Lecture Notes for MATH2230. Neil Ramsamooj

Transcription

1 Lecture Notes for MATH3 Neil amsamooj

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3 Table of contents Vector Calculus Parametric curves and arc length eview of partial differentiation Vector Fields Divergence and curl of a vector field Gradient of a function Line integrals and double integrals Line integrals Path independence and conservative vector fields Double integrals Green s theorem Surface integrals Parametric surfaces Surface integrals Surface integrals over vector fields Triple integrals and Divergence theorem Triple integrals Divergence theorem Laplace transforms Definition and existence of Laplace transforms Improper integrals Definition and examples of Laplace tranform Existence of Laplace transform Properties of Laplace transforms Linearity property The inverse Laplace transform Shifting the s variable; shifting the t variable Laplace transform of derivatives Applications and more properties of Laplace transforms Solving differential equations using Laplace transforms Solving simultaneous linear differential equations using the Laplace transform Convolution and Integral equations Dirac s delta function Differentiation of transforms The Gamma function Γ(x) Fourier series Definitions Convergence of Fourier Series Even and odd functions Half range expansions Partial Differential Equations Definitions The Heat Equation A derivation of the heat equation

4 4 Table of contents 4.. Solution of the heat equation by seperation of variables The heat equation with insulated ends as boundary conditions The heat equation with nonhomogeneous boundary conditions The Wave Equation A derivation of the wave equation Solution of the wave equation by seperation of variables Laplace s equation Solving Laplace s equation by seperation of variables Laplace s equation in polar coordinates

5 Chapter Vector Calculus. Parametric curves and arc length ecall that a function y f(x) describes a curve in the Cartesian plane which consists of points (x, f(x)) where x is the independent variable. Another method of defining a curve in the Cartesian plane is by the use of a parameter t. Definition.. A parametric curve C in the Cartesian plane is obtained by specifying x and y to be functions of a parameter t x f(t) y g(t) where a t b. Notice that any specific value of the parameter t t describes a particular point (x(t ), y(t )) of C; the curve C consists of the set of such points C {(x(t), y(t)) a t b}. Example.. Sketch the parametric curve C where t π 4. Answer: From the trigonometric identity we have xcost y sint cos t + sin t x + y, and therefore the curve C consists of points (x, y) that lie on a circle of radius. The values of the parameter t, t π at the ends of the interval t π correspond respectively to the points (, 4 4 ) and (, ). Therefore the curve C consists of the points on the circle of radius that lie between (, ) and ( ):, 5

6 6 Vector Calculus Notation.3. A parametric curve x f(t) y g(t) where a t b can also be written in vector notation where i and j are the standard unit vectors. r(t) f(t)i+ g(t)j where a t b Definition.4. A parametric curve r(t) f(t)i+ g(t)j where a t b is said to be smooth if the component functions f(t) and g(t) have continuous derivatives f (t), g (t) respectively on the interval a <t<b. The definition of a parametric curve in three dimensional space is analogous to the definition in the Cartestian plane. Definition.5. A parametric curve C in the three dimensional space is obtained by specifying x, y and z to be continuous functions of a parameter t x f(t) y g(t) z h(t) where a t b. Such a parametric curve C can also be written in vector notation r(t) f(t)i+ g(t)j +h(t)k where a t b where i, j and k are the standard unit vectors. Furthermore, C is said to be smooth if the component functions f(t), g(t) and h(t) have continuous derivatives f (t), g (t) and h (t) respectively on the interval a <t<b. Example.6. Sketch the parametric curve r(t)ti +tj + ( t)k where t. Answer: Notice that each of the functions x, y and z are linear functions of t x t y t z t so the curve C will be a segment of a straight line. The values of the parameter t, t at the ends of the interval t correspond respectively to the points (,, ) and (,, ). Therefore the curve C consists of the line that connects the points (,,) and (,,):

7 . Parametric curves and arc length 7 The following results express the arc length of a parametric curve as an integral. Theorem.7. i. The arc length s of a smooth parametric curve C in the Cartesian plane is given by r(t)x(t)i + y(t)j where a t b b s a ( dx dt ii. The arc length s of a smooth parametric curve C ) + ( ) dy dt dt. r(t)x(t)i+ y(t)j +z(t)k where a t b in three dimensional space is given by ( ) b dx ( ) dy ( ) dz s + + dt. a dt dt dt Example.8. Find the arc length of the parametric curve of Example.. Answer: r(t) costi + sintj where t π 4 s a b π 4 π 4 π 4 [t] π 4 π 4. ( ) ( ) dx dy + dt dt dt ( sint) + (cost) dt sin t + cos t dt dt

8 8 Vector Calculus Notice from the diagram in Example., this answer agrees with the formula for the arc length of a circle s rθ ( π ) () 4 π 4.

9 . eview of partial differentiation 9. eview of partial differentiation ecall for a function of one variable f(x), the derivative at x a is given by f f(a +h) f(a) (a) lim h h and f (a) has the geometric interpretation as the slope of the tangent line to f(x) at x a as may be seen in the following diagram. Example.9. If f(x) x then f (x) x. At the value x, we have f () and so the slope of the tangent to the point (, ) is equal to as illustrated in the following diagram (.) The limit definition (.) is not usually used to compute derivatives, in practice derivatives are computed by a set of rules power rule, product rule, quotient rule etc. Similarly, partial derivatives are defined using limits, but actually computed by using rules. The following are the definitions of partial derivatives of a function f(x, y) of two variables Definition.. i. If f(x, y) is a function of two variables then the partial derivative of f with respect to x at the point (a, b) is denoted as f x (a,b) or f (a, b) and is defined as x f(a +h,b) f(a, b) f x (a, b) lim h h ii. If f(x, y) is a function of two variables then the partial derivative of f with respect to y at the point (a,b) is denoted as f y (a, b) or f (a,b) and is defined as y f(a,b+h) f(a, b) f y (a, b) lim h h Partial derivatives of a function of two variables are usually computed by the following ule for finding partial derivatives of f(x, y) i. To find f x, treat y as a constant and differentiate with respect to x. ii. To find f y, treat x as a constant and differentiate with respect to y. Example.. Let f(x, y) x y 3. Find f f and x y. Answer: To find f x, treat y as a constant in f(x, y) x y 3. One can imagine that y is equal to some fixed constant, say y. Therefore f(x, y) x 3 and now differentiate x 3 as usual with respect to x to get (x 3 ) x 3. Now replace the by y to get the answer f x xy3.

10 Vector Calculus Similarly to find f y, treat x as a constant in f(x, y) x y 3. As before, imagine that x is equal to a fixed constant, say x 7. Therefore f(x, y) 7 y 3 and now differentiate 7 y 3 as usual with respect to y to get (7 y 3 ) 7 (3y ). Now replace the7 by x to get the answer f y x (3y ) 3x y. Example.. Let f(x, y) x cos(x + y). Find f f and x y. Answer: To find f x, treat y as a constant in f(x, y) x cos(x + y), say y 3. Then f(x, y) x cos(x + 3) and this is a product of two functions, therefore we use the usual product rule to differentiate with respect to x and replacing the 3 by y we have the answer (x cos(x+3)) (x ) cos(x+3) +x (cos(x + 3) ) x cos(x+3) x sin(x+3) f x x cos(x + y) x sin(x + y). To find f y, treat x as a constant. In this case f(x, y) x cos(x + y) is a product of a constant x and a function cos(x+ y) and so it is not necessary to use product rule. We have the answer f y x (cos(x + y)) y x ( sin(x+ y)) x sin(x + y). ecall that one obtains the second derivative d f of a function f(x) of one variable by differentiating the first derivative, for example f(x) sinx dx df dx cosx ( ) df dx d dx (cosx) sinx. d f dx d dx f or. To be pre- y Similarly, one obtains the second partial derivatives f, f, f x y x y and f y x y) of two variables by taking the partial derivative of a first partial derivative f x cise f x ( ) f x x f y ( ) f y y ( ) f f y x y x f x y ( ) f. x y of a function f(x,

11 . eview of partial differentiation Example.3. Find the second partial derivatives of f(x, y)x y + y 3. Answer: The first partial derivatives are The second partial derivative x The second partial derivative y The second partial derivative x The second partial derivative y f x f y f x y f y x f x xy f y x +3y is obtained by taking the partial derivative of f x ( ) f f x x x x (xy) y f is obtained by taking the partial derivative of y f y ( ) f y y ( x + 3y ) y 6y f is obtained by taking the partial derivative of y f x y ( ) f x y ( x +3y ) x x f is obtained by taking the partial derivative of x f y x ( ) f y x y (xy) x with respect to with respect to with respect to with respect to The partial derivatives of a function f(x, y, z) of three variables are defined in a similar manner to the partial derivatives of a function of two variables for example, to obtain f, one treats the variables x and z as constants and differentiates with respect to y y. Example.4. Find the first and second partial derivatives of f(x, y,z)x yz +xe z. Answer: f f xyz +ez x f x yz f y x xz f xy +ez z x y x z f y f x y xz f z y x f z x y +xe z f z xez f xy +ez x z f y z x

12 Vector Calculus We saw in Example.9 that the derivative f (a) can be interpreted as the slope of the tangent of the function f(x) at the point x a. In a similar manner, if f(x, y) is a function of two variables then the first partial derivatives f f (a, b) and (a, b) may also be interpreted as slopes of lines x y passing through the point corresponding to (x, y)(a,b). ecall that the graph of a function f(x, y) is obtained by plotting points (x, y, f(x, y)) in three dimensional space. In this way, the graph of a function of two variables forms a surface in 3 dimensions. For example, the graph of the function f(x, y) x forms a surface that is a plane: We illustrate the geometric interpretation of the partial derivatives f x, f by considering the y behaviour of the function f(x, y) x at (x, y) (, ). By taking partial derivatives, we have that ( ) ( ) f x, f y,. Now notice that f(, ). Therefore the point ( ) (,, f(,),, ) ( ) lies on the graph of f(x, y) x as we see in the above diagram. The z coordinate of,, ( ) is the function value f(, ) and corresponds to the height of the point,, f(, ) above the xyplane. We now keep x fixed and change the y-value. Plotting such points f(, y) gives the line B on the above diagram. Notice from the diagram that line B is parallel to the xy-plane, that is, the value of the function f does not change if we keep x fixed and change the y-value. This corresponds to ( ) f y,

13 . eview of partial differentiation 3 because the partial derivative f y (, ) measures the rate at which at the function f is increasing as the y-value changes while keeping x fixed. Now keep y fixed and change the x-value. Plotting such points f(x, ) gives the line A in the above diagram. Notice that the height of the points on line A is decreasing as the x-values increase. This corresponds to ( ) f x, because the partial derivative measures the rate at which the function f is increasing as the x-value increases while keeping y fixed.

14 4 Vector Calculus.3 Vector Fields Definition.5. A vector field on the Cartesian plane is a function F that assigns a two dimensional vector F(x, y) to each point (x, y). We may write F in terms of its component functions F(x, y) P(x, y)i+ Q(x, y)j. Furthermore, the vector field F is said to be continuous if and only if each its component functions is continuous. Example.6. The following diagram is a plot of the vector field F(x, y) yi+xj on the Cartesian plane. Notice that each point (x, y) on the Cartesian plane has a vector associated to it. Similarly, we may define vector fields in the three dimensional space 3. Definition.7. A vector field on the three dimensional space 3 is a function F that assigns a three dimensional vector F(x, y, z) to each point (x, y, z). We may write F in terms of its component functions F(x, y,z)p(x, y, z)i + Q(x, y,z)j + (x, y,z)k. Furthermore, the vector field F is said to be continuous if and only if each its component functions is continuous..3. Divergence and curl of a vector field Definition.8. Let F (x, y)p(x, y)i+ Q(x, y)j be a vector field on the Cartesian plane where the first partial derivatives of the component functions P and Q exist. Then the divergence of the vector field F is the function div(f) P x + Q y.

15 .3 Vector Fields 5 Example.9. Find the divergence of the following vector fields on the Cartesian plane i. F(x, y)3i ii. G(x, y)3x i Answer: i. ii. div(f) P x + Q x x (3)+ x () div(g) P x + Q x x (3x )+ x () 6x The divergence of a vector field has the following interpretation. Consider a infinitesimally small box of length x and width y located at the point (x, y) Then the divergence of a vector field F at the point (x, y) can be viewed as the net flow of F out of an infinitesimally small box located at the point (x, y). Consider the two vector fields F and G of Example.9. Notice that the vector field F(x, y) 3i is a constant vector field so the net flow of F out of the small box located at the point (x, y) is zero, which agrees with the answer div(f) obtained in Example.9.

16 6 Vector Calculus Notice however that the vector field G(x, y) 3x i is nonconstant, and that the length of the vectors increases as x increases. From the diagram below we see that the vectors at one vertical side of the small box are longer than the vectors at the other vertical side. Hence one can regard the net flow out of the box as positive, which agrees with the answer div(g)6x. obtained in Example.9. The following is the definition of the divergence of a three dimensional vector field. Definition.. Let F(x, y, z)p(x, y, z)i + Q(x, y, z)j + (x, y, z)k be a vector field on three dimensional space 3 where the first partial derivatives of the component functions P, Q and exist. Then the divergence of the vector field F is the function div(f) P x + Q y + z. The divergence of a three dimensional vector field can be expressed in an alternative form by the use of a differential operator Definition.. Let denote the vector differential operator Then the divergence of the vector field is given by the dot product of and F x i+ y j + z k. F(x, y, z)p(x, y, z)i + Q(x, y, z)j + (x, y, z)k div(f).f It is not hard to check that Definition. and Definition. are equivalent: div(f).f ( x i+ y j + ) z k (P(x, y, z)i + Q(x, y,z)j + (x, y,z)k) x (P)+ y (Q)+ z () P x + Q y + z... For a box located at (x, y) where x. If the box is located on the opposite side of the y axis, then the vectors reverse direction and our interpretation still holds.

17 .3 Vector Fields 7 Note that the divergence of three dimensional vector field has a similar interpretation to the divergence in two dimensions div(f) at the point (x, y, z) can be viewed as the net flow of F out of an infinitesimally small cube located at the point (x, y, z). Definition.. Let F(x, y, z)p(x, y, z)i + Q(x, y, z)j + (x, y, z)k be a vector field on three dimensional space 3 where the first partial derivatives of the component functions P, Q and exist. Then the curl of the vector field F is the vector field on 3 defined by curl (F) F i j k x y z P Q ( y Q z ) i+ ( P z ) ( Q j + x x P ) k. y Example.3. Find the curl of the vector field F(x, y,z)xzi+xyzj y k. Answer: curl(f) F i j k x y z xz xyz y ( y xy)i+(x )j + (yz )k ( y xy)i +xj + yzk. Note that at any specific point (x, y, z), curl (F) is a three dimensional vector and so corresponds to some specific direction in three dimensional space 3. Consider a small paddle that can rotate along an axis. If the vector field F is viewed as the velocity of a liquid then curl (F) at the point (x, y, z) would be the direction in which the axis of the paddle should be aligned in order to get the fastest counterclockwise rotation at the point (x, y, z)..3. Gradient of a function Definition.4. If f(x, y) is a function of two variables defined on the Cartesian plane where the first partial derivatives of f(x, y) exist, then the gradient of the function, denoted as grad(f), is the vector function defined as grad(f) f f i+ x y j. emark.5. Note that grad(f) assigns a two dimensional vector f x (x, y )i+ f y (x, y )j to each point (x, y ) of the Cartesian plane. It follows from Definition.5 that grad(f) is a vector field and because of this, grad(f) is also referred to as the gradient vector field of f.

18 8 Vector Calculus Example.6. Let f(x, y) x + y. Then grad(f) x (x + y )i+ y (x + y )j xi+yj. At any specific point (x, y), grad(f) is a vector that gives the direction of the maximum increase of the function f(x, y). Consider the following contour plot of the function f(x, y)x + y Figure.. Contour plot of f(x, y) x + y Note that all points (x, y) on the circle labelled f have function value equal to. The point (x, y) (/, 3 /) is one such point on the circle f. Notice that if we change the (x, y) values in the direction of arrow A, that is if we take (x, y) values along the circle f, then the function f(x, y) does not change value. In order to get the maximum increase in the function f(x, y) we must move in the direction perpendicular to the circle f ; this is the direction specified by in the diagram. grad(f)xi+yj i+ 3 j Definition.7. Let (x, y) be a point on a curve C in the Cartesian plane that has a welldefined tangent vector T. Then any vector n at the point (x, y) that is perpendicular to the tangent vector T is called a normal vector to the curve C at the point (x, y). Example.8. In Figure. above, the arrow A is a tangent vector to the curve f at the point (x, y)(/, 3 /). The vector grad(f)i+ 3 j is an example of a normal vector to the curve f at the point (/, 3 /). Definition.9. Let f(x, y) be a function of two variables and let c be a constant. The set of points that satisfy called a level curve of the function f(x, y). f(x, y)c

19 .3 Vector Fields 9 Example.3. Let f(x, y) x + y. Then the level curve is the circle f(x, y) x + y. Notice that the three level curves f(x, y), f(x, y).49 and f(x, y).5 are shown in Figure. above. Theorem.3. Let f(x, y) c be a level curve C in the Cartesian plane where f(x, y) is a differentiable function. Then if grad(f) at a point (x, y) is not the zero vector, then it is a normal vector to the curve C at the point (x, y). Example.3. Find a normal vector to the ellipse x + 4y at the point (, ). Answer: Notice that the curve x +4y is a level curve as it is in the form f(x, y) where f(x, y)x + 4y. The gradient of the function f is and at the point (x, y) (, grad(f) f f i+ x y j xi+8yj ), grad(f) is equal to grad(f)xi+8yj i+ 8 j i+ j which, by Theorem.3 is a normal vector to x +4y at the point (, ). ecall that the partial derivative f x (x, y ) measures the rate at which the function f is increasing as the x-value increases while keeping the y value fixed, in other words, f x (x, y ) gives the rate of change in the function f as the (x, y) points move from (x, y ) in the direction of the horizontal unit vector i; see Figure. below. Similarly the partial derivative f y (x, y ) gives the rate of change in the function f as the (x, y) points move from (x, y ) in the direction of the vertical unit vector j as is also illustrated in Figure. below.

20 Vector Calculus Figure.. Diagram of the domain of the function f If we wish to determine the rate at which a function changes as the (x, y) points move from a fixed point (x, y ) in the direction of a unit vector that is not horizontal or vertical, then we shall need the following definition. Definition.33. The directional derivative of a function f(x, y) in the direction of a unit vector u at a point (x, y ), denoted as D u f, is defined as the dot product D u f u grad(f) Example.34. Find the directional derivative of the function f(x, y) x + y at the point (, ) in the direction i j. Answer: In the definition of the directional derivative, the direction is specified by a unit vector. We first find the unit vector u parallel to i j The vector grad(f) at the point (x, y) (,) is and the required directional derivative is i j u length of i j i j +( ) i j grad(f) f x i+ f y j xi+yj i +4j D u f u grad(f) ( i ) j (i+4j).

21 .3 Vector Fields The gradient of a function f(x, y, z) of three variables is defined and has properties analogous to the gradient of a function of two variables. Definition.35. If f(x, y, z) is a function of three variables, then the gradient of the function, denoted as grad(f), is the vector function defined as grad(f) f f i+ x y j + f z k. Note that the vector differential operator of Definition. can be used to give an alternative expression of grad(f) grad(f) f. Definition.36. Let f(x, y, z) be a function of three variabless and let c be a constant. The set of points that satisfy f(x, y,z)c called a level surface of the function f(x, y,z). Example.37. Let f(x, y, z) x + y +z. Then the level surface S f(x, y,z) is the sphere x + y +z. of radius is illustrated in Figure.3 below. Figure.3. Definition.38. The tangent plane to a point P on a surface S is the plane that touches the surface at the point P. A normal vector to a surface S at the point P is a vector that is perpendicular to the tangent plane at the point P. Theorem.39. Let f(x, y, z) a be a surface S in three dimensional space where f(x, y, z) is a differentiable function. Then if grad(f) at a point (x, y, z) is not the zero vector, then it is a normal vector to the surface S at the point (x, y, z).

22 Vector Calculus Example.4. Find a normal vector to the sphere x + y +z at the point (,,). Answer: Notice that the surface x + y +z is in the form f(x, y,z) where f(x, y,z)x + y +z. The gradient of the function f is grad(f) f f i+ x y j + f z k xi+yj +zk and at the point (x, y)(,,), grad(f) is equal to grad(f)xi +yj + zk i+j +k i which, by Theorem.39 is a normal vector to the sphere x + y +z at the point (,, ). f x (x, y, z ), f y (x, y, z ) and As in the case of a function of two variables, the partial derivatives f z (x, y, z ) measure the rate at which a function f(x, y, z) changes as the (x, y, z) points move from a fixed point (x, y,z ) in the direction of a unit vectors i, j and k respectively. If we wish to determine the rate at which a function changes as the (x, y, z) points move from a fixed point (x, y, z ) in the direction of a unit vector that is not a standard unit vector i, j or k, then, as in the case of a function of two variables, we need the definition of a directional derivative. Definition.4. The directional derivative of a function f(x, y, z) in the direction of a unit vector u at a point (x, y, z ), denoted as D u f, is defined as the dot product D u f u grad(f).

23 .4 Line integrals and double integrals 3.4 Line integrals and double integrals.4. Line integrals Definition.4. Let F(x, y) be a continuous vector field defined on the Cartesian plane and let C be the smooth parametric curve r(t)x(t)i + y(t)j a t b. Then the line integral of the vector field F along the curve C is b F dr F (x(t), y(t)) (x (t)i + y (t)j ) dt C a A line integral F dr can be interpreted as the work done on a particle by a force field F as it C travels on a path C. Consider the following example. Example.43. Let F (x, y) 3x i be a vector field on the Cartesian plane. Calculate the following line integrals i. C F dr where C is the curve ii. r(t) ( + t)i + j t C F dr where C is the curve r(t) i+( +t)j t. Answer: i. ii. b F dr F(x(t), y(t)) (x (t)i+ y (t)j )dt C a F(( + t),) (( + t) i+() j )dt ( 3( + t) i ) (i+j )dt 3( +t) dt [ ( +t) 3] 8. b F dr F(x(t), y(t)) (x (t)i+ y (t)j )dt C a F(, ( +t)) (() i+( +t) j )dt ( 3() i ) (i+j )dt 3i (i+j )dt dt.

24 4 Vector Calculus Figure.4 below is an illustration of the vector field F(x, y)3x i and the two curves C : r(t) ( + t)i + j t C : r(t)i+( +t)j t. If we regard the vector field F as a force field then as C lies in the direction of F, we expect F to add energy to a particle travelling along the path C. This agrees with C F dr 8 in part i). However, the curve C is perpendicular to the direction of vector field and therefore F does not add any energy to a particle travelling along C which agrees with the answer C F dr. in part ii). Figure.4. The definition of a line integral in three dimensional space is similar to the definition in the Cartesian plane. Definition.44. Let F(x, y, z) be a continuous vector field defined in three dimensional space and let C be the smooth parametric curve r(t)x(t)i+ y(t)j + z(t)k a t b. Then the line integral of the vector field F along the curve C is b F dr F (x(t), y(t),z(t)) (x (t)i + y (t)j + z (t)k )dt. C a Example.45. Evaluate C F dr where F is the vector field in three dimensional space F (x, y,z)xyi + yzj +zxk and C is the parametric curve r(t)ti +t j +t 3 k t.

25 .4 Line integrals and double integrals 5 Answer: From Definition.44 above, we have b F dr F(x(t), y(t),z(t)) (x (t)i + y (t)j + z (t)) dt C a F(t, t,t 3 ) ((t) i+(t ) j +(t 3 ) k ) dt ( t 3 i+t 5 j +t 4 k ) (i +tj + 3t k ) dt (t 3 +t 6 + 3t 6 )dt [ ] t t Path independence and conservative vector fields Definition.46. If C is a parametric curve in the Cartesian plane of the form r(t) f(t)i+ g(t)j where a t b, then the initial point of C is given by the position vector r(a) and the terminal point of C is given by the position vector r(b). Similarly, one can define the initial and terminal points of a parametric curve in three dimensional space. r(t) f(t)i+ g(t)j +h(t)k a t b Example.47. Let F be the vector field F(x, y) y i +xj and let C and C be the parametric curves defined as C : r(t)(5t 5)i+(5t 3)j t C : r(t)(4 t )i +tj 3 t. a) Sketch the curves C and C and verify that C, C have the same initial and terminal point b) By evaluating the respective line integrals, show that F dr F dr. C C Answer: a) In the case of C, we have x 5t 5 y 5t 3. Note that both x, y are linear functions of t and that the parameter t takes values in a finite interval t. It follows that C is a line segment with initial point r() ( 5, 3) and terminal point r()(, ). In the case of C, the initial point is r( 3) ( 5, 3) and terminal point r() (, ). To determine the shape of C, we eliminate t from to obtain x 4 t y t x 4 y.

26 6 Vector Calculus It follows that C is a segment of the parabola x 4 y that has initial point ( 5, 3) and terminal point (, ). Figure.5. Sketch of curves C and C b) Using the definition of the line integral, we have b F dr F(x(t), y(t)) (x (t)i+ y (t)j )dt C a F((5t 5), (5t 3)) ((5t 5) i +(5t 3) j )dt ( (5t 3) i +(5t 5)j ) (5i+5j )dt ( 5(5t 3) +5(5t 5) ) dt ( 5 5t 5t + 4 ) dt and and clearly 5 6 b F dr F(x(t), y(t)) (x (t)i+ y (t)j )dt C a F((4 t ),t) ((4 t ) i +(t) j ) dt 3 ( t i+(4 t )j ) (( t)i+ j )dt 3 ( ( t)t +4 t ) dt 3 ( t 3 t +4 ) dt F dr F dr. C C Definition.48. A curve C is called a piecewise smooth curve if it is a finite union C C C C n

27 .4 Line integrals and double integrals 7 of smooth curves C,C,,C n such that the terminal point of C i is the initial point of C i+. Figure.6. A piecewise smooth C C C C 3 A line integral along a piecewise smooth curve is obtained by adding the line integrals of its individual pieces. Definition.49. Let C C C C n be a piecewise smooth curve. Then the line integral of the vector field F along the curve C is F dr F dr + F dr + + F dr C C C C n Definition.5. A line integral is said to be independent of path if C C F dr F dr F dr C for any piecewise smooth curve C which has the same initial point and terminal point as the curve C. Example.5. Consider curves C, C and the line integral C F dr as defined in Example.47. This line integral is not independent of path as C has the same initial and terminal points as C but F dr F dr. C C ecall from emark.5 that the gradient grad(f) of a function f(x, y) is in fact a vector field. Definition.5. Let F be a vector field defined on n.. Then F is said to be a conservative vector field if there exists a function f such that F grad(f). For such a case, f is called a potential function for the vector field F..3.. In this class we consider only the cases of n (the Cartesian plane) and n 3 (three dimensional space).3. A different definition F grad(f) is used in physics; with this alternative definition, the function f gives a more accurate physical interpretation of the work done by an outside force in moving against the vector field. In this class, we use the definition F grad(f).

28 8 Vector Calculus The above is the standard definition of a conservative vector field; however it is not a practical method of determining whether or not a vector field is conservative. A more useful criterion for vector fields in is the following Theorem.53. Let F(x, y) P(x, y)i+ Q(x, y)j. be a vector field defined on where the component functions P and Q have continuous first partial derivatives. Then F is conservative if and only if Example.54. Show that the vector field P y Q x. F(x, y)(6x +5y)i+(5x +4y)j is conservative and determine a potential function for F(x, y). Answer: In this case and P(x, y) 6x +5y Q(x, y)5x+ 4y P Q 5 y x so it follows that F is conservative. Let f be a potential function for F. Then grad(f)f f f i+ j (6x+ 5y)i +(5x+ 4y)j x y f 6x+ 5y x. (.) f y 5x +4y Using partial integration.4 to integrate the first equation of (.) with respect to x, we have f(x, y)3x +5xy + C(y). (.3) The function C(y) can be determined by differentiating equation (.3) with respect to y and using the second equation of (.) f y 5x +C (y)5x+ 4y C (y)4y C(y)y + K where K is a constant of integration. We therefore have f(x, y)3x +5xy +y + K and by choosing K we have that f 3x + 5xy + y is a potential function for the vector field F. The following result specifies exactly what conditions are required for a line integral in n to be independent of path..4. see Example.59 on page 3

29 .4 Line integrals and double integrals 9 Theorem.55. Let F be a vector field on n. Then the line integral F dr is independent of path if and only if F is a conservative vector field. C Furthermore, the value of a line integral that is independent of path can be determined from the endpoints of the curve and a potential function of the vector field Theorem.56. (Fundamental Theorem of Line Integrals) Let F be a vector field defined on n and F dr be a line integral that is independent of path. C Then F dr f(r ) f(r ) C where f is a potential function for the the vector field F and r, r are respectively the initial and terminal points of the curve C. Example.57. Let F be the vector field F(x, y, z) x 3 y 4 i+x 4 y 3 j and let C be the parametric curve defined as C: r(t) t i+( +t 3 )j a) Show that F is a conservative vector field b) Find a potential function for F t. c) Use the potential function of part (b) to evaluate the line integral Answer: a) The given vector field is of the form F P(x, y)i+ Q(x, y)j where As so it follows that F is conservative. b) Let f be a potential function for F. Then P(x, y) x 3 y 4 Q(x, y)x 4 y 3. P y 4x3 y 3 Q x grad(f) F f f i+ x y j x3 y 4 i+x 4 y 3 j C F dr. f x x3 y 4 f y x4 y 3. (.4) By partially integrating the first equation of (.4) with respect to x, we have f(x, y) x4 y 4 + C(y). (.5) 4 The function C(y) can be determined by differentiating equation (.5) with respect to y and using the second equation of (.) f y x4 y 3 +C (y) x 4 y 3 C (y) C(y)K

30 3 Vector Calculus where K is a constant of integration. We have f(x, y) x4 y 4 4 +K and by choosing K we have that f x4 y 4 is a potential function for the vector field F. 4 c) Using the parametric definition of C, we have that the initial and terminal points of C are r r()(, ) r r()(, ) and from Theorem.56 F dr f(r ) f(r ) C f(,) f(,) Double integrals We give the definition of a double integral over a rectangular region and then state a theorem that gives a procedure for evaluating double integrals over a rectangular region. We then note that a double integral may be interpreted as a volume. Finally we explain the evaluation of double integrals over Type I, Type II and circular regions. Definition.58. Let be a rectangular region in the Cartesian plane defined by {(x, y) a x b, c y d}. Divide the rectangular region into subrectangles by partitioning the intervals a x b and c y d : and define the subrectangle ij as a x <x < <x m b c y < y < < y n d ij {(x, y) x i x x i, y j y y j }

31 .4 Line integrals and double integrals 3 Choose points (x ij, y ij ) in each subrectangle ij. Let A ij denote the area of the subrectangle ij and let P denote the length of the largest diagonal of all subrectangles (note that as we take smaller subrectangles of we have that P ). Then the double integral of the function f(x, y) over the rectangular region is defined to be the limit if this limit exists. f(x, y)da lim P m n i j f(x ij, y ij ) A ij The above definition is not usually used to evaluate double integrals. We shall give a method of evaluating double integrals based on the following procedure of partial integration. Example.59. (of partial integration) [ ] xy xy dy y y x x 3x. Notice in the above procedure of partial integration we integrate a function of two variables with respect to y treat x as a constant when integrating the answer is a function of x. xy dy 3x We can also integrate partially with respect to x: Example [ ] sin(xy) cos(xy) dy y sin(4y) sin(3y) y y and notice in this case our answer is a function of y. x4 x3 Using the procedure of partial integration and the following theorem we can evaluate double integrals over rectangular regions. Theorem.6. (Fubini s Theorem) If the function f(x, y) is continuous at each point in a rectangular region {(x, y) a x b, c y d} then ( b ) d ( d ) b f(x, y)da f(x, y)dy dx f(x, y)dx dy. (.6) a c c a Note that the integrals within the brackets of (.6) are evaluated by partial integration. Also notice that (.6) implies that the double integral over a rectangular region can be obtained by integrating with respect to y and then x or integrating with respect to x and then y. Example.6. Evaluate the double integral x y da

32 3 Vector Calculus where is the rectangular region {(x, y) x 3, y }. Answer: By Fubini s Theorem Notation.63. i. The integral is usually denoted as ii. Similarly, the integral is usually denoted as x y da 3 ( ) x ydy dx 3 [ x y ] y dx 3 ( x 3 3x dx [ ] x 3 x3 x 7. ( b a ab c c ( d c c d a d a b d b y x f(x, y)dy ) f(x, y)dydx. f(x, y)dx ) f(x, y)dxdy. dx dy ) dx ecall that the graph of a function f(x, y) forms a surface in three dimensional space. Given a function f(x, y) for each point (x, y) in a rectangular domain, then the double integral f(x, y)da can be interpreted as the volume between the graph of f(x, y) and the rectangle lying in the xy plane

33 .4 Line integrals and double integrals 33 We now consider double integrals over regions in the Cartesian plane that are not rectangular. Definition.64. i. A Type I region is a region in the Cartesian plane that may be described as {(x, y) a x b, f(x) y g(x)} where f(x) and g(x) are continuous functions of x. ii. A Type II region is a region in the Cartesian plane that may be described as {(x, y) k(y) x h(y), c y d} where k(y) and h(y) are continuous functions of y. Figure.7. As in the evaluation of a double integral over a rectangular region, evaluating double integrals over Type I and Type II regions requires the use of partial integration. Note that for a Type I region, the integration is done with respect to the y variable first and then with respect to x variable. For a Type II region, the integration is done with respect to the y variable first and then with respect to x variable. Theorem.65. i. If the function f(x, y) is continuous at each point in a Type I region then {(x, y) a x b, g(x) y f(x)} ( b ) f(x) f(x, y)da f(x, y)dy dx a g(x) ii. If the function f(x, y) is continuous at each point in a Type II region then {(x, y) k(y) x h(y), c y d} ( d ) h(y) f(x, y)da f(x, y)dx dy. c k(y)

34 34 Vector Calculus Example.66. Evaluate the double integral where {(x, y) x, y x }. Answer: Notice that is a Type I region. Then xda [ ( ( 3. x da x xdy y x ) dx dx [xy] y ) x x x() dx x x ( x ) 3 3 dx ] x x Example.67. Evaluate the double integral xyda where is the region bounded by the line x y + and the parabola x y 3. Answer: By solving the equations x y + x y 3 y + y 3 y, 4 and we see that the line x y + and the parabola x y 3 intersect at the points (, ) and (5, 4):

35 .4 Line integrals and double integrals 35 From the above diagram we can write as {(x, y) y 3 x y +, y 4} and so the region is of Type II. Then 4 y+ xyda xydx dy y 3 4 [ x ] xy+ y dy x y 3 ( ) 4 y (y +) y ( y) 3 dy 4 ( ) y 3 +y + y y5 4 +3y3 9y dy 4 ( ) y5 4 +4y3 +y 8y dy 36. Some regions in the Cartesian plane are more easily described by the use of polar coordinates. Definition.68. A polar rectangle is a region in the Cartesian plane that may be described in polar coordinates (r, θ) as {(r, θ) θ θ θ, a r b} where a and b are real constants such that a<b. Figure.8. Example.69. Express the following region as a polar rectangle. Answer: The equations {(x, y) 9 x + y 5, x, y } x + y 9 x + y 5 describe circles of radius 3 and 5 respectively that have the origin as center. Therefore the inequalities 9 x + y 5, x and y

36 36 Vector Calculus describe points that lie between the circles and in the first quadrant and so we can write as a polar rectangle {(r, θ) θ π, 3 r 5}. We shall use the following theorem to evaluate double integrals over regions in the Cartesian plane that are polar rectangles. Notice that the following theorem is a conversion of a double integral in xy coordinates to a double integral in polar coordinates. Theorem.7. If the function f(x, y) is continuous at each point in a polar rectangle then {(r, θ) θ θ θ, a r b} ( θ ) f(x, y)da f(r cosθ, r sinθ)rdr dθ. θ ab Example.7. Evaluate the following double integral ( (x + y ) ) da where {(x, y) x + y } by converting to polar coordinates. Answer: The region is the unit disc and so we can write as a polar rectangle {(r, θ) θ π, r }.

37 .5 Green s theorem 37 From Theorem.7, we have ( (x + y ) ) da π ( ) ( (r cos θ +r sin θ))rdr dθ π ( ) ( r )rdr dθ π ( ) (r r 3 )dr dθ π [ r ] r4 dθ 4 π 4 dθ π..5 Green s theorem Definition.7. i. A parametric curve C is called closed if its terminal point coincides with its initial point. ii. A simple curve C is a curve that does not intersect itself anywhere between its endpoints. Figure.9. Examples of curves Convention: A positive orientation of a simple closed curve C refers to a counterclockwise traversal of C. Theorem.73. (Green s Theorem) Let C be a positively oriented, piecewise smooth, simple closed curve in the Cartesian plane. Let D be the region bounded by C. Let F be the vector field F (x, y)p(x, y)i+ Q(x, y)j where P(x, y) and Q(x, y) have continuous partial derivatives on an open region that contains D. Then ( Q F dr C D x P ) da y

38 38 Vector Calculus Figure.. egion D bounded by curve C Note that Green s Theorem states that the line integral around a simple closed curve C may be obtained by evaluating a double integral over the region D enclosed by C. Example.74. Let C C C C 3 be the simple closed curve enclosing the triangular region D. Use Green s Theorem to determine the line integral (x 4 i+ xyj).dr (.7) by evaluating an appropriate double integral. C Answer: By Green s Theorem C ( Q F dr D x P ) da y we have that the line integral (.7) is equal to a double integral ( (x 4 i+ xyj) dr C D x (xy) ) y (x4 ) da (y )da D y da D (.8) and this last integral is a double integral of the function f(x, y) y over the triangular region D. Notice that D is a Type I region

39 .6 Surface integrals 39 that is, we can describe the region D as the set and therefore D {(x, y) x, y x} Substituting (.9) into (.8) we have (x 4 i + xyj) dr C D y da x y dydx (.9) D y da x y dydx and notice that this is the point of Green s theorem for a simple closed curve C, a line integral around C is equal to a double integral over the region enclosed by C C (x 4 i+ xyj).dr x y dydx and we can evaluate this double integral by using partial integration C (x 4 i+ xyj).dr x y dy dx [ y ] y ( x) dx [ ] ( x)3 6 6 y x dx.6 Surface integrals.6. Parametric surfaces ecall that some curves C in three dimensional space can be described parametrically C: r(t)x(t)i+ y(t)j +z(t)k a t b, for example the curve shown below

40 4 Vector Calculus may be written parametrically as r(t)ti+tj + ( t)k t. We now describe surfaces in three dimensional space by the use of vector functions r(u, v) of two parameters u and v. Definition.75. A parametric surface S in three dimensional space is obtained by specifying x, y and z to be continuous functions of parameters u and v x f(u,v) y g(u,v) z h(u, v) where (u, v) and is a region in the uv plane. The parametric surface S can be written in vector form r(u,v) f(u,v)i+ g(u,v)j +h(u,v)k (u,v). Example.76. Let S be the (truncated) plane defined by y, x,z. Then we can describe S a parametric surface by where is the region {(u, v) u,v } r(u,v)ui+ j +vk (u,v). in the uv plane. Notice that each point (u,v) of defines a unique point in S. Example.77. Describe the cylinder S as a parametric surface. x + y 4, z 3.

41 .6 Surface integrals 4 Answer: We can use polar coordinates to describe the cylinder S as the parametric surface r(θ, z) cosθi +sinθj + zk (θ,z) where is the region {(θ, z) θ π, z 3} in the θz plane. Lemma.78. Let S be a parametric surface r(u,v) f(u,v)i+ g(u,v)j +h(u, v)k (u, v) that is smooth. Then a normal vector n to S at the point r(u,v ) is given by where r u and r v are the vectors provided that r u r v. nr u r v r u f u (u,v )i+ g u (u, v )j + h u (u, v )k r v f v (u,v )i+ g v (u, v )j + h v (u,v )k Example.79. Find a normal vector to the cylinder S: r(θ, z) cosθi +sinθj + zk (θ,z) (.) where is the region {(θ, z) θ π, z 3} at the point (,, ). Answer: In this case the parameters are θ and z. Note that the point (,, ) corresponds to (θ, z) (, ) as r(,)cosi+sinj + k i

42 4 Vector Calculus The vector functions r θ (θ, z) and r z (θ, z) are obtained by differentiating (.) partially with respect to θ and z respectively r θ (θ,z) sinθi+cosθj +k and at (θ,z)(,) r z (θ, z) i+j +k r θ (,)j r z (, )k and the normal vector at the point (,, ) is nr θ (,) r z (, ) j k i.6. Surface integrals Definition.8. Let S be a smooth parametric surface r(u,v)x(u,v)i+ y(u,v)j + z(u,v)k (u, v) where is a region in the uv plane and let f(x, y, z) be a continuous function defined on S. Then the surface integral of the function f(x, y,z) over the surface S is f(x, y,z)ds f(x(u, v), y(u,v),z(u, v)) r u r v da (.) S Note that equation (.) defines a surface integral to be equal to a double integral over a region in the uv plane where u and v are the parameters of the surface S. Example.8. Evaluate the surface integral S (x y +z ) ds where S is the part of the cylinder x + y 9 that lies between the plane z and z.

43 .6 Surface integrals 43 Answer: Our definition (.) of a surface integral requires that the surface S be given in parametric form: S: r(θ, z) 3cosθi+3sinθj + zk (θ,z) (.) where is the region {(θ, z) θ π, z }. Note in this case we use the parameters r and θ instead of u and v. The x, y and z components of S are The vectors r θ and r z are and we have and therefore x3cosθ y 3sinθ z z. r θ x y i+ θ θ j + z k 3sinθi +3cosθj +k θ r z x y i+ z z j + z k i +j +k z r θ r z ( 3sinθi+3cosθj) k i j k 3sinθ 3cosθ 3cosθi+3sinθj r θ r z 3cosθ i+3sinθj 9cos θ +9sin θ 3. From the definition of a surface integral (x y + z ) ds f(x(θ,z), y(θ, z),z(θ,z)) r θ r z da S ( (3cosθ) (3sinθ)+z ) 3dA ( 3 7cos θ sinθ +z ) da This last integral is a double integral of the function f(θ, z) 7cos θ sinθ + z region over the rectangular

44 44 Vector Calculus in the θz plane. Therefore S ( (x y +z )ds 3 7cos θ sinθ +z ) da 3 π ( 7cos θ sinθ + z ) dzdθ π [ ] 3 7cos θ sinθz + z3 dθ 3 3 π ( 54cos θ sinθ + 8 ) dθ 3 [ 3 54cos3 θ + 8θ ] π (use substitution u cosθ) 3 3 6π A surface integral of a function f over a surface S can be interpreted as follows. Suppose that a surface S is subdivided into n smaller surfaces S i. Choose a point (x i, y i, z i ) on each smaller surface S i. Let A i be the area of the small surface S i and multiply this area by the value f(x i, y i, z i ) of the function f at the point (x i, y i, z i ). Hence we have a weighted area f(x i, y i,z i ) A i for each of the n smaller surfaces S i of the entire surface S. Then the surface integral of the function f over the surface S is approximately the sum of each of the weighted areas for the S i S n f(x, y, z)ds i f(x i, y i,z i ) A i (.3) and if the number n of smaller surfaces S i gets larger then the approximation (.3) gets better, so we have n f(x, y, z) ds lim f(x i, y i, z i ) A i. (.4) n S Notice that if the function f(x, y, z) then from (.4) we have S i ds lim n n A i and the right side of (.4) is nothing but the sum of the areas of each of the smaller surfaces S i. This sum is obviously equal to the surface area of S. Therefore we have the following lemma i

45 .6 Surface integrals 45 Lemma.8. The surface integral of the constant function f over the surface S is equal to the surface area of S ds surface area of S. S Example.83. Find the surface area of the truncated plane S r(u,v)ui+ j +vk (u,v). where is the region {(u, v) u,v }. Answer: The surface S is shown in the following diagram (see Example.76). Clearly the surface area of S is equal to ; let us verify this by evaluating the surface integral ds. The x, y and z components of S are The vectors r u and r v are and we have xu S y z v. r u x u i + y u j + z k i+j +k u r v x y i+ v v j + z k i+j +k v r u r v i k i j k j and so r u r v j From the definition of a surface integral S ds r u r v da da

46 46 Vector Calculus This last integral is a double integral of the function f(u,v) over the rectangular region in the uv plane. Therefore S ds da dudv [u] dv ( )dv dv which agrees with our expected answer. ecall that the graph of a function g(x, y) forms a a surface in three dimensional space. Let be a region in the xy plane and let S be the surface formed as the image of the region in the graph of g(x, y): Then the following lemma expresses a surface integral of a function f(x, y, z) over such a surface S in terms of the function g(x, y). Lemma.84. Let be a region in the xy plane and let S be the image of in the graph of a function g(x, y). Then the surface integral of the function f(x, y, z) over the surface S is equal to a double integral over the region in the xy plane: S f(x, y, z) ds f(x, y, g(x, y)) + Proof. The surface S can be written parametrically as ( ) g + x ( g y ) da r(x, y)xi+ yj + g(x, y)k (x, y)

47 .6 Surface integrals 47 with parameters x and y. The vectors r x and r y are and therefore r x x (x)i+ x (y)j + g (g(x, y))k i+j + x x k r y y (x)i + y (y)j + g (g(x, y))k i+ j + y ( r x r y i+ g ) x k i j k g x g y g x i g y j + k ( j + g ) y k y k hence r x r y g x i g y j +k ( ) g ( ) g + + x y and from the parametric definition of a surface integral in Definition.8 we have f(x, y, z) ds f(x, y, g(x, y)) r x r y da S ( ) g ( ) g f(x, y, g(x, y)) + + da x y which is the desired result. Example.85. Find the surface area of the hemisphere of radius a z a (x + y ). Answer: Denote the hemisphere as S Notice that the hemisphere S is the image of the region in the graph of the function : x + y a z g(x, y) a (x + y ).

48 48 Vector Calculus ecall from Lemma.8 that the surface area of S is given by the surface integral of the constant function f over S, that is surface area of S dss. S Now because the hemisphere S is the image of a region in the graph of the function g(x, y), from Lemma.84 we have ( ) g ( ) g ds + + da S x y and therefore we have surface area of S ds S a where :x + y a is the disc of radius a ( ) ( ) g g + + da x y ( ) x y + +( a (x + y ) a (x + y ) x + + y a (x + y da ) a a (x + y da ) ( a (x + y ) ) da ) da and so we convert to polar coordinates surface area of S a a a a a π ( a (x + y ) ) da a ( a (r cos θ + r sin θ) ) rdrdθ a ( a r ) rdr ) dθ π ( π [ ( ] a r ) πa π a dθ ra r dθ.6.3 Surface integrals over vector fields Definition.86. A surface S is called an orientable surface if there exists a continuous function n on S that assigns a unit normal vector to each point of S. Such a function n is called an orientation of S. There are two possible orientations for any orientable surface.

49 .6 Surface integrals 49 An oriented surface S is a surface together with one of the two possible choices of orientation. Example.87. Let S be the cylinder S: r(θ, z) cosθi +sinθj + zk (θ,z) where is the region {(θ, z) θ π, z 3} defined in Example.79. ecall from that example, the vector r θ (θ,z) r z (θ, z) ( sinθi +cosθj) k i j k sinθ cosθ cosθi+sinθj is a normal vector at any point r(θ,z) on S. The function n(θ, z) given by n(θ,z) r θ r z r θ r z cosθi+sinθj 4cos θ + 4 sin θ cosθi+sinθj assigns the unit normal vector cosθ i+sinθj to each point on the cylinder S. Therefore n(θ, z) cosθi+sinθj is an example of an orientation on the surface S. Notice that is the other possible orientation of the cylinder S. n (θ,z) n(θ, z) (cosθi+sinθj) We now define the surface integral over a vector field. Definition.88. If F(x, y, z) is a continuous vector field defined on a smooth parametric surface S r(u,v)x(u,v)i+ y(u,v)j + z(u,v)k (u, v) where is a region in the uv plane. Let S have an orientation n given by n r u r v r u r v. Then the surface integral of the vector field F(x, y,z) over the surface S is F.n ds F(x(u,v), y(u, v), z(u,v)) (r u r v ) da. (.5) S Note that equation (.5) defines a surface integral of a vector field over a surface to be equal to a double integral over a region in the uv plane where u and v are the parameters of the surface S.

50 5 Vector Calculus Example.89. Evaluate the surface integral (xi+3xzj + y k) nds where S is the truncated plane.5 S r(u,v)ui+ j +vk (u,v) and is the region {(u,v) u,v }. Answer: The x, y and z components of S are xu y z v. The vectors r u and r v are and we have r u x u i + y u j + z k i+j +k u r v x y i+ v v j + z k i+j +k v r u r v i k i j k j In this case the vector field F is F(x, y, z) xi +3xzj + y k From the definition of a surface integral of a vector field over a surface S (xi+3xzj + y k).n ds F(x(u,v), y(u, v), z(u,v)).(r u r v ) da (ui +3uvj + k).( j)da 3uv da This last integral is a double integral of the function f(u,v) 3uv over the rectangular region.5. See Example.76

51 .6 Surface integrals 5 in the uv plane. Therefore (xi+3xzj + y k).nds 3uv da S 3uv dudv [ 3u ] v dv ( 3() v 3() v 3v dv 3 4 ) dv If the vector field F represents the velocity of a fluid moving in three dimensional space, then the surface integral of the vector field F over the surface S F.n ds can be interpreted as the S volume of fluid flowing through the surface S in unit time. Consider the following example. Example.9. Let F(x, y, z) yj be a vector field in three dimensional space. Evaluate the following surface integrals i. F.n ds where S S is the parametric surface r(u, v)i+uj + vk (u, v). ii. where is the region {(u, v) u,v } S F.n ds where S is the parametric surface r(u, v)vi + j +uk (u, v). where is the region {(u, v) u,v }. Answer: i. Note that the vector field F (x, y, z) yj is in the direction of j at each point (x, y, z) in three dimensional space. The surface S is the truncated plane shown in the diagram below and notice S is parallel to the vector field F. If we regard F as the velocity of a fluid and as S F.n ds can be interpreted as the volume of fluid flowing through the surface S in unit time, we expect S F.n ds

52 5 Vector Calculus as there is no fluid flowing through S (only along S ). We verify this answer: the x, y and z components of S are The vectors r u and r v are and we have x y u z v. r u x y i+ u u j + z k i + j + k u r v x y i+ v v j + z k i+j +k v r u r v j k i j k i and by therefore the surface integral of the vector field F over the surface S is F.ndS F(x(u, v), y(u,v),z(u, v)).(r u r v )da S (uj) ida da ii. In this case the vector field F is perpendicular to the surface S and if we again interpret S F.n ds as the volume of fluid flowing through the surface S in unit time, we expect S F.ndS >. We verify this answer: the x, y and z components of S are The vectors r u and r v are and we have x v y z u. r u x y i+ u u j + z k i +j + k u r v x y i+ v v j + z k i+j + k v r u r v k i i j k j

53 .6 Surface integrals 53 and by therefore the surface integral of the vector field F over the surface S is F.ndS F(x(u, v), y(u,v),z(u, v)).(r u r v )da S (j) j da da This last integral is a double integral of the function f(u, v) u over the rectangular region in the uv plane. Therefore F.ndS u da S dudv [u] dv dv. Consider a surface S that is a subset of the graph of a function g(x, y), to be specific, let be a region in the xy plane and let S be the oriented surface formed as the image of the region in the graph of g(x, y) where we assume that the orientation of S is chosen to point upward Then the following lemma expresses a surface integral of a vector field F(x, y, z) over such an oriented surface S in terms of the function g(x, y).

54 54 Vector Calculus Lemma.9. Let be a region in the xy plane and let S be the oriented surface formed as the image of in the graph of a function g(x, y) where the orientation of S is chosen to point upward. Then the surface integral of the vector field F(x, y, z) P(x, y,z)i+ Q(x, y, z)j +T(x, y,z)k over the oriented surface S is equal to a double integral over the region in the xy plane: ( F.n ds P g ) x Q g y +T da S Example.9. Evaluate the surface integral (xi + yj +zk).nds S where S is the part of the paraboloid z x y that lies above the plane z. Assume S has an upward orientation. Answer: Note that the paraboloid z x y intersects the plane z where x y that is, the paraboloid intersects the plane z at the circle x + y. From the diagram, we see that the surface S is the image of the region : x + y in the graph of the function g(x, y) x y. The components of the given vector field are P(x, y,z)x Q(z, y, z) y T(x, y,z)z and therefore from Lemma.9 the surface integral of the given vector field over the oriented surface S is ( (xi + yj +zk).nds P g ) S x Q g y + T da ( x( x) y( y)+z)da ( x( x) y( y) + x y ) da

55 .6 Surface integrals 55 because z x y. Hence where the region is the unit disc S ( (xi+ yj + zk).nds +x + y ) da and so we convert to polar coordinates (xi + yj + zk).nds S ( + x + y ) da π ( +r cos θ + r sin θ ) rdrdθ π ( ) ( +r )rdr dθ ( [ ] ) π r + r4 dθ 4 3π. π 3 4 dθ

56 56 Vector Calculus.7 Triple integrals and Divergence theorem.7. Triple integrals Definition.93. Let B be a rectangular box in three dimension defined by B {(x, y,z) a x b, c y d,r z s}. Divide the rectangular box B into suboxes by partitioning the intervals a x b, c y d and r z s: and define the subbox B ijk as a x <x < <x m b c y < y < < y n d r z < z < < z l s B ijk {(x, y,z) x i x x i, y j y y j,z k y z k } Choose points (x ijk, y ijk, z ijk ) in each subbox B ijk. Let V ijk denote the volume of the subrectangle B ijk and let P denote the length of the longest diagonal of all subboxes (note that as we take smaller subboxes of B we have that P ). Then the triple integral of the function f(x, y, z) over the rectangular box B is defined to be the limit if this limit exists. B f(x, y, z)dv lim P m n l i j k f(x ijk, y ijk, z ijk ) V ijk The above definition is not usually used to evaluate triple integrals. The method of evaluating triple integrals over rectangular boxes is based on the following theorem. Theorem.94. (Fubini s Theorem for triple integrals) If the function f(x, y, z) is continuous at each point in a rectangular box then B B {(x, y) a x b, c y d,r z s} f(x, y, z)dv ab d s f(x, y, z)dzdy dx. (.6) c r From the above theorem we see that the evaluation of a triple integral requires two instances of partial integration.

57 .7 Triple integrals and Divergence theorem 57 Example.95. Evaluate the triple integral where B is the rectangular box B xyz dv B {(x, y,z) x, y, z 3}. Answer: By Fubini s Theorem for triple integrals xyz dv B xyz 3 dzdy dx ( ) xyz 3 dz dydx [ xyz 3 ] z3 dy dx 3 z 9xy dydx [ 9xy ] y dx y 7x dx [ ] 7x We also wish to evaluate triple integrals over cylindrical solids that are centered around the z axis. The following diagram illustrates the solid V {(x, y,z) x + y a, m z l } that is bounded by the cylinder x + y a and the horizontal planes z l, z m. If we use the tranformation x r cosθ y r sinθ z z then the set V can be rewritten in cylindrical coordinates (r,θ,z) V {(r, θ,z) r a, θ π, m z l}.

58 58 Vector Calculus Theorem.96. If the function f(x, y,z) is continuous at each point in a cylindrical solid V {(x, y,z) x + y a, m z l } then by converting to cylindrical coordinates f(x, y, z)dv π a l f(r cosθ, r sinθ, z)rdzdrdθ. m V.7. Divergence theorem Theorem.97. (Divergence theorem) Let E be a solid region whose boundary surface S has a outward orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then F.ndS div(f)dv S E Note that the Divergence theorem states that a surface integral over a surface S that encloses a solid region E may be obtained by evaluating a triple integral the solid E. We illustrate the Divergence theorem by the following examples. Example.98. Verify the Divergence theorem for the vector field F F(x, y, z) 3xi +xyj + xzk and the solid region E that is the cube bounded by the six planes x y z x y z. Answer: We evaluate the surface integral (3xi+xyj +xzk).nds and the triple integral and verify that these are equal. The solid cube E S E div(3xi+xyj +xzk)dv. has boundary surface S consisting of the six truncated planes S,S, S 3,S 4, S 5 and S 6

59 .7 Triple integrals and Divergence theorem 59 Figure.. and each of these six planes has an orientation that points out of the solid E. We first evaluate the triple integral div(3xi +xyj +xzk)dv E [ 9 E E ( x (3x)+ y (xy)+ ) z (xz) dv (3 +x+x) dv (3 + 3x) dzdydx (by Theorem.94) ( [3z +3xz] z (3 +3x)dydx ( [3y +3xy] y (3 +3x) dx ] 3x+ 3x x x z ) dydx y ) dx Next we evaluate the surface integral of F over S. As the surface S consists of S, S, S 3, S 4, S 5 and S 6 we have F.ndS F.n ds + F.ndS + F.ndS S S S S 3 (.7) + F.n ds + F.n ds + F.ndSs S 4 S 5 S 6 and each S i F.n ds on the right hand side of (.7) is evaluated by first expressing S i as parametric surface r(u, v) and then using the definition of a surface integral F.ndS F (x(u, v), y(u,v),z(u, v)).(r u r v )da. (.8) S i Note that we should choose each parametrization r(u, v) of S i such that the normal vector r u r v has the same direction with the outward normal vector of S i as shown in Figure. (if the normal vector r u r v is in the opposite direction of the outward normal vector of S i then S i F.n ds on the right hand side of (.7) should be adjusted by a minus sign).

60 6 Vector Calculus S A parametrization for S is where is the rectangular region r(u,v)vi+ j + uk (u, v) Figure.. The vectors r u and r v are and r u x u i + y u j + z k i+j +k u r v x y i+ v v j + z k i+j +k v r u r v k i i j k j. Notice that the normal vector r u r v j points in the the same direction with the outward normal vector of S as shown in Figure.. Using (.8) (3xi +xyj +xzk).nds (3vi +vj +uvk).(r u r v ) da S (3vi +vj +uvk).(j)da v da v dv du (by Theorem.6) [ v ] du du and therefore S F.n ds

61 .7 Triple integrals and Divergence theorem 6 S A parametrization for S is r(u, v)ui+j + vk (u, v) (.9) where is the rectangular region shown in Figure.. Notice the difference with the parametrization for S the i and k components are exchanged. This causes the normal vector r u r v to point in the opposite direction r u x u i + y u j + z k i+j +k u and r v x y i+ v v j + z k i+j +k v r u r v i k i j k j. The parametrization (.9) is chosen so that the normal vector r u r v j points in the the same direction with the outward normal vector of S as shown in Figure.. Using (.8) and therefore (3xi+xyj + xzk).n ds (3ui+j +uvk).(r u r v )da S (3ui+j +uvk).( j)da da dv du S F.ndS S 3 A parametrization for S 3 is r(u,v)i+uj + vk (u, v) where is the rectangular region shown in Figure.. The vectors r u and r v are and r u x u i + y u j + z k i+ j +k u r v x y i+ v v j + z k i+j +k v r u r v j k i j k i.

62 6 Vector Calculus The surface integral is (3xi+xyj + xzk).n ds (3i +uj +vk).(r u r v )da S 3 (3i +uj +vk).(i)da 3dA 3dv du and therefore S 4 A parametrization for S 4 is 3 S 3 F.ndS 3 r(u, v)i+vj + uk (u,v) where is the rectangular region shown in Figure.. The vectors r u and r v are and r u x u i + y u j + z k i+j +k u r v x y i+ v v j + z k i+ j +k v r u r v i k i j k i. The surface integral is (3xi+xyj + xzk).n ds (i +j +k).(r u r v )da S 4 (i +j +k).( i)da da dv du and therefore S 5 A parametrization for S 5 is S 4 F.ndS r(u, v)ui+vj +k (u,v) where is the rectangular region shown in Figure.. The vectors r u and r v are r u x u i + y u j + z k i+j +k u r v x y i+ v v j + z k i+ j +k v

63 .7 Triple integrals and Divergence theorem 63 and r u r v i j i j k k. The surface integral is and therefore S 6 A parametrization for S 6 is (3xi +xyj +xzk).nds (3ui+uvj +uk).(r u r v )da S 5 (3ui+uvj +uk).(k)da u da u dv du S 5 F.ndS r(u, v)vi+uj +k (u,v) where is the rectangular region shown in Figure.. The vectors r u and r v are and r u x u i + y u j + z k i+ j +k u r v x y i+ v v j + z k i+j +k v r u r v j i i j k k. The surface integral is (3xi +xyj +xzk).nds (3vi +uvj + k).(r u r v )da S 6 (3vi +uvj + k).( k)da da dv du and therefore S 6 F.ndS

64 64 Vector Calculus So we have S F.n ds F.ndS + F.ndS + F.n ds S S S 3 + F.ndS + F.ndS + F.n ds S 4 S 5 S Hence we have 9. S F.n ds div(f)dv 9 E which verifies the Divergence theorem. One application of the Divergence theorem is in the evaluation of surface integrals F.n ds S where S is the boundary surface of a solid region E. As we saw in the previous example, the evaluation of such surface integrals can be tedious. It is sometimes easier to calculate the corresponding triple integral div(f)dv ; by the Divergence theorem E F.ndS div(f)dv this gives the answer of the required surface integral. S E Example.99. Use the Divergence Theorem to evaluate the surface integral (ye z i+ y j + cos(xy)k).n ds S where S is the surface of the solid bounded by the cylinder x + y 9 and the planes z and z. Answer: The closed cylinder S below encloses the solid region E {(x, y, z) x + y 9 and z } From the Divergence Theorem S F.ndS divf dv E

65 .7 Triple integrals and Divergence theorem 65 we have that our required surface integral I can be expressed as a triple integral ( I (ye z i+ y j + cos(xy)k).n ds S E x (yez )+ y (y )+ ) z (cos(xy)) dxdy dz y dxdydz From Theorem.96 we can convert to cylindrical coordinates I y dxdy dz E π 3 r sinθ rdzdrdθ π 3 [ r sinθz ] z z drdθ π 3 4r sinθdrdθ π [ 4r 3 ] r3 sinθ dθ 3 r π 36 sinθdθ and so the surface integral S E (ye z i+ y j + cos(xy)k).n ds.

66

67 Chapter Laplace transforms The Laplace transform changes a function y(t) in the t variable to a corresponding function Y (s) in the new independent variable s. The Laplace transform provides a method of solving differential and integral equations by converting these to algebraic expressions as we shall see in the following sections.. Definition and existence of Laplace transforms.. Improper integrals ecall that a definite integral has two finite limits of integration b f(t)dt. a If we require that one of these limits of integration be then we need an improper integral Definition.. The improper integral is defined as the limit a a g(t) dt (.) g(t)dt lim g(t)dt. (.) a The improper integral (.) is said to converge if the limit of (.) exists and is finite. Otherwise the improper integral is said to diverge. Example.. Determine if the following improper integrals converge or diverge i. e t dt ii. e t dt Answer: i. e t dtlim lim [ e t e t dt ] lim ( e e ) lim e lim e e lim e (as e is a constant) e (e as ) 67

68 68 Laplace transforms ii. and therefore e t dt converges. Note by defintion of the improper integral e t dt e. e t dtlim e t dt [ lim ] e t ( lim e e ) lim e lim e ( lim e ) e (as e is a constant) (e as ) and therefore e t dt diverges. The shaded areas of the graphs below represent the integrals of e t dt and e t dt respectively. Notice that the shaded area in the both graphs increase as t increases, however the rate at which the area in the first graph increases is much smaller than the rate at which the area in the second graph increases. This should suggest that e t dt diverges. Figure.. Notice in the above graph that e t tends to zero as t gets larger, in other words lim e t, t also notice that e t does not tend to zero as t gets larger From Example. we saw that seem that if then lim e t. t e t dt converges and e t dt diverges. Therefore it would lim g(t) t a g(t) dt diverges. This is in fact the case, we state this fact formally: Lemma.3. If lim t g(t) then the improper integral a g(t)dt diverges.

69 . Definition and existence of Laplace transforms 69 Proof. (Not required). The lim t g(t) is if, by definition, given any K > there exists a t value t such that g(t) K for all t >t. Fix such a K and t. Then ( t ) lim g(t)dt lim g(t)dt + g(t) dt a a t t g(t)dt + lim g(t) dt a t t g(t)dt + lim Kdt because g(t) K a t t g(t)dt + lim (K Kt ) a and hence a g(t)dt diverges... Definition and examples of Laplace tranform Definition.4. The Laplace transform of the function y(t), denoted by L[y(t)], is L[y(t)] e st y(t)dt (.3) and is defined for those values of s for which the improper integral (.3) converges. Note that L[y(t)] is a function of s and we denote L[y(t)] Y (s). Example.5. Use the definition of the Laplace transform to determine L[e kt ] where k is a real constant. and when s k we have L[e kt ] lim e st e kt dt e (s k)t dt e (s k)t dt lim e (s k)t dt lim lim dt (.4) (.5) and therefore L[e kt ] does not exist for s k. Now consider when s k [ ] lim e (s k)t e (s k)t dt lim (s k) ( lim s k e (s k) s k s k lim e (s k) s k )

70 7 Laplace transforms From the graphs of Figure. we see that and in general lim e and lim e { lim e a if a< if a > (.6) that is, lim e a depends only on the sign of a. Hence and therefore { lim e (s k) if (s k)< if (s k)> L[e kt ] s k lim e (s k) s) k s k ( s k lim (.7) e (s k) (.8) and from (.7) the limit in (.8) is zero only when s > k. So we have and this holds only when s >k. L[e kt ] s k (.9) Note by setting k in (.9) we get L[] s when s >. (.) Example.6. Use the definition of the Laplace transform to show that when s >. L[sinkt] k s +k Answer: L[sinkt] lim e st sinktdt e st sinktdt. Using the integration by parts formula with gives a b v du [uv] b b a u dv a v sinkt du e st dt dv k coskt dt u e st s [ e e st sinktdt st ] sinkt + k e st coskt dt (.) s s Integrate e st coskt by parts again with to get v coskt du e st dt dv k sinkt dt u e st s [ e e st cosktdt st ] coskt k e st sinktdt. (.) s s

71 . Definition and existence of Laplace transforms 7 Substituting (.) into (.) gives [ e e st sinktdt st sinkt s ] k s and solving for e st sinktdt in (.3) gives s +k s [ e st coskt s ] k s e st sinktdt [ e st sinkt ] k [ s e st s coskt ] and evaluating the right side of (.4) using the limits of integration gives s +k s e st sinktdt. (.3) (.4) e st sinktdt s e s sink k s e s cosk + k s (.5) and taking the limit lim of both sides of (.5) we have ( ) s +k s e st sinktdt lim Notice as sink implies that and taking limits gives ( lim s e s sink ks e s cosk + ks ). (.6) e s e s sink e s ( lim e s lim e s sink ) lim e s (.7) Now from equation (.6) above if s > (and only if s >) then lim e s. So when s > the inequality (.7) simplifies to ( lim e s sink ) and therefore Similarly one can show when s > that lim e s sink. (.8) lim e s cosk, (.9) and substituting (.8) and (.9) into the right side of (.6) we have ( ) s +k s lim e st sinktdt k s ( ) e st sinktdt k s +k when s >. lim L[sinkt] k s +k Example.7. Use the definition of the Laplace transform to show that when n,,, when s >. L[t n ] n! s n+ (.)

72 7 Laplace transforms Answer: From (.) we know that L[] s when s > L[t ]! s + when s > and so (.) holds when n. We therefore need to consider when n,,3, L[t n ] e st t n dt lim e st t n dt. Using the integration by parts formula with gives v t n dv nt n dt [ e e st t n dt st t n s ] du e st dt u e st s + n e st t n dt. (.) s Taking the limit lim of both sides of (.) we have ( ) ( [ e e st t n dt lim st t n ] ) ( + n s s lim lim e st t n dt ) and by using the definition of the Laplace transforms L[t n ] and L[t n ] we have ( [ e L[t n ] lim st t n ] ) + n s s L[tn ]. (.) Now lim ( [ e st t n s ] ) ( lim e s n s ( ) lim n se s and this limit is of the indeterminate form. Applying L Hospitals rule by differentiating the numerator and denominator n times with respect to, we have and therefore lim lim ( ( n se s [ e st t n s Substituting (.3) into (.) gives the formula ) lim n! s n+ e s when s > ) ] ) when s >. (.3) L[t n ] n s L[tn ] valid for n,, 3, and s >. (.4) Notice that by replacing n by n in (.4) we have L[t n ] n L[t n ] (.5) s

73 . Definition and existence of Laplace transforms 73 and similarly by replacing n with n, n 3,,, in (.4) we have, respectively, L[t n ] n L[t n 3 ] s L[t n 3 ] n 3 L[t n 4 ] s L[t ] s L[t ] (.6) L[t] s L[t ] Finally we get (.) by using the equations (.5) and (.6) to express L[t n ] in terms of L[t ] L[] s : which is our desired result. L[t n ] n s L[tn ] from (.4) n ( ) n L[t n ] s s n ( ) n n L[t n 3 ] s s s n n n s s s s L[t ] n n s s s n s n! s n L[t ] n! s n+ s L[t ] from (.5)..3 Existence of Laplace transform In Examples.5,.6 and.7 we were able to determine the Laplace transforms of the functions e kt, sin kt and t n respectively. We also showed that the Laplace tranforms of those functions existed for only certain value of s: y(t) L[y(t)] Laplace tranform exists when e kt s >k s k k sinkt s + k s > t n n! s n+ s > We now give an example of a function that does not have a Laplace tranform for any value of s. Example.8. Show that the Laplace transform of e t does not exist for any real value of s. Answer: ecall that from Definition.4 L[e t ], given by: L[e t ] e st e t dt (.7) is defined only for those values of s for which the improper integral on the right converges. Fix any value of s. If we can show that for this value of s lim e st e t t

74 74 Laplace transforms then by Lemma.3 the improper integral e st e t dt (.8) does not converge for this fixed value of s. And if (.8) does not converge for any such fixed s then L[e t ] does not exist for any s. So we consider (.7) lim e st e t lim e t st t t and for fixed s, t st is an increasing quadratic function in t which implies t st goes to infinity as t and therefore which is our desired result. lim e st e t lim e t st (.9) t t In Example.9 we saw that the Laplace transform of e t did not exist because for any value of s. Equation (.3) is equivalent to lim e t st. (.3) t e t lim t e st which indicates that the function e t is much larger than e st for large values of t as we can see in the following graph when s. 6 5 e t 4 3 e t.5.5 Example.9 indicates that a function such as e t that is much larger than the standard exponential function e st as t does not have a Laplace transform. On the other hand functions that are comparable (in the sense that they are less than or equal to) to exponential functions will have Laplace transforms. as we shall see in Theorem.5. The precise phrase for a function being comparable to an exponential function is that the function is of exponential order. Definition.9. A function y(t) is of exponential order if there exist constants A, b and a t value t such that y(t) <Ae bt when t t. Example.. Show that y(t)t 3 is of exponential order... Provided such functions satisfy certain continuity conditions.

75 . Definition and existence of Laplace transforms 75 ecall that e t has a Taylor series expansion about t, that is and clearly when t e t m t m m! +t+ t + t3 6 + t4 4 + ( ) t 3 < 6 +t+ t + t3 6 + t e t and as t 3 t 3 when t we have t 3 <6e t when t and hence t 3 is of exponential order where the constants of Definition.9 are A 6, b andt. We shall need the following definition Definition.. A function y(t) is piecewise continuous on a finite interval a t b if on this interval i. y(t) is defined and takes finite value at each point and ii. y(t) only has a finite number of discontinuities. Example.. The function t t < y(t) t t<3 t 3 t 4 is piecewise continuous on t 4 as we can see from its graph that y(t) has one point of discontinuity at t and is defined and finite everywhere on t 4. Example.3. Find 4 y(t)dt of the function y(t) defined in Example.. The integral of a piecewise defined function is computed by adding the integrals of each of the intervals y(t) dt y(t)dt + y(t)dt + y(t) dt 3 3 t dt+ t 4 dt + ( t)dt

76 76 Laplace transforms To find the Laplace transform of a piecewise defined function y(t) such as the one defined in Example., we use the same procedure of adding the integral over each interval as done in Example.3 : Example.4. Sketch the graph of the following function t < y(t) t t< t and determine L[y(t)]. Answer: The graph of y(t) is By definition 4 3 L[y(t)] e st y(t)dt Now as y(t) is a piecewise-defined defined function, it follows that e st y(t) is also piecewise-defined: t< e st y(t) e t t< t and e st y(t)dt e st y(t)dt + e st y(t)dt + e st y(t)dt dt + e st y(t)dt + dt + e st tdt+ e st tdt and using integration by parts with we have and therefore v t du e st dt dv dt u e st s [ ] e st t dt e st t + e st dt s s [ ] e st [ ] t + e st s s ( s + ) ( e s s s + ) e s s ( L[y(t)] s + ) ( e s s s + ) e s. s

77 . Definition and existence of Laplace transforms 77 We now state a theorem that guarantees the existence of the Laplace transform of functions that are of exponential order and that are piecewise continuous. Theorem.5. (Existence) Let y(t) be a function such that y(t) <Ae bt when t t for some constants A, b and t (that is, y(t) is of exponential order). Also let y(t) be piecewise continuous on any finite interval c t c where c, c. Then the Laplace transform of y(t) exists for s >b. Proof. Not required. Example.6. Show that the following functions are of exponential order and hence use Theorem.5 to prove that their Laplace transforms exist.. i. e 5t ii. sint t < iii. y(t) t t < t Answer: i. e 5t <.e 5t for t and so the constants of Definition.9 are A, b 5 and t. e 5t is a function that is continuous everywhere (see the graph on Figure.) and is hence continuous on every finite interval and is therefore piecewise continuous on every finite interval. So by Theorem.5 the Laplace transform of e 5t exists for s > 5, a fact we already know from Example.5. ii. sint < e t for t and so A, b and t. sin t is continuous everywhere and is therefore piecewise continuous on every finite interval. So by Theorem.5 the Laplace transform of sint exists for s >, which we already know from Example.6. iii. From the graph of y(t) in Example.4 we see that y(t) < for t and therefore y(t) < e t for t and so A, b and t. If we consider any finite interval c t c where c, c then y(t) has at most two points of discontinuity in the interval c t c depending on whether or not x, lie in the interval c t c. Hence y(t) is piecewise continuous on every finite interval. By Theorem.5 the Laplace transform of y(t) exists for s >. emark.7. It is possible to obtain the Laplace transform of functions that do not satify the hypotheses of Theorem.5, for example.3 and the Dirac delta function considered later. The t main use of Theorem.5 is to guarantee the existence of the Laplace transform of standard functions such as e kt, cos kt, sinh kt, t 5 + 4t 3 3t + and piecewise functions such as those in Examples. and.4... Assume that we do not know the Laplace transforms of these functions have not been determined in Examples.5,.6 and.4. Note we do not actually find Laplace transforms by Theorem.5, we just show that the transform exists..3. See Kreyszig [] Section 6.

78 78 Laplace transforms. Properties of Laplace transforms.. Linearity property The Laplace transform has the linearity property Theorem.8. Let y (t) and y (t) be functions whose Laplace transforms exist and let c and c be scalars. Then L[c y (t)+c y (t)]c L[y (t)] +c L[y (t)]. Proof. Using the definition of the Laplace transform ( L[c y (t)+c y (t)] e st c y (t) +c y )dt c e st y (t)dt +c e st y (t)dt c L[y (t)]+c L[y (t)] Example.9. Use the linearity property of the Laplace transform to determine i. L[t +3t +] ii. L[coshkt] wherek is a real constant. Answer: i. L[t +3t +] L[t ] +3L[t] +L[] ( ( ) )+ s 3 +3 s s s s + s ii. By definition coshkt e kt + e kt therefore [ e L[coshkt]L kt +e kt ] L[e kt ]+ L[ekt ] ( ) ( ) + s +k s k ( ) (s k)+(s +k) (s + k)(s k) s s k.. The inverse Laplace transform In the above sections, we saw that the Laplace transform changes a function y(t) in the t variable to a function L[y(t)]Y (s) in a new independent variable s. We now consider the reverse problem Definition.. Given some function Y (s), if there exists a function y(t) such that L[y(t)] Y (s) then we call y(t) the inverse Laplace transform of Y (s) and write L [Y (s)] y(t). Example.. Find the inverse Laplace transform of Y (s) s +4.

79 . Properties of Laplace transforms 79 Answer: This is done by recognizing the form of the function Y (s). We recognize that and therefore L[sinkt] k s +k L[sint] s +4 hence by the definition of the inverse Laplace transform [ ] L s sint. +4 A useful fact is that the inverse Laplace transform has the linearity property Lemma.. Let Y (s) and Y (s) be functions whose inverse Laplace transforms exist. Then for any scalars c,c L [c Y (s)+c Y (s)]c L [Y (s)]+c L [Y (s)]. Proof. Let the inverse Laplace transforms of Y (s) andy (s) be y (t) and y (t) respectively, that is L [Y (s)] y (t), L [Y (s)] y (t). (.3) Then by the definition of the Laplace inverse transform and using the linearity property of the Laplace transform By substituting (.3) we get and using the definition of the inverse transform Finally, using (.3) we have L[y (t)] Y (s), L[y (t)] Y (s) (.3) L[c y (t)+c y (t)]c L[y (t)] +c L[y (t)]. L[c y (t)+c y (t)] c Y (s)+c Y (s) c y (t)+c y (t) L [c Y (s)+c Y (s)]. c L [Y (s)]+c L [Y (s)] L [c Y (s)+c Y (s)]. Example.3. Find the inverse Laplace transform of Y (s) s s s 4 s + 5. By the linearity property of the inverse Laplace transform [ L [Y (s)]l s s 9 + s + ] 4 s + 5 [ [ [ ] (.33) L s ]+L s ]+L s 9 4 s + ( 5 ) and we showed in previous examples that y(t) cosh kt L[y(t)] s s k t n n! s n+ sinkt k s + k (.34)

80 8 Laplace transforms We use the linearity property again to rewrite the constants of (.33) [ [ [ ] L [Y (s)]l s ]+L s ]+L s 9 4 s +( 5 ) [ ] L s + [ ] [ 3! s 3 3! L + ] L 5 s 4 5 s + ( 5 ) so that the functions inside the square brackets of L [ table (.34) and hence we have ] match those in the right column of the L [Y (s)] cosh3t + 3 t3 + sin( 5 t). 5 The following well-known result on partial fractions is useful when finding inverse transforms: Lemma.4. (Partial Fraction Expansion)Let p(s) q(s) denominator q(s) factors as be a ratio of polynomials with real coefficients. If the q(s)k(s a ) n (s a ) n (s a k ) nk (s +b s +c ) m (s +b j s +c j ) mj where the quadratic factors are irreducible then the following identity holds p(s) q(s) A, s a + A, (s a ) + + A,n (s a ) n + A, s a + A, (s a ) + + A,n (s a ) n + A k, s a k + A k, (s a k ) + + A,n k (s a k ) nk + B,s +C, s + b s +c + B,s +C, (s +b s +c ) + B,m s +C,m (s + b s+ c ) m + B j,s +C j, s + B j,s +C j, +b j s +c j (s +b j s +c j ) + B j,m j s +C j,mj (s +b j s +c j ) mj + (s) (.35) where (s) is a polynomial that is equal to zero if the degree of the polynomial p(s) is less than the degree of the polynomial q(s) and A r,s, B r,s andc r,s are real constants for all subscripts r,s. Proof. Not required. Example.5. The partial fraction expansion of s + 5s + (s )(s +) (s 5) 3 (s +)(s +9) takes the form s +5s + (s )(s +) (s 5) 3 (s + )(s + 9) A s + B s + + C (s + ) + D s 5 + E (s 5) + F (s 5) 3 + Gs+ H s + Is +J Ks +L + s + +9 (s +9) where A, B, C,D,E,F,G, H,I,J,K and L are constants to be determined.

81 . Properties of Laplace transforms 8 Example.6. Find the inverse Laplace transform of Y (s) s s+9 s 3 +9s. as the Laplace transform of a func- As it is, we do not immediately recognize the expression s s+9 s 3 + 9s tion that we know. So we use partial fractions to write s s +9 s 3 +9s do recognize. We first factorize the denominator s s +9 s 3 + 9s s s +9 s(s +9) then by the formula (.35) the partial fraction expansion in this case is s s +9 s(s +9) A Bs +C + s s + 9 as a sum of expressons in s that we where A, B and C need to be determined. We do this by first clearing denominators s s+9 A(s +9) +s(bs +C). (.36) We need to find three constants (namely A, B and C) so shall need three equations with variables A,B andc which may be obtained by either substituting values for example putting s into (.36) gives 9 A( +9) +(B. +C) A comparing coefficients of like powers rearranging the right side of (.36) gives and comparing the coefficients of s in (.37) gives while comparing the coefficients of s gives s s+9 (A + B)s +Cs +9A (.37) A +B, C and therefore we have three equations in the variables A, B and C A A +B C and solving these give A, B and C. Hence [ s L [Y (s)]l ] s +9 s(s +9) [ L s ] s +9 [ [ ] L ] L s s + 9 [ L ] 3 [ ] 3 s L s +9 which is our answer. 3 sin(3t)

82 8 Laplace transforms..3 Shifting the s variable; shifting the t variable Given a function f(x), recall that the graph of the function f(x a) is shifted horizontally by a units. If a > then this horizontal shift is to the right; else if a < the shift is to the left. The following graph is an example of a shift to the right when f(x)sin(x) and a..5 sin(x) sin(x-) We consider shifting in the context of Laplace transforms. ecall that the Laplace transform changes functions y(t) in the t variable to corresponding functions Y (s) in the variable s so we can shift the s variable or shift with respect to the t variable. Shifting the s-variable Let Y (s) be a function of s with inverse Laplace transform L [Y (s)] y(t). If we shift Y (s) then we obtain a new function Y (s a) in the variable s. The inverse Laplace transform of this new function Y (s a) is related to the inverse transform of Y (s) by the formula We state and prove this result in the following L [Y (s a)] e at L [Y (s)]. (.38) Lemma.7. Let Y (s) have inverse Laplace transform L [Y (s)], and let Y (s a) be the function obtained by shifting Y (s). Then Proof. Let L [Y (s)] y(t). Then L[e at y(t)] L [Y (s a)] e at L [Y (s)]. e st( e at y(t))dt e (s a)t y(t)dt (.39) Y (s)l[y(t)] and by the definition of the Laplace transform Y (s) e st y(t)dt. (.4) eplacing s by s a on both sides of (.4) gives Y (s a) e (s a)t y(t)dt. (.4) But this is identical to the second integral of (.39) and therefore L[e at y(t)]y (s a) (.4)

83 . Properties of Laplace transforms 83 which implies L [Y (s a)] e at y(t) L [Y (s a)]e at L [Y (s)]. Equation (.4) above will also be useful so we state it in the following Corollary.8. If the function y(t) has Laplace transform Y (s) then L[e at y(t)] Y (s a). Lemma.7 and Corollary.8 are particularly useful in finding inverse Laplace transforms. Example.9. [ ] i. Find L s s + [ ] ii. Find L whenn,, (s +a) n+ 3 Answer: i. First complete the square of the quadratic.4 We need to determine s s + (s ) +9. [ [ ] L ]L s s + (s ). +9 [ ] 3 3 L (s ) +9 Notice that this is similar to shifting this by using (.4) we have and therefore ii. ecall from Example.7 From formula (.4) with a replaced by a L[sin3t] 3 s + 9, L[e t 3 sin3t] (s ) +9 [ L ] 3 [ ] 3 s s + L (s ) et sin3t. L[t n ] n! whenn,,3 sn+ L[e at t n ] n! (s +a) n+ and from the linearity property of Laplace transforms therefore L[ n! e at t n ] n! L[e at t n ] (s +a) n+ [ ] L (s+ a) n+ n! e at t n..4. The formula for completing the square is ax +bx+c a(x+ b a ) +c b 4a

84 84 Laplace transforms Shifting the t-variable When shifting the t variable, a function that is useful is the Heaviside step function. Definition.3. The Heaviside step function H(t t ) is defined as { when t<t H(t t ). when t t t t Figure.. Graph of H(t t ) Notice that H(t t ) is a piecewise defined function and using the definition of the Laplace transform as in Example.4 we can determine L[H(t t )]. Lemma.3. The Laplace transform of the Heaviside step function is L[H(t t )] e st. s Proof. L[H(t t )] e st H(t t )dt t e st ()dt + e st ()dt t from the definition of H(t t ). Therefore L[H(t t )] e st dt t lim e st dt t [ ] e st lim s t ( e st lim s s as lim e s. e st s e st s lim e s e s s )

85 . Properties of Laplace transforms 85 Example.3. Sketch the graphs of the following functions i. y (t)sint when t ii. y (t)sin(t ) when t iii. y 3 (t)h(t )sin(t ) when t Answer: The graphs of i) and ii) are y (t) y (t) The graph of y (t) is obtained from y (t) by shifting units to the right. Notice that y (t) is not defined for t <. y 3 (t) is the product of functions and we use the following table to determine its graph t interval H(t ) H(t )sin(t ) t< t sin(t ) so we see that { when t < H(t )sin(t ) sin(t ) when t and therefore the graph of y 3 (t) is obtained y (t) by shifting units to the right and taking the graph to be zero on the interval t <. Notice that, unlike y (t), y 3 (t) is defined for t < and the answer for part iii) is y 3 (t) We saw in Example.3, that H(t )sin(t ) is a t-shift of sin(t ). Below we give a result that expresses the Laplace transform of a shifted function H(t t )y(t t ) in terms of the Laplace transform of the original function Theorem.33. Let L[y(t)] Y (s). Then L[H(t t )y(t t )] e st Y (s). Proof. L[H(t t )y(t t )] e st H(t t )y(t t )dt t e st ()y(t t )dt + e st ()y(t t )dt t

86 86 Laplace transforms from the definition of H(t t ) in Definition.3. Therefore L[H(t t )y(t t )] e st y(t t )dt. Now use the change of variables T t t in the integral on the right to get L[H(t t )y(t t )] e s(t +t) y(t)dt e st e st y(t)dt But T is a dummy variable in the above integral that is, the integral takes the same value regardless of the name of the independent variable: e st y(t)dt e st y(t)dt. Therefore which is the desired result. L[H(t t )y(t t )]e st t e st L[y(t)] e st Y (s) e st y(t)dt Using the definition of the inverse Laplace transform and Theorem.33 we clearly have Corollary.34. Let L[y(t)] Y (s). Then Example.35. Sketch the graph of L [ e 3s Answer: Notice that and by Corollary.34 L [e st Y (s)]h(t t )y(t t ). s + ]. e 3s s + is of the form e st Y (s) with e st e 3s and Y (s) s +. Now Y (s) s + L [ e 3s s + y(t)l [Y (s)]sin(t) ] H(t 3)y(t 3) H(t 3)sin(t 3) Using the definition of H(t t ) in Definition.3 { when t <3 H(t 3)sin(t 3) sin(t 3) when t 3 and following Example.3 the graph of H(t 3)sin(t 3) is t -

87 . Properties of Laplace transforms Laplace transform of derivatives The Laplace transform of the derivative y (t) can be expressed in terms of the Laplace transform of the original function y(t). This is essentially done by using integration by parts L[y (t)] e st y (t)dt lim e st y (t)dt now let v e st du y (t)dt and so dv se st dt u y(t) L[y (t)] lim e st y (t)dt ( [e lim y(t) ] ) + s e st y(t)dt ( lim e s y() e y() ) (.43) +s lim e st y(t)dt ( ) lim e s y() y()+sl[y(t)]. and provided lim e s y() (as will be the case if y(t) is of exponential order) then we have that L[y (t)]sl[y(t)] y() (.44) which expresses the Laplace transform of y (t) in terms of the Laplace transform of y(t). Note that the argument (.43) is not a proof, we have ignored some continuity issues. The full proof, given in amkissoon [], is not required for this course. We give a statement of the hypotheses required for (.44) in the following theorem. Theorem.36. If y(t) is continuous and of exponential order and y (t) is piecewise continuous on every finite interval then L[y (t)] sl[y(t)] y(). Proof. Not required. By two applications of Theorem.36 we can determine the Laplace transform of y (t) in terms of the transform of y(t): [ (y L[y (t)] L (t)) ] sl[y (t)] y () using (.44) with y replaced by y s(sl[y(t)] y()) y () using (.44) as it is which gives us L[y (t)] s L[y(t)] sy() y () (.45) Similarly, one can apply Theorem.36 n times to obtain the Laplace transform of the n th derivative y (n) (t) in terms of the transform of y(t): L[y (n) (t)] s n L[y(t)] s n y() s n y () s n 3 y () sy (n ) () y (n ) () (.46) The formulae (.45) and (.46) require certain continuity conditions to hold. We give a statement of these conditions in the following theorem.

88 88 Laplace transforms Theorem.37. If y(t), y (t), y (t),, y (n ) (t) are continuous and of exponential order and y (n) (t) is piecewise continuous on every finite interval then L[y (n) (t)] s n L[y(t)] s n y() s n y () s n 3 y () sy (n ) () y (n ) (). Proof. Not required. Example.38. Use the formula for the Laplace transform of a derivative to determine i. L[cos(kt)] ii. L[sinh(kt)] Answer: i. Let y(t)sin(kt). Then y (t) k cos(kt) and applying Theorem.36 we have L[y (t)]sl[y(t)] y() L[k cos(kt)] sl[sin(kt)] sin() kl[ cos(kt)] sl[sin(kt)] sin() where we used the linearity property of Theorem.8 to move the k outside of the Laplace transform. Then k kl[ cos(kt)]s s + k as we know L[sinkt] k from Example.6. Therefore s +k L[ cos(kt)] s s +k ii. ecall that the hyperbolic cosine cosh(t) and hyperbolic sine sinh(t) are defined as coshkt e kt +e kt sinhkt e kt e kt and it is easy to check from these definitions that (coshkt) k sinhkt, (sinh kt) k coshkt cosh(), sinh() (.47) Let y(t)coshkt. Then y (t)k sinh(kt) and applying Theorem.36 we have L[k sinh(kt)] sl[cosh kt] cosh() and as we know L[coshkt] s from Example.9 s k ( s kl[sinh(kt)]s s k and so kl[sinh(kt)] s (s k ) s k L[sinh(kt)] k s k. )

89 .3 Applications and more properties of Laplace transforms 89.3 Applications and more properties of Laplace transforms.3. Solving differential equations using Laplace transforms We can use the results of the previous sections to solve differential equations, in particular initial value problems, that is, differential equations with specified initial values. We illustrate the application of Laplace transforms to initial value problems in the following examples. Example.39. Solve the differential equation subject to the initial conditions y(), y (). y 3y +y e 3t (.48) Answer: Take the Laplace transform of both sides of the given differential equation L[y 3y +y]l[e 3t ] L[y ] 3L[y ]+L[y] s 3 Using formulae (.44), (.45) for the transform of a derivative we have s L[y(t)] sy() y () 3(sL[y(t)] y()) +L[y(t)] s 3 Substituting the given initial conditions y(), y ()gives For convenience let L[y(t)]Y (s) Now solve for Y (s) s L[y(t)] 3sL[y(t)] +L[y(t)] s 3 s Y (s) 3sY (s) +Y (s) s 3 (s 3s +)Y (s) s 3 Y (s) (s 3s + )(s 3) (.49) Our required answer is y(t) L [Y (s)]. We use the partial fraction expansion method (as in Example.6) to find the inverse transform of (.49). We first factorize the denominator of Y (s) (s 3s +)(s 3) (s )(s )(s 3) then by the formula (.35) the partial fraction expansion is (s )(s )(s 3) A s + B s + C s 3 where A, B and C need to be determined. Clear denominators A(s )(s 3)+B(s )(s 3)+C(s )(s ) (.5) In this case it is easiest to substitute values into (.5) in order to determine A, B and C. Substituting values s,s and s 3 into (.5) gives

90 9 Laplace transforms s A( )( )+B()( )+C()( ) A s A()( )+B()( )+C()() B s 3 A()()+B()()+C()() C Therefore [ ] y(t)l [Y (s)]l (s 3s +)(s 3) [ ] L (s )(s )(s 3) [ ] L s + ( ) s + s 3 L [ s ] L [ s ] + L [ s 3 ] et e t + e3t and so the solution to this initial value problem is y(t) et e t + e3t It is possible to solve the initial value problem of Example.39 by other methods of solving differential equations such as D-operators and power series. One value of using Laplace transforms has over these other methods is that the inhomogeneous term, usually on the right side of the differential equation (for example the e 3t in equation (.48)) can be a piecewise defined function as we see in the following example. Example.4. Solve the differential equation x (t) +x(t) F(t) subject to the initial conditions x(),x () and where F(t) is the piecewise defined function when t< π F(t) when t π Answer: Take the Laplace transform of both sides of the given differential equation and so we need to determine L[F(t)]. The graph of F(t) is L[x +x] L[F(t)] (.5)

91 .3 Applications and more properties of Laplace transforms 9 t pi/ One way to determine the Laplace transform of a piecewise defined function is to do so from first principles, that is, using the definition of the Laplace transform as in Example.4. Another way is to use Heaviside functions. Consider the following table t interval H(t π ) H(t π ) F(t) t < π and therefore t π F(t) H(t π ) L[F(t)]L [ H(t π ] [ ) L[] L H(t π ] ) s e sπ s using Lemma.3. Substituting L[F(t)] into (.5) we have L[x +x] s L[x ]+L[x] s sπ e s sπ e s Using formula (.45) for the transform of the second derivative we have s L[x(t)] sx() x ()+L[x(t)] s Substituting the given initial conditions x(), x ()gives For convenience let L[x(t)] X(s) and solve for X(s) s L[x(t)] +L[x(t)] s s X(s) +X(s) At this point we use partial fractions to expand e sπ s e sπ s sπ (s e +)X(s) + s sπ e X(s) s(s +) + s + e sπ s(s + ) e sπ s (.5) (.53)

92 9 Laplace transforms In this case it may be best to write (.53) as ( s(s e sπ ) +) and apply partial fractions to (.54) s(s + ) : s(s + ) A s + Bs+ C s + A(s + )+s(bs +C) (A + B)s + Cs + A and by comparing coefficients of, s and s we obtain, respectively, three equations and so and therefore ( s(s e sπ + ) A C A +B s(s +) s ) ( s s s +. s s + )( s s s + e s e sπ sπ + ) se sπ s + and this is the required partial fraction expansion of (.54). Subtituting this into X(s) in (.5) we have sπ sπ X(s) s + s + s s + e se + s s + and therefore x(t)l [X(s)]L s + s + s s + e + s [ [ [ ] L ]+L ] s s L s + s L e sπ +L se sπ + s s + sπ s e sπ s + +sint cost H(t π )+H(t π )cos(t π ) and so the solution to this initial value problem is x(t) +sint cost H(t π )+H(t π )cos(t π ). (.55) So when the inhomogeneous part F(t) of our differential equation is piecewise defined, we obtain a solution to the differential equation that is also piecewise defined as we can see by using the definition of the Heaviside function to write (.55) as + sint cost when t < π x(t) + sint cost + cos(t π ) when t π It is interesting to note that while the inhomogeneous part F(t) of the differential equation of Example.4 is not continuous at t π (there is a break in the graph of F(t) at t π ), the solution x(t) given in (.55) is actually continuous and in fact differentiable at t π as we can see from the graph of x(t)

93 .3 Applications and more properties of Laplace transforms 93 x(t) - pi/ t - where we have drawn the interval t < π in solid and the interval t π in dashes. Notice both pieces meet smoothly at t π. Also notice from the graph that x(t) satisfies the initial conditions x(), x ()..3. Solving simultaneous linear differential equations using the Laplace transform We briefly discuss an example that motivates how simultaneous differential equations arise. This example serves only to help illustrate simultaneous differential equations and its details are not required for this course. Example.4. Consider a farm with mice and cats. The mouse population changes with time t and denote this population by M(t). Similarly denote the cat population by C(t). Now the rate M (t) at which the mouse population changes at time t depends on two things the number of mice alive at time t (that is M(t)), as more mice present means a greater mouse reproduction rate. the number of encounters between cats and mice as the population of mice decreases at these encounters. For simplicity we assume the number of encounters is proportional to the product M(t)C(t). Hence M (t) is proportional to M(t) and M(t)C(t) and we have the differential equation M (t)am(t) bm(t)c(t) where a and b are constants such that a, b. Notice that this differs from the differential equations we have considered so far as there are two dependent variables M, C and one independent variable t. Similarly the rate C (t) at which the cat population changes depends on the number of encounters between cats and mice as encounters means food for the cats and therefore an increased survival rate C (t) of cats the more cats present means less food for each individual cat and therefore an increased death rate C (t) of cats and we have the differential equation where d,e. The two equations C (t) dc(t)+em(t)c(t) M (t)am(t) bm(t)c(t) C (t) dc(t) +em(t)c(t) form an example of simultaneous differential equations in which the dependent variables interact This particular system of differential equations is called a Lotka-Volterra system and is used in predator-prey models.

94 94 Laplace transforms We now illustrate the solution of simultaneous linear differential equations by using the Laplace transform. Example.4. Use the Laplace transform to solve the following simultaneous linear differential equations y y + y y y y subject to the initial conditions y () and y (). Answer: Take the Laplace transform of both equations L[y ] L[y ]+L[y ] L[y ] L[y ] L[y ] Using formulae (.44) for the transform of a derivative we have sl[y ] y () L[y ]+L[y ] sl[y ] y () L[y ] L[y ] Substituting the initial conditions y () and y ()gives sl[y ] L[y ] +L[y ] sl[y ] L[y ] L[y ] For convenience let L[y (t)]y (s) and L[y (t)] Y (s) and we can rewrite these two equations as sy (s) Y (s)+y (s) sy (s) Y (s) Y (s) We eliminate Y (s). Multiplying equation (.56) by s+ we have and adding equations (.57) and (.58) gives From equation (.57) we know ecall that Shifting these along the s-axis gives and so (s +)Y (s)+y (s) (.56) Y (s) (s +)Y (s) (.57) (s +) (s+ ) Y (s)+(s + )Y (s) (.58) L[e t cost] (s +) ((s + ) +)Y (s) s + Y (s) (s +) + L[cost] Y (s) Y (s) s + Y (s) (s + ) + s s + s + (s +) + L[sint] s + L[e t sint] (s +) + y (t)l [Y (s)] y (t) L [Y (s)] [ ] [ ] L s + (s +) L + (s +) + e t cost e t sint

95 .3 Applications and more properties of Laplace transforms 95 and the solution to our simultaneous linear differential equations is y (t) e t cost y (t) e t sint..3.3 Convolution and Integral equations ecall from Lemma. that the inverse Laplace transform of a sum is the sum of inverse transforms L [Y (s) +Y (s)]l [Y (s)] +L [Y (s)]. However, a similar statement does NOT hold for products. In general L [Y (s)y (s)] L [Y (s)]l [Y (s)] (.59) as the following example demonstrates: consider Y (s) s and Y (s) s. Then ( ) L [Y (s)y (s)] L s s [ ] L s and clearly this is a case of (.59). L [Y (s)]l [Y (s)]l ( s. t ) ( ) L s In order to obtain a formula for L [Y (s)y (s)] where Y (s) and Y (s) are Laplace transforms, we shall need the following Definition.43. The convolution of f(t) and g(t) is denoted as f g and is defined as t (f g)(t) f(t β)g(β)dβ. Example.44. Find the convolution f g when f(t) t and g(t) t, that is, find t t. Answer: eplacing the variable t with t β in f(t) t gives and similarly we obtain g(β) β. Then f(t β)(t β) and so (f g)(t) t t t (t β) βdβ t (t tβ + β )β dβ t (t β tβ + β 3 ) dβ [ t β ] tβ3 3 + β4 t 4 t4 t4 3 + t4 4 t4 t t t4. Notice from Definition.43 and Example.44 that if f, g are functions of t then f g is again a function of t. We now give a formula for L [Y (s)y (s)] in the theorem below.

96 96 Laplace transforms Theorem.45. (Convolution Theorem) If Y (s)l[y (t)] and Y (s) L[y (t)] then or equivalently L [Y (s)y (s)] y y L[y y ] L[y (t)]l[y (t)]. Proof. Not required. Example.46. Find L [ s 5 ] by using the Convolution Theorem. Answer: We can split s 5 as a product in several ways. Let us choose one such way and using the formula L[t n ] n! s n+ Y (s) s 3 Y (s) s. y (t)l [Y (s)]t Then using the Convolution Theorem and from Example.44 therefore L [ s 5 ] L [ s 3 s y (t)l [Y (s)]t ] L [Y (s)y (s)] y y t t t t t4 L [ s 5 ] t4. [ Of course, we did not need to use the Convolution Theorem to determine L ], s 5 we can use the formula L[t n ] n! sn+ directly. On the other hand, examples in which the Convolution Theorem is needed are given by integral equations. Definition.47. An equation which contains the dependent unknown variable under an integral operator is called an integral equation. We will be interested in those integral equations that can be solved by using Laplace tranforms and the Convolution theorem. Example.48. Use the Laplace transform to solve the integral equation t y(t)4t 3 sin(t β)y(β)dβ. Answer: Take the Laplace transform of both sides of the integral equation [ t ] L[y(t)] L[4t] 3L sin(t β)y(β)dβ. Notice that the integral t sin(t β)y(β) dβ (.6)

97 .3 Applications and more properties of Laplace transforms 97 is in the form of a convolution, that is t sin(t β)y(β)dβ sint y(t) and from the Convolution Theorem [ ] t L sin(t β)y(β)dβ L[sint y(t)] L[sint]L[y(t)]. (.6) Substituting (.6) into (.6) we have and denoting L[y(t)] Y (s) and using partial fractions L[y(t)] L[4t] 3L[sin t]l[y(t)] Y (s) 4 ( ) s 3 s Y (s) + ( Y (s) 3 ) s 4 + s ( s Y (s) ) +4 s 4 + s Y (s) 4(s + ) s (s +4) 4(s +) s (s +4) A s + B Cs +D + s s + 4 4s +4 As(s +4)+B(s + 4)+(Cs +D)s 4s +4 (A +C)s 3 + (B +D)s + 4As +4B and by comparing coefficients of, s, s and s 3 we obtain, respectively, four equations 4B 4 4A B + D 4 A +C Solving these gives A, B,C,D 3 and therefore Y (s) s + 3 s + 4 and so the answer to our integral equation is y(t) where y(t)l [Y (s)]l [ s ] + L [ L [ s ] + 3 L [ t+ 3 sin(t). ] 3 s + 4 ] s Dirac s delta function Dirac s delta function δ(t) is not actually a function of the variable t. It is a distribution. The theory of distributions belongs to a branch of mathematics called functional analysis. We shall discuss the proof of the formula which can more generally be stated as L[δ(t)] (.6) L[δ(t t )]e st where t (.63)

98 98 Laplace transforms however this proof is not required for this course. Knowledge and application.6 of the two formulae (.6) and (.63) is the only requirement of this subsection. In functional analysis, a functional is a map from a set of functions to a set of numbers. Example.49. Let C denote the set of continuous functions and denote the set of real numbers. Then i. the map φ: C defined by b φ(f) f(t)dt a is a functional as integrating any continuous function f(t) over a finite interval [a, b] gives a number. ii. the map σ: C defined by σ(f) f() is also a functional. Notice that σ associates to any continuous function f(x) the value of that function at t. One can define functionals on many sets of functions, in the Example.49 above we described two different functionals on a particular set of functions the set of continuous functions C. To define the Dirac delta function we shall need another specific set of functions the set of test functions. Denote the set of test functions as D. Definition.5. A test function on is a function that is identically zero outside a sufficiently large interval c <t<c and that has derivatives of any order. ecall that as any differentiable function is necessarily continuous, it follows that D C, that is the set of test functions on is contained in the set of continuous functions on. Example.5. The function f(t) e t when <t< when t or t is an example of a test function. From the graph of f(t) f(t) - it is clear that f(t) is identically zero outside <t<. Notice that f(t) is a continuous function. Definition.5. Let D denote the set of all test functions on. Then a functional that has domain equal to D is called a distribution on. We can now give a precise definiton of the Dirac delta function δ(t)..6. see Example.63

99 .3 Applications and more properties of Laplace transforms 99 Definition.53. The Dirac delta function δ(t) is the distribution δ(t): D that is defined as δ(t)(φ) φ(). So the Dirac delta function δ(t) maps any test function φ in the set D to a corresponding number φ(). In other words, δ(t) takes any test function φ and evaluates that function at t. More generally, we can define the shifted Dirac delta function δ(t t ). Definition.54. The Dirac delta function δ(t t ) is the distribution δ(t t ): D that is defined as where t is a constant. δ(t t )(φ) φ(t ) Notice by letting t, Definition.54 reduces to Definition.53. We now give a more intuitive but less technically accurate description of the Dirac delta function δ(t t ). Consider the following function n when t f n (t) t t + n (.64) otherwise where n,,3, is a parameter that can vary. The graph of f n (t) is f n (t) n t t +/n and we see that the graph of f n (t) takes the form of a pulse. Notice that the area under this graph is n(t + n t ). Now as the parameter n gets larger, the pulse f n (t) gets narrower and higher. We see this in the graphs below illustrating the cases n,,3 respectively: t f (t) f (t) 3 f 3 (t) t t t t + t +/ Notice that the area under the graph of each f, f and f 3 is. t +/3 Intuitively, as n the functions f n tend to the Dirac delta function, that if it were to exist as a function, would take the form of a pulse with zero width, infinite height and an area of.

100 Laplace transforms The above intuitive description of the Dirac delta function is not precise as the pointwise limit of the functions f n (t) as n lim f n (t) n does not exist. If the limit lim n f n (t) were to exist then for any value t a, the sequence of numbers f n (a) would converge. However, when t t the sequence of numbers f (t ), f (t ), f 3 (t ), is equal to,, 3, which is a sequence that does not converge. Therefore the pointwise limit lim n f n (t) does not exist. However, it turns out that the theory of distributions was developed to handle such situations as taking the limit of the sequence f n above. In the theory of distributions, each function f n can be converted into a corresponding distribution which is denoted by T fn. Then one takes the limit of these distributions this is actually a limit of integrals. In this particular case, the limit of the T fn as n does exist and is in fact equal to the distribution δ(t t ) as defined in Definition.54. It is in this sense of distributions that the above sequence f n converge to the Dirac delta δ(t t ). The following definition describes how a function f n is converted to a distribution T fn. b Definition.55. Let f(t) be a function that is locally integrable, that is, a f(t)dt < for any finite interval a < x < b. Then the distribution T f : D associated to the function f(t) is defined as where φ is any test function in D. T f (φ) fφ dt Example.56. Let f n be the function defined in (.64). Then T fn (φ) f n φ dt t+ n n φdt. Let T n n,, 3, be a sequence of distributions. Then notice for a fixed test function φ, T n (φ) is a sequence of real numbers. We now describe the limit of a sequence of distributions. Definition.57. The sequence of distributions T n n,, 3, has limit equal to the distribution T only if for each test function φ the sequence of real numbers T n (φ) has limit equal to T (φ). In other words T n T as distributions only if T n (φ) T (φ) as real numbers for each test function φ. The following lemma will be useful in proving that the pulse functions f n defined in (.64) has limit equal to the Dirac delta function δ(t t ) in the sense of distributions, that is, T fn δ(t t ) as n. Lemma.58. Let φ(t) be a continuous function and let f n be defined as in equation (.64) above. Then lim f n (t)φ(t)dt φ(t ). n Proof. From equation (.64) f n (t)φ(t)dt n t t+ n φ(t)dt. t+ n n (φ(t) φ(t )+ φ(t ) )dt t t t+ t+ n n (φ(t) φ(t ))dt +n n φ(t )dt t t (.65)

101 .3 Applications and more properties of Laplace transforms Let us consider the first integral on the right. As φ(t) is continuous at t t, by the definition of continuity, given any ǫ > there exists a δ > such that ǫ <(φ(t) φ(t )) <ǫ when δ <t t <δ. (.66) Now as δ >, there exists an integer N such that < δ for all n N. Integrating the first inequality n in (.66) between t + and t n for any n N gives t+ t+ t+ n n ( ǫ)dt < (φ(t) φ(t ))dt < n ǫdt (.67) t t and as ǫ is independent of t, it is easy to evaluate the first and third inequalities in (.67) ǫ n < t t+ n (φ(t) φ(t ))dt< ǫ n and multiplying throughout by n we have t+ n ǫ <n (φ(t) φ(t ))dt <ǫ. (.68) t We have shown that for any ǫ > there exists some N such that for all n N the inequality (.68) holds. As this is true for any ǫ >, we can make ǫ in (.68) as small as we wish, this therefore implies that t+ n lim n (φ(t) φ(t ))dt. (.69) n t t Now consider the second integral on the right in the last line of (.65). As φ(t ) is a constant and therefore t+ n t + n φ(t )dt nφ(t )[t] n t φ(t ) t t + t n + f n (t)φ(t)dt n (φ(t) φ(t ))dt +n n φ(t )dt t t t+ n f n (t)φ(t)dt n (φ(t) φ(t ))dt + φ(t ) Taking the limit as n gives t+ n lim f n (t)φ(t)dt lim n (φ(t) φ(t ))dt + lim φ(t ) n n n t t From (.69) we have lim n f n (t)φ(t)dt + lim φ(t ) n and as φ(t ) is a constant we get our desired result f n (t)φ(t)dt φ(t ). lim n We now use Lemma.58 to show that the pulse functions f n defined in (.64) has limit equal to the Dirac delta function δ(t t ) in the sense of distributions. Lemma.59. Let f n be defined as in equation (.64) and let T fn be the corresponding distributions as in Definition.55. Then as n T fn δ(t t )

102 Laplace transforms as distributions. Proof. By Definition.57 T fn δ(t t ) as n if for each test function φ, Now from Definition.55 lim n T fn (φ)δ(t t )(φ) (.7) T fn (φ) f n φ dt and from the definition of the Dirac delta as a distribution in Definition.54 δ(t t )(φ) φ(t ). Substituting these two definition into (.7) we have that T fn δ(t t ) as n if for each test function φ lim f n φdt φ(t ). (.7) n However since any test function φ is continuous, (.7) is true from Lemma.58 and therefore the result follows. We now determine the Laplace transform of the Dirac delta function δ(t t ). ecall that δ(t t ) is a distribution. The following is the definition of the Laplace transform of a distribution (also see einhard [3]). Definition.6. Let { when t< φ s e st whent If T is a distribution and if T(φ s ) exists then T has Laplace transform L[T] T(φ s ) where T(φ s ) is the evaluation of the functional T at the function φ s. Using this definition we have Theorem.6. The Laplace transform of the Dirac delta function δ(t t ) where t is L[δ(t t )]e st. (.7) Proof. As stated in Definition.54 the Dirac delta is a distribution. As a functional δ(t t ) takes any function φ and evaluates that function at t t, hence the functional δ(t t ) takes the function φ s and evaluates it at t t, that is as t. By Definition.6 we have δ(t t )(φ s )e st L[δ(t t )]δ(t t )(φ s )e st. By letting t in the formula (.7) we have the following Corollary.6. The Laplace transform of the Dirac delta function δ(t) is L[δ(t)].

103 .3 Applications and more properties of Laplace transforms 3 In this course we shall treat δ(t t ) as a function. The above precise definition of δ(t t ) is only for the interest of the reader, it is not required material. Knowledge of the formulae L[δ(t t )]e st L[δ(t)] is required as is the application of these formulae as done in the following example. Example.63. Solve the differential equation subject to the initial conditions y(), y (). y + 9y δ(t ) (.73) Answer: Take the Laplace transform of both sides of the given differential equation L[y +9y] L[δ(t )] L[y ]+9L[y]e s Using formula (.45) for the transform of the second derivative we have s L[y(t)] sy() y ()+9L[y(t)] e s Substituting the given initial conditions y(), y ()gives Let L[y(t)] Y (s) Now solve for Y (s) s L[y(t)]+9L[y(t)]e s s Y (s)+9y (s)e s (s +9)Y (s) e s Y (s) e s (s +9) e s 3 3 (s +3 ) Notice that e s 3 3 (s +3 ) is of the form e st Y (s) with e st e s and Y (s) 3 3 (s + 3 ). Now Y (s) 3 3 (s +3 y(t)l [Y (s)] ) 3 sin(3t) and by the t-shift formula of Corollary.34 [ L e s 3 3 (s + 3 ) and so the solution to this initial value problem is ] H(t )y(t ) H(t ) sin(3(t )) 3 y(t)h(t ) sin(3(t )). 3 One can associate a physical interpretation to Example.63. The homogeneous part of the differrential equation (.73), that is y +9y is an equation that describes some type of oscillation about an equilibrium position, for example the simple harmonic motion of a mass loaded on a spring, where y measures the displacement from the equilibrium position

104 4 Laplace transforms y y (equilibrium position) The inhomegeneous part of the differential equation y + 9y δ(t ) (.74) that is, the right hand side of the equation (.74), may be interpreted in our example of a spring as a vertical outside force applied to the mass over time. In this case δ(t ) can be viewed as a sudden impulse at an instant of time when t. Now the initial condition y() means that the mass is at the equilibrium position at time t. The condition y () means that the velocity at time t is zero. The mass is therefore at rest and does NOT move until time t when the impulse δ(t ) is applied, causing the mass to oscillate as we can see from the following graph of the solution of the differential equation (.74) y(t)h(t ) sin(3(t )) (.75) 3 /3 y(t) t Also notice from the graph of y(t) that after time t the system oscillates freely about the equilibrium position as there is no outside force after t.

105 .3 Applications and more properties of Laplace transforms Differentiation of transforms The result given in the following is useful for determining tranforms and inverse transforms. Theorem.64. If Y (s) L[y(t)] then Y (s) L[ ty(t)] where Y (s) denotes the derivative of Y (s) with respect to the variable s. Proof. From the definition of the Laplace transform Y (s) e st y(t)dt and differentiating both sides of this equation with respect to s gives Y (s) d e st y(t)dt. (.76) ds It is possible to.7 exchange the differentiation and integration operations in equation (.76) Y d (s) ds (e st y(t))dt and differentiating e st y(t) partially with respect to s we have Y (s) te st y(t)dt e st ( ty(t)) dt L[ ty(t)]. The following two examples are applications of Theorem.64. Example.66. Find L[t sin t]. Answer: and by Theorem.64 and therefore by linearity L[sint] s + ( ) L[ tsint] s + s (s + ) L[t sint] s (s + ). Using Theorem.64 gives a more efficient method of proving L[t n ] n! sn+ than Example.7. Example.67. Use the principle of induction, the definition of the Laplace transform and Theorem.64 to show that L[t n ] n! s n+ when n,,,.7. This is done by an application of the following y(s, t) Lemma.65. Let y(s, t) and be continuous on the set {(s,t) c s d, t < }; let the integral s be uniformly convergent and denote Y (s) y(s, t)dt. Then Y (s) is differentiable and Y y(s, t) (s) dt. s y(s, t) dt s Knowledge of this lemma is not required for this course.

106 6 Laplace transforms Answer: The statement is true when n as L[t ] L[] e st dt lim e st dt [ ] e st lim s ( ) lim s e s s s! s + Assume the statement is true when n k, that is Then by using Theorem.64 and therefore L[t k ] k! s k+. when s > ( ) k! L[ t.t k ] s ( k+ ) L[ t k+ ] k! s ( k+ ) (k +) k! L[t k+ ] s k+ (k +)! s k+ (k +)! s k+ and so the statement is true when n k +, hence by the principle of induction is true for all n,,, L[t n ] n! s n+.3.6 The Gamma function Γ(x) In Examples.7 and.67 we obtained the formula L[t n ] n! s n+ when n,,,. We now wish to obtain a more general formula, that is, a formula for L[t x ] when x is a real number and x >. To do this, we need to define the gamma function Γ(x). Definition.68. The gamma function Γ(x) is defined when x > as an integral Γ(x) e t t x dt. (.77) The following result gives some properties of the gamma function. Lemma.69. Let Γ(x) denote the gamma function. Then i. Γ() ii. Γ(x +)xγ(x) when x > iii. Γ(n +)n! when n,,,

107 .3 Applications and more properties of Laplace transforms 7 ( iv. Γ ) π Proof. i. Let x in (.77) Γ() ii. eplacing x by x+ in (.77) Γ(x +) now use integration by parts with to get and the limit lim lim lim lim e t t dt. e t dt e t dt [ e t ] ( ) e e t t x dt e t t x dt v t x dv xt x dt Γ(x +) lim e t t x dt ( lim [ e t t x ] +x lim ( lim du e t dt u e t ( e. x e x) +x lim e x )+ xγ(x) lim e x lim x e ) e t t x dt e t t x dt (.78) is of the indeterminate form. Applying L Hospitals rule by differentiating the numerator and denominator x times with respect to, where x is the the smallest integer x, we have that lim e x and therefore Γ(x+) xγ(x) and clearly this result holds for x > as Γ(x) is defined for x >. iii. We prove Γ(n + ) n! when n,,, by the principle of induction.the statement is true when n as Γ() from part i. Assume the statement is true when n k, that is Then letting x k + in Γ(k + )k! (.79) Γ(x+) xγ(x)

108 8 Laplace transforms of part ii) we have and using (.79) we have Γ(k +) (k + )Γ(k + ) Γ(k +)(k +)k! (k + )! so the statement is true for n k +, hence by the principle of induction is true for all n,,, Γ(n +)n! iv. Let x in (.77) using the substitution ( ) Γ e t t dt (.8) in (.8) we get u t ( Γ du t dt ) e u du (.8) Using double integrals (a topic done later in this course) we can determine the right side of (.8) ( )( ) e x dx e y dy e x e y dxdy e (x +y ) dxdy Changing variables from Cartesian (x, y) to polar (r,θ).8 Therefore ( e (x +y ) dxdy π ( π [ e r π dθ π )( e x dx ( e dx) x π since x, y are dummy variables and therefore e x dx π As e x is symmetrical about the y axis e x dx π ) e r rdr dθ ] ) e y dy π and substituting (.8) into (.8) we get our desired result ( ) Γ π. dθ (.8).8. see Section 9.3 of Kreyszig []

109 .3 Applications and more properties of Laplace transforms 9 Having defined the gamma function, we can now determine L[t x ] when x >. Theorem.7. Let x be a real number such that x >. Then L[t x ] Γ(x +) s x+..9 Proof. Substitute x + for x in the definition of Γ(x) Γ(x+ ) e t t x dt Using the substitution t su dt sdu where we treat s as a constant and u as the variable, we get Γ(x+ ) e su (su) x s du s x+ e su u x du and as u is a dummy variable, we can replace u by another dummy variable t to get Γ(x+ )s x+ e st t x dt. and from the definition of the Laplace transform of t x which gives our desired result Γ(x +)s x+ L[t x ] L[t x ] The following is an application of Theorem.7. [ ] 5 Example.7. Show that L t 5 π 7. 8s By Theorem.7 and from Lemma.69 part ii) L Γ(x +) s x+. (.83) [ t 5 ] Γ ( 7 s 7 ) ( ) 7 Γ 5 ( ) 5 Γ and applying Lemma.69 part ii) two more times we have ( ) 7 Γ 5 ( ) 5 Γ 5 ( ( )) 3 3 Γ 5 ( ) 3 Γ.9. We require x> because in the definition of Γ(x) we havex >.. If s < then the integral diverges from Lemma.3. and clearly (.83) is undefined for s, so it is necessary that s > for(.83) to hold.

110 Laplace transforms and from Lemma.69 part iv) and therefore ( ) 7 Γ 5 3 π 5 π 8 [ ] 5 L t 5 π. 8s 7

111 Chapter 3 Fourier series 3. Definitions Definition 3.. Let f(x) be a function that is defined for all x. Then f(x) is called periodic if there exists a positive number p such that f(x+ p) f(x) for each x. The number p is called a period of the function f(x). Example 3.. The function sinx has period p π as one can check sin(x + π) sin(x) (3.) by using the trigonometric formula sin(x + y)sinxcos y + cosxsin y: sin(x + π) sin(x)cos(π) + cos(x)sin(π) sin(x).()+cos(x).() sin(x) Equation (3.) can be interpreted graphically as follows. The graph of sin(x + π) is a shift of the graph of sin(x) to the left by π units. Notice that this shifted graph sin(x + π) coincides with the original sin(x) and this implies equality in (3.). sin(x) - period pi Note 3.3. In the following, we shall always let the period p L. We do this to be consistent with notation used in a later section of partial differential equations. For example, in the case of Example 3., sin(x) has period Lπ and clearly L π. Definition 3.4. Let the functionf(x) be a periodic function of period L. a ( + a n cos nπx L + b nsin nπx ) L n Then (3.)

112 Fourier series is a Fourier series for f(x) if its coefficients are given by the formulae a n L f(x)cos nπx dx where n,,,3, (3.3) L L L b n L f(x)sin nπx dx where n,, 3, (3.4) L L L Equations (3.3) and (3.4) are called Euler s formulae and the numbers a n, b n are called the Fourier coefficients of f(x). Example 3.5. Let f(x) x where π <x π and f(x +π) f(x) (3.5) Sketch the graph of f(x) and find the Fourier series of f(x) on the interval ( π, π). Answer: Note that the function f(x) is defined to be periodic with period π by the statement f(x+ π) f(x) in (3.5). The graph of f(x) is obtained by first drawing f(x) x on the interval π < x π and then repeatedly shifting this drawing by multiples of π Figure 3.. Graph of f(x) Notice the closed dots of Figure 3. imply that f(x)π when x, π, π, 3π,5π, We need to determine the Fourier coefficients a n and b n. It is typical to split the case of a n into two subcases n and n,, this is because n causes the cosine term in (3.3) to become. a L x cos πx L L L dx π x cosdx π π π xdx π π [ ] x π π π as Lπ as cos

113 3. Definitions 3 and so a (3.6) Now determine a n when n,, a n L π π L L π π π π x cos nπx L dx x cos nπx π dx x cosnx dx notice the π s cancel. Integrating by parts with gives recall when n is an integer a n [ ] π x sinnx π n [ ] π x sinnx π n (( π sinnπ π n v x dv dx π π du cosnxdx u sinnx n π sinnx dx π π n + [ cosnx π π n ] π + cosnπ ) ( π sin( nπ) n + n expression value sin nπ sin( nπ) cosnπ ( ) n cos( nπ) ( ) n Table 3.. )) cos( nπ) n and therefore which gives a n π (( + ( ) ( )n n + ( )) )n n a n when n,,3, (3.7) And now determine the b n when n,, Integrating by parts with b n L π π v x dv dx L L π π π π x sin nπx L dx x sin nπx π dx x sinnx dx du sinnxdx u cosnx n

114 4 Fourier series gives b n [ x cosnx ] π π cosnx dx π n π π π n [ x cosnx ] [ ] π π sinnx + π n π π n π (( π cosnπ + sinnπ ) ( π cos( nπ) π n n + n )) sin( nπ) n and using the values from Table 3. b n (( π( )n π n ( π( )n π n ( )n n which gives ) ( + π ( )n n π ( )n n ) )) + b n ( )n+ n when n,, 3, (3.8) Substituting L π and the expressions for a,a n and b n given by (3.6),(3.7) and (3.8) into we get our answer a ( + a n cos nπx L + b nsin nπx ) L n the Fourier series of the function f(x) given in (3.5) is ( ) n+ sin nx. n n 3. Convergence of Fourier Series In Example 3.5 we proved that the Fourier series of f(x)x where π < x π and f(x+ π) f(x) (3.9) is ( ) n+ n n sin nx. (3.) We will see, from Theorem 3.8 and Example 3.9 below that the function f(x) given in (3.9) converges to its Fourier series (3.) for all values of x except when x, π, π, 3π,5π, Definition 3.6. Suppose a ( + a n cos nπx L +b nsin nπx ) L n (3.)

115 3. Convergence of Fourier Series 5 is the Fourier series of a function g(x). Let m a ( + a n cos nπx L +b nsin nπx ) L n (3.) be the sum of the first m terms of (3.). Then the Fourier series (3.) is said to converge to g(x) at xx if ( m a ( lim m + n a n cos nπx L +b nsin nπx ) ) g(x ). L Intuitively Definition 3.6 means that the sum of the first m terms (3.) of the Fouries series is an approximation of the function g(x), and this approximation gets better as m gets larger. Let us return to our example of f(x) given in (3.9), in which the Fourier series (3.) converges to f(x) except when x, π,π, 3π, 5π, Consider when m 5, that is the sum of the first five terms of (3.) 5 n ( ) n+ n sin nx (3.3) and consider the graph of this function obtained using the computer program Maple x Figure 3.. Graph of the first five terms of Fourier series (3.) Notice that this graph is approximately the same as the graph of f(x) given in Figure 3., hence the function f(x) x is approximately equal to (3.3), that is 5 f(x) n ( ) n+ n sin nx. Now consider when m 5, that is the sum of the first twenty-five terms of (3.) 5 n ( ) n+ n sin nx, (3.4) and the graph of the first twenty-five terms is

116 6 Fourier series x Figure 3.3. Graph of the first twenty-five terms of Fourier series (3.) and we see that the graph Figure 3.3 more closely resembles the graph of f(x) given in Figure 3.. When m, the graph of the first thousand terms of (3.) is x Figure 3.4. Graph of the first thousand terms of Fourier series (3.) which is almost identical to the graph of f(x) given in Figure 3., except for the presence of what appear to be vertical lines at x π, π, 3π,. These lines are in fact not vertical but are almost vertical with large negative slope 3.. These almost vertical lines are present in Figure 3.4 because ( ) n+ n n sinnx 3.. This slope can be determined. ecall we are graphing ( ) n+ sin nx (3.5) n n To get the slope of (3.5) at xπ, simply differentiate (3.5) w.r.t. x and substitute xπ; this will give ( ) n+ cosnπ ( ) n+ ( ) n ( ) n+ n n n

117 3. Convergence of Fourier Series 7 is a finite sum of continuous functions sin nx, therefore is itself continuous which means that its graph cannot have gaps. These almost vertical lines are therefore necessary to maintain continuity. It is very interesting that when m these almost vertical lines disappear and the graph of the Fourier series, in this case, becomes discontinuous as we see next. It is not possible to directly plot the case of m via computer. However from Theorem 3.8 we can predict what the graph of m, that is the graph of the Fourier series is ( ) n+ n n sin nx. Figure 3.5. Graph of the Fourier series (3.) and notice that this graph is identical to the graph of f(x) given in Figure 3. except when x π, π, 3π,, that is except at the points at which our original function f(x) is discontinuous. By examining the closed dots of Figures 3. and 3.5, we see that f(x) takes value π and the Fourier series (3.) takes value at these points of discontinuity. We now explain some terms necessary for the statement of Theorem 3.8. We do not define these terms precisely, but illustrate their meaning via an example; see Kreyszig [] for precise definitions. Definition 3.7. Consider the following piecewise defined function x when x < h(x) x when x >. The graph of this function is h(x) / x

118 8 Fourier series Then the left hand limit at x is lim h(x) x and is the value h(x) approaches as x tends to from the left; one may think of a small object moving along the parabola x heading right 3. to just before the point (, ); the height of the object is the value of lim x h(x). The right hand limit at x is lim x + h(x) and is the value h(x) approaches as x tends to from the right; again one may think of a small object sliding left along the line x to just before the point (, ); the height of the object is the value of lim x +h(x). The left hand derivative at x is defined as h( +ǫ) (lim lim x h(x)) ǫ ǫ ( + ǫ) lim ǫ ǫ lim ǫ+ ǫ and this can be interpreted as the slope of the function h(x) at a point just before (, ); in this case this is the same as the slope of the parabola at (, ). The right hand derivative at x is defined as h( +ǫ) (lim lim x +h(x)) ǫ + ǫ ( +ǫ) ǫ lim ǫ + lim ǫ + and this can be interpreted as the slope of the the function h(x) at a point just after (, ); in this case this is the same as the slope of the line at (, ). ecall that it is not possible for a function to have a derivative at a point of discontinuity 3.3 such as x in the example above. Note however that the definitions of left hand/right hand derivative are constructed to extend the notion of a derivative to points of discontinuity, one simply examines the slope of the curve just before the point of disconuity to get the left hand derivative (if it exists), likewise examining the slope just after the point of discontinuity gives the right hand derivative. We can now state a theorem that describes, for which values of x, a Fourier series of a function f(x) is equal 3.4 to the function f(x). Theorem 3.8. Let f(x) be a function that is periodic with period L and that is piecewise continuous on the interval L < x < L. Also assume that for each x that either f(x) is differentiable at x or the left hand derivatives and right hand derivatives of f(x) exist. Let a ( + a n cos nπx L +b nsin nπx ) L n (3.6) 3.. heading rightapproach from the left 3.3. because differentiablity implies continuity 3.4. more precisely, for which values of x the Fourier series of f(x) converges to its function f(x)

119 3. Convergence of Fourier Series 9 be the Fourier series of f(x). Then the series (3.6) converges to f(x) at each point x except possibly at points where f(x) is discontinuous. At any such point x of discontinuity, the Fourier series converges to ( ) lim f(x)+ lim f(x). + x x x x We now apply Theorem 3.8 to the answers in Example 3.5: Example 3.9. Show that the Fourier series of the function converges to function g(x) ( ) n+ sin nx (3.7) n n f(x) x where π <x π and f(x +π) f(x) { x when π < x <π when x π and g(x+ π) g(x). (3.8) Answer: We first show that f(x) satisfies the hypotheses of Theorem 3.8. By the definition of f(x) we have f(x+ π) f(x) and so f(x) is periodic with period π. From the graph of f(x) Figure 3.6. Graph of f(x) it is clear f (x) and is hence is differentiable at each point x, π, π, 3π, 5π,. At the points of discontinuity, that is for x (k + )π where k is an integer, the left hand derivative exists: ( f((k + )π +ǫ) limx (k+)π f(x)) lim ǫ ǫ π +ǫ π lim ǫ ǫ lim ǫ

120 Fourier series and is equal to. Alternatively one can clearly see that the slope of the curve just before the point x (k + )π is equal to. Similarly the right hand derivative exists and is equal to at each point of discontinuity x (k + )π. Hence we may apply Theorem 3.8 and let g(x) be the function that the Fourier series (3.7) converges to. Then according to Theorem 3.8, g(x) is equal to f(x) everywhere f(x) is continuous and so g(x) x when π < x <π g(x+ π) g(x) (3.9) At the point of discontinuity x π according to Theorem 3.8 and this repeats every π so g(x) ( ) lim f(x) + lim f(x) x π x π + (π +( π)) g(π) g(x +π) g(x) (3.) and combining (3.9) and (3.) we get our desired answer (3.8). Therefore the Fourier series (3.7) converges to (in other words, is equal to) g(x). The graph of g(x) is Figure 3.7. Graph of g(x) and notice this almost identical to the graph of f(x) in Figure 3.6. Example 3.9 illustrates that a Fourier series of a function f can be identical to the function f for almost all values of x.

121 3.3 Even and odd functions 3.3 Even and odd functions If a function is even or odd we can save some work when computing its Fourier series. Definition 3.. i. A function f(x) is said to be even if f( x) f(x) for all values of x. ii. A function f(x) is said to be odd if f( x) f(x) for all values of x. Example 3.. Determine if the following functions are even, odd or neither. i. f(x) x 3 ii. g(x)x 4 iii. h(x) x 3 +x 4 iv. k(x)cos(x)+5x x 4 + x Answer: i. eplacing the x in f(x)x 3 by x we have which implies f( x) ( x) 3 f( x) x 3 (3.) Comparing the right hand side of (3.) to the definition of the original function f(x) x 3 we see that so f(x) x 3 is an odd function. ii. eplace the x in g(x)x 4 by x to get which implies f( x) f(x) g( x)( x) 4 g( x) x 4 (3.) Comparing the right hand side of (3.) to the definition of the original function g(x) x 4 we see that so g(x) x 4 is an even function. iii. eplace the x in h(x)x 3 +x 4 by x to get g( x) g(x) which implies In this case, notice that h( x)( x) 3 + ( x) 4 h( x) x 3 + x 4 (3.3) h(x) x 3 +x 4 and h(x) x 3 x 4 and that the right hand side of (3.3) is equal to neither of these. Therefore h(x) is neither even nor odd.

122 Fourier series iv. eplace the x in by x to get k(x) cos(x) +5x x 4 + x k( x)cos( x)+5( x) ( x) 4 + x cos(x)+5x x 4 + x k(x) and so k(x) is an even function. It is slightly more difficult to determine if a periodic function that is piecewise defined is even or odd. To determine if such a function, for example g(x) { x when π <x <π when x π and g(x+ π) g(x). is even or odd, one may check the function on one period as the following lemma states. Lemma 3.. i. A periodic function of period L is even if f( x) f(x) on the interval L x L. ii. A periodic function of period L is odd if f( x) f(x) on the interval L x L Proof. Not required. Example 3.3. Show that the function is odd. g(x) { x when π < x <π when xπ and g(x +π) g(x) Answer: For any value of x in the interval π < x <π we have and when x π or π and hence g( x) x g(x) g( x) g(x) g( x) g(x) on the interval π x π and it follows from Lemma 3. that g(x) is odd. (the graph of g(x) is given in Figure 3.7). Lemma 3.4. Let f(x) be an odd function. Then for any L > L f(x)dx. L Proof. Using a property of definite integrals L L f(x)dx f(x)dx + f(x) dx (3.4) L L

123 3.3 Even and odd functions 3 From a change of variables u x on the first integral on the right of (3.4) we have f(x)dx f( u)( du) L L f(u)( du) L f(u)du L L f(u)du L f(x)dx and using this in (3.4) we have L f(x)dx L L L f(x)dx + f(x)dx. A more intuitive reason for Lemma 3.4 follows from the graph of an odd function f(x) L in which the total shaded area represents L f(x)dx. The area to the left represents f(x)dx L and is equal in magnitude but opposite in sign to the area on the right which represents L f(x)dx; clearly these two cancel each other in equation (3.4). Lemma 3.5. The Fourier series of an even function f(x) has no sine terms, that is, b n for each n,,3,. Proof. Let a ( + a n cos nπx L + b nsin nπx ) L n be the Fourier series of the even function f(x). From Euler s formulae b n L f(x)sin nπx L L dx where n,,3, As f(x) is even and sin nπx L therefore f(x)sin nπx L is odd L ( ) nπ( x) ( f( x)sin f(x) sin nπx ) L L f(x)sin nπx L is an odd function. From Lemma 3.4 L f(x)sin nπx L dx L

124 4 Fourier series and clearly this implies b n for each n,,3,. So we have The Fourier series of an even function takes the form a + a n cos nπx L n Similarly, we can prove Lemma 3.6. The Fourier series of an odd function f(x) has no cosine terms, that is, a n for each n,,, 3,. Therefore we also have Example 3.7. Let The Fourier series of an odd function takes the form n b n sin nπx L f(x) x where <x and f(x+ ) f(x) (3.5) Sketch the graph of f(x) and find the Fourier series of f(x). Answer: The graph of f(x) is obtained by first drawing f(x) x on the interval < x and then repeatedly shifting this drawing by multiples of. f(x) is clearly periodic with period L. Notice that f( x) ( x) x f(x) when x and from Lemma 3. f(x) is even. Therefore from Lemma 3.5, the Fourier series of f(x) takes the form a + a n cos nπx (3.6) L n

125 3.3 Even and odd functions 5 that is, b n when n,, 3,. We use Euler s formula to determine a n. Split the case of a n into two subcases n and n,, a L ( x ) cos πx L L L dx ( x )dx as L [ ] x x Now determine the a n when n,, Integrating by parts with gives Integrating by parts again with gives a n [ [ π L a n L L π ( x ) cosnπx dx ( x ) cos nπx L dx ( x ) cos nπx dx v x du cosnπxdx dv x dx u sinnπx nπ [ ( x a n ] ) sinnπx x sinnπx dx nπ nπ x sinnπx + dx nπ v x dv dx x sinnπx dx nπ ] x cosnπx n π ] x cosnπx n π ( cosnπ n π and recall when n is an integer and therefore Hence the Fourier series of f(x) is du sinnπx dx nπ u cosnπx n π cosnπx n π dx [ ] sinnπx + n 3 π 3 ( ) cos( nπ) n π expression value sin nπ sin( nπ) cosnπ ( ) n cos( nπ) ( ) n a n 4( )n n π 3 + n ) + sinnπ n 3 π 3 sin( nπ) n 3 π 3 when n,, 4( ) n n π cosnπx.

126 6 Fourier series 3.4 Half range expansions Let a function f(x) be defined on an interval x L. We want to be able to represent f(x) by two types of Fourier series a Fourier series that consists of cosine terms only or a Fourier series that consists of sine terms only. 3.5 Essentially, this may be done because we only want the Fourier series to represent the function for half of the period L x L. Example 3.8. Let f(x) x on the interval x π. Then we will show in Example 3.4 that we can represent this function f(x) by a Fourier series that consists of cosine terms only, which in this case is x π + ( ( ) n ) π n cosnx n and this equality holds when x π. We can also represent f(x) by a Fourier series that consists of sine terms only, which in this case is and this equality holds when x<π. Definition 3.9. x ( ) n+ sinnx n n i. A Fourier series that consists of cosine terms only is called a Fourier cosine series. ii. A Fourier series that consists of sine terms only is called a Fourier sine series. Hence, given a function f(x) defined on an interval x L, we want to be able to represent f(x) by either a Fourier cosine series or a Fourier sine series. ecall from Lemma 3.5 that the Fourier series of an even function consists of cosine terms only, in other words, the Fourier series of an even function is a Fourier cosine series. Also recall from Lemma 3.6 that the Fourier series of an odd function is a Fourier sine series. We use these facts to obtain our desired representation of f(x) by either a Fourier cosine series or a Fourier sine series. We obtain the representation of f(x) by a Fourier cosine series by first constructing an even periodic function that is equal to f(x) on the interval x L. Such a function is called an even periodic extension of f(x). Similarly we obtain the representation of f(x) by a Fourier sine series by constructing an odd periodic function that is equal to f(x) on the interval x L. In this case, such a function is called an odd periodic extension of f(x). Definition 3.. Let a function f(x) be defined on an interval x L. Then i. the function defined as E(x) { f(x) x L f( x) L < x < is the even periodic extension of f(x) of period L. ii. the function defined as { f(x) x L O(x) f( x) L <x< is the odd periodic extension of f(x) of period L., E(x +L) E(x), O(x +L)O(x) 3.5. We will use this when solving partial differential equations.

127 3.4 Half range expansions 7 Example 3.. Let f(x)x on the interval x π. Then i. the even periodic extension of f(x) is { x x π E(x) x π <x<, E(x+ π)e(x) and from the graph of E(x) we see that E(x) is an even periodic function of period π. Also note that E(x) f(x) when x π. ii. the odd periodic extension of f(x) is { x O(x) ( x) and from the graph of O(x) x π π <x<, O(x+ π)o(x) we see that O(x) is an even periodic function of period π. Also note that O(x) f(x) when x π. We now define the half-range expansions of a function f(x). Definition 3.. Let f(x) be defined on an interval x L. i. The Fourier series of the even periodic extension E(x) of f(x) is called the cosine halfrange expansion of f(x) or Fourier cosine series of f(x). ii. The Fourier series of the odd periodic extension O(x) of f(x) is called the sine half-range expansion of f(x) or Fourier sine series of f(x).

128 8 Fourier series We explain the idea of Definition 3.. Consider the case of the cosine half-range expansion of f(x). ecall from Theorem 3.8, that under certain conditions, the Fourier series of any periodic function g(x) is equal to g(x). As E(x) is, by definition a periodic function, the Fourier series of the even periodic extension E(x) is equal to E(x) on the interval L x L and hence the Fourier series of the even periodic extension E(x) is equal to f(x) on the interval x L as E(x) f(x) when x L. As E(x) is even, its Fourier series consist of cosine terms only. Summarizing, the Fourier series of the even periodic extension E(x) is a cosine series that is equal to f(x) 3.6 on the interval x L. By a similar reasoning, the Fourier series of the odd periodic extension O(x) is a sine series that is equal to f(x) on the interval x<l. The following result gives formulae for the cosine and sine half-range expansions of a function. Lemma 3.3. Let f(x) be defined on an interval x L. Then i. the cosine half-range expansion of f(x) is given by where a n L L a + a n cos nπx L n f(x)cos nπx L dx ii. the sine half-range expansion of f(x) is given by where b n L L n f(x)sin nπx L dx b n sin nπx L where n,,, 3, where n,,3, Proof. i. By definition, the cosine half-range expansion of f(x) is the Fourier series of the even periodic extension E(x) of f(x). As E(x) is an even function, by Lemma 3.5 the Fourier series of E(x) consists of cosine terms only and hence is of the form where for n,,, 3, a n L. But we know that E(x) a + a n cos nπx L n L L and so we may determine a n directly from f(x): L a n L ( L L ( L L E(x)cos nπx L dx { f(x) x L f( x) L <x< E(x)cos nπx L dx L E(x)cos nπx L dx + f( x)cos nπx L dx+ L ) E(x)cos nπx L dx L f(x)cos nπx L dx ) 3.6. provided f(x) is continuous on the interval x L, which we will assume in this class.

129 3.4 Half range expansions 9 Using the trigonometry identity cos( x)cosx we have that and therefore L f( x)cos nπx L dx which is our desired result. ( a n L ( L L L L L L L L L The proof of ii) is similar to that of i). ( f( x)cos nπx L ( nπu f(u)cos L ( nπu f(u)cos L f(x)cos nπx L dx ) du ) du ) dx by the substitution u x asu is dummy variable f( x)cos nπx L L dx + f(x)cos nπx L dx ) f(x)cos nπx L dx + f(x)cos nπx L dx L f(x)cos nπx L dx ) Example 3.4. Find the i. cosine half-range expansion ii. sine half-range expansion of the function f(x)x on the interval x π. Answer: i. In this case L π. From the formulae in Lemma 3.3 the cosine half-range expansion of f(x) is given by a + a n cosnx n where a n π f(x)cosnx dx π π x cosnx dx π Split into the case of n and the case of n,, a π x dxπ π a n π x cosnx dx π ([ ] π x sinnx π sinnx π n n ([ ] π x sinnx [ cosnx + π n n ] ( ) ( ) n π ) ) dx π n n

130 3 Fourier series and therefore the cosine half-range expansion of f(x) is π + ( ( ) n ) π n cosnx. n ii. From the formulae in Lemma 3.3 the sine half-range expansion of f(x) is given by where b n sinnx n b n π f(x)sinnx dx π π x sinnx dx π ( [ x cosnx ] π π cosnx + π n n ( [ x cosnx ] [ ] π sinnx + π n n ( )n+ n and therefore the sine half-range expansion of f(x) is π ) ) dx ( ) n+ sinnx. n n

131 Chapter 4 Partial Differential Equations 4. Definitions Definition 4.. An equation involving one or more partial derivatives of an unknown function of two or more variables is called a partial differential equation or p.d.e.. The unknown function is called the dependent variable. The order of the highest derivative is called the order of the partial differential equation. Example 4.. i. ii. u x u t is a first order p.d.e. u u is a second order p.d.e. x t Definition 4.3. A partial differential equation is said to be linear if it is of first degree in the unknown function and its partial derivatives. Example 4.4. i. ii. u x u t, ( u x ) u t, u u x t, u x. u t u x u + u are each linear p.d.e. x t are each non-linear p.d.e. iii. Let u(x, t) be a function of the two variables. Then the general form of a second order linear p.d.e. is a(x, t) u x + u b(x,t) t + c(x, t) u x t +d(x,t) u +e(x, t) u t x x + f(x,t) u + g(x,t)u h(x, t) t where a, b,c, d, e, f, g and h are given functions of x and t. Definition 4.5. If each term of a partial differential equation contains the dependent variable or its partial derivatives then the equation is said to be homogeneous, otherwise it is said to be inhomogeneous. Example 4.6. i. ii. u + u is a homogeneous p.d.e. x t u + u x t is an inhomogeneous p.d.e. x t 3

132 3 Partial Differential Equations Consider a function u(x, y) of the two independent variables x and y. The domain of the function u(x, y) will be a subset of the xy-plane. For example the real-valued function u(x, y) (x + y ) has domain equal to {(x, y) x + y } which is a closed disc of radius : The closed disc above is an example of a subset in the space of independent variables x and y of the function u(x, y). Note in this example, the space of independent variables x and y is simply the xy-plane. Definition 4.7. A solution of a partial differential equation in a subset of the space of independent variables is a function for which all the partial derivatives that appear in the equation exist and satisfies the equation everywhere in. Example 4.8. i. Show that the function u(x, y) x y is a solution of the p.d.e. xy-plane. Answer: When u(x, y) x y, the second partial derivatives u x and u y are and so for u(x, y)x y u x u + u in the set x y u y (4.) u x + u y +( ) and notice that the second partial derivatives in (4.) exist for all points in the xy-plane and also these second partial derivatives satisfy the p.d.e for all points in the xy-plane. Therefore, by Definition 4.7, u(x, y) x y is a solution of the p.d.e. u + u at each point (x, x y y) in the xy-plane. ii. Show that the function u(x, y) ln(x + y ) is a solution of the p.d.e. u + u in the x y set {xy-plane (, )}, in other words, u(x, y) ln(x + y ) is a solution of the given p.d.e. at all points on the xy-plane except at the origin.

133 4. Definitions 33 Answer: When u(x, y) ln(x + y ), the second partial derivatives u x and u y are u x y x u (x + y ) y x y (x + y ) (4.) and so for u(x, y) ln(x + y ), ( u x + u y y x ) ( x (x + y ) + y ) (x + y ) however, notice that the second partial derivatives in (4.) exists for all points except when (x, y) (, ), therefore by Definition 4.7, u(x, y) ln(x + y ) is a solution of the p.d.e. u + u at each point (x, y) in the xy-plane except when (x, y) (,). x y Theorem 4.9. (Superposition principle I) If u and u are solutions of a linear homogeneous partial differential equation in some subset in the space of independent variables then, for any scalars c and c u c u +c u is also a solution of that partial differential equation in the subset. Proof. (Not required). Example 4.. From Example 4.8 we have that u x y and u ln(x + y ) are solutions of the p.d.e. u + u in the set {xy-plane (, )}. Therefore by the Superposition principle x y I u 5(x y )+9ln(x + y ) is a solution of u x + u y in the set {xy-plane (,)}. We shall see later that we will usually obtain an infinite set of solutions u, u, u 3, to a given linear homogeneous p.d.e.. The following theorem states, that under certain conditions, the sum of such an infinite set of solutions is also a solution. Theorem 4.. (Superposition Principle II) If each function of an infinite set u, u, u 3, is a solution of a linear homogeneous partial differential equation in some subset in the space of independent variables then the function u n c n u n c u + c u + (4.3) is also a solution of that partial differential equation in the subset provided the series (4.3) converges at each point in. Proof. (Not required).

134 34 Partial Differential Equations Note 4.. In this class, we shall use a method of solving p.d.e. that determine constants c n that will ensure that the series u c n u n c u + c u + n converges. This method is based on Theorem 3.8 on page 8. We will not directly test for convergence, our method will automatically guarantee convergence. Example 4.3. One can check each of the functions is a solution of e t cosx, e 9t cos3x, e 5t cos5x,,e (n+)t cos(n +)x, in the subset {(x,t) t }. It is possible to prove 4. that the series u t u x (4.4) n (n ) e (n )t cos(n )x e t cosx+ 9 e 9t cos3x + 5 e 5t cos5x+ (4.5) converges at each point in the subset {(x, t) t }, therefore by the Superposition Principle II the series (4.5) is a solution of the partial differential equation (4.4) in the subset {(x, t) t }. We shall be interested in the following linear homogeneous partial differential equation of the second order: ( ( r r u t u c x u t c u x u x + u x u + u r r r + u r θ u + u r r r + u + u r θ z r u r ) + sin φ φ ( sin φ u φ one dimensional heat equation one dimensional wave equation two dimensional Laplace s equation Laplace s equation in polar coordinates Laplace s equation in cylindrical coordinates ) ) + u Laplace s equation in spherical coordinates sin φ θ 4.. A direct proof of this convergence is beyond the scope of this class

135 4. The Heat Equation The Heat Equation 4.. A derivation of the heat equation We illustrate how the one-dimensional heat equation models a certain physical situation. u t c u x (4.6) Consider a bar of metal of length L. We wish to describe how the temperature of this bar changes with time t. To simplify our model we assume that our rod is one-dimensional, that is, it has no thickness. Assume also that the metal bar is insulated along the sides. We place our bar of length L along the x-axis as follows. Notice that as the bar is insulated along the sides, heat may escape the bar only at points x that is, heat can only escape at the ends of the bar. and xl, Let u(x, t) denote the temperature of the bar at a point x and at time t. Note that the domain of this temperature function u(x, t) is x L and t <. We wish to model the following problem. The metal bar has an initial temperature distribution, which is not necessarily uniform along the bar. That is, at time t, the bar has a temperature that varies along the length of the bar, so this initial temperature distribution varies with x and can be stated as u(x,) f(x) where x L (4.7) where the function f(x) describes the initial temperature distribution along the bar. The equation (4.7) is an example of an initial condition. For simplicity, we assume in this particular problem that the ends of the bar will be at zero temperature for all time t. This may be stated as Equations (4.8) and (4.9) are examples of boundary conditions. u(,t) for any t (4.8) u(l, t) for any t (4.9)

136 36 Partial Differential Equations Summarizing so far, our physical problem is as follows. We have a bar that has an initial temperature that is given by equation (4.7). We insulate the bar along the sides and allow the heat to escape from the ends. Therefore the temperature of the bar will change with time. We wish to find the temperature function u(x, t) that describes this change and will show that such a u(x, t) will be a solution of the heat equation (4.6). To derive the heat equation, we consider a small segment of the metal bar that lies between the point x and x + x where x represents a small positive number. We represent this segment by the grey area in the following diagram of the metal bar. 4. As the bar is insulated along the sides, the heat flow out of the segment is only in the indicated direction of the arrows A and B. ecall that heat flows from one point to another point if these points are at different temperatures. Now the temperature at a point x on our metal bar at time t is given by the function u(x,t). Note that a difference in temperature u(x, t) from one point to the next is measured by the change of the function u(x, t) with respect to x. By the definition of partial derivatives, such a change in u with respect to x is given by the partial derivative u x. And as heat flow occurs when we have temperature change, we expect that heat flow at the point x + x on the metal bar at time t should be proportional to the partial derivative u (x + x, t). This is in fact true the Law of Heat Con- x duction from physics tells us that the rate of heat flow at the point x + x at time t (indicated by arrow A) is given by k u (x+ x, t) (4.) x where k is the thermal conductivity of the metal. The sign appears in (4.) because heat flows from points of high temperature to points of low temperature while u is positive when temperature x increases with distance x. Similarly the rate of heat flow at the point x at time t (as indicated by arrow B) is given by k u (x,t). (4.) x Notice that the sign of (4.) is opposite to that of (4.) because arrow B points in the opposite direction of arrow A. Now as heat flows out of the segment only in the direction of arrows A and B, we have that rate of heat loss in segment k u x u (x,t) k (x + x, t) (4.) x 4.. We draw our one dimensional metal bar as having width and height ONLY for illustrative purposes.

137 4. The Heat Equation 37 We can determine the rate of heat loss in the segment by another means. ecall that the more the temperature of an object increases, the more heat it will contain. In fact if an object has mass m, specific heat capacity C and temperature T(t) then, also from physics: rate of heat increase of object mass specific heat capacity rate of temperature increase mc dt dt. Using the above physical law, it is possible to show 4.3 that rate of heat increase in segment ρc u(ξ,t) x t where x ξ x+ x where ρ is the mass per unit length of the metal bar and C is the specific heat capacity of the metal bar. Therefore rate of heat loss in segment ρc and so from equations (4.) and (4.3) we have ρc u(ξ, t) t earranging equation (4.4) gives u(ξ, t) t xk u x and taking the limit of (4.5) as x gives u(ξ, t) lim k x t ρc lim x u(ξ, t) x where x ξ x+ x (4.3) t u (x, t) k (x + x,t). (4.4) x ( k u u (x + x,t) (x,t) x x ρc x ( u x u (x+ x, t) x x Now because x ξ x + x we have that ξ x as x and therefore By definition of the partial derivative lim x u(ξ, t) lim x t ( u u (x + x,t) x x (x,t) x and so from equation (4.6) we obtain u(x, t) t u(x,t). t ) ) ( u x x (x,t) (4.5) ) (x, t). (4.6) ) u(x,t) x k u(x,t) ρc x (4.7) and as k, ρ and C are positive constants we may write equation (4.7) in the standard form of the heat equation u t u c x where c is a constant by using a Mean Value theorem to obtain x+ x ρc u(s,t) ds ρc u(ξ,t) x where x ξ x+ x t t x

138 38 Partial Differential Equations We have shown that the temperature u(x, t) of our metal bar satisifes the partial differential equation u t u c x subject to the initial condition and boundary conditions u(x,) f(x) u(,t) u(l,t) where x L where t. In the next section we give a technique for obtaining a solution of such partial differential equations. 4.. Solution of the heat equation by seperation of variables We illustrate the use of the seperation of variables technique in the solution of the following onedimensional heat equation. Example 4.4. Let u(x, t) denote the temperature of a thin metal bar of length that is insulated along the sides. Let the temperature along the length of the bar at time t be given by u(x, )4x( x) where x. The ends of the bar are kept at a constant temperature of, that is u(,t) u(,t) where t. Determine the temperature u(x, t) of the bar by solving the following heat equation u t u c x (4.8) where c is a fixed constant that depends on the physical properties of the bar. Answer: The method of seperation of variables begins with the assumption that our solution u(x, t) takes the form u(x, t) X(x)T(t), (4.9) that is, the function of two variables u(x, t) is seperated into a product of a single variable function X(x) and a single variable function T(t). Substituting the solution (4.9) into heat equation (4.8) gives t (X(x)T(t))c x(x(x)t(t)). (4.) ecall that to differentiate partially with respect to t, we treat the variable x as a constant. Therefore, when differentiating X(x)T(t) with respect to t, the function X(x) behaves as a constant and we have t (X(x)T(t)) X(x)T (t) where T (t) denotes the (ordinary) derivative of T(t). Similarly x (X(x)T(t))X (x)t(t)

139 4. The Heat Equation 39 and so from equation (4.) we have X(x)T (t)c X (x)t(t) T (t) c T(t) X (x) X(x) (4.) The left side T (t) c T(t) of equation (4.) is a function of the variable t; the right side X (x) of (4.) is X(x) a function of a different variable x. If T (t) c T(t) constant then the value of the function T (t) will vary with t. Similarly, if c T(t) X (x) X(x) constant then the value of the function X (x) will vary with x and, in fact, equation (4.) will be false X(x) because we can choose values of t and x such that T (t) c T(t) and X (x) take different values. The only X(x) T way for equation (4.) to hold is if (t) c T(t) and X (x) are both constant functions and are equal to X(x) the same constant k. So we have T (t) c T(t) X (x) k. (4.) X(x) We now need to determine if this constant k is zero, positive or negative. That is, we need to determine if k, a or a where a. Using the boundary conditions u(, t) u(, t) where t (4.3) we show that the only non-zero solutions of the heat equation (4.8) with these particular boundary conditions 4.4 occurs when k a : k If k then from equation (4.) we obtain two ordinary differential equations T (t) Solving these gives where A, B and C are constants. X (x). T(t)A X(x)B x +C (4.4) Substituting equations (4.4) into equation (4.9) gives where E A B and F A C. u(x,t)x(x)t(t)ex +F (4.5) From the first boundary condition of (4.3) we get u(, t) E()+F F 4.4. In Example 4.5 we will see that the value k will give a non-zero solution because the boundary conditions of that example are different from the boundary conditions of this example.

140 4 Partial Differential Equations The second boundary condition gives u(,t)e() + E As E F in (4.5), we have that the solution (4.5) of our heat equation corresponding to k must be identically equal to zero. k a If k a then from equation (4.) we have X (x) X(x) a X (x) a X(x) X(x)C cosh(ax)+dsinh(ax) where C and D are constants. From equation (4.9) we have From the first boundary condition of (4.3) we get u(x,t)(c cosh(ax) +D sinh(ax))t(t). (4.6) u(,t)(c cosh()+d sinh())t(t) (C()+D())T(t) CT(t) and this implies C. The second boundary condition gives u(, t)d sinh(a)t(t) (4.7) and as sinh a when a, equation (4.7) implies that D. As C D, we have that the solution (4.6) obtained when k a is identically equal to zero. The last possibility of k a will lead to non-zero solutions of the heat equation (4.8). Subsituting k a into equation (4.) gives from which we obtain T (t) c T(t) X (x) X(x) a T (t)+c a T(t) (4.8) X (x)+a X(x) (4.9) and this is the point of the seperation of variables method, the assumption converts the partial differential equation u(x, t)x(x)t(t) u t u c x into two ordinary differential equations (4.8) and (4.9) which can be easily solved. By using the D-operator method, the general solution of (4.8) is and the general solution of (4.9) is T(t) Ce c a t X(x) D cos(ax) +E sin(ax).

141 4. The Heat Equation 4 And so we have solutions of the heat equation (4.8) of the form u(x,t)e c a t (A cos(ax)+b sin(ax)) (4.3) where the arbitrary constants A CD and B CE. Note that a is also an arbitrary constant. We now impose the boundary conditions on equation (4.3). From the first boundary condition of (4.3) we get u(,t)e c a t (A cos()+b sin()) Ae c a t A. Substituting A into (4.3) implies that our solution of the heat equation (4.8) is of the form u(x,t)be c a t sin(ax). (4.3) Imposing the second boundary condition of (4.3) on equation (4.3) gives and so the second boundary condition forces u(,t) Be c a t sin(a) sin(a) a kπ where k,, 3, (4.3) (note we do not consider k as this leads to the zero solution; k,, do not give solutions that are linearly independent of the solutions corresponding to k,, ). Substituting each of these different values of a into equation (4.3) implies that e c π t sin(πx), e c (π) t sin(πx), e c (3π) t sin(3πx), are each solutions of the heat equation (4.8). So from by the Superposition Principle II on page 33 we have that u(x, t) n b n e c (nπ) t sin(nπx) (4.33) is a solution of the heat equation (4.8) which, by construction, satisfies the boundary conditions. Finally, we use the given initial condition u(x, )4x( x) where x to determine the constants b n of equation (4.33). Setting t in (4.33) gives u(x,) n b n e c (nπ) sin(nπx)4x( x) when x n b n sin(nπx)4x( x) when x (4.34) The left side of equation (4.34) is nothing but a Fourier sine series that represents the function 4x( x) on the interval x. ecall that such a sine series is called the sine half-range expansion of the function 4x( x) on the interval x and from Lemma 3.3 on page 8, the coefficients b n of the left side of (4.34) are given by b n L f(x)sin nπx L L dx b n 4x( x)sin(nπx) dx

142 4 Partial Differential Equations Integrating by parts with gives Integrating by parts again with v 4x( x) dv (4 8x)dx du sin(nπx)dx u cos(nπx) nπ [ ] 4x( x)cos(nπx) (4 8x)cos(nπx) b n + dx nπ nπ (4 8x)cos(nπx) + dx nπ gives v (4 8x) dv 8dx [ ] (4 8x)sin(nπx) b n n π [ + 8cos(nπx) ] n 3 π 3 ( 8( )n n 3 π ) n 3 π 3 when n is even 3 n 3 π 3 when n is odd du cos(nπx) dx nπ u sin(nπx) n π + 8sin(nπx) n π dx Substituting these values for b n into (4.33) and writing the odd values of n as k + gives our solution u(x,t) k 3 (k +) 3 π 3e c ((k+)π) t sin((k +)πx) (4.35) that satisfies the heat equation (4.8) subject to the given boundary and initial conditions. In the above problem, the metal bar initially had a non-zero temperature distribution given by the initial condition u(x, )4x( x) where x and the ends of the bar are held at zero temperature. As heat flows from a high temperature to a lower temperature, we expect the initial heat contained in the bar to flow out of the bar through the ends which are held at zero temperature. Therefore the bar should eventually lose its heat and hence each point in the bar should have zero temperature. This physical expectation actually agrees with the solution (4.35) above. First notice that where ( 3 π lim t sin(πx) t π 3e c lim t (. lim t ) 3 ( π t lim π 3e c ) sin(πx) lim e c π t t sin(πx) t )

143 4. The Heat Equation 43 as the coefficient c π of t in e c π t is negative. Similarly the limit of each term in the solution (4.35) is zero as each term in u(x,t) k 3 (k +) 3 π 3e c ((k+)π) t sin((k +)πx) 3 π 3e c π t sin(πx)+ 3 7π 3e c (π) t sin(πx)+ 3 5π 3e c (3π) t sin(3πx)+ contains a exponential with a negative coefficient. Therefore we have lim u(x,t) lim t t and note lim t lim t 3 ((k+)π) t (k +) 3 sin((k + )πx) π 3e c ( k 3 π t sin(πx) + 3 (π) t sin(πx)+ 3 ) (3π) t sin(3πx)+ π 3e c 7π 3e c 5π 3e c 3 π t 3 (π) sin(πx)+ lim t 3 (3π) sin(πx) + lim t sin(3πx) + π 3e c t 7π 3e c t 5π 3e c +++. lim u(x,t) t can be interpreted as meaning that the temperature of the bar u(x, t) eventually becomes zero. (4.36) 4..3 The heat equation with insulated ends as boundary conditions We now consider a similar heat conduction problem in a thin metal bar that is insulated along the sides of the bar. Again we assume the bar has a initial non-zero temperature distribution. However, unlike Example 4.4, we do not hold the ends of the bar at temperature zero. In this case we assume the ends of the bar to be insulated (so that the bar is entirely insulated). ecall from equation (4.) on page 36, that the heat flow at point x on the bar is given by k u (x, t). x If the ends x and x L of a bar of length L are insulated then there is no heat flow at these points and therefore k u u (, t) and k x x (L,t), and hence the insulation of the ends of a bar of length L can be mathematically stated as The equations (4.37) is another example of boundary conditions. u (, t) x (4.37) u x (L,t). Example 4.5. Let u(x, t) denote the temperature of a thin metal bar of length π that is insulated along the sides. Let the temperature along the length of the bar at time t be given by u(x,) x π where x π. The ends of the bar are insulated for time t and therefore u x (,t) u x (π,t) where t.

144 44 Partial Differential Equations Determine the temperature u(x, t) of the bar by solving the following heat equation u t c u x (4.38) where c is a fixed constant that depends on the physical properties of the bar. Answer: We use the method of seperation of variables and therefore assume that our solution u(x, t) takes the form Substituting the solution (4.39) into heat equation (4.38) gives u(x, t) X(x)T(t). (4.39) t (X(x)T(t))c x (X(x)T(t)) X(x)T (t) c X (x)t(t) T (t) c T(t) X (x) X(x) (4.4) and the only way for equation (4.4) to hold is if are equal to the same constant k. So we have T (t) c T(t) and X (x) X(x) where the constant k can be zero, negative or positive, that is are both constant functions and T (t) c T(t) X (x) k (4.4) X(x) k, a or a where a. We now use the boundary conditions u (, t) x where t (4.4) u x (π, t) to show that the only non-zero solutions of the heat equation (4.38) with the boundary conditions (4.4) occur when k and k a. k a If k a then from equation (4.4) we have X (x) X(x) a X (x) a X(x) X(x)C cosh(ax)+dsinh(ax) where C and D are constants. From equation (4.39) we have which implies u(x, t)(c cosh(ax)+d sinh(ax))t(t). (4.43) u (x, t) (ac sinh(ax)+ad cosh(ax))t(t) (4.44) x From the first boundary condition of (4.3) we get u (, t) (ac sinh()+ad cosh())t(t) x (ac()+ad())t(t) a DT(t)

145 4. The Heat Equation 45 and this implies D. The second boundary condition gives u (π,t)ac sinh(aπ)t(t) (4.45) x and as sinh (aπ) when a, equation (4.45) implies that C. As C D, we have that the solution (4.43) obtained when k a is identically equal to zero. k If k then from equation (4.4) we obtain two ordinary differential equations T (t) Solving these gives where A, B and C are constants. X (x). T(t)A X(x)B x +C (4.46) Substituting equations (4.46) into equation (4.39) gives where E A B and F A C. Taking the partial derivative gives From the first boundary condition of (4.4) we get The second boundary condition also gives u(x, t) Ex+ F (4.47) u (x,t)e (4.48) x u (,t)e. x u (π,t)e. x Notice that the boundary conditions (4.4) do not give any information about the constant F in equation (4.47), so it is possible that F, and therefore when k the heat equation (4.38) may have a non-zero solution of the form u(x, t)f. (4.49) k a The last possibility of k a will also lead to non-zero solutions. Subsituting k a into equation (4.4) gives T (t) c T(t) X (x) X(x) a from which we obtain the two ordinary differential equations By using the D-operator method, the general solution of (4.5) is T (t)+c a T(t) (4.5) X (x)+a X(x) (4.5) T(t) Ce c a t

146 46 Partial Differential Equations and the general solution of (4.5) is X(x) D cos(ax) +E sin(ax). And so we have solutions of the heat equation of the form u(x,t)e c a t (A cos(ax)+b sin(ax)) (4.5) where the arbitrary constants A CD and B CE. Note that a is also an arbitrary constant. By taking the partial derivative of (4.5) we get u x (x,t) e c a t ( aa sin(ax)+ab cos(ax)) (4.53) and we now impose the boundary conditions on equation (4.53). From the first boundary condition of (4.4) we get u x (, t)e c a t ( aa sin()+ab cos()) abe c a t B. Substituting B into (4.3) implies that the solutions corresponding to k a is of the form u(x,t)ae c a t cos(ax) u x (x,t)e c a t ( aa sin(ax)) (4.54) Imposing the second boundary condition of (4.4) on equation (4.54) gives and so the second boundary condition forces u x (π,t)e c a t ( aa sin(aπ)) a k sin(aπ) where k,,3, (note we do not consider k as this leads to the zero solution; k,, do not give solutions that are linearly independent of those corresponding to k,, ). Substituting each of these different values of a into equation (4.53) implies that e ct cos(x), e c t cos(x), e c 3 t cos(3x), are each solutions of the heat equation with the given boundary conditions. So from by the Superposition Principle II on page 33 we have that u(x, t) n a n e c n t cos(nx) (4.55) is a solution of the heat equation that corresponds to k a. From equation (4.49) u(x,t) a (4.56) is also a solution of the heat equation (that corresponds to k ) where we rename the constant F as a. Since both (4.55) and (4.56) are solutions, by the Superposition Principle their sum is a solution of the heat equation. u(x, t) a + a n e c n t cos(nx) (4.57) n

147 4. The Heat Equation 47 Finally, we use the given initial condition u(x,) x π where x π to determine the constants a n of equation (4.57). Setting t in (4.57) gives u(x, ) a + a n e c (nπ) cos(n x) x π n when x π a + a n cos(n x) x π n when x π (4.58) The left side of equation (4.34) is nothing but a Fourier cosine series that represents the function x on the interval x π. Such a cosine series is called the cosine half-range expansion of π the function x π on the interval x π and from Lemma 3.3 on page 8, the coefficients a n of the left side of (4.58) are given by a n L a n π When n we have L f(x)cos nπx L dx where n,,, π ( x ) cos(nx) dx where n,,, π a π [ π π ( x x π x ] π π ) dx When n,, then applying integration by parts to a n π ( x ) cos(nx) dx π π with gives v x π dv π dx a n [ ( x ) ] π sin(nx) + π π n π [ ] π cos(nx) π n when n is even 4 n π when n is odd du cos(nx)dx sin(nx) u n π sin(nx) n Substituting these values for a n into (4.58) and writing the odd values of n as k + gives our solution u(x,t) + 4 (k+) t (k +) cos((k +)x) (4.59) π e c k that satisfies the heat equation (4.38) subject to the given boundary and initial conditions. dx In the previous problem, the metal bar initially had a non-zero temperature distribution given by the initial condition u(x,) x π where x π.

148 48 Partial Differential Equations At time t the ends of the bar were insulated. As the bar is also insulated along the sides, this implies that the bar is entirely insulated for time t. Hence we expect no heat loss from the bar; furthermore we expect the initial heat of the bar to eventually distribute itself evenly throughout the length of the bar. So we expect as time t, that each point of the bar should have the same temperature. This physical expectation agrees with the solution (4.59) above, because and note lim u(x,t) lim t t + lim t ( + k (k +) π e c (k+) t cos((k + )x) 4 π e ct cos(x)+ lim t ) 4 3 π e 9ct cos(3x) + lim t lim t u(x, t) 4 5 π e 5ct cos(5x)+ can be interpreted as meaning that the temperature of each point on the bar u(x, t) eventually becomes equal to. Also notice that is the average of the initial temperature distribution as u(x,) x where x π π ( ) π x ( ) dx π π x dx π. length of bar π 4..4 The heat equation with nonhomogeneous boundary conditions The boundary conditions of the Example 4.4 u(,t) u(,t) where t and the boundary conditions of Example 4.5 u x (,t) u x (π,t) are examples of homogeneous boundary conditions. where t Definition 4.6. If the boundary conditions of the one-dimensional heat equation can be written in the form u t u c x where <x<a and t > αu(, t)+ β u (,t) x γu(a,t)+δ u where (a, t) x t where α, β, γ and δ are constants, then such boundary conditions are said to be homogeneous. Otherwise, the boundary conditions are called nonhomogeneous.

149 4. The Heat Equation 49 Consider the following heat equation with nonhomogeneous boundary conditions. Example 4.7. Let u(x, t) denote the temperature of a thin metal bar of length that is insulated along the sides. Let the temperature along the length of the bar at time t be given by u(x,)x where x. Let the temperature of the end of the bar corresponding to x be held at degrees; let the other end be held at degrees that is u(, t) u(,t) where t Determine the temperature u(x, t) of the bar by solving the following heat equation u t c u x (4.6) where c is a fixed constant that depends on the physical properties of the bar. Answer: We will use a substitution of the form u(x, t) u s (x,t) +U(x,t) to convert this heat equation with nonhomogeneous boundary conditions to a heat equation with homogeneous boundary conditions as in Example 4.4. We do this because the seperation of variables method fails in the case of nonhomogeneous boundary conditions. Consider the following function u s (x,t)x. (4.6) This function u s can be interpreted as a linear increase in temperature from the end x which is at temperature to the end x at temperature. We define our substitution U(x, t) by the equation U(x, t) u(x,t) u s (x,t), (4.6) the idea being to use u s to subtract the boundary condition from u(x, t) so that U(x, t) has zero temperature at both ends, that is, the boundary conditions for U are homogeneous: U(,t) u(, t) u s (,t) U(,t)u(,t) u s (, t) () where t

150 5 Partial Differential Equations The function u s was also chosen to be a solution of the heat equation. By differentiating (4.6) partially it is easy to see that u s u s t c x (4.63) and because (4.63) holds, we have that U(x, t) also satisfies the heat equation: substituting into gives and subtracting (4.63) from (4.64) gives u(x, t) U(x,t) +u s (x,t) (U +u s ) t u t u c x c (U + u s ) x U t + u s t c u s x + c U x (4.64) U t c U x. (4.65) (If the function u s were not a solution of the heat equation, then the partial differential equation in U obtained would not take the standard form of the heat equation (4.65) ). From equation (4.6) we can also determine the initial condition for U(x, t) u(x,)x +U(x, ) U(x, )x x for x. Therefore we have that the substitution (4.6) converts the heat equation u t u c x subject to the nonhomogeneous boundary conditions u(, t) u(,t) where t and initial condition u(x,)x where x to another heat equation U t U c x subject to the homogeneous boundary conditions and initial condition U(,t) U(,t) U(x, ) x x where t (4.66) where x

151 4. The Heat Equation 5 Now the heat equation (4.66) has boundary conditions exactly of the type considered in Example 4.4 and so the technique of solving (4.66) is exactly as the technique used to solve Example 4.4. We use the method of seperation of variables and therefore assume that the solution U(x, t) of (4.66) takes the form Substituting this into heat equation (4.66) gives U(x,t) X(x)T(t). t (X(x)T(t))c x (X(x)T(t)) T (t) c T(t) X (x) X(x) k and, exactly as in Example 4.4, the boundary conditions for U(x, t) imply that the only non-zero solutions are obtained when k a T (t) c T(t) X (x) X(x) a from which we obtain the two ordinary differential equations By using the D-operator method, the general solution of (4.67) is and the general solution of (4.68) is T (t)+c a T(t) (4.67) X (x)+a X(x) (4.68) T(t) Ce c a t X(x) D cos(ax) +E sin(ax). And so the solutions of the heat equation take the form U(x, t)e c a t (A cos(ax)+b sin(ax)) (4.69) where the arbitrary constants A CD and B CE. We impose the boundary conditions on equation (4.69). From the first boundary condition of (4.66) we get U(, t)e c a t (A cos() +B sin()) Ae c a t A. Substituting A into (4.69) implies that our solutions are of the form Imposing the second boundary condition on equation (4.7) gives U(x, t) Be c a t sin(ax). (4.7) U(,t)Be c a t sin(a) and so the second boundary condition forces a kπ sin(a) where k,, 3, Substituting each of these different values of a into equation (4.7) implies that e c( π ) t sin( πx ), π ( )t e c sin( πx ), 3π ( )t e c sin( 3πx ), are each solutions of the heat equation. So from by the Superposition Principle we have that U(x,t) b n e c( nπ ) t ( nπx ) sin n (4.7)

152 5 Partial Differential Equations is a solution of the heat equation (4.66) which, by construction, satisfies the boundary conditions. We use the initial condition U(x, ) x x where x to determine the constants b n of equation (4.7). Setting t in (4.7) gives U(x,) n b n e c( nπ ) sin ( nπx ) x x when x ( nπx ) b n sin x x when x (4.7) n The left side of equation (4.7) is a Fourier sine series that represents the function x x on the interval x. From Lemma 3.3 on page 8, the coefficients b n are given by b n L f(x)sin nπx L L dx b n ( nπx ) (x x)sin dx Integrating by parts with gives ( nπx ) v x x du sin dx dv (x )dx u ( nπx ) nπ cos ) ] b n [ (x x) ( nπx cos 5 nπ + ( nπx (x ) cos nπ + (x ) cos nπ ) dx ( nπx Integrating by parts again with ( nπx ) v x du cos dx dv dx u ( nπx ) nπ sin gives ( [ b n (x ) ( nπx ) ] ) ( nπx ) sin nπ nπ nπ sin dx [ ( nπx ) ] nπ n π cos ( ( ) n nπ n π ) n π when n is even 8 n 3 π 3 when n is odd Substituting these values for b n into (4.7) and writing the odd values of n as k + gives our solution for U(x,t) U(x, t) ( ) 8 (k+)π tsin ( ) (k +)πx (k +) 3. π 3e c k and since u(x, t) u s (x,t) +U(x,t) we obtain the solution of our nonhomogeneous problem ) dx

153 4. The Heat Equation 53 u(x, t)x k 8 (k +) 3 π 3e c ( (k+)π ) tsin ( ) (k +)πx (4.73) In Example 4.7, holding one end at temperature zero in effect creates a heat sink. Holding the other end at temperature creates a heat source, and we should expect that as time t that the presence of this source and sink to dominate the temperature distribution along the length of the bar. As this source and sink does not change with time (because the temperature at the ends of the bar are fixed) we should also expect that the temperature distribution along the bar to stabilize, that is, each point in the bar should as time t to have a temperature that is independent of time. This physical expectation agrees with the solution (4.73) of Example 4.7 as ( lim u(x, t) lim x ( ) 8 (k+)π tsin ( ) ) (k + )πx t t (k +) 3 π 3e c k( ( ) 8 (k+)π tsin ( ) ) (k +)πx lim x lim t t (k +) 3 π 3e c and note k 8 lim x lim ( t ) t π 3 e c x lim t x ( π ) t sin( πx ) lim 8 t lim u(x,t) x t 3 3 π 3e c ( π )t sin( πx ) can be interpreted as meaning that the temperature at any point x on the bar (where x ) will eventually reach the value of x. Also note that the temperature distribution is independent of time. x

154 54 Partial Differential Equations 4.3 The Wave Equation 4.3. A derivation of the wave equation Consider an elastic string that is stretched taut along the x-axis, with one end of the string fixed at the point x and the other end fixed at the point x L. The string is then distorted at time t and because the string is under tension, when the string is released it will oscillate about the x axis with the endpoints of the string being stationary. We make the assumption that each point on the string travels vertically under this oscillation. Therefore we can identify each point on the string by an x value and we let the displacement of this point from the horizontal axis at time t be given by the function value u(x, t). Note that as the ends of the string (which correspond to the points x and x L) are fixed we have the boundary conditions u(,t) for all t. (4.74) u(l, t) If the distortion of the string at time t is given by the function f(x) where x L, then we have an initial condition u(x,) f(x) where x L. (4.75)

155 4.3 The Wave Equation 55 Now if then u(x,t)displacement of point x at time t u (x,t) velocity of pointx at time t. t Therefore if the velocity of each point x on the string is given by the function g(x) where x L, then we have a second initial condition u (x,) g(x) where x L. (4.76) t Summarizing, our physical problem is as follows. We have a string that is stretched horizontally between two points that are at a distance L apart. The string has an initial distortion given by equation (4.75) and an initial velocity given by (4.76). Each point on the string can be identified with a point x where x L. We wish to find a function u(x, t) that measures the displacement of each point on the string from the horizontal axis at time t. This function u(x, t) will be a solution of the wave equation. To derive the wave equation we consider, at time t, a small segment AB of the string which consists of the points that correspond to the x values that lie between x and x + x where x represents a small positive number: In the above diagram T denotes the tension in the string at the point A; T denotes the tension at point B. When x is small enough we may assume that the segment of string AB does not move horizontally, this implies that the net horizontal force on the segment A B is zero. By considering the horizontal components of the tensions T and T we have T cosα T cosβ and so T cosα and T cosβ have a common value which we call T, therefore T cosαt cosβ T. (4.77) For x small enough we may also assume that the segment of string AB moves as a particle. The displacement of this particle from the horizontal axis is given by u(x, t), therefore the velocity of this particle in the vertical direction is u t (x, t) and the acceleration in the vertical direction is u (x, t). t By considering the vertical components of the tensions T and T, from Newton s Second Law we have T sin β T sinα(ρ x) u t (4.78)

156 56 Partial Differential Equations where ρ is the mass per unit length of the string. Dividing equation (4.78) by the value T from equation (4.77) we have T sin β T T sinα T ( ρ x u T ) t and using the expressions for T from equation (4.77) we have T sin β T cosβ T sinα T cosα (ρ x u T ) t tan β tanα ( ρ x T ) u t. (4.79) In the following diagram the line segment BC is tangent to the curve u(x, t) at the point B (note that we regard the t variable as being fixed). By the definition of slope we have slope of tangent BC CD tan β. BD However, from the definition of the partial derivative and therefore the slope of the tangent at B is Similarly the slope of the tangent at A is and from the equations of (4.79) we have earranging this equation gives u x and we take the limit of (4.8) as x to get lim x slope of tangent BC u (x + x, t) x u (x + x,t)tan β. x u (x, t) tanα x u u (x + x,t) x x (x,t)(ρ x u T ) t. ( u x u (x + x, t) (x, t) x ( ρ u x T ) t (4.8) u (x + x, t) x x ) (x, t) ( ρ u T ) t. (4.8)

157 4.3 The Wave Equation 57 By definition of the partial derivative lim x ( u u (x + x,t) (x,t) x x x and so from equation (4.8) we obtain u(x,t) x ) ( u x x (x,t) ) u(x,t) x ( ρ T ) u t (4.8) and as ρ and T are positive constants we may write equation (4.8) in the standard form of the wave equation where c is a constant. u u t c x 4.3. Solution of the wave equation by seperation of variables The solution of the wave equation by the method of seperation of variables is similar to the solution of the heat equation by seperation of variables that was done in Section 4... In solving the wave equation we shall assume a solution of the form u(x,t)x(x)t(t) substitute this solution into the wave equation to obtain T (t) c T(t) X (x) X(x) k consider the cases of k a,, a and use boundary conditions to determine which of these cases of k lead to nonzero solutions of the wave equation use the Superposition Principle to write these non-zero solutions in expressions that resemble 4.5 u(x,t) ( ( ) ( )) nπct nπct ( nπx ) a n cos +b n sin sin L L L n use initial conditions and the half-range cosine/sine formulae a n L f(x)sin nπx L L dx nπcb n L L g(x)sin nπx L L dx to determine the constants a n and b n, where f(x) and g(x) describe the initial displacement and velocity solutions may not always take this form

158 58 Partial Differential Equations We illustrate the solution of the wave equation by the following example. Example 4.8. Solve the wave equation subject to the boundary conditions and initial conditions u t u x (4.83) u(,t) u(π, t) u(x, ) sinx sinx u t (x,) where x π. Answer: We use the method of seperation of variables and therefore assume that our solution u(x, t) takes the form Substituting the solution (4.84) into wave equation (4.83) gives u(x, t) X(x)T(t). (4.84) t(x(x)t(t)) x (X(x)T(t)) X(x)T (t) X (x)t(t) T (t) T(t) X (x) X(x) (4.85) and the only way for equation (4.85) to hold is if T (t) T(t) are equal to the same constant k. So we have where the constant k can be zero, negative or positive, that is and X (x) X(x) are both constant functions and T (t) T(t) X (x) k (4.86) X(x) k, a or a where a. We now use the boundary conditions u(,t) u(l,t) where t (4.87) to show that the only non-zero solutions of the wave equation (4.83) with the boundary conditions (4.87) occur when k a. k If k then from equation (4.84) we obtain two ordinary differential equations T (t) Solving these gives X (x). T(t)A t + B X(x)C x+d (4.88) where A, B,C and D are constants.

159 4.3 The Wave Equation 59 Substituting equations (4.88) into equation (4.84) gives u(x,t) (C x+d )(A t + B ) (4.89) From the first boundary condition of (4.87) we get u(,t)d (A t + B ) for all values of t, which implies that D. The second boundary condition gives u(π,t) πc (A t +B ) for all values of t, which implies that As the boundary conditions imply that C. C D we have that when k, the corresponding solution (4.89) is identically equal to zero. k a If k a then from equation (4.84) we obtain two ordinary differential equations T (t) a T(t) Solving these gives X (x) a X(x). T(t)A cosh(at)+b sinh(at) X(x)C cosh(ax)+d sinh(ax) (4.9) where A, B,C and D are constants. Substituting equations (4.9) into equation (4.84) gives u(x,t)(c cosh(ax) +D sinh(ax))(a cosh(at)+b sinh(at)) (4.9) From the first boundary condition of (4.87) we get u(, t)(c () +D ())(A cosh(at)+b sinh(at)) for all values of t, which implies that The second boundary condition gives C. u(π, t) (D sinh(aπ))(a cosh(at) +B sinh(at)) for all values of t, and as sinh(aπ) when a, this implies that As the boundary conditions imply that D. C D we have that when k a, the corresponding solution (4.9) is identically equal to zero.

160 6 Partial Differential Equations k a The last possibility of k a will lead to non-zero solutions. Subsituting k a into equation (4.86) gives T (t) T(t) X (x) X(x) a from which we obtain the two ordinary differential equations By using the D-operator method, the general solution of (4.9) is and the general solution of (4.93) is T (t)+a T(t) (4.9) X (x)+a X(x) (4.93) T(t) Acos(at)+B sin(at) X(x)C cos(ax) +D sin(ax). So we have solutions of the wave equation of the form u(x, t)(a cos(at) +B sin(at))(c cos(ax) +D sin(ax)). (4.94) From the first boundary condition we get and therefore the solutions take the form where a AD and b BD. u(,t)(a cos(at)+b sin(at))(c() +D()) C(A cos(at)+b sin(at)) C u(x,t) (a cos(at) +b sin(at))sin(ax) (4.95) Imposing the second boundary condition on equation (4.95) gives and so the second boundary condition forces u(π,t) (a cos(at) +b sin(at))sin(aπ) a k sin(aπ) where k,,3, (note we do not consider k as this leads to the zero solution; k,, do not give solutions that are linearly independent of those corresponding to k,, ). Substituting each of these different values of a into equation (4.95) implies that (a cos(t)+b sin(t))sin(x), (a cos(t) +b sin(t))sin(x), (a 3 cos(3t)+b 3 sin(3t))sin(3x), are each solutions of the wave equation (4.83). So from by the Superposition Principle II on page 33 we have that u(x, t) n (a n cos(nt)+b n sin(nt))sin(nx) (4.96) is a solution of the wave equation (4.83) which, by construction, satisfies the boundary conditions. Finally, we use the given initial conditions u(x, ) sinx sinx u x (x,) where x π.

161 4.3 The Wave Equation 6 to determine the coefficients a n, b n of the solution (4.96). u(x,) n Setting t in (4.96) gives (a n cos()+b n sin())sin(n x)sinx sinx when x π n a n sin(n x)sinx sinx when x π (4.97) The left side of equation (4.97) is a Fourier sine series that represents the function sin x sin x on the interval x π. Such a sine series is called the sine half-range expansion of the function sin x sin x on the interval x π and from Lemma 3.3 on page 8, the coefficients a n of the left side of (4.97) are given by a n π Now if n is an integer, then a n L π L f(x)sin nπx L dx where n,, π ( sinx sinx ) sin(nx) dx where n,, π sin(nx)sin(nx) dx sin (nx) dx π cos(nx) dx π and when n m are both integers π π cos(m n)x cos(m + n)x sin(nx)sin(mx) dx. Therefore we have a π ( sinx ) π sinx sinx dx π sinx sinx dx π sinx sinx dx π π and similarly π.π π. a π ( sinx ) π sinx sinx dx π sinx sinx dx π sinx sinx dx π π dx π. π.π. For a k where k > we have a k π ( sinx ) π sinx sinkx dx π sinx sinkx dx π sinx sinkx dx π π π. π.

162 6 Partial Differential Equations and therefore we have determined the a n a a and a n for all n >. (4.98) Differentiating the solution (4.96) partially with respect to t gives and from the second initial condition we have which implies u t (x, t) ( na n sin(nt)+nb n cos(nt))sin(nx) n u t (x,) ( na n sin()+nb n cos())sin(nx) when x π n nb n sin(nx) (4.99) n The left side of equation (4.99) is a Fourier sine series that represents the function on the interval x π and from Lemma 3.3 on page 8, the coefficients nb n of the left side of (4.99) are given by nb n L nb n π L f(x)sin nπx L dx π where n,, ()sin(nx) dx where n,, nb n b n where n,, (4.) Substituting the values of the coefficients a n and b n given by (4.98) and (4.) into (4.96) we obtain our solution u(x,t)cos(t)sin(x) cos(t)sin(x) of the wave equation (4.83) that satisfies the given boundary and initial conditions.

163 4.4 Laplace s equation Laplace s equation Laplace s equation is a partial differential equation that does not involve the time variable t. In Cartesian coordinates, the form of Laplace s equation depends on dimension: Laplace s equation in dimension Laplace s equation in dimensions Laplace s equation in 3 dimensions d u dx u x + u y u x + u y + u y Laplace s equation arises when one considers solutions of physical problems that are independent of time. Solutions that are independent of time are called steady state solutions. Example 4.9. In Example 4.7 we showed on page 5 that is a solution of the heat equation u s (x,t)x u t u c x subject to the nonhomogeneous boundary conditions u(, t) u(,t) where t Notice that the solution u s does not involve the t variable, so u s is an example of a steady-state solution. As u s is a solution of this heat equation we clearly have u s t c u s x. (4.) Furthermore as u s is independent of the time variable t, when we differentiate partially with respect to t we have u s (4.) t and substituting (4.) into (4.) we have that, after dividing by c, that and as u s is a function of x only we have that u s x d u s dx which is Laplace s equation in one dimension. So we see that u s is an example of a steady-state solution of a one-dimensional heat equation being a solution of Laplace s equation in one-dimension. To be precise, u s is a solution of the Laplace s equation together with specified boundary conditions:

164 64 Partial Differential Equations d u dx subject to the boundary conditions u() u() Finally, recall that the limit of the solution of the heat equation in Example 4.7 ( lim u(x,t) lim x ( ) 8 (k+)π tsin ( ) ) (k +)πx t t (k +) 3 π 3e c k x u s (x,t) So we see that the steady-state solution u s of the the heat equation in Example 4.7 gives the temperature distribution that the metal bar eventually stabilizes to. As u s is a solution of a Laplace s equation, we have that the limiting temperature distribution of the solution of a heat equation can sometimes be obtained as a solution of a corresponding Laplace s equation with the appropriate boundary conditions. We see an example of this in the following Example 4.. From Example 4.4 we saw the solution of the following heat equation problem Let u(x, t) denote the temperature of a thin metal bar of length that is insulated along the sides. Let the temperature along the length of the bar at time t be given by u(x, )4x( x) where x. The ends of the bar are kept at a constant temperature of, that is u(,t) u(,t) where t. is given by u(x,t) k 3 (k +) 3 π 3e c ((k+)π) t sin((k +)πx) and this temperature function eventually stablilizes to lim u(x,t) (4.3) t as we would expect because all the initial heat in the bar flows out of the ends of the bar which are at zero temperature. Notice that this limiting temperature distribution of zero (4.3) is identical to the solution of a corresponding one-dimensional Laplace s equation with the same boundary conditions as the heat equation d u dx subject to the boundary conditions u() u()

165 4.4 Laplace s equation 65 because the solution of the above Laplace s equation is easily shown to be zero: and the boundary conditions imply that hence d u dx u(x)ax+b u() u() A B u. which is the limiting temperature distribution of (4.3). The previous two examples provide evidence that Laplace s equation in one dimension can be interpreted as a steady-state heat flow problem in one dimension, that is, a problem of finding a steady-state solution of a one dimensional heat equation that satisfies some given boundary conditions. In this section we shall be interested in solving the two dimensional Laplace s equation over a rectangular domain. The following lemma implies that Laplace s equation in two dimensions can also be interpreted as a steady-state heat flow problem in two dimensions, that is, a problem of finding a steady-state solution of a two dimensional heat equation that satisfies certain conditions on the boundary of the domain. Lemma 4.. Let u s be a steady-state solution of the two dimensional heat equation ( u ) u t c x + u y. (4.4) Then u s is a solution of Laplace s equation in two dimensions u x + u y. Proof. As u s is a solution of this heat equation we clearly have ( u s ) u s t c x + u s y (4.5) As u s is steady-state, it is independent of the time variable t and therefore u s t (4.6) and by substituting (4.6) into (4.5) and dividing by c we have that u s x + u s y and therefore u s is a solution of Laplace s equation in two dimensions.

166 66 Partial Differential Equations 4.4. Solving Laplace s equation by seperation of variables We illustrate the solution of Laplace s equation over a rectangular domain with prescribed boundary conditions by the following example. Example 4.. By solving Laplace s equation in the square domain u x + u y (4.7) {(x, y) x and y }, find the steady-state temperature u(x, y) of a thin square metal plate of length which has each of its four sides at temperatures that are given by the following boundary conditions u(, y) u(, y) u(x,) u(x, ) x( x) where x and y. (4.8) Answer: The following diagram illustrates the boundary conditions around the square plate: We solve this Laplace s equation by using seperation of variables and therefore assume that our solution u(x, y) takes the form u(x, y) X(x)Y (y). (4.9) Substituting the solution (4.9) into Laplace s equation (4.7) gives x(x(x)y (y))+ y(x(x)y (y)) X (x)y (y)+x(x)y (y) X (x) X(x) Y (y) Y (y) (4.)