Applied Differential Equation. October 22, 2012

Transcription

1 Applied Differential Equation October 22, 22

2 Contents 3 Second Order Linear Equations 2 3. Second Order linear homogeneous equations with constant coefficients Solutions of Linear Homogeneous Equations, the Wronskian Complex Roots of the characteristic Equations Repeated Roots Nonhomogeneous Equation, Method of undetermined coefficients Variation of Parameters Mechanical and Electrical Vibrations Review of Chapter The Laplace Transform Definition of the Laplace transform Solution of Initial Value Problems Step functions Differential Equations with Discontinuous Forcing Functions Impulse functions Dirac delta function The Convolution Integral Review of Chapter

3 Chapter 3 Second Order Linear Equations Two important areas of application of 2nd order linear equations are in the fields of mechanical and electrical oscillations. Example Mechanical system of spring. l l + L m KL mg gravitational force: F g = mg m: mass of the ball, g acceleration due to gravity. Spring force (Hooke s Law): F s = kl L: displacement. k: spring constant. Hooke s Law: Spring force is proportional to the displacement of the spring (L). The direction of the force is opposite to the direction of the motion. when the ball is in equilibrium, mg = kl, k = mg L. 2

4 l still l + L l + L + u(t) m still m vibrated u(t) F s F g = mg Newton s Law of motion (neglect damping or resistive force such as air resistance) Suppose the spring is stretched by a displacement of.l and released later. l: original length of the spring. L: displacement of the spring caused by gravity in still state. u(t): displacement caused by vibration. F g = mg: gravity force. F s = k(l + u(t)): elastic force. Choose the downward direction as positive direction, that is, u(t) is positive, when the spring is stretched. F = ma, a = d2 u dt = u (t) mu (t) = mg k(l + u(t)): downward direction is positive, so mg has a positive sign. No matter what we choose as the positive direction, the elastic force is always opposite to direction of u(t), its sign is always negative. mg = kl, k = mg : see the second graph, when the spring is still, it is obvious mg = kl. L mu = mg mg (L + u), L u = g L u u() =.L, v = u () = : initial condition. 3

5 u = g L u u() =.L u () = Generally speaking, 2nd order DE: 2nd order linear DE: d 2 y dy = f(x, y, dx2 dx ) y + p(x)y + q(x)y = g(x) y(x ) = y y (x ) = y 3. Second Order linear homogeneous equations with constant coefficients Example : y = y Possible solutions: e x, e x, C e x, C 2 e x ; General solution y = C e x + C 2 e x. ay + by + cy = (a ) For ay + by + cy =, seek for solution in form y = e rx, r to be determined. (ar 2 + br + c)e rx = ar 2 + br + c = r,2 = b ± b 2 4ac 2a (characteristic equation) If b 2 4ac >, then r, r 2 real and different, then e r x and e r 2x are two different solutions. y = C e r x + C 2 e r 2x C and C 2 can be determined by y(x ) = y, y (x ) = y. Ex. y + 4y + 3y =, y() = 2, y () = r 2 + 4r + 3 = (r + 3)(r + ) = r = 3, r = y(x) = C e 3x + C 2 e x 4

6 y() = C e + C 2 e = C + C 2 = 2 y (x) = 3C e 3x C 2 e x y () = 3C C 2 = C = 2, C 2 = 5 2 y(x) = 2 e 3x e x Ex 2. Find the general solution of 2y 3y + y = 2r 2 3r + = (r 2)(r ) = r = 2 r = y(x) = C e 2x + C 2 e x Ex 3. y = 9.8y, y() =., y () = r = r = ± 9.8 y(x) = C e 9.8x + C 2 e 9.8x x,. y, C + C 2 =. y (x) = 9.8C e 9.8x 9.8C 2 e 9.8x x, y, 9.8C 9.8C 2 = C = C 2 =.5 y(x) =.5e 9.8x +.5e 9.8x 3.2 Solutions of Linear Homogeneous Equations, the Wronskian Recall { ay + by + cy = ar 2 + br + c = r, r 2 real, different. Then y(x) = C e r x + C 2 e r 2x Linear homogeneous equation: y + p(x)y + q(x)y = y(x ) = y y (x ) = y where p(x), q(x) are defined in an interval I = (α, β) and x (α, β). Theorem 2. (Existence and Uniqueness). y + p(t)y + q(t)y = g(t), y(t ) = y, y (t ) = y 5

7 If p(x), q(x) and g(x) are continuous in I : t I, then there is exactly one solution y = φ(x) of the IVP (Initial Value Problem), and the solution exists throughout the interval I. Theorem 2.2 (Superposition Principle) If y (x) and y 2 (x) are two solutions of y + p(x)y + q(x)y = Then the linear combination C y (x) + C 2 y 2 (x) is also a solution for any values of the constants C and C 2. Proof: General Solution: by Crammer s rule, once y and y 2 are known, and given initial value y(x ) = y, y (x ) = y, we can determine C and C 2 as following: y(x) = C y (x) + C 2 y 2 (x), C y (x ) + C 2 y 2 (x ) = y C y (x ) + C 2 y 2(x ) = y y y 2 (x ) y y 2(x ) y (x ) y y C = y (x ) y, C (x ) y 2 (x ) 2 = y (x ) y 2(x ) y (x ) y 2 (x ) y (x ) y 2(x ) W [y, y 2 ](x ) = y (x ) y 2 (x ) y (x ) y 2(x ) W = W [y, y 2 ](x) = y (x) y (x) Theorem 2.3 (General solution) y 2 (x) y 2(x) Wronskian determinant of y and y 2 If y and y 2 are two solution of y + p(x)y + q(x)y = and if there is a point x where the wronskian of y and y 2 is nonzero, i.e., W [y, y 2 ](x ) = y (x ) y 2 (x ) y (x ) y 2(x ), then the general solution of the DE is y = C y (x) + C 2 y 2 (x) where C and C 2 are arbitrary constants. y and y 2 are said to form a fundamental set of solutions. 6

8 Example y + 4y + 3y =. r 2 + 4r + 3 = (r + 3)(r + ) = r = 3, r = Two particular solutions y (x) = e 3x and y 2 (x) = e x. At x =, y (x ) = e =, y 2 (x ) = e =, y () = 3, y 2() =, So W [y, y 2 ]() = 3 = + 3 = 2 is the fundamental solution. Def: Linear Independence of functions y = C e 3x + C 2 e x Two functions f and g are said to be linearly dependent on interval I if there exist two constants k and k 2, not both zero, such that k f(t) + k 2 g(t) =, t I f and g are said to be linearly independent on I if they are not linearly dependent. Example : f(t) = 3t, g(t) = t, (a) I = (, ), (b) I = (, ). k 3t + k 2 t =, t I (a)i = (, ): k 3t + k 2 t = k 3t + k 2 t = (3k + k 2 )t = k 2 = 3k So as long as k 2 = 3k, for example k 2 = 3 and k =, then k 3t + k 2 t =, implies f(t) and g(t) are linear dependent on (, ). t I, which (b)i = (, ) : { k 3t + k 2 t =, t [, ) k 3t k 2 t =, t (, ) k 3t + k 2 t =, t [, ) k 2 = 3k and k 2 = 3k k 3t k 2 t =, t (, ) k 2 = 3k k = k 2 = 7

9 which implies f(t) and g(t) are linear independent on (, ). Example 2: f(t) = t, g(t) = t 2, I = (, ) k t + k 2 t 2 =, k = k 2 t, t = k k 2 t (, ) Linear independence The Wronkskian Theorem 3. (Abel s Theorem) If y and y 2 are two solutions of y + p(x)y + q(x)y = where p and q are continuous on an open interval I, then the Wronskian W [y, y 2 ](x) is given by W [y, y 2 ](x) = Ce p(x)dx where C is a certain constant that depends on y and y 2, but not on x. Furthermore, w[y, y 2 ](x) is either zero for all x I (if C = ) or else is never zero in I (if C ) Proof. +p(x) y 2 y 2 y + p(x)y + q(x)y ] y 2 = }{{} () = [y 2 + p(x)y 2 + q(x)y 2 ] y = }{{} (2) = (2) () y y 2 y 2 y }{{} (y } {{ ) = } =W =W W = y y 2 y 2 y by definition W = y y 2 + y y 2 y 2y y 2 y = y y 2 y 2 y W + p(x)w =, a separable equation [y W W = p(x), (ln W ) = p(x) ln W = p(x)dx + C, W = Ce p(x)dx Example Find the Wronskian of x 2 y (x + 2)y + (x + 2)y = without solving this DE. First, we need to divide the differential equation by the first coefficient of the second derivative term:x 2. This gives us y x + 2 x 2 y + x + 2 x 2 =. It turns out p(x) = x + 2 x 2. x + 2 x x 2 dx = x x 2 dx = ln x 2 x W [y, y 2 ](x) = Ce p(x)dx = Ce ln x 2 x = Cxe 2 x 8

10 Euler s Formula Motivation: e 3+6i Taylor series of e x e x x n = n! n= e ix (ix) n = = i 2n x 2n + n! (2n)! n= n= n= ( ) n x 2n = + i (2n)! n= n= i 2n+ x 2n+ (2n + )! ( ) n+ x 2n+ (2n + )! = cos x + i sin x Euler s formula Example : e λ+iµx = e λ e iµx = e λ (cos(µx) + i sin(µx)) e 3+6i = e 3 (cos(6) + i sin(6)) Property: derx dx = rerx 3.3 Complex Roots of the characteristic Equations Consider ay + by + cy =. y = e rx, r to be determined. Characteristic equation: ar 2 + br + c =. If b 2 4ac <, then b ± b 2 4ac = b ± ( ) b 2 4ac 2a 2a r = λ + iµ, r 2 = λ iµ λ = b b 2a, µ = 2 4ac 2a y (x) = e (λ+iµ)x = e λx (cos µx + i sin µx) = b ± i b 2 4ac 2a y 2 (x) = e (λ iµ)x = e λx (cos µx i sin µx) 9

11 e (λ+iµ)x e (λ iµ)x W [y, y 2 ](x) = (λ + iµ)e (λ+iµ)x (λ iµ)e (λ iµ)x = (λ iµ)e 2λx (λ + iµ)e 2λx = 2iµe 2λx General solution: y(x) = C y (x) + C 2 y 2 (x) y(x) = C e λx (cos µx + i sin µx) + C 2 e λx (cos µx i sin µx) = (C + C 2 )e λx cos µx + i(c C 2 )e λx sin µx Define: C = C + C 2, C2 = i(c C 2 ). r = λ ± iµ y(x) = C e λx cos µx + C 2 e λx sin µx, ỹ (x) = e λx cos µx ỹ 2 (x) = e λx sin µx Here ỹ and ỹ 2 are two real solutions. Example y = y y + y = r 2 + = general solution b 2 4ac = 4 < λ = b b 2a =, µ = 2 4ac = 2a y (x) = e ix = cos x + i sin x, y 2 (x) = e ix = cos x i sin x y(x) = C e ix + C 2 e ix C = C + C 2, C2 = i(c C 2 ) ỹ (x) = cos x, ỹ 2 = sin x. Example 2 y + y +.25y =, y() = 3, y () = r 2 + r +.25 = b 2 4ac = 5 = 4 λ = b 4 2a = 2, µ = = 2 y = e 2 x+ix = e 2 x (cos x + i sin x), y 2 = e 2 x ix = e 2 x (cos x i sin x) y(x) = C y (x) + C 2 y 2 (x) C = C + C 2 C2 = i(c C 2 ) ỹ (x) = e 2 x cos x, ỹ 2 (x) = e 2 x sin x.

12 3.4 Repeated Roots Consider: ay + by + cy = ar 2 + br + c = r,2 = b + b 2 4ac 2a b 2 4ac > : r and r 2 real and different. b 2 4ac < : r and r complex and different. b 2 4ac = : r = r 2 = b 2a real. In the last case, y (x) = e r x = y 2 = e r 2x. W [y, y 2 ](x) = General solution? To find the general solution, we have to find y 2 (x) which is different from y (x). Assume that y 2 (x) has the form where υ(x) is to be determined. y 2 = υ e b 2a x b 2a υe b 2a x y 2 = υ e b 2a x b a v e b 2a x + b2 4a 2 υe b 2a x υ(x)y (x) = υ(x)e r x = υ(x)e b 2a x a[υ e b 2a x b a υ e b 2a x + b2 4a 2 υe b 2a x ] + b[υ e b 2a x b 2a υe b 2a x ] + cυe b 2a x = aυ + c b2 υ = }{{ 4a } = υ = υ(x) = C x + C 2 y 2 (x) = C xe r x + C 2 e r x, r = b 2a

13 { y (x) = e r x y 2 (x) = xe r x Verify y 2 (x) = xe r x is also a solution. y 2 = e r x + r xe r x y 2 = r e r x + r e r x + r 2 xe r x = 2r e r x + r 2 xe r x Check a[2r e r x + r 2 xe r x ] + b[e r x + r xe r x ] + cxe r x = 2ar + ar 2 x + b + br x + cx b 2a r, b + b2 4a x + b b2 2a x + b2 4a x = General solution: y(x) = C e b 2a x + C 2 xe b 2a x Summary General solution of ay + by + cy = ar 2 + br + c = r,2 = b ± b 2 4ac 2a b 2 4ac > : y(x) = C e r x + C 2 e r 2x b 2 4ac < : λ = b b 2a, µ = 2 4ac 2a y(x) = e λx (C cos µx + C 2 sin µx) b 2 4ac = : y(x) = C e b 2a x + C 2 xe b 2a x Examples: y 2y + y = y = 4y 9y + 6y + y = y + 4y + 2y = y = 4y 2

14 3.5 Nonhomogeneous Equation, Method of undetermined coefficients Homogeneous DE: ay + by + cy = Nonhomogeneous DE: ay + by + cy = g(x)? Theorem: The general solution of nonhomogeneous equation can be written in the form y + p(x)y + q(x)y = g(x) y = C y (x) + C 2 y 2 (x) + Y (x) where y and y 2 are a fundamental set of solutions of the corresponding homogeneous equation y + p(x)y + q(x)y = C and C 2 are arbitrary constants, and Y is some specific solution of the nonhomogeneous equation, such that Y + p(x)y + q(x)y = g(x) Procedure: Find y and y 2 Find Y (x) y = C y (x) + C 2 y 2 (x) + Y (x) How to find Y (x): The Method of undetermined coefficents ay + by + cy = g (x) + g 2 (x) Splitting If Y (x) is the solution of ay + by + cy = g (x), Y 2 (x) is the solution of ay + by + cy = g 2 (x) then Y (x) + Y 2 (x) is a solution of ay + by + cy = g (x) + g 2 (x). Examples (a) y 2y 3y = 3e 2x r 2 2r 3 =, (r 3)(r + ) =, r =, 3 3

15 y = C e x + C 2 e 3x + Y (x) Guess: Y = Ae 2x Y = 2Ae 2x, Y = 4Ae 2x (4A 2A 3)e 2x = 3e 2x A = 3, Y = 3e 2x y = C e x + C 2 e 3x + 3e 2x (b) y + 4y = 5 sin 3x, y() = 2, y () = r =, r = ±2i, λ =, µ = 2 y = C cos 2x + C 2 sin 2x + Y Guess Y = A sin 3x Y = 3A cos 3x Y = 9A sin 3x 9A sin 3x + 4A sin 3x = 5 sin 3x, y = y = C cos 2x + C 2 sin 2x sin 3x (c) y 4y 2y = sin(2t) r 2 4y 2 =, r = 2, 6 y = C e 2t + C 2 e 6t + Y Guess Y = A sin(2t) + B cos(2t) Y = 2A cos(2t) 2B sin(2t) Y = 4A sin(2t) 4B cos(2t), A = ( 6A + 8B) sin(2t) (6B + 8A) cos(2t) 6A + 8B =, (6B + 8A) =, A = 2, B = 4 (d) y 3y 4y = 8e x cos 2x r 2 3r 4 =, (r 4)(r + ) =, r =, 4 y = C e x + C 2 e 4x + Y Guess Y = Ae x cos 2x + Be x sin 2x Y Y Y e x cos 2x e x cos 2x 2e x sin 2x 3e x cos 2x 4e x sin 2x e x sin 2x e x sin 2x + 2e x cos 2x 3e x sin 2x + 4e x cos 2x ( A 2B) e x cos 2x + (2A B) e x sin 2x = 8e x cos 2x }{{}}{{} = 8 = A = 5 3, B = 2 3 4

16 (e) y 3y 4y = 2t 2 t + r 2 3r 4 =, r =, 4 y = C e x + C 2 e 4x + Y Guess Y = At 2 + Bt + C Y = 2At + B Y = 2A 2A 6At 3B 4At 2 4Bt 4C = t 2 t 4A = (6A + 4B) = (2A 3B 4C) = A B C = A =.25, B =.25, C = Variation of Parameters Find a specific solution of the non-homogeneous equation. Example Find the general solution of y + 4y = sin x Solution Find y and y 2 of y + 4y =, y (x) = cos 2x and y 2 (x) = sin 2x, y(x) = C y (x) + C 2 y 2 (x)(*) The basic idea in the method of variation of parameters is to replace the constants C and C 2 in (*) by functions u (x) and u 2 (x), resp., and then to determine these functions so that Y (x) = u (x) cos 2x + u 2 (x) sin 2x is a particular solution of the non-homogeneous equation. Y = u cos 2x + u 2 sin 2x 2u sin 2x + 2u 2 cos 2x Y = u cos 2x + u 2 sin 2x 2u sin 2x + 2u 2 cos 2x 2u sin 2x + 2u 2 cos 2x 4u cos 2x 4u 2 sin 2x = u cos 2x + u 2 sin 2x 4u sin 2x + 4u 2 cos 2x 4u cos 2x 4u 2 sin 2x Since Y + 4Y = sin x [u cos 2x + u 2 sin 2x 4u sin 2x + 4u 2 cos 2x 4u cos 2x 4u 2 sin 2x] + 4[u cos 2x + u 2 sin 2x] = sin x u cos 2x + u 2 sin 2x 4u sin 2x + 4u 2 cos 2x = sin 2x 5

17 which is one equation for unknown functions u and u 2. To determine them, we need to seek for another equation or relation. Let us impose one more restriction u cos 2x + u 2 sin 2x =, then Y = 2u sin 2x + 2u 2 cos 2x Y = 2u sin 2x + 2u 2 cos 2x 4u cos 2x 4u 2 sin 2x Since Y + 4Y = sin x 2u sin 2x + 2u 2 cos 2x = sin x u sin 2x = cos 2x u 2 2u (sin 2x) 2 2 cos 2x + 2u 2 cos 2x = sin x u 2 = sin x cos 2x 2 u 2 = sin x cos 2xdx 2 = [sin(x + 2x) sin(x 2x)]dx 2 2 = 2 cos 3x 4 cos x + C 3 u = sin x sin 2x 2 u = sin x sin 2xdx 2 = 4 sin x + 2 sin 3x + C 4 Y (x) = [ 4 sin x + 2 sin 3x] cos 2x + [ 2 cos 3x cos x] sin 2x 4 y(x) = C cos 2x + C 2 sin 2x + Y (x) Summary To solve y + p(x)y + q(x)y = g(x) (*) Suppose y (x) and y 2 (x) are solutions of y + p(x)y + q(x)y = Y (x) is a particular solution of (*), in the form of Y (x) = u (x)y (x) + u 2 (x)y 2 (x). Take the first derivative of Y, we have Y = u y + u 2y 2 + u y + u 2 y 2. 6

18 Suppose u y + u 2y 2 =, then Y = u y + u 2 y 2. Y = u y + u 2y 2 + u y + u 2 y 2 Since Y + py + qy = g u y + u 2y 2 + u y + u 2 y 2 + p[u y + u 2 y 2] + [u y + u 2 y 2 ] = g u y + u 2y 2 + u [y + py + qy ] +u 2 [y 2 + py 2 + qy 2 ] = g }{{}}{{} = = u y + u 2y 2 = g { u By y + u 2y 2 = u y + u 2y 2 = g y 2 u g y 2 y = y 2 g = y 2 y y 2 W [y, y 2 ](x) y u y g y 2 = y g = y 2 y y 2 W [y, y 2 ](x) gy 2 u = W [y, y 2 ](x) dx, u 2 = Y (x) = y gy 2 W [y, y 2 ](x) dx + y 2 gy W [y, y 2 ](x) dx y = C y (x) + C 2 y 2 (x) y gy W [y, y 2 ](x) dx Theorem If y and y 2 are a fundamental set of solutions of then a particular solution of is and general solution is Y (x) = y gy 2 W [y, y 2 ](x) dx + y 2 y + p(x)y + q(x)y = y + p(x)y + q(x)y = g(x) gy 2 W [y, y 2 ](x) dx + y 2 y = C y (x) + C 2 y 2 (x) + Y (x) 7 gy W [y, y 2 ](x) dx gy W [y, y 2 ](x) dx

19 Example y + 4y = sin x y + 4y = gives us y = cos 2x, y 2 = sin 2x W [y, y 2 ](x) = cos 2x sin 2x 2 sin 2x 2 cos 2x = 2 gy 2 sin 2x sin x W [y, y 2 ](x) dx = dx = 2 4 sin x 2 sin 3x + C gy cos 2x sin x W [y, y 2 ](x) dx = dx = 2 4 cos x 2 cos 3x + C 2 A particular solution is Y (x) = cos 2x[ 4 sin x 2 sin 3x] + sin 2x[ 4 cos x cos 3x] 2 General solution is y(x) = C cos 2x + C 2 sin 2x + Y (x) Example 2: y y 2y = 2e x To find a set fundamental solutions of y y 2y =, r 2 r 2 = (r 2)(r + ) = r = 2, y = e 2x, r = y 2 = e x W [y, y 2 ](x) = y y 2 y y 2 = e 2x e x 2e 2x e x = ex 2e x = 3e x gy 2 2e x W [y, y 2 ](x) dx = e x 3e x dx = 2 e 3x dx = e 3x gy 2e x W [y, y 2 ](x) dx = e 2x 3e x dx = 2 3 dx = 2x 3 A particular solution is Y = e 2x 2 9 e 3x e x 2x ( 2 3 = 9 + 2x ) e x 3 General solution is y(x) = C e 2x + C 2 e x + Y (x) 8

20 3.7 Mechanical and Electrical Vibrations l still l + L l + L + u(t) m still m vibrated u(t) F s F g = mg undamped free vibration m: mass. k: spring constant. g: gravity constant. L: length of spring stretched by mass m. mg = kl. mu + ku = u = A cos k k m t + B sin m t = A cos w t + B sin w t w = = R cos(w t δ) k m Here A = R cos δ, B = R sin δ or R = A 2 + B 2, tan δ = B A. w = k : in radians per unit time, natural frequency of a vibration. m T = 2π w : period of a vibration. R: amplitude. 9

21 R R cos δ I δ 2π + δ w t δ: phase or phase angle. Damped free vibration air resistance force = γv(t), γ : damped constant mu + γu + ku =, γ > ( = γ ± 2m r,2 = γ ± γ 2 4mk 2m 4km γ 2 ) γ 2 4km > : u = Ae r t + Be r 2t γ 2 4km = : u = (A + Bt)e γt 2m γ 2 4km < : u = e γ 2m t (A cos µt + B sin µt) = Re γ 2m t cos(µt + δ), µ = R = A 2 + B 2, tan δ = A B. µ: quasi frequency, T d = 2π µ : quasi-period. 4km γ 2 2m >, 2

22 R cos δ Re γt δ δ Damped vibration with external force mu + γ }{{} damped constant u + ku = F cos wt }{{} external periodical force (a) γ =, w w u = C cos w t + C 2 sin w t + = R cos(w t δ) + F m(w 2 cos wt w2 ) F m(w 2 cos wt w2 ) u() =, u () = (no initial displacement, no initial velocity) F R cos δ + m(w 2 w2 ) = F R = m(w 2 w2 ) Rw sin δ = δ = F u = m(w 2 w2 ) (cos wt cos w t) ( 2F = m(w 2 w2 ) sin (w ) w)t sin w + w t 2 2 Example u + u =.5 cos.8t u() =, u () = u = sin.t sin.9t 2

23 Graph of u = sin.t sin.9t sin.t (b) w = w u = C cos w t + C 2 sin w t + F 2mw t sin w t t u. Example 2 u + u =.5 cos t, u() =, u () = then u =.25t sin t. y =.25t This phenomenon is called resonance. 22

24 3.8 Review of Chapter 3 Constant coefficients homogeneous equations ay + by + cy =, ar 2 + br + c = b 2 4ac particular solutions general solution > y = e rx, y 2 = e r 2x y = C e rx + C 2 e r 2x < y = e λx cos(µx), y 2 = e λx sin(µx) y = e λx [C cos(µx) + C 2 sin(µx)] = y = e b 2a x, y 2 = xe b 2a x y = [C + C 2 x]e b 2a x Constant coefficients non-homogeneous equations general solutions: ay + by + cy = g(x) y = C y (x) + C 2 y 2 (x) + Y (x) where y and y 2 are a set of fundamental solutions of ay + by + cy =, and Y (x) is a particular solution satisfying ay + by + cy = g(x) Method : undetermined coefficients a b c and d are known constants, and A, B, C and D are unknown constants to be determined. g(x) e ax cos(ax) sin(ax) ax 3 + bx 2 + cx + d (ax + b)e ax (ax + b) cos(ax) (ax + b) sin(ax) sin(ax)e bx Y (x) Ae ax A cos(ax) + B sin(ax) A cos(ax) + B sin(ax) Ax 3 + Bx 2 + Cx + D (Ax + B)e ax (Ax + B) cos(ax) + (Cx + D) sin(ax) e bx [A sin(ax) + B cos(ax)] Note: if g(x) is also a solution to the homogeneous equation, then multiply x to Y (x) in the above table. Method 2: Variation of parameters gy 2 Y (x) = y W [y, y 2 ](x) dx + y 2 Existence and Uniqueness theorem gy W [y, y 2 ](x) dx y + p(t)y + q(t)y = g(t), y(t ) = y, y (t ) = y If p(x), q(x) and g(x) are continuous in I : t I, then there is exactly one solution y = φ(x) of the IVP (Initial Value Problem), and the solution exists throughout the interval I. 23

25 Wronskian W [y, y 2 ](x) = y y 2 y y 2 Abel s Theorem If y and y 2 are two solutions of y + p(x)y + q(x)y = where p and q are continuous on an open interval I, then the Wronskian W [y, y 2 ](x) is given by W [y, y 2 ](x) = Ce p(x)dx Fundamental set of solutions: If y and y 2 are two solution of y + p(x)y + q(x)y = and if there is a point x where the wronskian of y and y 2 is nonzero, i.e., W [y, y 2 ](x ) = y (x ) y 2 (x ) y (x ) y 2(x ), then the general solution of the DE is y = C y (x) + C 2 y 2 (x) where C and C 2 are arbitrary constants. y and y 2 are said to form a fundamental set of solutions. Spring mechanical system (a) Undamped free vibration: mu + ku = (b) Damped free vibration: mu + γu + ku = (c) Damped free vibration with external periodic force: mu + γu + ku = F cos(wt) 24

26 Chapter 4 The Laplace Transform Integral transforms are a class of most useful tools for solving linear differential equations. We are going to study one of these transforms: the Laplace transform. 4. Definition of the Laplace transform Consider a function f(t) defined for t. Define a new function Improper integrals: divergence, convergence, existence L{f(t)} F (s) = a f(t)dt = lim A A a e st f(t)dt f(t)dt Piecewise Continuous: A function is said to be piecewise continuous on an interval α t β if the interval can be partitioned by a finite number of points α = t < t < < t n β such that. f is continuous on each open subinterval t i < t < t i (i =,, n). 2. f has a finite limit at the endpoints of each subinterval. Examples: f(t) = tan t, t π is not a piecewise continuous function 25

27 {, for t [, ) f(t) =, for t [, ] f(t) =, for t [, ), 2 for t [, 2), n for t [n, n) 26

28 The Laplace transform: if f(t) is a piece-wise continuous function, Examples L{} = L{e at } = L{sin(at)} = e st dt = e at e st dt = L{f(t)} F (s) = s e st = s [e }{{} e ] = s = sin(at)e st dt = Property: α, β numbers, f(t), g(t) e st f(t)dt e (a s)t dt = a s e(a s)t a a 2 + s 2 (s > ) L{αf(t) + βf(t)} = αl{f(t)} + βl{g(t)}. Integration Table, s a = s a, s > a f(t) af(t)+bg(t) F (s) = L{f(t)} af (s) + bf (s) /s a a/s t /s 2 t n /n! /s n+, n is integer e αt /(s + α) sin(t) /( + s 2 ) sin(at) a/(a 2 + s 2 ) cos(t) s/( + s 2 ) cos(at) s/(a 2 + s 2 ) 27

29 4.2 Solution of Initial Value Problems Theorem 4.2. Suppose that f(t) Ke at for t M. Then, L{f (t)} = sl{f(t)} f(). (s > a) Proof. f (t)e st dt = e st df = e st f(t) = [ f()] + s = sl{f} f(). f(t)de st f(t)e st dt f (t)e st dt = e st df = e st f (t) = [ f ()] + s = sl{f } f () f (t)d st f (t)e st dt } {{ } =L{f } = s 2 L{f} sf() f (). Generalization: L{f (n) (t)} = S n L{f(t)} s n f() s n 2 f () f (n ) () L{f (t)} = sl{f(t)} f() L{f (t)} = s 2 L{f(t)} sf() f () Examples:. y 2y 2y =, y() = 2, y () = 28

30 Sol: y (t) 2y (t) 2y(t) = L{y (t) 2y (t) 2y(t)} = L{} = s = L{y (t)} 2L{y (t)} 2L{y(t)} = Define: Y (s) = L{y(t)} L{y (t)} = sy (s) y() = sy (s) 2 L{y (t)} = s 2 Y (s) sy() y () = s 2 Y (s) 2s [s 2 Y (s) 2s] 2[s Y (s) 2] 2Y (s) = [s 2 2s 2]Y (s) = 2s 4 Y (s) = 2s 4 s 2 2s 2 The inverse Laplace transform: y(t) = L {Y (s)} 2s 4 [s ( + 3)][s ( 3)] = A s ( + 3) + B s ( 3) = (A + B)s A( 3) B( + 3) [s ( + 3)][s ( 3)] { A + B = 2 A( 3) + B( + 3) = 4 [ ] [ ] [ 3 + A 2 = 3 B 4 A =.4226, B =.5774 { } { } y(t) = L s ( + + L 3) s ( 3) { } { }.4226 L s ( + =.4226 L 3) s ( + =.4226e (+ 3)t 3) { } { }.5774 L s ( =.5774 L 3) s ( =.5774e ( 3)t 3) ] 29

31 method 2: 2s 4 s 2 2s 2 = 2s 4 (s ) 2 3 2(s ) 2 = (s ) 2 ( 3) 2 = 2 s (s ) 2 ( 3) (s ) 2 ( 3) 2 s F (s) = s 2 ( 3) 2 f(t) = cosh( 3t) s (s ) 2 ( 3) 2 e t cosh( 3t) s 2 (s ) 2 ( 3) 2 e t cosh( 3t) 2. y + y = sin 2t, y() = 2, y () = Let Y (s) = L{y(t)} 3 F (s) = s 2 ( f(t) = sinh( 3t) 3) 2 3 (s ) 2 ( e t sinh( 3t) 3) (s ) 2 ( 2 3 e t sinh( 3t) 3) 2 L{y + y} = L{sin 2t} L{y 2 } + L{y} = s 2 s 2 Y sy() y () + Y = s 2 (s )Y = s 2 + 2s + 2 Y = (s 2 + 4)(s 2 + ) + 2s + s y = L { (s 2 + 4)(s 2 + ) } + L { 2s + s 2 + } 3

32 2 (s 2 + 4)(s 2 + ) = A s B s 2 + = A(s2 + ) + B(s 2 + 4) (s 2 + 4)(s 2 + ) { A + B = A + 4B = 2 A = 2/3, B = 2/3 { } { } L 2 2/3 (s 2 + 4)(s 2 = L + ) s = { } 2 3 L s { } 3 L s 2 + = 3 sin(2t) sin t { } 2/3 + L s 2 + { } { } { } 2s + s L s 2 = 2L + s 2 + L + s 2 + = 2 cos t + sin t y = 3 sin(2t) + 2 sin t + 2 cos t + sin t 3 3. y (4) y =, y() =, y () =, y () =, y () = Let Y (s) = L{y(t)} L{y (4) } L{y} = L{} = s 4 Y s 3 y() s 2 y () sy () y () Y = (s 4 )Y s 2 = Y = s2 s 4 = s 2 (s 2 )(s 2 + ) = A s 2 + B s 2 + A = B = /2 y = L { /2 s 2 } + L = 2 L { s 2 = 2 sinh t + 2 sin t { } /2 s 2 + } + 2 L { s 2 + } 4.3 Step functions Definition of the unit step function: u c (t) = {, for t c, for t > c 3

33 or u c (t) y = u(t) 2 u c (t) = u(t c) c Hat function, for t a f(t) = c, for a < t b, for t > b = c[u a (t) u b (t)] c[u a (t) u b (t)] c a b Examples: Express the following functions in terms of unit step function., for t 2 f(t) =, for 2 < t 6, for t > 6 Sol: = u 2 (t) u 6 (t). 32

34 , t < f(t) = 2, t < 2 3, t 2 sol: f(t) = [u(t) u (t)] + ( 2) [u (t) u 2 (t)] + 3 u 2 (t) t, t < f(t) = t, t < 2 t 2, t 2 Sol: f(t) = t [u(t) u (t)] + (t ) [u (t) u 2 (t)] + t 2 u 3 (t). 2, t < 2 (t f(t) = ) 2 2 t < 4 t 4 t < t 33

35 Sol: f(t) = 2[ u 2 (t)] + (t ) 2 [u 2 (t) u t (t)] + t[u 4 (t) u 5 (t)] + 4u 5 (t) Translation of a function Given b > f(x b) is the translation of f(t) to the right by b units. f(x + b) is the translation of f(t) to the left by b units. 5 f(x) = x 2 y = (x 2) u c (t)f(t) is the cut-off of f(t) at c. u c (t)f(t b): move f by b units ((a)to the right, if b > ; (b) to the left if b < ). And then cut off f(t b) at c. 34

36 6 5 4 f(x) = x u 2 (x) f(x 2) Laplace transforms: Suppose F (s) = L{f(t)} L{u c (t)f(t c)} = e cs F (s) u c (t)f(t c) = L {e cs F (s)} Example : Given sin t, t < π f(t) = 4 sin t + cos(t π/4), t π 4 find L{f(t)}. 35

37 Note that f(t) = sin t + g(t), where, t < π g(t) = 4 cos(t π 4 ), t π 4 = u π/4 cos(t π/4) L{f(t)} = L{sin t} + L{u π/4 cos(t π/4)} = L{sin t} + e πs 4 L{cos t} = s s e πs/4 s Differential Equations with Discontinuous Forcing Functions Example Find the solution of the differential equation 2y + y + 2y = g(t), y() =, y () = where {, 5 t < 2 g(t) = u 5 (t) u 2 (t) =, t < 5 and t 2 Soln: Let Y = L{y} L{2y + y + 2y} = L{g} 2[s 2 Y sy() y ()] + sy y() + 2Y = L{u 5 } L{u 2 } (2s 2 + s + 2)Y = e 5s /s e 2s /s Y = e 5s s(2s 2 + s + 2) e 2s s(2s 2 + s + 2) 36

38 s(2s 2 + s + 2) = A s + Bs + C 2s 2 + s + 2 = (2A + B)s2 + (A + C)s + 2A s(2s 2 + s + 2) A = /2, B =, C = /2 = s /2 + 2 s 2s 2 + s + 2 s /2 2s 2 + s + 2 = s + /2 2(s 2 + /2s + ) = 2 s + /2 s 2 + /2s + /4 + 3/4 = 2 s + /4 + /4 (s + /4) 2 + ( 3/2) 2 = 2 s + /4 (s + /4) 2 + ( 3/2) /8 2 (s + /4) 2 + ( 3/2) 2 { } L = 2 s 2 t L { } 2 s + /4 (s + /4) 2 + ( = 3/2) 2 2 e /4t cos( 3/2t) { } { } L /8 (s + /4) 2 + ( 3/2 = 3/2) 2 8( 3/2) L (s + /4) 2 + ( 3/2) 2 = 4 3 e /4t sin( 3/2t) Example 2 soln: y + 4y = g(t), y() =, y () =, t < 5 g(t) = (t 5)/5, 5 t <, t g(t) = (t 5)/5 [u 5 (t) u (t)] + u (t) 37

39 L{y + 4y} = L{g(t)} s 2 Y sy() y () + 4Y = 5 L{u 5 (t 5)} 5 L{u (t 5)} + L{u } we can not calculate L{u (t 5)} directly u (t 5) = u (t + 5) = u (t ) + 5u (s 2 + 4)Y = e 5s 5s 2 e s 5s 2 e 5s Y = 5s 2 (s 2 + 4) e s 5s 2 (s 2 + 4) s 2 (s 2 + 4) = A s 2 + B s = (A + B)s2 + 4A s 2 (s 2 + 4) A + B =, 4A = A = /4, B = /4 Let H(s) = 5s 2 (s 2 + 4) h(t) = L {H(s)} = 2 L {/s 2 } 4 L {2/(s )} = /2t /4 sin(2t) L {e 5s H(s)} = u 5 h(t 5) = u 5 [/2(t 5) /4 sin(2(t 5))] L {e s H(s)} = u h(t ) = u [/2(t ) /4 sin(2(t ))] 4.5 Impulse functions Dirac delta function Def: Dirac delta function, or unit impulse function δ(t) =, t Properties δ(t)dt =. 2. proof. δ(t t )dt = f(t)δ(t t )dt = f(t ) sifting property, or sampling operator. f(t)δ(t t )dt = f(t )δ(t t )dt = f(t ) δ(t t )dt = f(t ) 38

40 3. L{δ(t t )} = e st Example Soln: Let Y (s) = L{y(t)} L{y + 2y + 2y} = L{δ(t π)} s 2 Y sy() y () + 2[sY y()] + 2Y = e πs y + 2y + 2y = δ(t π), y() =, y () = (s 2 + 2s + 2)Y = e πs + s + 2 ( ) Y = e πs s 2 + 2s s + 2 s 2 + 2s + 2 s 2 + 2s + 2 = (s + ) 2 + { } L (s + ) 2 = e t sin(t) + { } L e πs s 2 = u π e (t π) sin(t π) + 2s + 2 s + 2 s 2 + 2s + 2 = s + 2 (s + ) 2 + = s + (s + ) (s + ) 2 + { } L s + (s + ) (s + ) 2 = e t [cos(t) + sin(t)] + y(t) = u π e (t π) sin(t π) + e t [cos(t) + sin(t)] { (s 2 + 2s + 2)Y Actually, we can rewrite (*) as = s + 2 (s 2 + 2s + 2)Y 2 = e πs Then Y (s) is the laplace transform of the solution to the homogeneous equation y + 2y + 2y =, y() =, y () =, and Y 2 (s) is the laplace transform of the solution to y + 2y + 2y = δ(t π), y() = y () =. And Y = Y + Y 2. Damping system my + γy + ky = f m: mass, k: spring constant, γ damping constant, f: external force. y + 2y + 2y =, y() =, y () =. The displacement y() = from the equilibrium position generates the potential energy. Once released, the potential energy and the kinetic energy are transformed into each other alternatively. Soln: y = e t [cos t + sin t] = 2e t [/ 2 cos t + / 2 sin t] = 2e t [cos t cos(π/4) + sin t sin(π/4)] = 2e t cos(t π/4) 39

41 y = 2e t y = 2e t cos(t π) y + 2y + 2y = δ(t π), y() = y () = : impulsive force was applied to the system at the instant moment of time π. Soln: y = u π e (t π) sin(t π) y = e t sin t y = e t y = e (t π) sin(t π) y = e (t π) 4

42 Graph of y(t) Example 2 y + y = k= δ(t k), y() =, y () = L{y + y} = (s 2 + )Y = L{δ(t k)} k= e ks k= Let H(s) = s 2 + h(t) = L {H(s)} = sin t L {e ks H(s)} = u(t k) sin(t k) y(t) = u(t k) sin(t k) k= time-shifting of sin(t) by k units 4.6 The Convolution Integral Def: convolution Properties (f g)(t) = t f(t τ)g(τ)dτ. f g = g f (commutative law) 4

43 Figure 4.: Graph of y(t) 2. f (ag + bg 2 ) = af g + bf g 2 (distributive law) 3. f (g h) = (f g) h (associative law) 4. f = f = 5. L{f g} = F (s)g(s) where F (s) = L{f}, G(s) = L{g}. Example Find the Laplace transform of the given function (a) f(t) = t (t τ) 2 cos(2τ)dτ Soln: f(t) = t 2 cos(2t), F (s) = 2 s 3 (b) f(t) = t e (t τ) sin τdτ. s s SolnL f(t) = e t sin t, F (s) = s s 2 + Example 2 Find the inverse laplace transform of sol: F (s) = /s 2, G(s) = a/(s 2 + a 2 ) f(t) = t, g(t) = sin(at) h(t) = (f g)(t) = t (t τ) sin(aτ)dτ = H(s) = a s 2 (s 2 + a 2 ) at sin(at) a 2. 42

44 Integral Equation. Volterra integral equations: Find φ(t) of φ(t) + t k(t τ)φ(τ)dτ = f(t) t k(t τ)φ(τ)dτ = k φ L{φ(t) + (k φ)(t)} = L{f(t)} Φ(s) + K(s)Φ(s) = F (s) Φ(s) = F (s) + K(s) { } F (s) φ(t) = L + K(s) Example Find φ(t) of φ(t) + t (t τ)φ(τ)dτ = sin(2t) t (t τ)φ(τ)dτ = t φ(t) L{φ(t) + t φ(t)} = L{sin(2t)} Φ(s) + s 2 Φ(s) = 4 s Φ(s) = 4s 2 (s 2 + 4)(s 2 + ) = A s B s 2 + A = 6 3, B = s = s sin(t) y(t) = 8 3 sin(2t) 4 3 sin(t) 2 s sin(2t) Volterra integral equations find application in demography, the study of viscoelastic materials, and in insurance mathematics through the renewal equation. 2. Integro-differential equations, find φ(t) of φ (t) + t k(t τ)φ(τ)dτ = f(t), φ() = 43

45 L{φ (t) + t k(t τ)φ(τ)dτ} = L{f(t)} sφ(s) + φ() + K(s)Φ(s) = F (s) Φ(s) = F (s) s + K(s) Impulse response function: Consider ay + by + cy = δ(t), y() = y () = (i) L{ay + by + cy} = (as 2 + bs + c) Y (s) }{{} characteristic function L{δ(t)} = Y (s) = as 2 + bs + c Let H(s) = as 2. It is known as the transfer function. + bs + c The solution to (i), h(t) = L {H(s)} is called the impulse response of the system (i). Consider the following system with forcing function g(t) Applying Laplace transform ay + by + cy = g(t), y() = y () = (ii) (as 2 + bs + c)y (s) = G(s) Y (s) = G(s) as 2 + bs + c = G(s)H(s) y(t) = L {G(s)H(s)} = (g h)(t) In the time domain, the transfer function transfers any input g(t) to the output y(t) = (g h)(t); in the frequency domain, it transfers the input G(s) to the output Y (s) = G(s)H(s). Once we know the impulse response function h(t), in the time domain, the solution to the system (ii) with forcing function g(t) is g h. 44

46 4.7 Review of Chapter 4. Definition of Laplace transform and the table of common Laplace transform. L{f} = e st f(t)dt. 2. L{f (n) } = s n F (s) s n f() s n 2 f () f (n ) () 3. Step function u c (t) = u(t c): L{u(t)} = /s, L{u c (t)} = e cs /s 4. Delta function δ(t c): L{δ(t)} =, L{δ(t c)} = e cs 5. Convolution integral (f g)(t) = τ 6. Transfer function and Impulse response f(t τ)g(τ)dτ: L{f g} = F (s)g(s) The Laplace transform of following IVP homogeneous equation a n y (n) + a n y (n ) + a y + a y = δ(t), y() = y () = = y (n ) () = is (a n s n + a n s n + a s + a )Y (s) =. Here D(s) = a n s n + a n s n + a s + a is the characteristic function of a n y (n) + a n y (n ) + a y + a y H(s) = a n s n + a n s n is called the transfer function. + a s + a h(t) = L {H(s)} is called the impulse response of the system. Given h(t), solution to a n y (n) + a n y (n ) + a y + a y = g(t), y() = y () = = y (n ) () = is y(t) = (g h)(t). 7. Comparison of method of characteristic equation and Laplace transform on constant coefficient Characteristic equation Laplace transform Order st, 2nd (so far) any order. INV no constraint initial value at x = General solution yes no Step function no yes Delta function no yes Nonhomogeneous equation undetermined coefficient laplace transform variation of parameter 45