MA22S2 Lecture Notes on Fourier Series and Partial Differential Equations.

Transcription

1 MAS Lecture Notes on Fourier Series and Partial Differential Equations Joe Ó hógáin Main Text: Kreyszig; Advanced Engineering Mathematics Other Texts: Nagle and Saff, Zill and Wright, Farlow, Brown and Churchill Online Notes: Ruth Baker (Oxford), Paul s online notes 1

2 FOURIER SERIES

3 Some Definitions Definition: A function f : [a, b] R is piecewise continuous if [a, b] can be divided into a finite number of subintervals on the interiors of which f is continuous and has finite left-hand and right-hand limits at all the endpoints of these subintervals. Example: f is piecewise continuous on [, 3]. Let C p [a, b] denote the set of all piecewise continuouos functions on [a, b]. Then C p [a, b] is a vector space under function addition and scalar multiplication. If x is an endpoint of some subinterval we write lim f(x) = f(x +) and lim f(x) = x x + x x f(x ). If f is continuous at x, then both of those limits are the same. Definition: f : R R is periodic if there exists some T such that f(x+t ) = f(x) for all x R. The smallest positive 3

4 such T is called the period of f. Example: (i) f(x) = sin x is periodic with period π, so f(x) = sin nx is periodic with period π n since sin n(x + π n ) = sin(nx + π) = sin nx. (ii) f(x) = c, a constant, is periodic but does not have a period. (iii) f(x) = x is not periodic. Note: The sum and the product of periodic functions is periodic. Definition: f : R R is even if f( x) = f(x) for all x and f is odd if f( x) = f(x) for all x. Example: cos is even and sin is odd. Note that (even)(even) = even, (odd)(odd) = even and (even)(odd) = odd. Theorem: If f is even, then if f is odd, then Proof: a a a a f(x)dx = f(x)dx =. a a a f(x)dx + f(x)dx = substitution u = x now gives both results. a a f(x)dx and f(x)dx. Making the 4

5 We will need to consider the vector space nature of C p [a, b], so we recall the abstract definition of a vector space. Definition: A set V along with an operation + : V V V ( addition) and an operation. : R V V (scalar multiplication) is called a real vector space if the following properties hold (i) (ii) (iii) u + v = v + u u, v V (commutivity) (u + v) + w = u + (v + w) u, v, w V ( associativity) an element V such that u + = + u u V ( zero element) (iv) Given any v V v V such that v + ( v) = ( v) + v = ( inverse element) (v) (vi) (vii) α.(u + v) = α.u + α.v α R, u, v V (α + β).u = α.u + β.u α, β R, u V α.(β.u) = (αβ).u α, β R, u V ( (v), (vi), (vii) imply distributivity) (viii) 1.u = u u V. 5

6 R 3 Recall the structure of R 3 : R 3 = {(a, b, c) a, b, c R}, where we write v = (a, b, c). If i = (1,, ), j = (, 1, ) and k = (,, 1), then v = (a, b, c) = a(1,, ) + b(, 1, ) + c(,, 1) = a i + b j + c k, where addition and scalar mutiplication is defined componentwise. We say that R 3 is a vector space with a basis { i, j, k}. Thr scalar (dot) product on R 3 is defined by (a, b, c).(d, e, f) = ad + be + cf so that i. i = j. j = k. k = 1. We define u v if and only if u. v =, so i, j and k are mutually perpendicular. The norm or length of v is defined by v = v. v = a + b + c and the distance from u to v is u v. Note that if v = (a, b, c) = a i + b j + c k, then a = v. i, b = v. j and c = v. k. Since every basis of R 3 has three elements we say that R 3 is three-dimensional. Other vector spaces are infinite-dimensional e.g. C p [a, b]. 6

7 The Vector Space C p [a, b] We define an inner product on C p [a, b] by (f, g) = Then f g if (f, g) =. b a f(x)g(x)dx. In this case we say that f is orthogonal to g. We write the norm of f to be f = b (f, f) = ( f (x)dx) 1 and the distance from f to g as f g = ( f = 1 i.e. a b a b a (f(x) g(x)) dx) 1. We say that f is normal if f (x)dx = 1. A set of functions {φ, φ 1,..., φ n,...} is orthonormal if φ n = 1 for all n and (φ n, φ m ) = for all n m. Given a set of orthogonal functions {ψ, ψ 1,..., ψ n,...} we get an orthonormal set by taking φ n = ψ n ψ n for all n. Example: Consider [, π] and {1, cos x, cos x,..., cos nx,...}. ( 1 dx) 1 = π. ( cos nxdx) 1 = ( 1 (1+cos nx)dx) 1 = ( 1 (x + 1 n sin nx) π ) 1 = ( 1 [(π ) ( )])1 = π. Also we get cos nx cos mxdx = for all n m. Hence {1, cos x, cos x,..., cos nx,...} is an orthogonal set and { 1 π, π cos nx n = 1,, 3...} is an orthonormal set in C p[, π]. Similarly the set {sin x, sin x,..., sin nx,...} is an orthogonal 7

8 set and { π sin nx n = 1,, 3...} is an orthonormal set in C p [, π]. 8

9 Generalized Fourier Series Let {φ n } be an orthonormal set in C p [a, b]. Consider any f C p [a, b] and let c n = (f, φ n ) for all n. Is it possible to write f as an infinite series f = c n φ n, where convergence is in n= the norm, or better still, f(x) = n= c n φ n (x), for all x [a, b], where convergence is now the usual convergence in R? This is analagous to any vector v R 3 being written as v = a i + b j + c k, where a = v. i, b = v. j and c = v. k. We call the series b c n φ n (x), where c n = (f, φ n ) = f(x)φ n (x)dx n= for all n, the generalized Fourier series of f(x). We don t yet know whether this series actually converges for any x [a, b] or, if it does converge, whether the limit is equal to f(x). We write f(x) c n φ n (x) to mean that the right-hand side is n= the series so obtained from f(x) leaving aside the questions of convergence for the moment. The theory of Fourier series tells us that for some subsets of C p [a, b] and for some orthonormal sets {φ n (x)} we do in fact have convergence and equality. Consider C p [, π] and the set {cos nx, sin mx}, n, m N. a 9

10 We have and sin mx cos nxdx = for all m, n; sin mx sin nxdx cos mx cos nxdx both = π, if m = n and if m n. Also for n = we get Hence the set { 1 1 π, cos xdx = π. an orthonormal set in C p [, π]. Letting c = 1 π f(x)dx, c n = 1 π d n = 1 π f(x) sin nxdx we have π 1 cos nx, π sin nx}, n = 1,, 3, forms f(x) c π + (c n cos nx π f(x) cos nxdx and + d n sin nx π ), so that f(x) a + (a n cos nx + b n sin nx), where a = c π, so a = 1 π f(x)dx, a n = c n π = 1 π f(x) cos nxdx and b n = d n π = 1 π f(x) sin nxdx. This is called the Fourier series associated with f(x). Example: f(x) = x + π, x <, x π 1

11 a = 1 π f(x)dx = 1 π 1 π (1 π π ) = π. a n = 1 π f(x) cos nxdx = 1 π 1 π x cos nxdx+ + ( 1 n sin nx) = 1 1 (x + π)dx = 1 π (1 x + πx) = (x + π) cos nxdx = cos nxdx = 1 π [( 1 n x sin nx) 1 n πn (1 cos nπ). πn b n = 1 π f(x) sin nxdx = 1 π 1 π x sin nxdx+ sin nxdx] sin nxdx = 1 πn ( 1 n cos nx) = (x + π) sin nxdx = sin nxdx = 1 π [( 1 n x cos nx) + 1 n ( 1 n cos nx) = 1 π [ n cos nπ) + ( 1 n sin nx) ] + ( 1 n + 1 n cos nπ) = 1 n. Therefore f(x) π 4 + Note: real number, then [ 1 cos nxdx] (1 cos nπ) cos nx 1 πn n sin nx]. Recall that if f(x) is an odd function and c is any c c function, we get that f(x)dx =. Similarly, if f(x) is an even c c f(x)dx = Example: f(x) = x, x [, π]. c f(x)dx. 11

12 Since x is even we get a = π xdx = π. Also a n = π x cos nxdx = (cos nπ 1) and b πn n = for all n 1. Therefore x π + 1 π (cos nπ 1) cos nx. n In general, if f(x) is even, then b n = for all n 1 and a n = π f(x) cos nxdx for all n. Hence f(x) a + a n cos nx. If f(x) is odd, then a n = for all n and b n = π f(x) sin nxdx for all n 1. Hence f(x) b n sin nx. Example: f(x) = 1, x < 1, x π Note that f(x) is an odd function on [, π]. Hence a n = for all n. Also b n = π f(x) sin nxdx for all n 1, so b n = π 1. sin nxdx = π ( 1 n cos nx) π = πn (1 cos nπ). Therefore f(x) πn (1 cos nπ) sin nx. 1

13 Convergence of Fourier Series We now discuss the convergence of Fourier series. There are two types of convergence (i) (ii) Convergence in the norm or convergence in the mean and Pointwise convergence. (i) Suppose the f C p [, π]; Then we write f(x) a + (a n cos nx + b n sin nx), as before. Let S N (x) = a + N (a n cos nx + b n sin nx). Convergence in the norm (mean) means that S N f as N in i.e. f S N as N or (f(x) S N (x)) dx as N. We have Theorem: If f C p [, π] (in fact f can belong to a larger space called square-integrable functions), then S N f in the norm. This result is theoretically interesting but not very useful for our purposes. (ii) Now suppose that both f and f C p [, π]. Such a function is called piecewise smooth. We have Theorem: If f is piecewise smooth on [, π], then 13

14 S N (x) 1 (f(x +) + f(x )) or 1 (f(x +) + f(x )) = a + (a n cos nx + b n sin nx) for all x [, π]. If f is continuous at x, then 1 (f(x +)+f(x )) = f(x), so that f(x) = a + (a n cos nx + b n sin nx). If f : R R is piecewise smooth and periodic of period π, then this result applies for all x R. Example: f(x) = 1, x < 1, x π For all x [, π] we have 1 (f(x +) + f(x )) = 1 1 πn (1 cos nπ) sin nx. So x = gives = and at x = π we get 1 = πn (1 cos nπ) sin nπ 1 m+1 sin(m + 1)π = 4 ( 1) m π m+1, so π 4 = ( 1) m m+1. = 4 π m= m= m= 14

15 Functions of any Period Suppose that f : R R is piecewise smooth and periodic of period L. Let u = π L x and set g(u) = f(x) = f(l πu). We have g : [, π] [ L, L] R, where g = f L π on R. Then g(u + π) = f( L π (u + π)) = f(l π u + L) = f(l πu) = g(u), so g has period π. Therefore g(u) = a + (a n cos nu + b n sin nu), with a = 1 π g(u)du = 1 π f( L π u)du, a n = 1 π g(u) cos nudu = 1 π f( L π u) cos nudu and similarly b n = 1 π g(u) sin nudu = 1 π f( L πu) sin nudu. Making the change of variable x = L πu we get that L L a = 1 L f(x)dx, a n = 1 L f(x) cos nπ L xdx and b n = 1 L L L L f(x) sin nπ L xdx, so L f(x) = a + (a n cos nπ L x + b n sin nπ L x). A piecewise smooth function defined on an interval can be extended periodically to all of R and then we can apply the theorem to any interval. 15

16 Example: Find the Fourier series of the periodic extension of the function f(x) =, x < 1 k, 1 x 1, 1 < x The period is L = 4 so L =. Then a = k nπ sin nπ kdx = k, a n = 1 f(x) cos nπ xdx = f(x)dx = k cos nπ xdx = and b n = for all n since f is an even function. Therefore f(x) = k + k 1 π n sin nπ cos nπ x. Exercise: Find the Fourier series of the L-periodic extension of the function f(x) =, L x < x, x L 16

17 Sine and Cosine Series If a function f is defined on [, L] only we can extend the definition of f to an even (odd) function on all of [ L, L] and then extend to a L-periodic function on all of R. The even extension of f is given by f e (x) = f(x) for x L and f e (x) = f( x) for L x. Obviously f e ( x) = f e (x) for all x [ L, L]. The odd extension of f is given by f o (x) = f(x) for x L and f o (x) = f( x) for L x. Then we get f o ( x) = f o (x) for all x [ L, L]. Now f e (x) = a + for all x [ L, L], where a = 1 L a n = 1 L L L f e (x) cos nπ L xdx = L a n cos nπ L x and f o(x) = L L L f e (x)dx = L f(x) cos nπ L xdx and b n sin nπ L x L f(x)dx, 17

18 b n = 1 L L L f o (x) sin nπ L xdx = L L f(x) sin nπ L xdx. In particular this is true for all x [, L], so we have sine and cosine Fourier series for f(x) on [, L]. Example: Find the Fourier sine and cosine series for f(x) = x on [, L]. L a = L xdx = L, a n = L L x cos nπ L xdx = L [(x L nπ sin nπ L x) L L L nπ sin nπ L xdx] = L [ L cos nπ n π L ] L = L (cos nπ 1) and n π L L b n = L x sin nπ L xdx = L [( x L nπ cos nπ L x) L L + nπ cos nπ L xdx] = L nπ cos nπ. Hence the sine series for f(x) is x = L π the cosine series is x = L + L π cos nπ n (cos nπ 1) n cos nπ L x. sin nπ L x and 18

19 DIFFERENTIAL EQUATIONS 19

20 Ordinary Differential Equations We first need to briefly consider second order linear ordinary differential equations. A second order linear ordinary differential equations is an equation of the form y (x) + p(x)y (x) + q(x)y(x) = r(x) or simply y +p(x)y +q(x)y = r(x), where y is a real or complex valued function of a real variable x and p(x), q(x), r(x) are real-valued functions of x. It is homogeneous if r(x) =. If we insist that y(x ) = k and y (x ) = k 1 for some x, where k and k 1, are real or complex constants, then we get the Theorem: If p(x) and q(x) are continuous on some open interval I and x I, then the initial-value problem y + p(x)y + q(x)y =, y(x ) = k, y (x ) = k 1 has a unique solution y(x) on I. This is an example of an existence and uniqueness result. We can use any method we like to find this unique solution. Defining the differential operator L by L = d dx + p(x) d dx + q(x)

21 gives the equation in the form L(y) = r(x) with L(y) = in the homogeneous case. We have the obvious result, called the principle of superposition: Theorem: If y 1 and y satisfy L(y) =, then so does αy 1 + βy for any real or complex numbers α, β. Theorem: The solution space of L(y) = on the open interval I is two-dimensional. Proof: Let x be some point in I. Let y 1 (x) be the unique solution satisfying y 1 (x ) = 1 and y 1(x ) = and let y (x) be the unique solution satisfying y (x ) = and y (x ) = 1. Then y 1 and y are linearly independent (consider the Wronskian). Suppose that y is any other solution. Letting k 1 = y(x ) and k = y (x ) we get that y and k 1 y 1 + k y are solutions with the same initial conditions and hence, by uniqueness, are the same. We conclude that y 1 and y form a basis for the solution space. We shall be interested in the case where p(x) = a and q(x) = b where a and b are real constants i.e. y + ay + by =. 1

22 Recalling that a first order linear differential equation y +ky = has a solution y = e kx we try a solution y = e λx for the second order equation. Substituting into the equation gives (λ + aλ + b)e λx = and hence λ + aλ + b =. This is called the characteristic equation. Its roots are λ 1 = a+ a 4b and λ = a a 4b with corresponding solutions e λ 1x and e λ x. We have three cases: (1) a 4b >, two real roots () a 4b =, one real double root (3) a 4b <, two fully complex roots. In case (1) y 1 = e λ 1x and y = e λ x are linearly independent and so constitute a basis for the solution space on any interval I and the general solution is y = c 1 y 1 + c y i.e. y = c 1 e λ 1x + c e λ x. In case () we only get one root λ 1 = λ = a and hence only one solution y 1 = e a x. To find a second linearly independent solution we try a solution of the form y = uy 1, where u(x) is some function to be determined. Differentiating gives y = u y 1 + uy 1 and y = u y 1 + u y 1 + uy 1

23 and now substituting into the equation we get (u y 1 + u y 1 + uy 1) + a(u y 1 + uy 1) + buy 1 =. Rearranging gives u y 1 + u (y 1 + ay 1 ) + u(y 1 + ay 1 + by 1 ) =. Now the two terms in brackets are, so u y 1 = i.e. u e a x =. Hence u = and so u = d 1 x + d for any constants d 1 and d. Taking d 1 = 1 and d = gives u(x) = x. We conclude that y (x) = xy 1 (x) = xe a x is a second solution. y 1 and y are easily seen to be linearly independent on any interval I and so the general solution is y = c 1 y 1 + c y i.e. y = (c 1 + c x)e a x. In case (3) we have λ 1 = a + a 4b = a + i 4b a = a + i b a 4. Writing ω = b a 4 gives roots λ 1 = a + iω and λ = a iω. Then eλ 1x and e λ x are complex solutions. Now e s+it = e s e it = e s (cos t + i sin t), so e λ 1x = e a x (cos ωx + i sin ωx) and e λ x = e a x (cos ωx i sin ωx). Now 1 (eλ 1x + e λ x ) and 1 i (eλ 1x e λ x ) are also solutions i.e. e a x cos ωx and e a x sin ωx are solutions and are obviously linearly independent. Hence the general solution is 3

24 y = e a x (A cos ωx + B sin ωx). Example: Solve the initial-value problem y + y y =, y() = 4, y () = 5. Characteristic equation is λ + λ = with roots 1 and. General solution is y = c 1 e x + c e x. The initial conditions imply that c 1 + c = 4 and c 1 c = 5. Hence c 1 = 1 and c = 3 and the solution is y = e x + 3e x. Example: Solve the initial-value problem y 4y + 4y =, y() = 3, y () = 1. Characteristic equation is λ 4λ+4 = with a single real root. General solution is y = (c 1 + c x)e x. The initial conditions imply that c 1 = 3 and c 1 +c = 1. Hence c 1 = 3 and c = 5 and the solution is y = (3 5x)e x. Example: Solve the initial-value problem y + y + 5y =, y() = 1, y () = 5. Characteristic equation is λ +λ+5 = with roots 1+i and 1 i. General solution is y = e x (A cos x+b sin x). The initial conditions imply that A = 1 and A + B = 5. Hence A = 1 and B = 3 and the solution is y = e x (cos x+3 sin x). 4

25 General Partial Differential Equations A partial differential equation (PDE) is an equation involving partial derivatives. The order of the equation is the highest partial derivative in the equation. Example: u t = u x is first order, u u t x = f(x, t) is second order etc. A PDE can have any number of variables. Example: u t = u xx has variables, namely t and x, u t = u rr + 1 r u r + u θθ has 3 variables, namely t, r and θ, etc. A second order equation in variables is linear if it is of the form Au xx + Bu xy + Cu yy + Du x + Eu y + F u = G, where A, B, C, D, E, F and G are functions of x and y i.e. there exists a linear operator L such that Lu = G, where Example: L = A x + B x y + C y + D x + E y + F. u tt = e x u xx + sin t (linear) uu xx + u t = (non-linear) u xx + yu yy = (linear) xu x + yu y + u = (non-linear). We shall be interested in the linear case only. The general 5

26 linear equation above is homogeneous if G =. Otherwise it is inhomogeneous. A solution of a PDE in some region R of the space of the variables involved is a function of the variables that has all the partial derivatives appearing in the equation in some domain containing R and satisfies the equation everywhere in R. Usually there are many such solutions. However if we impose conditions that the solutions must satisfy on the boundary of R (boundary conditions) or, if one of the variables is time t, at t = (initial conditions), then hopefully we can get a unique solution. This is our objective. The boundary and initial conditions arise from physical considerations in each particular case. Important PDEs from physics: u t = c u x (one-dimensional heat equation) u t = c u x (one-dimensional wave equation) u + x u = (two-dimensional Laplace equation) y u = (three-dimensional Laplace equation). These equations arise in many areas of physics. Our task is 6

27 to solve these equations given various boundary and initial conditions. Note: Some PDEs can be solved by integrating. Example: Solve u xx u =, where u = u(x, y). For each y we consider u as a function of x and use ODE techniques to solve: d u dx u = has characteristic equation λ 1 =, so any solution has the form A(y)e x + B(y)e x. Example: Solve u xy = u x, where u = u(x, y). u x y = u x ; letting u x = p gives p y = p, so dp p = dy; hence ln p = y+c, so p = De y for each x i.e. u x = D(x)e y. We conclude that u = e y D(x)dx + g(y). 7

28 First Order Equations: Method of Characteristics A first order equation of the form a(x, y) u x + b(x, y) u y = c(x, y, u) is called semi- linear. If c = c(x, y), then it is called linear. Suppose that u(x, y) is a solution of the equation above in some domain D in R. The equation z = u(x, y) defines an explicit surface S in R 3 given by Φ(x, y) = (x, y, u(x, y)). The normal to S at any point (x, y, z) in S is ( u x, u y, 1). Consider the vector field given by (a(x, y), b(x, y), c(x, y, z)). We have ( u x, u y, 1).(a, b, c) = a u x b u y + c =, so (a, b, c) is perpendicular to the normal to S at every point of S or (a, b, c) lies in the tangent plane to S at every point on S. Hence solving the differential equation is equivalent to finding a surface S given by z = u(x, y) such that (a, b, c) lies in the tangent plane at every point (x, y, z) on S. Such a surface is called a characteristic surface. How do we find characteristic 8

29 surfaces? We begin by finding characteristic curves : Definition: A characteristic curve of the differential equation is a curve in R 3 whose tangent vector at each point (x, y, z) on the curve is (a(x, y), b(x, y), c(x, y, z)). If σ : I R 3 s (x(s), y(s), z(s)) is a characteristic curve, then dx and dz = a(x(s), y(s)), dy = c(x(s), y(s), z(s)). If we take initial conditions = b(x(s), y(s)), x(s ) = x, y(s ) = y, z(s ) = z i.e. σ(s ) = (x, y, z ), then this system of three ordinary differential equations has a unique solution X(s), Y (s), Z(s), say. Hence, if we have a solution of the equation, u(x, y), then through every point on the surface S, given by z = u(x, y), we have a characteristic curve. We show that these curves lie in S and so S is the union of these curves. Theorem: Suppose that C, given by X(s), Y (s), Z(s), is the unique characteristic curve through the point (x, y, z ) of S. Then C is contained in S. Proof: Let Z 1 (s) = u(x(s), Y (s)). 9

30 Then dz 1 = u dx x + u y dy = a(x(s), Y (s)) u x +b(x(s), Y (s)) u y = c(x(s), Y (s), Z 1 (s)). Also X(s ) = x and Y (s ) = y, so Z 1 (s ) = u(x(s ), Y (s )) = u(x, y ) = z. Now, if C 1 is the curve given by X(s), Y (s), Z 1 (s), then obviously C 1 is contained in S and, by uniqueness of solutions, Z = Z 1 and so C = C 1. Our problem is to find such a surface S. From now on we take the variables as x and t, for time. We will be interested in a solution of a(x, t) u x + b(x, t) u t = c(x, t, u) subject to a boundary condition of the form u(x, ) = φ(x), for some C 1 function φ. Such a problem is called an initial-value or Cauchy problem. Recall that a surface S is given by Φ(x, t) = (x, t, u(x, t)), so taking t = gives Φ(x, ) = (x,, u(x, )) = (x,, φ(x)), a curve in the xz plane. We parametrize this curve by x = r, t =, z = φ(r) and call it Γ. Specifying the initial condition u(x, ) = φ(x) is equivalent to saying that Γ must lie in the required surface. For each value of r we have a unique char- 3

31 acteristic curve through the point (x(r), t(r), z(r)) = (r,, φ(r)). Let these curves be parametrized by s. In this way we get a surface S defined on some domain of the r and s by Φ(r, s) = (x(r, s), t(r, s), z(r, s)), where for each r, x(r, ) = r, t(r, ) =, z(r, ) = φ(r) and dx = a, dt = b, dz = c. The Inverse Function Theorem now tells us that if (x,t) (r,s) on Γ, then we can solve to get r and s in terms of x and t and hence z = u(x, t), for some function u. Finally, this u(x, t) satisfies the initial-value problem since Γ lies in the surface (for s = ) and along each characteristic curve we have c = dz = d u dx u(x(s), t(s)) = x + u dt t = a u x + b u t. 31

32 Example: a u x + u t = (a constant), Curve Γ is x (x,, φ(x)). Parametrize Γ by r u(x, ) = φ(x). i.e. x = r, t =, z = φ(r) and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) =, z(r, ) = φ(r). We have dx dz = a, with x(r, ) = r, dt =, with z(r, ) = φ(r). Solving we get x = as + r, t = s, z = φ(r). = 1, with t(r, ) =, These equations give r = x at and s = t so that z = φ(x at) is our solution. Example: x u x + u t =, u(x, ) = φ(x). Curve Γ is x (x,, φ(x)). Parametrize Γ by r i.e. x = r, t =, z = φ(r) and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) =, z(r, ) = φ(r). We have dx dz = x, with x(r, ) = r, dt =, with z(r, ) = φ(r). = 1, with t(r, ) =, Solving we get t = s, ln x = s + ln r, so that x = re s, z = φ(r). 3

33 These equations give r = xe t and s = t so that z = φ(xe t ) is our solution. Example: a u x + u t = u (a constant), u(x, ) = cos x. Curve Γ is x (x,, cos x). Parametrize Γ by r i.e. x = r, t =, z = cos r and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) =, z(r, ) = cos r. We have dx = a, with x(r, ) = r, dt dz = z, with z(r, ) = cos r. = 1, with t(r, ) =, Solving we get t = s, x = as + r, 1 z = s + c. Initial conditions give c = 1 cos r. Hence we get r = x at and s = t so that 1 z = t 1 cos(x at) and z = Example: xt u x + u t = u, u(x, ) = x. Curve Γ is x (x,, x). Parametrize Γ by r cos(x at) 1 t cos(x at). i.e. x = r, t =, z = r and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) =, z(r, ) = r. We have dx dz = xt, with x(r, ) = r, dt = z, with z(r, ) = r. = 1, with t(r, ) =, 33

34 Solving we get t = s, so that dx = xs, giving ln x = s + ln r, or x = re s, z = re s. Hence r = xe t and s = t so that z = xe t e t = xe (t t). Example: x u x + t u t = u, u(x, 1) = sin x. Curve Γ is x (x, 1, sin x). Parametrize Γ by r i.e. x = r, t = 1, z = sin r and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) = 1, z(r, ) = sin r. We have dx dz = x, with x(r, ) = r, dt = z, with z(r, ) = sin r. = t, with t(r, ) = 1, Solving we get t = e s, ln x = s + ln r, or x = re s, z = sin re s. Hence r = x t and s = ln t so that z = sin x t t. Example: u x + x u t = u, u(, t) = φ(t). Curve Γ is t (, t, φ(t)). Parametrize Γ by r i.e. x =, t = r, z = φ(r) and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) =, t(r, ) = r, z(r, ) = φ(r). We have dx = 1, with x(r, ) =, dt = x, with t(r, ) = r, 34

35 dz = z, with z(r, ) = φ(r). Solving we get x = s, t = 1 s + r, ln z = s + ln φ(r). Hence s = x and r = t 1 x so that z = φ(r)e s = φ(t 1 x )e x. Example: u x + u t = u, u(x, ) = sin x. Curve Γ is x (x,, sin x). Parametrize Γ by r i.e. x = r, t =, z = sin r and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) =, z(r, ) = sin r. We have dx dz = 1, with x(r, ) = r, dt = z, with z(r, ) = sin r. = 1, with t(r, ) =, Solving we get x = s + r, t = s, ln z = s + ln sin r. Hence s = t and r = x t so that z = sin re s = sin(x t)e t. Example: x u x + u t = tu, u(x, ) = cos x. Curve Γ is x (x,, cos x). Parametrize Γ by r i.e. x = r, t =, z = cos r and parametrize the characteristic curves by s to get x(r, s), t(r, s), z(r, s) with x(r, ) = r, t(r, ) =, z(r, ) = cos r. 35

36 We have dx dz = x, with x(r, ) = r, dt = tz, with z(r, ) = cos r. = 1, with t(r, ) =, Solving we get ln x = s + ln r, so x = re s, t = s, ln z = 1 s + ln cos r. Hence s = t and r = xe t so that z = cos(xe t )e 1 t. 36

37 Second Order Equations: Separation of Variables The Heat Equation The temperature u(x, t) of a slender metal bar of length L satisfies the diffusion equation u t = c u x, where c is a positive constant. The bar is embedded in a perfect insulator so that the boundary conditions are given by u(, t) = and u(l, t) =. Initially the temperature of the bar is given by u(x, ) = f(x), for some function f(x). Find the temperature at a distance x from one end of the bar at any time t. This situation also applies to an infinite vertical slab. We have u t = c u, < x < L, t > x with u(, t) =, u(l, t) = for all t and u(x, ) = f(x) for all x. To solve we use the method of Separation of Variables. 37

38 Try a solution of the form u(x, t) = F (x)g(t). Then so that u t = F (x)dg dt and u x = d F dx G(t), F (x) dg dt = c d F dx G(t) or G cg = F F, where means differentiation with respect to the relevant variable. Now the left-hand of this equation is a function of t only while the right-hand side is a function of x only, so both sides G must be constant i.e. cg = F F get two ODEs = k, for some constant k. We dg dt = ckg and d F dx = kf or dg dt ckg = and d F dx kf =. Now u(, t) = implies that F ()G(t) = for all t, so F () = or G(t) = for all t. If G(t) = for all t, then u =, the trivial solution (with f(x) = also). Hence F () =. In the same way we get F (L) =. We now have the second order ODE d F dx kf = with F () = and F (L) =. For k = the general solution of this initial-value problem is F (x) = ax + b. The boundary conditions imply that a = b = 38

39 , again the trivial solution. Next consider k = p >. We have d F dx p F = with general solution F (x) = Ae px + Be px. The boundary conditions now give A + B = and Ae pl + Be pl =, so that B = A and Ae pl Ae pl =. Therefore A(e pl 1 e pl ) =, so A( epl 1 e pl ) = and hence A = since pl. Again we get the trivial solution. The final possibility is k = p <. We have d F dx general solution + p F = with F (x) = A cos px + B sin px. Applying the boundary conditions gives A = and B sin pl =, so A = and B = or sin pl =. If B = we again get the trivial solution. Hence B and sin pl =, so p = nπ L for n Z. Taking B = 1 gives solutions F n (x) = sin nπ L x for n = 1,, 3...(For n < we get sin nπ L x.) Next consider the other ODE dg dt ckg =, where now k = p = n π. For convenience write λ L n = nπ L, so that dg dt + cλ ng =. Try G = e µt to get (µ + cλ n)e µt = or µ = cλ n. We have 39

40 solutions G n = B n e cλ nt, n = 1,, 3... Now for each n = 1,, 3... we have a solution u n (x, t) = F n (x)g n (t) = B n e cλ nt sin nπ L x for our PDE. However the solution must also satisfy the initial condition u(x, ) = f(x). In general none of the u n (x, t), or any finite sum of them, will satisfy this condition. Let s try an infinite sum, called a formal sum, leaving aside questions of convergence for the moment, u(x, t) = u n (x, t) = B n e cλnt sin nπ L x. If u(x, ) = f(x), then f(x) = B n sin nπ L x i.e. the B n must be the coefficients of f(x) in its Fourier sine series expansion on the interval [, L]; in other wor L B n = L f(x) sin nπ L xdx. With the B n so chosen then u(x, t) above will be a solution of the heat equation satisfying all the boundary and initial conditions. Example: f(x) =, x L 1, L x L 1, L 1 < x < L 4

41 and so L B n = L f(x) sin nπ L xdx = L 1 sin nπ L xdx L 1 = L [ L nπ cos nπ L x]l L 1 = nπ (cos nπl 1 L cos nπl L ) u(x, t) = π L 1 n (cos nπl 1 L cos nπl nπx L ) sin L e cλ nt. Note that if L 1 = L and L = L, then u(x, t) = 1 π n (cos nπ nπx cos nπ) sin L e cλnt. 41

42 Different boundary conditions for heat equation: Consider u t = c u x, L < x < L, t > with u( L, t) = u(l, t), u u x ( L, t) = x (L, t) and u(x, ) = f(x). As before we write u(x, t) = F (x)g(t) to get dg dt ckg = and d F dx kf =, where k is a constant. The boundary conditions become F ( L) = F (L) and df df dx ( L) = dx (L). Again, k = gives F (x) = ax + b and F ( L) = F (L) implies that a( L) + b = a(l) + b, so a = and F (x) = b, a constant. For k = p > we have d F dx p F = with general solution F (x) = Ae px + Be px. F ( L) = F (L) now gives Ae pl + Be pl = Ae pl + Be pl, so that A + Be pl = Ae pl + B or (A B)(1 e pl ) =. Hence A = B since p and F (x) = A(e px + e px ). Also df df dx ( L) = dx (L) gives Ap(e pl e pl ) = Ap(e pl e pl ), so that Ap(e pl e pl ) = and hence A =, the trivial solution. The only possibility is k = p < to give, as before, F (x) = A cos px + B sin px. 4

43 F ( L) = F (L) implies that A cos( pl) + B sin( pl) = A cos pl + B sin pl, so B sin pl =. Hence B = or sin pl =. Also df df dx ( L) = dx (L) gives Ap sin pl+bp cos pl = Ap sin pl + Bp cos pl, so Ap sin pl =. Hence A = or sin pl =. If sin pl then both A and B are and we get again the trivial solution. We conclude that sin pl = so that p = nπ L, n = 1,, 3... We have solutions F n (x) = A n cos nπ L x + B n sin nπ L x, n = 1,, 3... and F (x) = A as the constant solution. Now for dg dg dt ckg =. k = gives dt =, so G = a constant. k = p gives solutions G n (t) = B ne cλ nt, where λ n = nπ L. Putting it all together we get solutions u n (x, t) = (A n cos nπ L x + B n sin nπ L x)e cλ nt for n =, 1,, 3...; n = giving the constant solution. Again to satisfy u(x, ) = f(x) consider u(x, t) = (A n cos nπ L x + B n sin nπ L x)e cλ nt and so n= f(x) = A + (A n cos nπ L x + B n sin nπ L x). Hence A, A n and B n must be the Fourier coefficients of f(x) on [ L, L] i.e. 43

44 A = 1 and B n = 1 L L L L L L f(x)dx, A n = 1 L L f(x) cos nπ L xdx L f(x) sin nπ L xdx. With these A, A n and B n we get that u(x, t) is a solution of the equation satisfying the boundary and initial conditions. 44

45 The Wave Equation The deflection of an elastic string of length L fixed at the endpoints is governed by the wave equation u t = c u x, where c is a positive constant. Suppose that the initial deflection is given by f(x) and the initial velocity is given by g(x). We have u t = c u x, < x < L, t > with u(, t) =, u(l, t) = for all t and u(x, ) = f(x), u t (x, ) = g(x) for all x. As before try a solution of the form u(x, t) = F (x)g(t). We get F G = c F G or G c G = F F. Again the only possibility is both sides must be a constant k, We have two ODEs, namely G c G = F F = k. 45

46 G c kg = and F kf =. The boundary conditions become F () = and F (L) =. Consider d F dx kf =, with F () = and F (L) =. This is identical to the heat equation with the only non-trivial solution given by k = p >. We have d F dx + p F =, with general solution F (x) = A cos px + B sin px. Again the boundary conditions give A = and B sin pl = so that p = nπ L F n (x) = sin nπ L, n = 1,, 3,... Taking B = 1 gives x is a solution for n = 1,, 3,... Now consider Writing λ n = cnπ L + λ ng = with general solution d G dt + c p G =, where p = nπ L. > gives d G dt G n (t) = B n cos λ n t + B n sin λ n t, where B n and B n are constants for each n. We now have u n (x, t) = F n (x)g n (t) = (B n cos λ n t + B n sin λ n t) sin nπ L x, n = 1,, 3... are solutions of the wave equation satisfying the 46

47 boundary conditions, called normal modes. To get a solution satisfying the initial conditions also we consider the formal sum u(x, t) = u n (x, t) = (B n cos λ n t + Bn sin λ n t) sin nπ L x. Then f(x) = u(x, ) = B n sin nπ L x, which is true if the B n are the coefficients of the Fourier sine series expansion of f(x) on the interval [, L]. Similarly g(x) = u t (x, ) = Bnλ n sin nπ L x, true if the B nλ n are the coefficients of the Fourier sine series expansion of g(x) on [, L]. In other wor, L L if B n = L f(x) sin nπ L xdx and B nλ n = L g(x) sin nπ L xdx, then u(x, t) is a solution satisfying the initial conditions also. Note: u(x, t) = B n cos λ n t sin nπ L x+ Bn sin λ n t sin nπ L x = B n cos λ n t sin nπ L x, if g(x) =. In this case u(x, t) = 1 B n sin nπ L (x ct) + 1 B n sin nπ L (x + ct) = 1 ((f(x ct) + f(x + ct)). Example: Suppose that f(x) = sin 3x 4 sin 1x and g(x) = sin 4x + sin 6x with L = π, c = Then sin 3x 4 sin 1x = B n sin nx, so B 3 = 1, B 1 = 4 and all other B n =. 47

48 Also sin 4x + sin 6x = λ n Bn sin nx = nbn sin nx, so 8B 4 =, 1B 6 = 1 and all other B n =. Hence the solution is u(x, t) = cos 6t sin 3x+ 1 4 sin 8t sin 4x+ 1 1 sin 1t sin 6x 4 cos t sin 1x. Example: Suppose that the midpoint of the string is pulled up a distance h and then released from rest, giving Example: f(x) = h L x, x L, h L (L x), L x L g(x) =. f(x) = L.h L L B n sin nπ L x, where B n x sin nπ L xdx + L.h L L = L L (L x) sin nπ L g(x) = means that all Bn =. We have u(x, t) = 8h 1 sin nπ π n sin nπ nπc L x cos L t. L f(x) sin nπ L xdx = xdx =... 8h π n sin nπ. 48

49 Two Dimensional Laplace s Equation Let u(x, y) be the steady-state temperature in a rectangular metal sheet x L, y H. It is known that u(x, y) satisfies the equation u x + u y =. Suppose the sheet is insulated along the sides y =, y = H and x = L and the temperature of the side x = is given by f(y). Hence the boundary conditions are u(, y) = f(y), u(l, y) =, u(x, ) = and u(x, H) =. Let u(x, y) = F (x)g(y), giving F (x)g(y) + F (x)g (y) =, so F F = G G = k, a constant. Again we get two ODEs F (x) kf (x) =, F (L) = and G (y) + kg(y) =, G() = G(H) =. Consider G (y) + kg(y) =. If k =, then G(y) = ay + b. Now G() = implies b = and then G(H) = gives a =, the trivial solution. Next consider k = p <. Then 49

50 G (y) p G(y) = with general solution G(y) = Ae py + Be py. G() = gives B = A, so G(y) = A(e py e py ). Now G(H) = implies that A(e ph 1) =, so A = also and again we get the trivial solution. The final situation is k = p >. We have G + p G(y) = with general solution G(y) = A cos py + B sin py. G() = gives A = and G(y) = B sin py; now G(H) = implies that B = or sin ph =. B = again gives the trivial solution, so sin ph = or p = nπ H, n = 1,, 3... Hence, for each n = 1,, 3... we have a solution G n (y) = sin nπ H y. Now consider F (x) kf (x) =, where k = p = ( nπ H ). Let λ n = nπ H. Then we have F (x) λ nf (x) = with general solution F (x) = Ae λ nx + Be λ nx. F (L) = gives B = Ae λnl and we get a solution F n (x) = A n (e λ nx e λ nx e λ nl ), for n = 1,, 3... Rearranging, we get F n (x) = A n e λ nl (e λ nx e λ nl e λ nx e λ nl ) = A n e λ nl (e λ n(x L) e λ n(x L) ) = A n sinh λ n (x L), where 5

51 A n = A n e λ nl. Now for each n = 1,, 3... we have a solution u n (x, y) = A n sinh λ n (x L) sin λ n y. None of the u n (x, y) will, in general, satisfy the boundary condition u(, y) = f(y), so we consider u(x, y) = u n (x, y) = A n sinh λ n (x L) sin λ n y. If this satisfies u(, y) = f(y), thenf(y) = A n sinh( λ n L) sin λ n y, which is true if the A n sinh( λ n L) are the Fourier sine series coefficients of f(y) on [, H] i.e. A n sinh( λ n L) = H H f(y) sin nπ H ydy. Example: Suppose f(y) = y(h y). H H H y(h y) sin nπ H ydy = H y sin nπ H ydy Let I = H H y sin nπ H ydy and J = H y sin nπ H ydy. y sin nπ H ydy. Integration by parts gives I = H[ H nπ nπy cos H y]h + H nπ = H3 nπ cos nπ. Also J = [ H nπ y cos nπ H y]h = H3 nπ + H nπ cos nπ + H n π [y sin nπ H y]h H y cos nπ H ydy H n π H H sin nπ H ydy = cos nπ H ydy 51

52 H3 nπ cos nπ + H3 n 3 π 3 [cos nπ H y]h = H3 nπ cos nπ + H3 n 3 π 3 (cos nπ 1). Hence A n sinh( λ n L) = 4H (1 cos nπ), so A n 3 π 3 n = 4H and u(x, y) = 4H (cos nπ 1) π 3 n 3 sinh(λ n L) sinh λ n(x L) sin λ n y. Note: Consider u x + u y =, x L, y H, (1 cos nπ) n 3 π 3 sinh( λ n L) where u(x, ) = f 1 (x), u(x, H) = f (x), u(, y) = g 1 (y) and u(l, y) = g (y). Separation of variables depen on some bounday conditions being homogeneous i.e. =. To solve the above we consider solutions of u + x u =, y with u(x, ) = f 1 (x) and the others = etc. If the solutions are u 1, u, u 3, u 4, respectively, then u = u 1 + u + u 3 + u 4 will be a solution satisfying all the boundary conditions above. Example: The voltage V (x, y) at any point in a square metal plate of side length π satisfies Laplace e equation V + x V =. y The plate is earthed at x =, x = π and y =, so that V (, y) =, V (π, y) = and V (x, ) =. A voltage f(x) is 5

53 applied along the fourth side y = π so that V (x, π) = f(x), where Solve for V (x, y). f(x) = x, x π, π x, π x π V (x, y) = F (x)g(y) so that F (x) kf (x) = and G (y) + kg(y) =, with F () =, F (π) = and G() =. Consider F (x) kf (x) =, with F () =, F (π) =. k = implies F =. k = p also gives F = as before. Therefore k = p and so F (x) + p F (x) =, with general solution F (x) = A cos px + B sin px. The boundary conditions now give A = and sin pπ = so that p = n, n = 1,, 3,... and sin nx is a solution for each n = 1,, 3... Now consider G (y) p G(y) =, with G() =, p = n. For each n we have a general solution G n (y) = A n e ny + B n e ny. G() = implies that B n = A n so G n (y) = A n (e ny e ny ). We have solutions V n (x, y) = A n sin nx(e ny e ny ) = A n sin nx sinh ny. 53

54 Consider V (x, y) = A n sin nx sinh ny. Then V (x, π) = f(x) implies that f(x) = A n sin nx sinh nπ, so that A n sinh nπ is the Fourier coefficient of the sine series expansion of f(x) on the iterval [, π] i.e. A n sinh nπ = π f(x) sin nxdx = π [ x sin nxdx+π sin nxdx x sin nxdx] π = 4 πn sin nπ, on integrating. Finally we have V (x, y) = 4 π Example: π sin nπ sin nx sinh ny. n sinh nπ The voltage V (x, y) at any point in a square metal plate of side length π satisfies Laplace e equation V x + V y =. The plate is earthed at x =, x = π and y = π, so that V (, y) =, V (π, y) = and V (x, π) =. A voltage sin x is applied along the fourth side y = so that V (x, ) = sin x. Solve for V (x, y). V (x, y) = F (x)g(y) so that F (x) kf (x) = and G (y)+kg(y) =, with F () =, F (π) = and G(π) =. Consider F (x) kf (x) =, with F () =, F (π) =. π 54

55 k = implies F =. k = p also gives F = as before. Therefore k = p and so F (x) + p F (x) =, with general solution F (x) = A cos px + B sin px. The boundary conditions now give A = and sin pπ = so that p = n, n = 1,, 3,... and sin n x is a solution for each n = 1,, 3... Now consider G (y) p G(y) =, with G() =, p = n. For each n we have a general solution G n (y) = A n e n y + B n e n y. G(π) = now gives A n e nπ + B n e nπ =, so B n = A n e nπ. Therefore G n (y) = A n (e n y e nπ e n y ) = A n e nπ (e (n y nπ) e (n y nπ) ) = A n e nπ sinh n( y π). Write G n (y) = A n sinh n( y π) and then V n (x, y) = A n sinh n( y π) sin n x is a solution for each n = 1,, 3... Now consider V (x, y) = A n sinh n( y π) sin n x. If this is a solution, then V (x, ) = sin x implies that sin x = A n sinh n() sin n x. We conclude that A 4 sinh( 4π) = 1 and all other A n =. We get our solution 55

56 V (x, y) = 1 sinh( 4π) sinh 4(y π) sin x = sinh(4π y) sinh 4π sin x. 56