7 Separation of Variables Chapter 5, An Introduction to Partial Differential Equations, Pichover and Rubinstein In this section we introduce the technique, called the method of separations of variables, for solving initial boundary value-problems. 7.1 Heat Equation We consider the heat equation satisfying the initial conditions u t = ku xx, x [,], t > u(x, = φ(x, x [,] (7.1 We seek a solution u satisfying certain boundary conditions. The boundary conditions could be as follows: (a Dirichlet u(,t = u(,t =. (b Neumann u x (,t = u x (,t. (c Periodic u(,t = u(,t and u x (,t = u x (,t. We look for solutions of the form u(x,t = X(xT(t where X and T are function which have to be determined. Substituting u(x,t = X(xT(t into the equation, we obtain X(xT (t = kx (xt(t from which, after dividing by kx(xt(t, we get T (t kt(t = X (x X(x. The left side depends only on t whereas the right hand side depends only on x. Since they are equal, they must be equal to some constant λ. Thus T + λkt = (7.2 and X + λx =. (7.3 53
The general solution of the first equation is given T(t = Be λkt for an arbitrary constant B. The general solutions of the second equation are as follows. (1 If λ <, then X(x = αcosh λx + β sinh λx. (2 If λ =, then X(x = αx + β. (3 If λ >, then X(x = αcos λx + β sin λx. (7.4 In addition, the function X which solves the second equation will satisfy boundary conditions depending on the boundary condition imposed on u. The problem X + λx = (7.5 X satisfies boundary conditions is called the eigenvalue problem, a nontrivial solution is called an eigenfunction associated with the eigenvalue λ. 7.1.1 Heat equation with Dirichlet boundary conditions We consider (7.1 with the Dirichlet condition u(,t = u(,t = for all t. In this case the eigenvalue problem (7.5 becomes X + λx = X( = X( =. (7.6 We have to find nontrivial solutions X of the eigenvalue problem (7.6. If λ =, then X(x = αx+β and = X( = α +β implies that β = and = X( = α implies that α =. If λ <. Then = X( = α cosh + β sinh = α and = X( = β sinh shows that also β =. We conclude that λ is not an eigenvalue of the problem (7.5. Finally, consider λ >.Then = X( = αcos λ = α and = X( = β sin λ. Since X is nontrivial solution, β and hence sin λ =. Consequently, λ = ( nπ 2, n 1 54
and the corresponding eigenfunction is given by X n (x = sin nπx x After substituting λ = (nπ/ 2 to (7.2, we get the family of solutions T n (t = B n e k( nπ 2t. Thus we have obtained the following sequence of solutions u n (x,t = X n (xt n (x = B n sin nπx nπ e k( 2t. We obtain more solutions by taking linear combinations of the u n s ( recall the superposition principle u(x,t = N u n (x,t = N B n sin nπx nπ e k( 2t, and then by passing to the limit N, u(x,t = B n sin nπx nπ e k( 2t. (7.7 Finally, we consider the initial condition. At t =, we must have u(x, = B n sin nπx = φ(x. (7.8 The coefficients, B n can be computed as follows. Fix m N. Multiplying the above equality by sin mπx and then integrating over [,], we get φ(xsin mπx dx = B n sin nπx mπx sin dx = B n sin nπx sin mπx dx. (above we ignore the question about interchanging with. Since sin nπx mπx sin dx = n m (7.9 2 n = m, we find that B m = 2 φ(xsin mπx dx. (7.1 55
Example 7.1. Consider the problem, u t u xx =, < x < π, t > u(,t = u(π,t =, t x x π/2 u(x, = φ(x = π x π/2 x π. (7.11 Here k = 1 and [,] = [,π]. In view of (7.8, the solution u(x,t is given by u(x,t = B n (sin nxe n2t. Using (7.1, the coefficients B n are equal to B n = 2 π = 2 π π π/2 φ(x sin nx dx xsin nx dx + 2 π π π/2 (π xsin nx dx. Integrating by parts we find that the right-hand side is equal to [ 2 xcos nx + sin x ] π/2 π n n 2 + 2 [ (π xcos nx sinx ] π π n n 2 π/2 = 4 nπ sin πn2 2. Since for k 1, we get sin nπ 2 = n = 2k ( 1 k+1 n = 2k 1 u(x,t = = 4 π B n (sin nxe n2 t ( 1 n+1 (2n 1 2 sin[(2n 1x]e (2n 12t. (7.12 We claim that u is indeed the classical solution of the equation (7.11. To see this, denote by u n (x,t = ( 1n+1 (2n 1 2 sin[(2n 1x]e (2n 12 t 56
so that u(x,t = 4 π n u n(x,t. Since u n (x,t = ( 1 n+1 (2n 1 2 sin[(2n t 1x]e (2n 12 1 (2n 1 2 and the series 1 n (2n 1 converges. Hence, in view of the Weierstrass M- 2 test (see Review, the series on the right hand side converges uniformly to a continuous function u in the region D = (x,t x π, t }. It remain to show that u is differentiable with respect to t and twice differentiable with respect to x, and satisfies the heat equation. Take ε >, and let D ε = (x,t < x < π, t > ε}. Differentiating u n with respect to t we get u n (x,t t = (2n 1 2 (2n 1 2 sin[(2n t 1x]e (2n 12 e (2n 12ε. The series n e (2n 12ε converges (apply the ratio test. Hence applying the Weierstrass M-test again we find that the series of u n (x,t t converges uniformly to a continuous function on D ε. Similarly, differentiating with respect to x twice we get u n (x,t xx = (2n 1 2 (2n 1 2 sin[(2n t 1x]e (2n 12 ε e (2n 12 and again n u n(x,t xx converges uniformly on D ε. Hence, using that (u n t (u n xx =, u t u xx = n (u n t n (u n xx = n [(u n t (u n xx ] =. So, u is a solution on D ε and since ε was arbitrary, u is a classical solution on D. 7.1.2 Heat equation with Neumann boundary conditions We consider the heat equation (7.1 but with Neumann boundary conditions u x (,t = u x (,t = for all t. In this case the eigenvalue problem (7.5 becomes X + λx = X ( = X ( = (7.13 57
As before the problem doesn t have negative eigenvalues. If λ =, the general solution is X(x = αx + β so that = X ( = β, implies that λ = is an eigenvalue with the unique (up to multiplication by a constant eigenfunction X (x 1. If λ >, then the general solution of the problem is X(x = α cos λx+β sin λx form which we conclude that = X ( = β and = X ( = λα sin λ implies that λ > is an eigenvalue if and only if ( nπ 2 λ = λ n =, n 1 and the corresponding eigenfunction X n (x is given by X n (x = cos nπx. Then the corresponding solutions of T + λkt = are T (t = B T n (t = B n e k( nπ 2t, n 1. Thus we obtain a sequence of solutions nπ «2 u n (x,t = B n e k t nπx cos, n which we combine to form the series u(x,t = n B n e k( nπ 2t cos nπ x. At t =, we have φ(x = u(x, = n B n cos nπx. To compute B m s, multiply this equality by cos mπx and integrate over [,]. Then φ(xcos mπx dx = n B n cos nπx cos mπx dx. Since cos mπx dx = m = m 1, (7.14 58
and it follows that cos nπx mπx cos dx = n m 2 n = m, (7.15 B = 1 φ(x dx and B m = 2 φ(xcos mπx dx, m 1. 7.1.3 Heat equation with periodic boundary conditions Next we consider (7.1 with the periodic boundary conditions u(,t = u(,t and u x (,t = u x (,t for all t. In this case the eigenvalue problem takes the form X + λx = X( = X(, X ( = X ( (7.16 This follows from X( T(t = X(T(t and X ( T(t = X(T(t. To find eigenvalues, we first consider λ <. From (7.4 and since sinh is odd and cosh is even, the condition X( = X( implies that β sinh λ = so that β =. The condition X ( = X ( implies that α sinh λ = so that α =. If λ =, then X( = α + β = α + β = X( so that α =. So λ = is an eigenvalue with the corresponding eigenfunction X (x 1. Finally, let λ >. Then X( = X( gives either β = or λ = nπ and the condition X ( = X ( gives either α = or λ = nπ. Hence the positive eigenvalues are λ n = ( nπ 2, n 1 and the corresponding eigenfunctions are X n (x = B n cos + C n sin. 59
Thus the product solutions of the periodic boundary problem are u (x,t = A u n (x,t = u(x,t = A + n 1 ( B n cos + C n sin which can be combined to form the series ( B n cos At t =, we have φ(x = u(x, = A + n 1 Integrating over [,], we get ( B n cos + C n sin e ( nπ 2 t + C n sin e ( nπ 2t.. A = 1 φ(x 2 since cos ( nπx dx = sin ( nπ dx =. Next multiplying both sides by cos ( mπx and integrating over [,] leads to since B m = 1 cos sin cos cos φ(x cos ( mπx ( mπx ( mπx dx n m dx = n = m dx =, all n,m 1. Finally, multiplying both sides by sin ( mπx and integrating over [,] leads to C m = 1 ( mπx φ(x sin dx since sin ( mπx n m sin dx = n = m. 6
7.2 Separation of variable for the wave equation The above techniques can be used to solve the wave equation. We focus on the wave equation satisfying Dirichlet boundary condition (Neumann boundary conditions are treated in the book. We consider the initial boundary problem u tt c 2 u xx =, < x <, t > u(,t = u(,t =, t u(x, = φ, < x < u t (x, = ψ, < x <. If we take u(x,t = X(xT(t, then (7.17 becomes Dividing by c 2 X(xT(t, we get X(xT (t = c 2 X (xt(t. T (t c 2 T(t = X (x X(x (7.17 from which we conclude that both sides of this equality must be equal to a constant λ. Thus, we obtain two second order differential equations The boundary conditions imply that X + λx = (7.18 T + λc 2 T. (7.19 X( = X(. so that the function X should be a solution of the eigenvalue problem, X + λx =, X( = X( =. < x < Just as in the case of the heat equation the eigenvalues are and the associated eigenfunctions are ( nπ 2 λ n =, Xn (x = sin nπx n 1. Then the solution of (7.19 with λ = λ n is of the form ( ( nπct nπct T n (t = A n cos + B n sin 61
where A n and B n are constants. The product solutions of the boundary value problem are given by ( ( ( nπct nπct u n (x,t = A n cos + B n sin sin nπx which can be combined in the series u(x,t = ( ( ( nπct nπct A n cos + B n sin sin nπx. (7.2 n 1 Setting t =, we get φ(x = u(x, = n 1A ( nπct n cos sin nπx. Multiplying by sin mπx and integrating over [,] we find that A n = 2 φ(xsin nπx dx. Next, differentiate (7.2 with respect to t at t = to get ψ(x = u t (x, = nπc B n sin nπx. n 1 To compute B n s, multiply both sides by sin mπx to get B m = 2 mπc φ(xsin mπx dx = 2 mπc and integrate over [,] φ(xsin mπx Example 7.2. Consider u tt 4u xx = (x,t (,π (, u(,t = u(,t, t u(x, = sin2x, u t (x, =, x π Here c = 2, [,] = [,π]. The formal solution is of the form u(x,t = n 1(A n cos 2nt + B n sin 2nt sinnx. From the above formulae, B n = since ψ and A n = 2 1 n = 2 sin 2xsin ndx = otherwise. Consequently, u(x,t = cos 4t sin 2x. dx. 62
7.3 The energy method and uniqueness In this section we use the so called energy method to prove the uniqueness of initial boundary value problems. The method is based on the physical principle of energy conservation. Theorem 7.3. Consider the Dirichlet problem u t ku xx = f(x,t x [,],t > u(,t = a(t,u(,t = b(t t u(x, = φ(x x [,]. Then the problem has at most one solution. Proof. Suppose that the problem has two solutions u 1 and u 2. Setting v = u 1 u 2, the function v solves v t kv xx = x [,],t > v(,t =,v(,t = t v(x, = x [,]. Define the energy E(t = 1 v(x,t 2 dx. 2 Differentiating with respect to t, we get E (t = v t (x,tv(x,t dx = k v(x,tv xx (x,t dx. Integrating by part and using the boundary conditions, we find that E (t k v(x,tv xx (x,t dx = [kv(x,tv x (x,t] kv x (x,t 2 dx. Hence the energy E is decreasing. Since E( = and E(t, we get that E. This implies that v(,t for every t. Consequently, u 1 (x,t = u 2 (x,t for all x [,] and t. Theorem 7.4. Consider the Neumann problem u tt c 2 u xx = f(x,t x [,],t > u x (,t = a(t,u x (,t = b(t t u(x, = φ(x u t (x, = φ(x Then the problem has at most one solution. 63 x [,]. x [,].
Proof. Suppose that the problem has two solutions u 1 and u 2. Setting v = u 1 u 2, the function v solves v tt c 2 v xx = x [,],t > v x (,t =,v x (,t = t Define the energy v(x, = v t (x, = E(t = 1 2 Differentiating with respect to t, we get Note that E (t = x [,]. x [,]. ( vt (x,t 2 + c 2 v x (x,t 2 dx. ( vt (x,tv tt (x,t + c 2 v x (x,tv xt (x,t dx. (7.21 c 2 v x v xt = c 2 [(v x v t x v xx v t ] = c 2 (v x v t x v tt v t where we have used v tt c 2 v xx =. Substituting this into (7.21 and using the boundary conditions, we get E (t = c 2 x (v x(x,tv(x,tdx = c 2 [v x (x,tv t (x,t] =. Hence E(t = constant for all t. At t, we have v(x, = so that v x (x, =. Since v t (x, =, we conclude that E( = and so E(t = for all t. This implies that v t (x,t 2 + c 2 v x (x,t 2 = for all x [,T] and t. Thus, v t (x,t = v x (x,t = for all x [,T] and t. Consequently, v constant. But then v(x, =, implies that v. Review Infinite series of functions Formally, an infinite series n 1 a n is a pair consisting of a sequence (a n and a sequence (s n of partial sums, s n = n k=1 a k. The series a n converges, provided that a sequence (s n converges, that is, there is s such that 64
lim s n = s. The number s is called the value of the series a n and is denoted by s = a n. The series a n is said to converge absolutely, if an converges. If a n converges, then a n converges. However, the convergence of a n does not guarantee convergence of a n. For example, ( 1 n /n converges but 1/n diverges. If a n converges but a n diverges, then series a n is said to converge conditionally. et (f n be a sequence of functions defined on some interval [a,b]. We can form a series f n (x for every x [a,b]. One says that the series f n converges pointwise on [a,b] if for the series f n (x converges for every point x [a,b]. That is, for each x [a,b], n f(x f k (x as n. If k=1 n max f(x f k (x as n, x [a,b] k=1 then, the series f n is said to converge uniformly on [a,b]. Theorem 7.5 (Weierstrass M-test. et (f n be a sequence of functions on [a,b] and let (M n be a sequence of positive constants satisfying: (a f n (x M n for all x [a,b] and all n, (b the series M n converges. The the series f n converges absolutely and uniformly on [a,b]. Theorem 7.6. et (f n be a sequence of functions on [a,b] converging pointwise to the function f, i.e., f(x = f n(x. (a If the f n s are continuous and the series f n converges uniformly on [a,b], then its sum f is a continuous function on [a,b], and b a f n (x dx = b a f n (x dx = b a f(x dx. (b If the f n s are of class C 1 and the series f n of derivatives converges uniformly on [a,b], then its sum f is of class C 1 and ( f (x = f n (x = f n(x. 65