6 Wave Equation on an Interval: Separation of Variables

Transcription

1 6 Wave Equation on an Interva: Separation of Variabes 6.1 Dirichet Boundary Conditions Ref: Strauss, Chapter 4 We now use the separation of variabes technique to study the wave equation on a finite interva. As mentioned above, this technique is much more versatie. In particuar, it can be used to study the wave equation in higher dimensions. We wi discuss this ater, but for now wi continue to consider the one-dimensiona case. We start by considering the wave equation on an interva with Dirichet boundary conditions, u tt c u xx = < x < u(x, = φ(x < x < (6.1 u t (x, = ψ(x < x < u(, t = = u(, t. Our pan is to ook for a soution of the form u(x, t = X(xT (t. Suppose we can find a soution of this form. Pugging u into the wave equation above, we see that the functions X, T must satisfy X(xT (t = c X (xt (t. (6. Dividing (6. by c XT, we see that for some constant λ, which impies T c T = X X = λ X (x = λx(x T (t = λc T (t. Our boundary conditions u(, t = = u(, t impy X(T (t = = X(T (t for a t. Combining this boundary condition with the ODE for X, we see that X must satisfy { X (x = λx(x (6.3 X( = = X(. If there exists a constant λ satisfying (6.3 for some function X, which is not identicay zero, we say λ is an eigenvaue of x on [, ] subject to Dirichet boundary conditions. The corresponding function X is caed an eigenfunction of x on [, ] subject to Dirichet boundary conditions. Caim 1. The eigenvaues of (6.3 are λ n = (nπ/ with corresponding eigenfunctions X n (x = (nπx/. 1

2 Proof. We need to ook for a the eigenvaues of (6.3. First, we ook for any positive eigenvaues. Suppose λ is a positive eigenvaue. Then λ = β >. Therefore, we need to find soutions of { X (x + β X(x = Soutions of this ODE are given by X( = = X(. X(x = C cos(βx + D (βx for arbitrary constants C, D. The boundary condition The boundary condition X( = = C =. X( = = D (β = = β = nπ for some integer n. Therefore, we have an infinite number of eigenvaues λ n = βn = (nπ/ with corresponding eigenfunctions X n (x = D n x where D n is arbitrary. We have proven that λ n = (nπ/ are eigenvaues for the eigenvaue probem (6.3. We need to check that there are no other eigenvaues. We check if λ = is an eigenvaue. If λ = is an eigenvaue, our eigenvaue probem becomes { X (x = (6.4 X( = = X(. The soutions of this ODE are for C, D arbitrary. The boundary condition The boundary condition X(x = Cx + D X( = = D =. X( = = C =. Therefore, the ony function X which satisfies (6.4 is X(x =. By definition, the zero function is not an eigenfunction. Therefore, λ = is not an eigenvaue of (6.3. Next, we check if there are any negative eigenvaues. Suppose λ = γ is an eigenvaue of (6.3. If we have any negative eigenvaues, our eigenvaue probem becomes { X (x γ X(x = (6.5 X( = = X(.

3 The soutions of this ODE are given by for C, D arbitrary. Our boundary condition Our boundary condition X(x = C cosh(γx + D h(γx X( = = C =. X( = = D =. Therefore, the ony soution of (6.5 is X =. Again, by definition, the zero function is not an eigenfunction. Therefore, there are no negative eigenvaues of (6.3. Consequenty, we have found an infinite sequence of eigenvaues and eigenfunctions for our eigenvaue probem (6.3. For each of these pairs {λ n, X n }, we ook for a function T n satisfying T n (t = λ n c T n (t. (6.6 Then etting u n (x, t = X n (xt n (t, we wi have found a soution of the wave equation on [, ] which satisfies our boundary conditions. Of course, we have not yet ooked into satisfying our initia conditions. We wi consider that shorty. First, we see that for each n the soution of (6.6 is given by Therefore, for each n, T n (t = A n cos u n (x, t = T n (tx n (x [ = A n cos ct + B n ct. ct + B n ] ct x is a soution of the wave equation on the interva [, ] which satisfies u n (, t = = u n (, t. More generay, ug the fact that the wave equation is inear, we see that any finite inear combination of the functions u n wi aso give us a soution of the wave equation on [, ] satisfying our Dirichet boundary conditions. That is, any function of the form u(x, t = [ ] A n cos ct + B n ct x soves the wave equation on [, ] and satifies u(, t = = u(, t. Now we aso want the soution to satisfy our initia conditions. Our hope is that we can choose our coefficients A n, B n appropriatey so that u(x, = φ(x and u t (x, = ψ(x. That is, we woud ike to choose A n, B n such that u(x, = u t (x, = n= A n x = φ(x B n nπc 3 x = ψ(x. (6.7

4 Our desire to express our initia data as a inear combination of trigonometric functions eads us to a coupe of quesitons. For which functions φ, ψ can we find constants A n, B n such that we can write φ and ψ in the forms shown in (6.7? How do we find the coefficients A n, B n? In answer to our first question, for genera functions φ, ψ we cannot express them as finite inear combinations of trigonometric functions. However, if we aow for infinite series, then for nice functions, we can express them as combinations of trigonometric functions. We wi be more specific about what we mean by nice functions ater. We wi answer the second question, regarding finding our coefficients, shorty. First, however, we make a remark. Remark. Stated above, we caim that we can represent nice functions φ and ψ in terms of infinite series expansions invoving our eigenfunctions x. By this, we mean there exist constants A n, B n such that φ(x ψ(x A n x B n nπc x, (6.8 where means that the infinite series converge to φ and ψ, respectivey, in some appropriate sense. We wi discuss convergence issues shorty. Assuming for now that we can find sequences {A n }, {B n } such that φ, ψ satisfy (6.8, then defining u(x, t = [ ] A n cos ct + B n ct x, we caim we wi have found a soution of (6.1. We shoud note that if u(x, t was a finite sum, then it woud satisfy the wave equation, as described earier. To say that the infinite series satisfies the wave equation is a separate question. This is a technica point which we wi return to ater. To recap, so far we have shown that any function of the form [ u n (x, t = A n cos ct + B n ct ] x is a soution of the wave equation on [, ] which satisfies Dirichet boundary conditions. In addition, any finite inear combination of functions of this form wi aso satisfy the wave equation on the interva [, ] with zero boundary conditions. We caim that u(x, t = [ ] A n cos ct + B n ct x 4

5 wi aso sove the wave equation on [, ]. Before discusg this issue, we ook for appropriate coefficients A n, B n which wi satisfy our initia conditions. As stated above, we woud ike to find coefficients A n such that φ(x = A n x. (6.9 Ignoring the convergence issues for a moment, if φ can be expressed in terms of this infinite sum, what must the coefficients A n be? For a fixed integer m, mutipy both sides of (6.9 by ( mπx and integrate from x = to x =. This eads to ( mπ ( mπ x φ(x dx = x A n x dx. (6.1 Now we use the foowing property of the e functions. In particuar, we use the fact that ( mπ x x dx = m = n m n. (6.11 Proof of (6.11. Recaing the trigonometric identities cos(a ± B = cos(a cos(b (A (B, we have (A (B = 1 cos(a B 1 cos(a + B. Therefore, ( mπ x x dx = 1 ( (m nπ cos x dx 1 ( (m + nπ cos x dx. First, Simiary, for m n, ( (m + nπ cos x dx = = ( (m nπ cos x dx = = ( (m + nπ (m + nπ x= x x= [((m + nπ (] =. (m + nπ ( (m nπ (m nπ x= x x= [((m nπ (] =. (m nπ 5

6 But, for m = n, Therefore, for m n, we have whie for m = n, we have ( (m nπ cos x dx = cos( dx = as caimed. Now ug (6.11, (6.1 becomes ( mπ x x dx =, ( mπ x x dx =, Therefore, our coefficients A m are given by A m = ( mπ x φ(x dx = A m. ( mπ x φ(x dx. dx =. With these ideas in mind, for a given function φ, we define its Fourier e series on the interva [, ] as A n x (6.1 where A n = x φ(x dx. (6.13 Remark. We shoud mention that a our estimates above have been rather forma. We have not shown yet that any function φ can be represented by its Fourier series. Rather, we have shown that if a function φ can be represented by the infinite series then its coefficients A n shoud be given by A n = A n x, x φ(x dx. 6

7 The work done above has ed us to the definition of the Fourier e series associated with a given function φ. It sti remains to determine when a Fourier e series for a given function φ wi actuay converge to φ. We now have enough information to put together a soution of the initia-vaue probem for the one-dimensiona wave equation with Dirichet boundary conditions (6.1. Combining the infinite series expansions of our initia data φ and ψ (6.8 with (6.13, we define coefficients A n nπc B n and then our soution wi be given by u(x, t = x φ(x dx x ψ(x dx. (6.14 [ ] A n cos ct + B n ct x. ( Orthogonaity and Symmetric Boundary Conditions In ug the method of separation of variabes in the previous section, one of the keys in finding the coefficients A n, B n (6.14 in the infinite series expansion (6.15 was ug the fact that the eigenfunctions x satisfy x ( mπ x dx = for m n. In this section, we consider the wave equation on [, ] with different boundary conditions and discuss when our eigenfunctions wi satisfy the same type of condition, thus aowing us to determine the coefficients in the infinite series expansion of our soution. First, we reca some facts from inear agebra. Let v = [v 1,..., v n ] T, w = [w 1,..., w n ] T be two vectors in R n. The dot product of v and w is defined as v w = v 1 w v n w n. The norm of v is given by v = v v n = ( v v 1/. We say v and w are orthogona if their dot product is zero; that is, v w =. We now extend these ideas to functions. Let f, g be two rea-vaued functions defined on the interva [a, b] R. We define the L inner product of f and g on [a, b] as f, g = b a f(xg(x dx. 7

8 We define the L norm of f on the interva [a, b] as ( b 1/ f L ([a,b] = f(x dx = f, f 1/, and define the space of L functions on [a, b] as a L ([a, b] = {f : f L ([a,b] < + }. We say f and g are orthogona on [a, b] if their L inner product on [a, b] is zero; that is, f, g = b a f(xg(x dx =. With these definitions, we note that the functions { x } are mutuay orthogona. We were abe to use this fact to find the coefficients in the infinite series expansion of our soution u of the wave equation on [, ] with Dirichet boundary conditions. If our eigenfunctions were not orthogona, we woud have had no way of soving for our coefficients. Suppose we consider the wave equation on [, ] with a different set of boundary conditions. Wi our eigenfunctions be orthogona? For exampe, consider the initia-vaue probem for the wave equation on [, ] with more genera boundary conditions, u tt c u xx = < x < u(x, = φ(x < x < u t (x, = ψ(x < x < u satisfies certain boundary conditions at x =, x = for a t. Ug the method of separation of variabes, we are ed to the foowing eigenvaue probem, { X = λx < x < (6.16 X satisfies certain boundary conditions at x =, x =. Let {X n } be a sequence of eigenfunctions for this eigenvaue probem. Under what conditions wi {X n } be a sequence of mutuay orthogona eigenfunctions? That is, when can we guarantee that X n, X m = X n X m dx = for m n? In order to answer this question, we reca some facts from inear agebra. Let A be an n n matrix. One condition which guarantees the orthogonaity of the eigenvectors of A is symmetry of A; that is, A = A T. (Note: A matrix A being symmetric is a sufficient but not necessary condition for its eigenvectors being orthogona. Now, we don t have a notion of symmetry for operators yet, but et s try to use the properties of symmetric matrices to 8

9 extend the idea to operators. We note that if A is a symmetric matrix, then for a vectors u, v in R n, A u v = (A u T v = u T A T v = u T A v = u A v. The converse of this statement is aso true. That is, if A u v = u A v for a u, v in R n, then A is symmetric. It is eft to the reader to verify this fact. Consequenty, we concude that if A u v = u A v for a u, v in R n, then the eigenvectors of A are orthogona. We now pan to extend the idea of symmetry of matrices to operators to find sufficient conditions for the orthogonaity of eigenfunctions. Suppose we have an operator L such that L(u, v = u, L(v for a u, v in an appropriate function space. It turns out that the eigenfunctions associated with the eigenvaue probem { L(X = λx x Ω R n (6.17 X satisfies certain boundary conditions on Ω are orthogona. We state this more precisey in the foowing emma. Lemma. Let X n, X m be eigenfunctions of (6.17 with eigenvaues λ n, λ m respectivey such that λ n λ m. If L(X n, X m = X n, L(X m, (6.18 then X n and X m are orthogona. Proof. Combining (6.18 with the fact that X n, X m are eigenfunctions of (6.17 with eigenvaues λ n, λ m respectivey, we have Therefore, λ n X n, X m = L(X n, X m = X n, L(X m = λ m X n, X m. (λ n λ m X n, X m =. Now ug the assumption that λ n λ m, we concude that as caimed. X n, X m =, 9

10 Now we return to our eigenvaue probem (6.16. We woud ike to find sufficient conditions under which the boundary conditions ead to orthogona eigenfunctions. We make use of the above emma. In particuar, for (6.16 our operator L = x. By the above emma, we know that if X n and X m are eigenfunctions of (6.16 which correspond to distinct eigenvaues λ n, λ m and X n, X m = X n, X m, (6.19 then X n and X m are orthogona. Therefore, if we can find a sufficient condition under which (6.19 hods, then we can prove orthogonaity of eigenfunctions corresponding to distinct eigenvaues. We state such a condition in the foowing emma. Lemma 3. If [f(xg (x f (xg(x] x= x= = (6. for a functions f, g satisfying the boundary conditions in (6.16, then for a eigenfunctions X n, X m of (6.16. Proof. Integrating by parts, we see that X n, X m = X n, X m X n, X m = X nx m dx = X nx m x= x= X nx m dx = X nx m x= x= X nx m x= x= + = [X nx m X n X m] x= x= + X n, X m. X n X m dx Therefore, if (6. hods for a functions f, g which satisfy the boundary conditions in (6.16, then necessariy [X nx m X n X m] x= x= =, and the emma foows. Putting together the two emmas above, we concude that condition (6. is sufficient to guarantee orthogonaity of eigenfunctions corresponding to distinct eigenvaues. Consequenty, we define this condition as a symmetry condition associated with the operator x. In particuar, for the eigenvaue probem (6.16, we say the boundary conditions are symmetric if [f(xg (x f (xg(x] x= x= = for a functions f and g satisfying the given boundary conditions. Coroary 4. Consider the eigenvaue probem (6.16. If the boundary conditions are symmetric, then eigenfunctions corresponding to distinct eigenvaues are orthogona. 1

11 Remark. The above coroary states that eigenfunctions corresponding to distinct eigenvaues are orthogona. In fact, eigenfunctions associated with the same eigenvaue can be chosen to be orthogona by ug the Gram-Schmidt orthogonaization process. (See Strauss, Sec. 5.3, Exercise 1 Now, et s return to soving the wave equation on an interva. Consider the initia-vaue probem u tt c u xx = < x < u(x, = φ(x < x < (6.1 u t (x, = ψ(x < x < u satisfies symmetric B.C. s Beow we wi show formay how to construct a soution of this probem. We wi indicate how symmetric boundary conditions aow us to use separation of variabes to construct a soution. Specificay, we wi find a formua for the coefficients in the infinite series expansion in terms of the eigenfunctions and the initia data. Ug separation of variabes, we are ed to the eigenvaue probem, { X = λx < x < (6. X satisfies symmetric B.C.s. Let {(λ n, X n } be a the eigenvaues and corresponding eigenfunctions for this eigenvaue probem. For each eigenvaue, we sove the foowing equation for T n, T n (t + c λ n T n (t =. The soutions of this equation are given by A n cos( λ n ct + B n ( λ n ct if λ n > T n (t = A n + B n t if λ n = A n cosh( λ n ct + B n h( λ n ct if λ n <, (6.3 where A n and B n are arbitrary. Putting each soution T n of this equation together with the corresponding eigenfunction X n, we arrive at a soution of the wave equation u n (x, t = X n (xt n (t which satisfies the boundary conditions. Due to the inearity of the wave equation, we know any finite sum of soutions u n is aso a soution. We now ook to find an appropriate combination of these soutions u n so that our initia conditions wi be satisfied. That is, we woud ike to choose A n, B n appropriatey for each T n such that by defining u(x, t = n X n (xt n (t (6.4 with that choice A n, B n, our initia conditions are satisfied. In particuar, we want to find coefficients A n, B n such that u(x, = n u t (x, = n X n (xt n ( = n X n (xt n( = n A n X n (x = φ(x C n X n (x = ψ(x, 11

12 where cb n λn λ n > C n B n λ n = cb n λn λ n <. If we can find coefficents A n, C n such that we can write our initia data φ, ψ as φ(x = n ψ(x = n A n X n (x C n X n (x, (6.5 we caim that we have found a soution of (6.1 given by u(x, t = n X n (xt n (t where T n is defined in (6.3 with A n defined by (6.5 and C n /c λ n λ n > B n C n λ n = C n /c λ n λ n <, (6.6 for C n defined in (6.5. Now whether we can represent our initia data in terms of our eigenfunctions is a key issue we need to consider if we hope to use this method. Luckiy, it turns out that for nice functions φ, ψ and any symmetric boundary conditions, we can represent φ, ψ in terms of our eigenfunctions. In fact, as we wi show ater, for any symmetric boundary conditions, there is an infinite sequence of eigenvaues {λ n } for (6. such that λ n + as n +. Corresponding with these eigenvaues, we wi have an infinite sequence of orthogona eigenfunctions. It turns out that any L function can be represented by these eigenfunctions. In addition, this infinite sequence of eigenfunctions aows us to represent our soution as u(x, t = n X n (xt n (t for appropriatey chosen constants A n, B n in the definition of T n. We shoud mention that in genera this soution wi be an infinite series soution. It remains to prove that an infinite sum of soutions wi sti give us a soution of (6.1. We wi return to these issues ater. First, however, et us try to find formuas for our coefficients A n, C n in (6.5. Assuming we can represent our initia data in terms of our eigenfunctions as in (6.5, we can use Coroary 4 to determine what our coefficients shoud be. Specificay, = φ(x = X m, A n X n (x A n X n = X m, φ. 1

13 Now by Coroary (6.5, we know that symmetric boundary conditions impy eigenfunctions corresponding to distinct eigenvaues are orthogona, and, consequenty, X m, A m X m = X m, φ. Therefore, the coefficients A m for φ in the infinite series expansion φ(x = n A n X n (x must be given by Simiary, the coefficients C m in the infinite series expansion A m = X m, φ X m, X m. (6.7 ψ(x = n C n X n (x must be given by C m = X m, ψ X m, X m. (6.8 Therefore, we caim that our soution of (6.1 is given by u(x, t = n X n (xt n (t where T n is defined in (6.3 with A n defined by (6.7 and B n defined by (6.6 for C n defined by (6.8. Let s ook at an exampe. Exampe 5. Consider the initia-vaue probem for the wave equation on an interva with Neumann boundary conditions, u tt c u xx = < x < u(x, = φ(x, < x < (6.9 u t (x, = ψ(x < x < u x (, t = = u x (, t. Ug the separation of variabes technique, we are ed to the eigenvaue probem { X = λx X ( = = X (. (6.3 We note that the Neumann boundary conditions are symmetric, because for a functions f, g such that f ( = = g ( and f ( = = g (, we have [f(xg (x f (xg(x] x= x= =. 13

14 Therefore, we wi be abe to use the orthogonaity of the eigenfunctions to determine the coefficients of our soution in the infinite series expansion. First, we ook for a our eigenvaues and eigenfunctions of (6.3. We start by ooking for positive eigenvaues λ = β >. In this case, our eigenvaue probem becomes { X + β X = The soutions of our ODE are given by The boundary condition The boundary condition X ( = = X (. X(x = C cos(βx + D (βx. X ( = = D =. X ( = = Cβ (β = = β = nπ. Therefore, we have an infinite sequence of positive eigenvaues given by λ n = βn = (nπ/ with corresponding eigenfunctions X n (x = C n cos x. Next, we check if λ = is an eigenvaue. If λ =, our eigenvaue probem (6.3 becomes { X = The soutions of this ODE are given by X ( = = X (. X(x = Cx + D, where C, D are arbitrary. The boundary condition The boundary condition X ( = = C =. X ( = is automaticay satisfied as ong as C = (i.e. - the first boundary condition is satisfied. Therefore, X(x = D is an eigenfunction with eigenvaue λ =. Last, we ook for negative eigenvaues. That is, we ook for an eigenvaue λ = γ. In this case, our eigenvaue probem (6.3 becomes { X γ X = The soutions of the ODE are given by X ( = = X (. X(x = C cosh(γx + D h(γx. 14

15 The boundary condition The boundary condition X ( = = D =. X ( = = Cγ h(γ = = C =. Therefore, the ony function X which satisfies our eigenvaue probem for λ = γ < is the zero function, which by definition is not an eigenfunction. Therefore, we have no negative eigenvaues. Consequenty, the eigenvaues and eigenfunctions for (6.3 are given by λ n = ( nπ, Xn (x = cos x, n = 1,,... λ =, X (x = 1. For each n =, 1,,..., we need to ook for a soution of our equation for T n. In particuar, we need to sove T n (t + c λ n T n (t =. As described earier, the soutions of this equation are given by T n (t = A n cos ct T (t = A + B t. + B n ct n = 1,,... for A n, B n arbitrary. Putting these functions X n, T n together, we propose that our soution is given by u(x, t = A + B t + [ ] A n cos ct + B n ct cos x for appropriatey chosen constants A n, B n. Ug our initia conditions, we want to choose A n, B n such that u(x, = A + u t (x, = B + A n cos x = φ(x nπc B n cos x = ψ(x. That is, we want constants A n, B n such that φ(x = A n X n (x n= ψ(x = B X (x + 15 nπc B n X n (x. (6.31

16 Note: For ψ, we ook for constants C n such that ψ(x = C n X n (x n= and then define B n such that { C n = B n = C nπc n n = 1,,... Ug the fact that the Neumann boundary conditions are symmetric, we know that eigenfunctions corresponding to distinct eigenvaues must be orthogona. Consequenty, we can use the formuas (6.7, (6.8 derived earier for A n, C n. In particuar, Now and where and X, φ = X n, φ = X, X = X n, X n = X φ dx = X n φ dx = dx = A n X n, φ X n, X n C n X n, ψ X n, X n φ(x dx cos x dx = Therefore, our soution of (6.9 is given by Remarks. u(x, t = A + B t + A = X, φ X, X = 1 A n = X n, φ X n, X n = cos x φ(x dx n = 1,,... n = 1,,... [ ] A n cos ct + B n ct cos x B = C = X, ψ X, X = 1 nπc B n = C n = X, ψ X n, X n = φ(x dx cos x φ(x dx n = 1,, ψ(x dx cos x ψ(x dx n = 1,,...

17 Ug the fact that our boundary conditions are symmetric, we know from Coroary 4 that eigenfunctions corresponding to distinct eigenvaues are orthogona. As a resut, we gain for free the orthogonaity property of coe functions: cos ( mπ x cos x = for m n. The series where A + A n cos x A 1 A n φ(x dx cos x φ(x dx n = 1,,... is known as the Fourier coe series of φ on the interva [, ]. We cose this section by giving some exampes of symmetric boundary conditions, Exampe 6. The foowing boundary conditions are symmetric for the eigenvaue probem (6.. Dirichet: X( = = X( Neumann: X ( = = X ( Periodic: X( = X(, X ( = X ( Robin: X ( = ax(, X ( = bx( 6.3 Fourier Series In the previous section we showed that soving an initia-vaue probem for the wave equation on an interva can essentiay be reduced to studying an associated eigenvaue probem. That is, ug separation of variabes, the IVP u tt c u xx = < x < u(x, = φ(x < x < (6.3 u t (x, = ψ(x < x < u satisfies certain B.C.s 17

18 can be reduced to studying the eigenvaue probem { X = λx < x < X satisfies certain B.C.s One of the key ingredients in ug the soutions of the eigenvaue probem to sove the initia-vaue probem (6.3 was being abe to represent our initia data in terms of our eigenfunctions. That is, to be abe to find coefficients A n, C n such that φ(x = n ψ(x = n A n X n (x C n X n (x. In this section, we discuss this issue in detai, showing when such a representation makes sense. First, we start with some definitions. For a function φ defined on [, ] we define its Fourier e series as φ(x A n x where A n (nπx/, φ (nπx/, (nπx/. Note: As we have not discussed any convergence issues yet, the notation shoud just be thought of as meaning φ is associated with the Fourier series shown. We saw this series earier in the case of Dirichet boundary conditions. In the case of Neumann boundary conditions, we were ed to the Fourier coe series of φ on [, ], where A n φ(x n= A n cos x cos(nπx/, φ cos(nπx/, cos(nπx/ The other cassica Fourier series is known as the fu Fourier series. The fu Fourier series for φ defined on [, ] is given by where φ(x A + [ ] A n cos x + B n x, cos(nπx/, φ A n cos(nπx/, cos(nπx/ (nπx/, φ B n (nπx/, (nπx/ 18 n =, 1,,... n = 1,,.... (6.33

19 Note: The inner products for the fu Fourier series are taken on the interva [, ]. The fu Fourier series arises in the case of periodic boundary conditions. Reationship between Fourier e, Fourier coe and fu Fourier series. Let φ be an odd function defined on [, ]. Then its fu Fourier series is an odd function because the coefficients A n defined in (6.33 are zero. Therefore, for φ odd, its fu Fourier series is given by φ(x B n x, where B n is defined in (6.33. In particuar, for an odd function φ defined on [, ], the fu Fourier series of φ is the Fourier e series of the restriction of φ to [, ]. Equivaenty, for a function φ defined on [, ] the Fourier e series of φ is the fu Fourier series of the odd extension of φ to [, ]. Simiary, for an even function defined on [, ], the fu Fourier series is an even function because the coefficients B n defined in (6.33 are zero. Therefore, for φ even, its fu Fourier series is φ(x A + A n cos x, where A n is defined in (6.33. For an even function φ defined on [, ], the fu Fourier series of φ is the Fourier coe series of the restriction of φ to [, ]. Equivaenty, for a function φ defined on [, ], the Fourier coe series of φ is the fu Fourier series of the even extension of φ to [, ]. We wi use these reationships beow when we discuss convergence issues. Genera Fourier Series. More generay than the cassica Fourier series discusses above, we introduce the notion of a generaized Fourier series. Let {X n } be a sequence of mutuay orthogona eigenfunctions in L ([a, b]. Let φ be any function defined on [a, b]. Let A n X n, φ X n, X n. Then the A n are the generaized Fourier coefficients for A n X n (x, a generaized Fourier series of φ on [a, b]. We now state and prove some resuts regarding the convergence of Fourier series. First, we define some different types of convergence. Types of Convergence. There are severa types of convergence one can consider. We define three different types. Let {s n } be a sequence of functions defined on the interva [a, b]. We say s n converges to f pointwise on (a, b if s n (x f(x as n + 19

20 for each x (a, b. We say s n converges to f uniformy on [a, b] if max s n(x f(x a x b as n +. We say s n converges to f in the L -sense on (a, b if s n f L as n +. Some Convergence Resuts for Fourier Series. We wi now state some convergence resuts for Fourier series with respect to the different types of convergence. First, we need to give some definitions. We say f(x + = im f(x x x + f(x = im f(x. x x Theorem 7. (Pointwise Convergence of Cassica Fourier Series (Ref: Strauss, Sec. 5.4 Let f, f be piecewise continuous functions on [, ]. Then the fu Fourier series converges to 1 [f ext(x + + f ext (x ] at every point x R, where f ext is the -periodic extension of f. Remark. Due to the reationship between the Fourier e, Fourier coe and fu Fourier series, this theorem impies the convergence of the Fourier e and Fourier coe series as we. For a function φ defined on [, ], the Fourier e series of φ converges if and ony if the fu Fourier series of the odd extension of φ to [, ] converges. The foowing two theorems aow for generaized Fourier series. Let f be any function defined on [a, b]. Let {X n } be a sequence of mutuay orthogona eigenfunctions for the eigenvaue probem { X = λx a < x < b X satisfies symmetric B.C.s Let n A nx n (x be the associated generaized Fourier series of f. Theorem 8. (Uniform Convergence (Ref: Strauss, Sec. 5.4 The Fourier series n A nx n (x converges to f uniformy on [a, b] provided that (1 f, f, and f exist and are continuous for x [a, b]. ( f satisfies the given boundary conditions. Remark. In the case of any of the cassica Fourier series, the assumption that f exists may be dropped.

21 Theorem 9. (L convergence (Ref: Strauss, Sec. 5.4 The Fourier series n A nx n (x converges to f in the L sense in (a, b as ong as f L ([a, b]. Proofs of Convergence Resuts. In this section, we prove Theorem 7. In order to do so, we need to prove some preiminary resuts. Theorem 1. (Ref: Strauss, Sec. 5.4 Let {X n } be a sequence of mutuay orthogona functions in L. Let f L. Fix a positive integer N. Then the expression N E N = f a n X n L is minimized by choog a n = f, X n X n, X n. Remark. This means that in the series n a nx n, the coefficients a n which provide the best east-squares approximation to f are the Fourier coefficients. Proof. Ug the properties of inner product and the orthogonaity of the functions X n, we have N E N = f a n X n L = f a n X n, f a n X n = f, f = f = f = f f, f, f, N a n X n + a n X n, a n X n + a n X n + f, X m a m + m=1 Now competing the square in a m, we have m=1 [ X m, X m a m f, X ] m X m, X m a m = m=1 a m X m, m=1 a n X n a n X n a m X m, a m X m m=1 X m, X m a m. m=1 [ ( X m, X m a m f, X ] m f, X m X m, X m X m, X m. 1

22 Substituting this expression into the expression for E N above, we see that [ ( E N = f + X m, X m a m f, X ] m f, X m X m, X m X m, X m. (6.34 m=1 Therefore, the choice of a m that wi minimize E N is ceary as caimed. a m = f, X m X m, X m, Coroary 11. (Besse s Inequaity Let {X n } be a sequence of mutuay orthogona functions in L. Let f L and et A n be the Fourier coefficients associated with X n ; that is, A n = f, X n X n, X n. Then A n b a X n (x dx b a f(x dx. (6.35 Proof. Let a m = A m in (6.34. Combining this choice of coefficients with the fact that E N is nonnegative, we have E N = f m=1 Therefore, we concude that f, X m X m, X m X m, X m = f A m X m, X m f. m=1 Taking the imit as N +, the coroary foows. A m X m, X m. (6.36 Remark. Besse s inequaity wi pay a key roe in the proof of Theorem 7 on pointwise convergence. Theorem 1. (Parseva s Equaity Let {X n } be a sequence of mutuay orthogona functions in L. Let f L. The Fourier series n A nx n (x converges to f in L if and ony if m=1 A n X n L = f L. (6.37

23 Proof. By definition, the Fourier series n A nx n for f converges in L if and ony if N f A n X n as N +. L But, this means E N as N +. Combining this with (6.36, the theorem foows. We now have the necessary ingredients to prove Theorem 7 on pointwise convergence, but first we introduce some notation. Without oss of generaity, we may assume = π. Then the fu Fourier series on [ π, π] for the function f is given by f(x A + [A n cos(nx + B n (nx] where the coefficients are given by A = A n = B n = π 1, f 1, 1 = 1 f(x dx π π cos(nx, f cos(nx, cos(nx = 1 π (nx, f (nx, (nx = 1 π π π π π cos(nxf(x dx n = 1,,... (nxf(x dx n = 1,,.... Let S N (x be the N th partia sum of the fu Fourier series. That is, S N (x = A + [A n cos(nx + B n (nx]. We need to show that S N converges pointwise to 1 [f(x+ + f(x ] π < x < π. Substituting the formuas for the coefficients into S N, we see that The term S N (x = 1 [ π 1 + π π = 1 [ π 1 + π π ] (cos(nx cos(ny + (nx (ny f(y dy ] cos(n(y x f(y dy. K N (θ = 1 + cos(nθ is caed the Dirichet kerne. With this definition, we write S N (x = 1 π π π K N (y xf(y dy. We ist two properties of the Dirichet kerne which wi be usefu in proving Theorem 7. 3

24 K n (θ dθ = π π K n (θ dθ = π for a positive integers N. ] θ K N (θ = ([ N + 1 ( 1 θ. The first property is an easy cacuation. The second property can be proven as foows. K N (θ = 1 + cos(nθ = 1 + = n= N (e inθ + e inθ e inθ = e inθ e i(n+1θ 1 e iθ = e i(n+ 1 θ e i(n+ 1 θ e i 1 θ e i 1 θ = ([N + 1]θ ( 1 θ. Proof of Theorem 7. We need to show that for a x [ π, π] that S N (x converges to 1 [f(x+ f(x ]. Fix a point x [ π, π]. Taking the π-periodic extension of f, we have S N (x = 1 π = 1 π = 1 π π π π x π x π π K N (y xf(y dy K N (θf(θ + x dθ K N (θf(θ + x dθ. Now ug the first property of K N (θ given above, we have Therefore, f(x + = 1 π f(x = 1 π π π K N (θf(x + dθ K N (θf(x dθ. S N (x 1 [f(x+ + f(x ] = 1 π K N (θ[f(θ + x f(x + ] dθ π + 1 K N (θ[f(θ + x f(x ] dθ. π 4 π

25 Now we caim that as N + both of the terms on the right-hand side tend to zero, giving us the desired resut. We prove this for the first term. The second term can be handed simiary. Defining g(θ = f(x + θ f(x+ ( 1 θ, and ug the second property of K N (θ above, we have Defining 1 π π K N (θ[f(θ + x f(x + ] dθ = 1 π π g(θ ([ φ N (θ = N + 1 ] θ, ([ N + 1 ] θ dθ. it remains to show that g, φ N as N +. In order to prove this, we wi make use of Besse s inequaity (6.35. We do so as foows. First, we caim that {φ N } is a mutuay orthogona sequence of functions in L ([, π]. It can easiy be verified that φ N L ([,π] = π, and, therefore, φ N L ([, π]. The orthogonaity can be proven in the same manner that we proved the orthogonaity of {(N πx/} on [, ]. Second, we caim that g L ([, π]. By assumption f is piecewise continuous. Therefore, the ony potentia reason why g woud not be in L is the guarity in g at θ =. Therefore, we need to ook at im θ + g(θ. We have f(x + θ f(x + im g(θ = im θ + θ + ( 1θ f(x + θ f(x + θ = im im θ + θ θ + ( 1θ = f (x +. By assumption, f is piecewise continuous. Therefore, g is piecewise continuous and consequenty g L ([, π]. Ug the fact that g L ([, π] and {φ N } is a mutuay orthogona sequence of functions in L ([, π], we can make use of Besse s inequaity. In particuar, we have which impies N=1 π g, φ N φ N, φ N g L, g, φ N g L. N=1 Now ug the fact that g L, we concude that N=1 g, φ N is a convergent sequence, and, therefore, g, φ N as N +. This proves the theorem. 5

26 For proofs of the other convergence theorems, see Strauss, Section 5.5. We now show that the infinite series we caimed were soutions of the wave equation are actuay soutions. We demonstrate this ug the foowing exampe. Exampe 13. We return to considering the initia-vaue probem for the wave equation on [, ] with Dirichet boundary conditions, u tt c u xx = < x < u(x, = φ(x < x < (6.38 u t (x, = ψ(x < x < u(, t = = u(, t. As shown earier, any function of the form [ u n (x, t = A n cos ct + B n ] ct x satisfies the wave equation and the Dirichet boundary conditions. Now et u(x, t = [ ] A n cos ct + B n ct x, (6.39 where A n nπc B n x φ(x dx x ψ(x dx. We caim that this infinite series u is a soution of (6.38. We see that u satisfies the boundary conditions and the initia conditions (assuming nice initia data to aow for our convergence theorems. Therefore, it remains ony to verify that this infinite series satisfies our PDE. We prove so as foows. Let φ ext, ψ ext be the odd periodic extensions of φ and ψ, respectivey. Consider the initia-vaue probem for the wave equation on the whoe rea ine, u tt c u xx = < x < u(x, = φ ext (x (6.4 u t (x, = ψ ext (x. By d Aembert s formua, the soution of (6.4 is given by v(x, t = 1 [φ ext(x + ct + φ ext (x ct] + 1 c x+ct x ct ψ ext (y dy. Let ũ(x, t = v(x, t restricted to x. First, we caim that ũ(x, t is a soution of (6.38. Second, we caim that ũ(x, t is given by the series expansion in (6.39, thus proving that the infinite series (6.39 is a soution of (6.38. It is easy to verify that ũ is a soution 6

27 of (6.38. Therefore, it remains ony to show that ũ(x, t is given by the series expansion in (6.39. We do so as foows. Now for φ defined on [, ], the Fourier e series of φ is given by where à n = φ(y ( nπ x, φ ( nπ x, x = à n y x φ(x dx. Now for a nice function φ, the Fourier e series converges to φ on [, ]. In addition, ug the fact that φ ext is the odd periodic extension of φ, we know that the Fourier e series above converges to φ ext on a of R. We can do a simiar anaysis for ψ. That is, the Fourier e series of ψ is given by ψ(y B n y where B n = x ψ(x dx. Now pugging these series representations into the formua for ũ(x, t, we have [ ũ(x, t = 1 ] à n (x + ct + Ãn (x ct + 1 x+ct B n c x ct y dy [ = 1 ] à n x cos ct B n cos nπc y y=x+ct y=x ct = à n x cos ct + B n nπc x ct. But, this means ũ(x, t can be written as [ ] ũ(x, t = A n cos ct + B n ct x where A n = Ãn = B n = nπc B n = nπc x φ(x dx x ψ(x dx. Therefore, we have shown that the infinite series (6.39 is actuay a soution of (

28 6.4 The Inhomogeneous Probem on an Interva We now consider the inhomogeneous wave equation on an interva with symmetric boundary conditions, u tt c u xx = f(x, t < x < u(x, = φ(x < x < (6.41 u t (x, = ψ(x < x < u satisfies symmetric BCs. We wi sove this ug Duhame s principe. Reca from our earier discussion that the soution of an IVP for an inhomogeneous evoution equation of the form { Ut + AU = F x R n is given by U( x, = Φ( x U( x, t = S(tΦ( x + t S(t sf ( x, s ds where S(t is the soution operator for the homogeneous equation { Ut + AU = U( x, = Φ( x. Therefore, writing the inhomogeneous wave equation as the system, [ ] [ ] [ ] [ ] u 1 u + v c = t x v f [ ] [ ] u(x, φ(x =, v(x, ψ(x etting S(t denote the soution operator for this evoution equation, and defining S 1 (t, S (t such that [ ] S1 (tφ S(tΦ =, S (tφ we see that the soution of the inhomogeneous wave equation on R is given by [ ] φ t [ ] u(x, t = S 1 (t + S ψ 1 (t s ds. f(s We caim that we can use the same idea to sove the inhomogeneous wave equation on an interva [, ]. We state this formay for the case of Dirichet boundary conditions, but this can be extended more generay to any symmetric boundary conditions. Caim 14. Consider the inhomogeneous wave equation on an interva (6.41. denote the soution operator for the homogeneous equation, u tt c u xx = < x < u(x, = φ(x < x < u t (x, = ψ(x < x < u(, t = = u(, t. 8 Let S 1 (t (6.4

29 That is, et S 1 (t be the operator such that the soution of (6.4 is given by [ ] φ(x v(x, t = S 1 (t. ψ(x Then the soution of (6.41 is given by [ ] φ u(x, t S 1 (t + ψ t [ ] S 1 (t s ds. (6.43 f(s Proof. As shown by our earier discussion of Duhame s principe, defining u(x, t by (6.43, u wi satisfy the PDE and the initia conditions. The ony thing that remains to be verified is that u wi satisfy the boundary conditions. Let [ ] φ(x Φ(x. ψ(x By assumption, S 1 (tφ satisfies (6.4 for a Φ. Therefore, [ ] φ( t u(, t = S 1 (t + ψ( Simiary, u(, t =. = + t ds =. Exampe 15. Sove the initia-vaue probem, u tt u xx = f(x, t u(x, = φ(x u t (x, = ψ(x u(, t = = u(π, t. Therefore, S 1 (tφ( = = S 1 (tφ(. [ S 1 (t s f(, s < x < π < x < π < x < π We know that the soution of the homogeneous equation, v tt v xx = < x < π v(x, = φ(x < x < π v t (x, = ψ(x < x < π v(, t = = v(π, t is given by where v(x, t = A n = nb n = ] ds [A n cos(nt + B n (nt] (nx, (nx, φ(x (nx, (nx = π (nx, ψ(x (nx, (nx = π 9 π π (nxφ(x dx (nxψ(x dx (6.44

30 for n = 1,,.... Therefore, the soution operator associated with the homogeneous equation by [ ] φ S 1 (t = [A ψ n cos(nt + B n (nt] (nx, with A n, B n as defined above. Therefore, [ ] S 1 (t s = [C f(s n (s cos(n(t s + D n (s (n(t s] (nx where C n (s = nd n (s = (nx, f(x, s (nx, (nx = π for s t, n = 1,,.... Therefore, the soution of (6.44 is given by u(x, t = [A n cos(nt + B n (nt] (nx + with A n, B n, D n (s as defined above. 6.5 Inhomogeneous Boundary Data π t (nxf(x, s dx D n (s (n(t s (nx ds We now consider the wave equation with inhomogeneous boundary data, u tt c u xx = f(x, t < x < u(x, = φ(x < x < u t (x, = ψ(x < x < u(, t = g(t, u(, t = h(t. Method of Shifting the Data Ref: Strauss: Sec Our pan is to introduce a new function U such that U(, t = g(t U(, t = h(t. (6.45 Then, defining v(x, t = u(x, t U(x, t, we wi study the new initia-vaue probem which now has zero boundary data, v tt c v xx = f(x, t U tt + c U xx < x < v(x, = φ(x U(x, < x < (6.46 v t (x, = ψ(x U t (x, < x < v(, t = = v(, t. 3

31 In order for U to satisfy the conditions stated, we define U as a inear function of x such that [ ] h(t g(t U(x, t = x + g(t. Now U xx =. Therefore (6.46 becomes v tt c v xx = f(x, t U tt v(x, = φ(x U(x, v t (x, = ψ(x U t (x, v(, t = = v(, t. < x < < x < < x < This probem we can sove ug Duhame s principe discussed above. (6.47 Remark. If the boundary conditions g, h and the inhomogeneous term f do not depend on t, then the method of shifting the data is especiay nice because we can immediatey reduce the probem to a homogeneous IVP with homogeneous boundary conditions. In particuar, consider u tt c u xx = f(x < x < u(x, = φ(x < x < u t (x, = ψ(x u(, t = g, u(, t = h Now considering (6.46, if we find U(x such that c U xx = f(x U( = g U( = h < x < then, etting v(x, t = u(x, t U(x, we see immediatey that v wi be a soution of v tt c v xx = < x < v(x, = φ(x U(x < x < (6.48 v t (x, = ψ(x < x < v(, t = = v(, t, a homogeneous initia-vaue probem with Dirichet boundary conditions. 6.6 Uniqueness In the previous sections, we used the method of refection and separation of variabes to find a soution of the wave equation on an interva satisfying certain boundary conditions. We now prove that the soutions we found are in fact unique. Caim 16. Consider the initia-vaue probem for the inhomogeneous wave equation on an interva with Dirichet boundary conditions, u tt c u xx = f(x, t < x < u(x, = φ(x < x < (6.49 u t (x, = ψ(x < x < u(, t = = u(, t. 31

32 There exists at most one (smooth soution of (6.49. Proof. Suppose there are two soutions u, v of (6.49. Let w = u v. Therefore, w is a soution of w tt c w xx = < x < w(x, = < x < (6.5 w t (x, = < x < w(, t = = w(, t. Define the energy function We see that E( = 1 E(t = 1 w t + c w x dx. w t (, t + c w x(, t dx =. We caim that E (t =. Integrating by parts, we have E (t = = = w t w tt + c w x w xt dx w t w tt c w xx w t dx + c w x w t x= x= w t [w tt c w xx ] dx +, ug the fact that w(, t = = w(, t for a t impies that w t (, t = = w t (, t. As w is a soution of the homogeneous wave equation, we concude that E (t =. Therefore, ug the fact that E( =, we concude that E(t =. Then, ug the fact that w is a smooth soution, we concude that w x (x, t = = w t (x, t. Therefore, w(x, t = C for some constant C. Ug the fact that w(x, =, we concude that w(x, t =, and, therefore, u(x, t = v(x, t. 3