Math 124B January 17, 2012

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1 Math 124B January 17, 212 Viktor Grigoryan 3 Fu Fourier series We saw in previous ectures how the Dirichet and Neumann boundary conditions ead to respectivey sine and cosine Fourier series of the initia data. In the same way the periodic boundary conditions u(, t) = u(, t), u x (, t) = u x (, t), (1) ead to the fu Fourier expansion, i.e. the Fourier series that contains both sines and cosines. Such boundary conditions arise naturay in appications. For exampe heat conduction phenomena in a circuar rod wi be described by the heat equation subject to boundary conditions (1). To see how the fu Fourier series comes about, et us consider the eigenvaue probem associated with the boundary conditions (1), X = λx X( ) = X(), (2) X ( ) = X (). To find the eigenvaues and eigenfunctions for (2), we consider the quaitativey distinct cases λ <, λ = and λ > separatey. We first assume λ = γ 2 <. In this case the equation takes the form X = γ 2 X, and the soutions are thus X(x) = Ce γx + De γx. The boundary conditions then impy Ce γ + De γ = Ce γ + De γ, Cγe γ Dγe γ = Cγe γ Dγe γ. Mutipying the first equation by γ and adding to the second equation gives 2Cγe γ = 2Cγe γ 2Cγe γ (1 e 2γ ) = C =, since γ,. Simiary, D =, which impies that there are no negative eigenvaues. We next assume that λ =, which resuts in the equation X =. The soution is then X(x) = C+Dx, and the boundary conditions impy C D = C + D 2D = D =. D = D So X(x) 1 is an eigenfunction corresponding to the eigenvaue λ =. Finay, we consider the case λ = β 2 >. The equation then takes the form X = β 2 X, and hence, has the soution X(x) = C cos βx + D sin βx. Checking the boundary conditions gives C cos β D sin β = C cos β + D sin β Cβ sin β + Dβ cos β = Cβ sin β + Dβ cos β 2D sin β = 2Cβ sin β = Since C and D cannot both be equa to zero, and β, we must have sin β = β = nπ, for n = 1, 2,.... Then the positive eigenvaues and the corresponding eigenfunctions are λ n = ( nπ ) 2, Xn (x) = C + D sin nπx, for n = 1, 2,.... Notice that for each of the eigenvaues λ n, n = 1, 2,... we have two ineary independent eigenfunctions, cos(nπx/) and sin(nπx/). This differs from the case of the Dirichet and Neumann boundary conditions, where we had ony one ineary independent eigenfunction for each of the same eigenvaues, 1

2 namey sin(nπx/) for Dirichet, and cos(nπx/) for Neumann. We wi see in the next ecture that in the cases where there are more than one ineary independent eigenfunctions corresponding to the same eigenvaue, one can aways choose a pairwise orthogona set of eigenfunctions, which is necessary for the method of computing the Fourier coefficients to go through. Having soved the eigenvaue probem (2), it is now cear that the series soutions of PDEs subject to the periodic boundary conditions (1) wi have both sines and cosines in their expansions. For exampe the soution to the heat equation satisfying the boundary conditions (1) wi be given by the series u(x, t) = A e (nπ/)2kt (A n + B n sin nπx ), where the coefficients A n, B n come from the expansion of the initia data φ(x) = A A n + B n sin nπx. (3) The series (3) is caed the fu Fourier series of the function φ(x) on the interva (, ). It is now cear that to sove boundary vaue probems with periodic boundary conditions (1) via separation of variabes, one needs to find the coefficients in the expansion (3). This procedure is simiar to finding the coefficients of the Fourier cosine and sine series, and reies on the pairwise orthogonaity of the eigenfunctions. Notice that (3) is a inear combination of the eements of the set 1,, sin nπx : n = 1, 2,... }. (4) Now, if we can prove that the eements of this set are pairwise orthogona, then the coefficients in (3) can be found by taking the dot product of the series (3) with the corresponding eigenfunction. Since the interva in this case is (, ), we define the dot product of two functions to be (f, g) = ˆ f(x)g(x) dx. Let us now check the orthogonaity of the set (4). We need to show that ˆ ˆ ˆ ˆ ˆ sin nπx 1 1 sin mπx sin mπx sin mπx dx =, for a n, m = 1, 2,..., (5) dx =, for n m, (6) dx =, for n m, (7) dx =, for a m = 1, 2,..., (8) dx =, for a m = 1, 2,.... (9) Notice that, since sin(nπx/) and cos(nπx/) sin(mπx/) are odd functions, (5) and (9) are trivia, since the integration is performed over a symmetric interva. The integrands in (6), (7) and (8) are even functions, so the integras are equa to twice the corresponding integras over the interva (, ). Then, say for (6), we have ˆ ˆ dx = 2 dx =, for n m, 2

3 as was computed in the ast ecture, when considering Fourier cosine series over the interva (, ). Proofs of(7) and (8) are simiar. Thus, the set (4) is indeed orthogona. To compute the coefficients in (3), we take the dot product of the series with the corresponding eigenfunctions. For exampe, ˆ φ(x) dx = ˆ ( A 2 + A n + B n sin nπx ) and simiary for the eigenfunctions 1 = cos(πx/), and sin(mπx/). But ˆ ˆ ˆ sin 2 mπx 1 2 dx = 2. So the coefficients can be computed as ˆ dx = 2 dx = 2 ˆ sin 2 mπx dx = 2 2 =, dx = 2 2 =, = A m ˆ dx, A m = 1 B m = 1 ˆ ˆ φ(x) φ(x) sin mπx dx, n =, 1, 2,..., dx, n = 1, 2,.... (1) Exampe 3.1. Find the fu Fourier series of the function φ(x) = x on the interva (, ). We compute the coefficients in the expansion using formuas (1). x = A A n A n = 1 ˆ x + B n sin nπx dx =, since the function x cos(nπx/) is odd, and the integration interva is symmetric. Using evenness of the function x sin(nπx/), we have B n = 1 ˆ x sin nπx dx = 2 ˆ x sin nπx n+1 2 dx = ( 1) nπ, where the ast integra is exacty the coefficient of the Fourier sine series of function φ(x) = x, computed in the ast ecture. Thus the fu Fourier series of x on the interva (, ) is x = ( 1) n+1 2 nπ nπx sin, which exacty coincides with the Fourier sine series of the function x on the interva (, ). 3

4 3.1 Even, odd and periodic functions In the previous exampe we coud take any odd function φ(x), and the coefficients of the cosine terms in the fu Fourier series woud vanish for exacty the same reason, eading to the Fourier sine series. Thus the fu Fourier series of an odd function on the interva (, ) coincides with its Fourier sine series on the interva (, ). Anaogousy, the fu Fourier series of an even function on the interva (, ) coincides with its Fourier cosine series on the interva (, ). This is not surprising, taking into account the fact that the sine and cosine functions are themseves respectivey odd and even. So how can the same function φ(x), be it odd, even, or neither, have both a nonzero Fourier cosine and a nonzero Fourier sine expansion on the interva (, )? The answer to this question is trivia, if one recas the refection method used to sove boundary vaue probems on the finite interva (, ). In the refection method one takes either an odd (Dirichet BCs), or an even (Neumann BCs) extension of the data. Then the resuting extended data is either odd or even on the symmetric interva (, ), and is extended to be periodic with period 2 to the rest of the number ine. But then one can find the fu Fourier series of this extended data, which wi contain ony sines or cosines, depending on the evenness or the oddness of the extension, thus giving either the Fourier cosine, or Fourier sine series of the function φ(x) on the interva (, ). In short, one has the foowing correspondence between the boundary conditions, extensions, and Fourier series. Dirichet: u(, t) = u(, t) = odd extension Fourier sine series. Neumann: u x (, t) = u x (, t) = even extension Fourier cosine series. Periodic: u(, t) = u(, t), u x (, t) = u x (, t) periodic extension fu Fourier series. 3.2 The compex form of the fu Fourier series Instead of writing the Fourier series in terms of cosines and sines, one can use compex exponentia functions. Reca the Euer s formua, e iθ = cos θ + i sin θ, (11) from which one can get the formua for e iθ in terms of cosine and sine by pugging in θ. Soving from these two equations, we have the expressions of sine and cosine in terms of the compex exponentias cos θ = eiθ + e iθ 2, sin θ = eiθ e iθ. (12) 2i So instead of (3), one can write the Fourier expansion in terms of the functions e inπx/ and e inπx/, themseves eigenfunctions corresponding to the eigenvaue λ n = n 2 π 2 / 2. The convenience of the compex form is in that the Fourier series can be written compacty as φ(x) = n= C n e inπx/, (13) which is a inear combination in terms of the eements of the set 1, e iπx/, e iπx/, e i2πx/, e i2πx/,... } = 1, e ±inπx/ : n = 1, 2,... }, (14) repacing the set (4). In order to find the coefficients in the expansion (13), we first need to show that the set (14) is orthogona, after which the coefficients can be found by taking the dot product of the series with the corresponding eigenfunction. Notice that the functions in the set (14) are compex vaued. For compex vaued functions the dot product is defined as (f, g) = ˆ f(x)g(x) dx, 4

5 where the bar stands for the compex conjugate. Then the dot product of two distinct eements of the set (14) is ˆ e inπx/ e imπx/ dx = 1 i(n m)π ei(n m)πx/ = e i(n m)π i(n m)π (ei2(n m)π 1) =, for n m. So taking the dot product of the series (13) with one of the eigenfunctions in (14) gives ˆ φ(x)e imπx/ dx = ˆ n= Thus, the coefficients can be computed as 3.3 Concusion ˆ C n e inπx/ e imπx/ dx = C m e imπx/ e imπx/ dx = C m 2. C n = 1 2 ˆ φ(x)e inπx/ dx. (15) We saw how the separation of variabes in the case of periodic boundary conditions eads to the fu Fourier series (3) in the same way as the Dirichet and Neumann boundary conditions ead to the Fourier sine and cosine series respectivey. The coefficients in the fu Fourier expansion were computed by making use of the pairwise orthogonaity of eigenfunctions, resuting in the coefficients formuas (1). We aso expored the connection between the Fourier cosine, respectivey sine, expansions on the interva (, ) and the fu Fourier expansion on the interva (, ), using the idea of even and odd extensions. Finay using Euer s formua (11), we rewrote the fu Fourier series in the compex form (13), the coefficients of which were again computed by taking the dot product of the series with the eigenfunctions, and reying on the fact that the distinct eigenfunctions are orthogona. We wi see in the next ecture that one aways has orthogona eigenfunctions, provided the boundary conditions are symmetric in an appropriate sense. This wi ead to the notion of generaized Fourier series. 5