Math 124B January 31, 2012

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1 Math 124B January 31, 212 Viktor Grigoryan 7 Inhomogeneous boundary vaue probems Having studied the theory of Fourier series, with which we successfuy soved boundary vaue probems for the homogeneous heat and wave equations with homogeneous boundary conditions, we woud ike to turn to inhomogeneous probems, and use the Fourier series in our search for soutions. We start with the foowing boundary vaue probem for the inhomogeneous heat equation with homogeneous Dirichet conditions. ut ku xx = fx, t, for < x <, t >, ux, = φx, 1 u, t = u, t =. A naïve separation of variabes wi not amount to much, since the source function fx, t may not be separabe. Instead, we can use the eigenfunctions of the corresponding homogeneous eigenvaue probem, and ook for a soution in the series form ux, t = u n t sin nπx. 2 This is, of course, the Fourier sine series of ux, t for each fixed t >, the coefficients of which wi vary with the variabe t. Such an expansion aways exists due to competeness of the set of eigenfunctions sinnπx/}. The coefficients wi then be given by the Fourier sine coefficients formua u n t = 2 We can simiary expand the source function, fx, t = ˆ ux, t sin nπx dx. f n t sin nπx, where f n t = 2 ˆ fx, t sin nπx dx. 3 Now, since we are ooking for a twice differentiabe function ux, t that satisfies the homogeneous Dirichet boundary conditions, we can differentiate the Fourier series 2 term by term to obtain u xx x, t = u n t nπ nπx sin. 4 Notice that to be abe to differentiate twice, we need to guarantee that u x satisfies homogeneous Neumann conditions, which can be arranged by taking the odd extension of u to the interva,, and then the periodic extension to the rest of the rea ine. We can aso differentiate the series 2 with respect to t to obtain since the Fourier coefficients of u t x, t are u t x, t = ˆ 2 u t x, t sin nπx dx = t u nt sin nπx, 5 ˆ 2 ux, t sin nπx dx = u nt. Differentiation under the above integra is aowed since the resuting integrand is continuous. 1

2 Substituting 5 and 4 into the equation, and using 3, we have u nt + k But then, due to the competeness, u n + k nπ un t sin nπx = f n t sin nπx. nπ un = f n t, for n = 1, 2,..., which are ODEs for the coefficients u n t of the series 2. Using the integrating factor e knπ/2t, we can rewrite the ODE as e knπ/2t u n = e knπ/t f n t. Hence, we get But the initia condition impies thus, u n t = ue knπ/2t + e knπ/2 t ux, = φx = u n = φ n = 2 So the soution can be written in the series form as ux, t = φ n e knπ/2t + ˆ e knπ/2s f n s ds. u n sin nπx, φx sin nπx dx. 6 e knπ/2 s t f n s ds sin nπx, 7 where φ n, and f n are the Fourier coefficients of the initia data φx and the source term fx, t, and can be found from 6 and 3 respectivey. Notice that the first coefficient term in the above series comes from the homogeneous heat equation, whie the second term can be thought of as coming from the variation of parameters for the inhomogeneous equation. The case of the Neumann boundary conditions for the inhomogeneous heat equation is simiar, with the ony difference that one ooks for a series soution in terms of cosines, rather than the sine series 2. The boundary vaue probem for the inhomogeneous wave equation, utt c 2 u xx = fx, t, for < x <, ux, = φx, u t x, = ψx, 8 u, t = u, t =, can aso be soved by expanding the soution into the series 2. In this case the ODEs for the coefficients wi be nπ u n + c 2 un = f n t, for n = 1, 2,..., with the initia conditions u n = φ n, u n = ψ n. These ODEs can be soved expicity using variation of parameters to obtain the coefficients u n t, with which one can form the series soution 2. 2

3 7.1 Inhomogeneous boundary conditions We next study the case of inhomogeneous boundary conditions, and consider the foowing boundary vaue probem for the inhomogeneous heat equation. ut ku xx = fx, t, for < x <, t >, ux, = φx, 9 u, t = ht, u, t = jt. We wi use the method of shifting the data to reduce this probem to one with homogeneous boundary conditions, which wi be equivaent to 1. The idea of this method is to subtract a function satisfying the inhomogeneous boundary conditions from the soution to the above probem. Let us define the function Ux, t = 1 x ht + x jt, for which triviay U, t = ht, and U, t = jt. But then for the new quantity vx, t = ux, t Ux, t, we have v t kv xx = u t ku xx U t ku xx = fx, t 1 x h t x j t, vx, = ux, Ux, = φx 1 x v, t = u, t U, t = ht ht =, v, t = u, t U, t = jt jt =. h x j, So vx, t satisfies the foowing boundary vaue probem with homogeneous boundary conditions, v t kv xx = fx, t, for < x <, t >, vx, = φx, v, t = v, t =, 1 where fx, t = fx, t 1 x h t x j t, φx = φx 1 x h x 11 j. Probem 1 is equivaent to 1, so the function vx, t can be found by formua 7, in which φ n and f n s must be repaced by respectivey φ n and f n s. Ceary, φ n and f n t can be found in terms of the Fourier coefficients of φx and fx, t using 11. Having found the function vx, t in the series form, vx, t = φ n e knπ/2t + we can write the soution to probem 9 as e knπ/2 s t fn s ds sin nπx, ux, t = vx, t + Ux, t = 1 x ht + x jt + φ n e knπ/2t + In the case of Neumann boundary conditions, e knπ/2 s t fn s ds sin nπx. 12 u x, t = ht, u x, t = jt, 3

4 one shoud use the function ˆ V x, t = Ux, t dx = x x2 ht + x2 2 2 jt to shift the data, since V x, t = U, t = ht, and V x, t = U, t = jt. For the inhomogeneous boundary vaue probem for the wave equation, utt c 2 u xx = fx, t, for < x <, ux, = φx, u t x, = ψx, u, t = ht, u, t = jt, 13 one can shift the data in the same way to reduce the probem to one equivaent to 8. An aternative method to shifting the data, is to use expansion 2 to find the soution of probem 9 directy. The difference between this case and the case of probem 1 is in that the soution does not satisfy the homogeneous boundary conditions, so the series 2 can not be differentiated term by term when substituting into the equation. The way out is to separatey expand u xx x, t as But then using Green s second identity, we can compute w n t = 2 ˆ ˆ b a u xx x, t = u xx x, t sin nπx dx = 2 ˆ nπ nπx ux, t sin w n t sin nπx. 14 f g gf dx = f g fg b a, dx + 2 u x sin nπx nπ u cos nπx Noticing that the first boundary term vanishes, and using the boundary conditions for u from 9 in the second term, we have nπ w n t = un t 2nπ 2 1n jt + 2nπ ht Then substituting 14 and 5 into the equation, and using 15, we get nπ u nt k un t 2nπ 2 1n jt + 2nπ ht sin nπx 2 Using the competeness, we obtain the foowing ODEs for the coefficients u n t. u n + k nπ un = f n t 2nπ 2 1n jt ht, with u n = φ n. =. f n t sin nπx. 4

5 One can sove these ODEs using the same integrating factor as before to get u n t = φ n e knπ/2t + e knπ/2 s t f n s 2nπ 2 1n js hs ds. The soution to 9 wi then be given by the series 2. It is eft as an exercise to convince yourseves that this soution is equivaent to 12. For the inhomogeneous wave probem 13 this method wi yied the foowing ODEs for the coefficients with the initia conditions u n + c 2 nπ un = f n t 2nπ 2 1n jt ht, u n = φ n, u n = ψ n. Soving these ODEs by variation of parameters, one can find the soution to 13 as the series Concusion Using the eigenfunctions corresponding to the associated homogeneous boundary vaue probems as buiding bocks, we searched for a soution to the inhomogeneous heat and wave equations in the form of a series in terms of these eigenfunctions. This ead to ODEs for the coefficients of the series, which is akin to the method of variation of parameters, where one varies the coefficients in the inear combination of soutions to the homogeneous equation to obtain a soution of the inhomogeneous one. We studied two methods of soving probems with inhomogeneous boundary conditions, the first of which amounted to merey shifting the boundary data to reduce the probem to one with homogeneous boundary conditions. The second method was to directy ook for a soution in the form of a series in terms of the eigenfunctions of the associated homogeneous probem. In this case one can no onger directy substitute the series into the equation due to the issue of boundary terms when differentiating Fourier series. However, after taking care of these boundary terms, we arrived at ODEs for the coefficients of this expansion as we, soving which wi give the series soution 2. 5