Math 356 Lecture Notes Intro to Pdes Eigenfunction Expansions for Ibvps
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MATH 356 LECTURE NOTES INTRO TO PDES EIGENFUNCTION EXPANSIONS FOR IBVPS J. WONG (FALL 2019) Topics covered • Eigenfunction expansions for PDEs ◦ The procedure for time-dependent problems ◦ Projection, independent evolution of modes ◦ Separation of variables • The wave equation ◦ Standard solution, standing waves ◦ Example: tuning the vibrating string (resonance) • Interpreting solutions (what does the series mean?) ◦ Relevance of Fourier convergence theorem ◦ Smoothing for the heat equation ◦ Non-smoothing for the wave equation 3 1. The eigenfunction method to solve PDEs We are now ready to demonstrate how to use the components derived thus far to solve the heat equation. The procedure here for the heat equation will extend nicely to a variety of other problems. For now, consider an initial boundary value problem of the form ut = −Lu + h(x; t); x 2 (a; b); t > 0 hom. BCs at a and b (1.1) u(x; 0) = f(x) We seek a solution in terms of the eigenfunction basis X u(x; t) = cn(t)φn(x) n by finding simple ODEs to solve for the coefficients cn(t): This form of the solution is called an eigenfunction expansion for u (or `eigenfunction series') and each term cnφn(x) is a mode (or `Fourier mode' or `eigenmode'). Part 1: find the eigenfunction basis. The first step is to compute the basis. The eigenfunctions we need are the solutions to the eigenvalue problem Lφ = λφ, φ(x) satisfies the BCs for u: (1.2) By the theorem in ??, there is a sequence of eigenfunctions fφng with eigenvalues fλng that form an orthogonal basis for L2[a; b] (i.e. one with all the required properties). 1 2 J. WONG (FALL 2019) If possible, we compute solutions explicitly via the standard procedure. Note that the BCs imposed on the eigenvalue problem must be homogeneous. Now at each fixed time t, the function u(x; t) is a function of x defined on [a; b]. It fol- lows that there are coefficients cn(t) such that X u(x; t) = cn(t)φn(x): (1.3) n For each t,fcn(t)g is the set of coefficients for expressing u(x; t) in terms of the basis fφng: Part 2: get equations for the coefficients: Our objective now is to reduce the prob- lem to ODEs for coefficients cn(t): The order of steps can be changed here. Step 2a (Write known functions in the basis): We express every known function in the problem (PDE and ICs) in the eigenfunction basis using the orthogonality rule (??). In this case, there are two such functions, the source term and the initial condition: X hf; φni f(x) = f φ (x); f = ; (1.4) n n n hφ ; φ i n n n 1 X hh(x; t); φn(x)i h(x; t) = h (t)φ (x); h (t) = : (1.5) n n n hφ ; φ i n=0 n n R b 2 Here hf; gi = a f(x)g(x) dx is the L inner product in x; often, it is not worth calculating the integrals explicitly (they can be a mess). Note that in (1.5), the t-variable is not part of the integral, e.g. Z b h(x; t) = tx =) hh; φni = t xφn(x) dx: a Step 2b (Plug series in to the PDE): We now `plug the series in' to the PDE - which expands the PDE in terms of the eigenfunction basis. Taking this one part at a time1, ! @ X u = c (t)φ (x) t @t n n n X 0 = cn(t)φn(x); (t-derivs. only act on coeffs.) n X Lu = cn(t)Lφn n X = λncn(t)φn(x)(φn is an eigenfunction) n . Plugging this into the PDE, we get ut = −Lu + h(x; t) X 0 X X =) cn(t)φn(x) = − λncn(t)φn(x) + hn(t)φn(x): n n n 1After getting used to the process, you can shortcut the work here and skip some steps. They tend to be the same for most problems, but you should be careful and recognize when the steps must be changed. EIGENFUNCTION EXPANSION 3 The eigenfunction property means that the coefficient of φn in Lu depends only on cn(t). That is, the n-th term in the series is independent of the others. Now since fφng is a basis, the coefficients on the left and right sides must be equal term-by- term. Alternatively, X 0 X (cn(t) + λncn(t) − hn(t))φn(x) = 0 = 0 · φn n n and since zero must have a unique representation in the basis, 0 cn(t) + λncn(t) = hn(t) for all n: (1.6) The difficult part is now over. Step 2c (solve the ODE): Next, solve the ODE (1.6) for each n. Note that sometimes, this will involve some case work (as we saw with Fourier series). P The solution will have some arbitrary constants. At this point, the series cn(t)φn(x) is a `general solution' to the PDE with the given BCs with the initial condition not yet imposed. Step 2d (apply initial conditions): This can be done earlier. We turn the ODE into an IVP with a unique solution by applying the initial conditions (any `leftover' equations not yet used). As in the PDE, plug in the series: X X u(x; 0) = f(x) =) cn(0)φn(x) = fnφn(x) n which gives initial conditions for the cn's: cn(0) = fn: (1.7) Solving (1.6) with (1.7) yields unique solutions for cn(t), and we are done. Step 3 (summarize the solution): As the steps can be long, it is useful to make the solution clear. You should check that all functions and constants are well-defined by some formula. We have obtained a solution X u(x; t) = cn(t)φn(x) n to the IBVP (1.1) where the cn's and φn's are given by cn(t) = ··· ; φn = ··· (replace with formula computed in the procedure) and fn; hn(t) are given by (1.4) and (1.5). Note that it is often best to leave expressions in terms of explicit integrals e.g. like Z π hf; φni 2 2 fn = = x sin nx dx hφn; φni π 0 when computing would not give any new insight. The inner product shorthand can be used, but its exact definition should be clear (the interval etc.). 4 J. WONG (FALL 2019) 1.1. Example 1: no source; Dirichlet BCs. The simplest case. We solve ut =uxx; x 2 (0; 1); t > 0 (1.8a) u(0; t) = 0; u(1; t) = 0; (1.8b) u(x; 0) = f(x): (1.8c) The eigenvalues/eigenfunctions are (as calculated in previous sections) 2 2 λn = n π ; φn = sin nπx; n ≥ 1: (1.9) Assuming the solution exists, it can be written in the eigenfunction basis as 1 X u(x; t) = cn(t)φn(x): n=0 R 1 2 There is only one function to write in the eigenfunction basis (note that 0 φn dx = 1=2 here) 1 Z 1 X hf; φni f(x) = f φ (x); f = = 2 f(x) sin nπx dx: (1.10) n n n hφ ; φ i n=1 n n 0 00 Substitute into the PDE (1.8a) and use the fact that −φn = λnφ to obtain 1 X 0 (cn(t) + λncn(t))φn(x) = 0: n=1 Doing the same for the initial condition, we get the IC for cn: 1 1 X X u(x; 0) = f(x) =) cn(0)φn(x) = fnφn(x) =) cn(0) = fn: n=1 n=1 Equating coefficients of each basis function (to zero), we find that 0 cn(t) + λncn(t) = 0; cn(0) = fn: (1.11) We are now done. The solution to the IBVP (1.8) is 1 X −n2π2t u(x; t) = bne sin nπx with bn given by (1.13): (1.12) n=1 Alternatively, we could summarize by writing that the solution is 1 X −λnt u(x; t) = fne φn(x) n=1 with eigenfunctions/values φn; λn given by (1.9) and fn by (1.10). Long-time behavior: Note that every term in the solution (1.12) has a negative exponential (since all the eigenvalues are positive). Furthermore, terms further down in the series decay much faster since λn grows quadratically with n. It follows (informally) that lim u(x; t) = 0 t!1 EIGENFUNCTION EXPANSION 5 independent of the initial condition f(x) (which just affects the bn's and not the eigenvalues). Moreover, by approximating u by its first term, −λ1t −λ1t u(x; t) ≈ f1e φ1(x) =) decays to zero at least as fast as f1e : Thus, the smallest eigenvalue of a non-zero term gives the convergence rate for the solution (which is exponential). As an explicit example, suppose the initial condition is f(x) = x(1 − x): After some laborious integration by parts, we get ( 2(1 − (−1)n) 0 n even bn = 3 3 = 2 (1.13) π n π3n3 n odd: The first few terms of the solution are 2 2 2 2 u(x; t) = e−π t sin πx + e−9π t sin 3πx + ··· π3 27π3 Note that u(x; t) is missing a φ2 = sin 2πx term; this does not affect the convergence rate. However, suppose instead that f(x) = x − 1=2: Then we have Z 1 Z 1 b1 = 2 (x − 1=2) sin πx dx = 0; b2 = 2 (x − 1=2) sin 2πx dx = −1=pi: 0 0 the n = 1 term vanishes, so 1 2 u(x; t) ≈ − e−λ2π tφ (x) + ··· as t ! 1 π 2 2 and the convergence rate is given by the second eigenvalue λ2 = 4π (faster convergence!).