MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � 119 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 21. Tuesday, April 28: Dirichlet problem on an interval I want to do a couple of ‘serious’ applications of what we have done so far. There are many to choose from, and I will mention some more, but let me first consider the Diriclet problem on an interval. I will choose the interval [0, 2π] because we looked at it before. So, what we are interested in is the problem of solving d2u(x)(21.1) − dx2 + V (x)u(x) = f(x) on (0, 2π), u(0) = u(2π) = 0 where the last part are the Dirichlet boundary conditions. I will assume that (21.2) V : [0, 2π] −→ R is continuous. Now, it certainly makes sense to try to solve the equation (21.1) for say a given f ∈ C0([0, 2π]), looking for a solution which is twice continuously differentiable on the interval. It may not exist, depending on V but one thing we can shoot for, which has the virtue of being explicit is the following: Proposition 28. If V ≥ 0 as in (21.2) then for each f ∈ C0([0, 2π]) there exists a unique twice continuously differentiable solution to (21.1). You will see that it is a bit hard to approach this directly – especially if you remember some ODE theory from 18.03. There are in fact various approaches to this but we want to go through L2 theory – not surprisingly of course. How to start? Well, we do know how to solve (21.1) if V ≡ 0 since we can use (Riemann) integration. Thus, ignoring the boundary conditions we can find a solution to −d2v/dx2 = f on the interval by integrationg twice: x y (21.3) v(x) = − f(t)dtdy satifies − d2v/dx2 = f on (0, 2π). 0 0 Moroever v really is twice continuously differentiable if f is continuous. So, what has this got to do with operators? Well, we can change the order of integration in (21.3) to write v as � x � x � 2π (21.4) v(x) = − f (t)dydt = a(x, t)f(t)dt, a(x, t) = (t − x)H(x − t) 0 t 0 where the Heaviside function H(y) is 1 when y ≥ 0 and 0 when y < 0. Thus a(x, t) is actually continuous on [0, 2π] × [0, 2π] since the t − x factor vanishes at the jump in H(t − x). Thus v is given by applying an integral operator to f. Before thinking more seriously about this, recall that there are also boundary conditions. Clearly, v(0) = 0 since we integrated from there. However, there is no particular reason why � 2π (21.5) v(2π) = (t − 2π)f(t)dt 0 should vanish. However, we can always add to v any linear function and still satify the differential equation. Since we do not want to spoil the vanishing at x = 0 we can only afford to add cx but if we choose c correctly, namely consider � 2π1(21.6) c = − 2π 0 (t − 2π)f(t)dt, then (v + cx)(2π) = 0.120 LECTURE NOTES FOR 18.102, SPRING 2009 So, finally the solution we want is � 2π (21.7) w(x) = b(x, t)f(t)dt, b(x, t) = (t − x)H(x − t) − t − 2πx.2π0 This is the unique, twice continuously differentiable, solution of −d2w/dx2 = f in (0, 2π) which vanishes at both end points. Lemma 16. The integral operator (21.7) extends by continuity from C0([0, 2π]) to a compact, self-adjoint operator on L2(0, 2π). Proof. Since w is given by an integral operator with a continuous real-valued kernel which is even in the sense that (check it) (21.8) b(t, x) = b(x, t) we might as well give a more general result. � Proposition 29. If b ∈ C0([0, 2π]2) then � 2π (21.9) Bf(x) = b(x, t)f(t)dt 0 defines a compact operator on L2(0, 2π) if in addition b satisfies (21.10) b(t, x) = b(x, t) then B is self-adjoint. Proof. � Now, recall from one of the Problem sets that uk = c sin(kx/2), k ∈ N, is an orthonormal basis for L2(0, 2π). Moreover, differentiating we find straight away that d2uk k2 (21.11) − dx2 =4 uk. Since of course uk(0) = 0 = uk(2π) as well, from the uniqueness above we conclude that 4(21.12) Buk = uk ∀ k. k2 Thus, in this case we know the orthonormal basis of eigenfunctions for B. They are the uk, each eigenspace is 1 dimensional and the eigenvalues are 4k−2 . So, this happenstance allows us to decompose B as the square of another operator defined directly on the othornormal basis. Namely 2(21.13) Auk = kuk = ⇒ B = A2 . Here again it is immediate that A is a compact self-adjoint operator on L2(0, 2π) since its eigenvalues tend to 0. In fact we can see quite a lot more than this. Lemma 17. The operator A maps L2(0, 2π) into C0([0, 2π]) and Af(0) = Af(2π) = 0 for all f ∈ L2(0, 2π).� � � 121 LECTURE NOTES FOR 18.102, SPRING 2009 Proof. If f ∈ L2(0, 2π) we may expand it in Fourier-Bessel series in terms of the uk and find (21.14) f = ckuk, (ck) ∈ L2 . k Then of course, by definition, � 2ck(21.15) Af = uk. k k Here each uk is a bounded continuous function, with the bound on uk being in-dependent of k. So in fact (21.15) converges uniformly and absolutely since it is uniformly Cauchy, for any q > p, ⎛ ⎞ 1 2q q q� 2ck � � (21.16) | kuk| ≤ 2|c| |ck|k−1 ≤ 2|c|�f�L2 ⎝ k−2⎠ k=p k=p k=p where Cauchy-Schwartz has been used. This proves that A : L2(0, 2π) −→ C0([0, 2π]) is bounded and by the uniform convergence uk(0) = uk(2π) = 0 for all k implies that Af(0) = Af(2π) = 0. � So, going back to our original problem we try to solve (21.1) by moving the V u term to the right side of the equation (don’t worry about regularity yet) and hope to use the observation that (21.17) u = −A2(V u) + A2f should satisfy the equation and boundary conditions. In fact, let’s hope that u = Av, which has to be true if (21.17) holds with v = −AV u + Af, and look instead for (21.18) v = −AV Av + Af = ⇒ (Id +AV A)v = Af. So, we know that multiplication by V, which is real and continuous, is a bounded self-adjoint operator on L2(0, 2π). Thus AV A is a self-adjoint compact operator so we can apply our spectral theory to Id +AV A. It has a complete orthonormal basis and is invertible if and only if it has trivial null space. An element of the null space would have to satisfy u = −AV Au. On the other hand we know that AV A is positive since (21.19) (AV Aw, w) = (V Av, Av) =