# Lecture 9: Regular Sturm-Liouville Theory $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\tG}{\mathcal{G}}$ ###### tags: 224a In the last lecture we saw that Fourier sine series may be derived by looking at solutions to a particular second order BVP. In this lecture, we generalize this to show that a large class of second order ODE come equipped with an analogous expansion. Consider the nonhomogeneous BVP $Lu=f$ where $$L = -\frac{d^2}{dx^2} + q(x)$$ for $q\in C[a,b]$ and $f\in C[a,b]$, both real-valued, with boundary conditions $$B_1u = \alpha_0 u(a) + \alpha_1 u'(a)=0, \quad B_2u = \beta_0 u(b) + \beta_1 u'(b)=0,$$ with real coefficients $\alpha_0^2+\alpha_1^2>0, \beta_0^2+\beta_1^2>0$. (these are known as *separated* boundary conditions, since they act on different points). The solutions of such a problem live in the vector space $C^2_B[a,b]:=\{u\in C^2: B_1u=B_2u=0\}.$, and we view $L$ as a linear transformation $L:C_B^2[a,b]\rightarrow C[a,b]$. ## Formal Self-Adjointness The corresponding eigenvalue problem is: $$Lu = \lambda u\quad u\in C^2_B[a,b].$$ The main question we want to address is: do the eigenfunctions form an ONB of $L^2[a,b]$? We begin by observing that $L$ is *formally selfadjoint* with respect to the standard inner product $$(u,v):=\int_a^b \overline{u(x)}v(x)dx.$$ **Lemma 1.(Lagrange Identity)** If $u,v\in C^2_B[a,b]$ then $(u,Lv)-(Lu,v) = (uv'-u'v)|_a^b = 0.$ *Proof.* Just integrate, apply $Lu=Lv=0$, and cancel terms using the BC.$\square$ This already has some nice consequences. **Lemma 2.** Every eigenvalue of $L$ is real. Eigenvectors corresponding to distinct eigenvalues are orthogonal. The multiplicity of each eigenspace is exactly one. *Proof.* Observe that if $Lu=\lambda u$ and $Lv = \mu v$ then $\mu(u,v) = (u,Lv) = (Lu,v) = \bar{\lambda}(u,v)$. For the statement about multiplicity, recall that by the IVP existence and uniqueness theorem, the solution space of the homogeneous equation $(L-\lambda)u=0$ for real $\lambda$ has dimension two and is isomorphic to $\R^2$ via the evaluation map. Thus, applying one boundary condition reduces this dimension by exactly one. $\square$ However, this does not imply anything about *existence* of eigenvectors, let alone an orthonormal basis of them, and we can't apply Hilbert space theory since $C_2[a,b]$ isn't a Hilbert space and $L$ is not bounded on $C_B^2[a,b]$ (consider e.g. the case $q(x)=0$ and the eigenvectors $u_n=\sin(n\pi x)$ on $[0,\pi]$). ## Green's Function We will show that enough eigenvectors exist by considering the *inverse* of $L$, to which we will be able to apply spectral theory. A prerequisite for this is that $L$ is 1-1, so we make the following **Assumption.** $\ker(L)=0$, which we will remove at the end of the lecture. Let $W(u,v):=uv'-u'v \in C^1[a,b]$ denote the *Wronskian* of $u,v$. **Lemma 3.** There are real nonzero functions $u_a\in C^2_{B_1}[a,b]$, $u_b\in C^2_{B_2}[a,b]$ which are linearly independent and satisfy $Lu_a=Lu_b=0$. The Wronskian of these solutions is nonzero and constant on $[a,b]$. *Proof.* Again, by the IVP theorem the unconstrainted equation $Lu=0$ has a two dimensional real subspace of $C^2[a,b]$ of solutions; adding a single boundary condition leaves a one-dimensional space, which must contain a nonzero vector. This shows existence of $u_a$ and $u_b$. For the Wronskian, we compute $$W'(x) = (u_au_b'-u_a'u_b)' = u_au_b'' - u_a''u_b = qu_au_b-qu_au_b=0,$$ so $W$ must be constant on $[a,b]$. It cannot be zero since otherwise the vectors $[u_a(a), u_a'(a)]$ and $[u_b(a),u_b'(a)]$ are linearly dependent, whence $B_2u_a=0$ which would imply $u_a\in C_B^2[a,b]$, ruled out by the assumption. This further implies that $u_a$ and $u_b$ are linearly independent as vectors in $C^2[a,b]$.$\square$ Define the bivariate function $$K(x,\xi) = -W^{-1} (u_a(x)u_b(\xi)\{x\le \xi\} + u_a(\xi)u_b(x)\{x\ge \xi\}),$$ as well as the integral transformation $$(Gf)(x) := \int_a^b K(x,\xi) f(\xi)d\xi.$$ Observe that $K$ is continuous on $C[a,b]^2$ and that $K(x,\xi)=K(\xi,x)$. **Lemma 4 (Green's Function).** $$G:C[a,b]\rightarrow C_B^2[a,b].$$ Moreover $LGf=f$ for every $f\in C[a,b]$ and $GLu = u$ for every $u\in C_B^2[a,b]$, i.e., $G$ is the inverse of $L$. *Proof.* Fix $f\in C[a,b]$ and let $u=Gf$. Splitting the integral, we have $$u(x) = -W^{-1}(\int_x^b u_a(x)u_b(\xi)f(\xi)d\xi + \int_a^x u_b(x)u_a(\xi)f(\xi)d\xi )=: u_a(x) F_b(x) +u_b(x) F_a(x),$$ by pulling $u_a(x)$ and $u_b(x)$ outside the integrals in $\xi$. Differentiating once, we have \begin{align*} u' & = u_a'F_b + u_aF_b' + u_b'F_a + u_bF_a' \\ &= u_a'F_b + u_a(W^{-1}u_bf) + u_b'F_a + u_b'(-W^{-1}u_af)\\ &= u_a'F_b + u_b'F_a, \end{align*} where we used the fundamental theorem of calculus and the crucial cancellation of the terms containing $f$ in the last step is because the first integral is oriented negatively. This is a differentiable function since $F_a,F_b$ are differentiable by the FTC; differentiating again and using $Lu_a=Lu_b=0$ we have \begin{align*} u'' & = u_a''F_b + u_a'F_b' + u_b''F_a + u_b'F_a' \\ &= (qu_a)F_b + u_a'(W^{-1}u_bf) + (qu_b)F_a + u_b'(-W^{-1}u_af)\\ &= qu + W^{-1}f(u_a'u_b-u_au_b')\\ &= qu-f, \end{align*} whence $Lu=LGf=f$. Let us now check that $u\in C_B^2[a,b]$. The left boundary condition is $$B_1u=\alpha_0 (u_aF_b+u_bF_a)(a) + \alpha_1(u_a'F_b+u_b'F_a)(a) = F_b(B_1u_a)+F_a(B_1u_b).$$ The first term is zero since $B_1u_a=0$ and the second is zero since $F_a(a)=0$ by examining the limits of the integral defining it. Since both $u$ and $GLu$ are solutions of the ODE $Lu = (Lu)$ and $\ker(L)=\{0\}$ implies there is only one solution in $C_B^2[a,b]$, we must have $GLu=u$ as well for every $u\in C_B^2[a,b]$.$\square$ Lemma 4 implies that if $\phi_n\in C[a,b]$ satisfies $$G\phi_n = \lambda_n \phi_n$$ then it must also be an eigenvector of $L$ with eigenvalue $1/\lambda_n$ since $$\phi_n = LG\phi_n = \lambda_n L\phi_n.$$ Note that such a $\phi_n$ is automatically in $C_B^2[a,b]$ since it is in the range of $G$. ## Extension to selfadjoint $\tG$ Let $\tG:L^2[a,b]\rightarrow L^2[a,b]$ be defined by the same integral formula as $G$; since $\tG$ agrees with $G$ on $C[a,b]$ it is an *extension* of $G$. $\newcommand{\ran}{\mathrm{ran}}$ **Lemma 5.** $\tG$ is compact and selfadjoint. Moreover, $\ran(\tG)\subset C[a,b]$ and $\ker(\tG)=\{0\}$. *Proof.* The first two properties follow because $K\in L^2[a,b]^2$ and $K$ is symmetric, and we proved them for integral kernel operators with these properties. Fix $f\in L^2[a,b]$, $x\in [a,b]$ and let $\epsilon>0$. Since $K$ is continuous on a compact set it is uniformly continuous; choose $\delta$ so that $|K(y,\xi)-K(x,\xi)|<\epsilon$ for all $\xi\in [a,b]$ whenever $|x-y|<\delta$. We now have $$|(\tG f)(x)-(\tG f)(y)|\le \int_a^b \epsilon |f(x)|dx\le \epsilon|b-a|\|f\|,$$ establishing contiunity. Assume $\tG f = 0$. Then, for every continuous $\phi$, $$0=(\phi, \tG f) = (\tG \phi, f) = (G \phi f),$$ so $f$ is orthogonal to every vector in $\ran(G)=C_B^2[a,b]$. Since this set is dense in $L^2[a,b]$ (homework), we conclude that $f=0$.$\square$ Finally, we invoke the spectral theorem for compact operators. **Theorem.** $L$ has countably many eigenvalues, all real of multiplicity one. The corresponding eigenvectors are an ONB of $L^2[a,b]$. *Proof.* By Lemma 5 $\tG$ has an ONB of eigenvectors $\tG\phi_n=\lambda_n \phi_n$, with $\lambda_n\neq 0$. Since $\ran(\tG)\subset C[a,b]$, we must have $\phi_n\in C[a,b]$ whence $$G\phi_n = \lambda_n \phi_n.$$ By the argument in the previous section, $1/\lambda_n$ are the desired eigenvalues of $L$, as desired.$\square$ See the class notes for a simple argument showing how to remove the **Assumption** by the estimate $(\phi, L\phi)> C\|\phi\|^2$ on the quadratic form and replacing $L$ by $L-C$. A more detailed argument is presented in https://sites.math.washington.edu/~hart/m556/notes3.pdf.