Lecture 9: Regular Sturm-Liouville Theory

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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

$$L=-\frac{d^2}{d{x}^2}+q(x)$$
for

$q\in C[a,b]$ and

$f\in C[a,b]$, both real-valued, with boundary conditions

$$B_{1}u={\alpha }_{0}u(a)+{\alpha }_1{u}^{\prime }(a)=0,B_{2}u={\beta }_{0}u(b)+{\beta }_1{u}^{\prime }(b)=0,$$
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

$C_B^2[a,b]:=\{u\in C^2:B_{1}u=B_{2}u=0\}.$, and we view

$L$ as a linear transformation

$L:C_B^2[a,b]\to C[a,b]$.

Formal Self-Adjointness

The corresponding eigenvalue problem is:

$$Lu=\lambda uu\in C_B^2[a,b].$$
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

$u,v\in C_B^2[a,b]$ then

$(u,Lv)-(Lu,v)=(u{v}^{\prime }-u^{\prime }v){|}_a^b=0.$
Proof. Just integrate, apply

$Lu=Lv=0$, and cancel terms using the BC.

$◻$

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

$\mu (u,v)=(u,Lv)=(Lu,v)=\overline{\lambda }(u,v)$.

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

${\mathbb{R}}^2$ via the evaluation map. Thus, applying one boundary condition reduces this dimension by exactly one.

$◻$

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

$W(u,v):=u{v}^{\prime }-u^{\prime }v\in C^1[a,b]$ denote the Wronskian of

$u,v$.

Lemma 3. There are real nonzero functions

$u_a\in C_{B_1}^2[a,b]$,

$u_b\in C_{B_2}^2[a,b]$ which are linearly independent and satisfy

$L{u}_a=L{u}_b=0$. 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

$$W^{\prime }(x)=(u_a{u}_b^{\prime }-u_a^{\prime }{u}_b{)}^{\prime }=u_a{u}_b^{″}-u_a^{″}{u}_b=q{u}_a{u}_b-q{u}_a{u}_b=0,$$
so

$W$ must be constant on

$[a,b]$. It cannot be zero since otherwise the vectors

$[u_a(a),u_a^{\prime }(a)]$ and

$[u_b(a),u_b^{\prime }(a)]$ are linearly dependent, whence

$B_2{u}_a=0$ 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

$C^2[a,b]$.

$◻$

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).

$$G:C[a,b]\to C_B^2[a,b].$$ 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

$$\begin{array}{rl}{u}^{\prime }& =u_a^{\prime }{F}_b+u_a{F}_b^{\prime }+u_b^{\prime }{F}_a+u_b{F}_a^{\prime }\\ & =u_a^{\prime }{F}_b+u_a(W^{-1}{u}_{b}f)+u_b^{\prime }{F}_a+u_b^{\prime }(-W^{-1}{u}_{a}f)\\ & =u_a^{\prime }{F}_b+u_b^{\prime }{F}_a,\end{array}$$
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

$L{u}_a=L{u}_b=0$ we have

$$\begin{array}{rl}{u}^{″}& =u_a^{″}{F}_b+u_a^{\prime }{F}_b^{\prime }+u_b^{″}{F}_a+u_b^{\prime }{F}_a^{\prime }\\ & =(q{u}_a)F_b+u_a^{\prime }(W^{-1}{u}_{b}f)+(q{u}_b)F_a+u_b^{\prime }(-W^{-1}{u}_{a}f)\\ & =qu+W^{-1}f(u_a^{\prime }{u}_b-u_a{u}_b^{\prime })\\ & =qu-f,\end{array}$$
whence

$Lu=LGf=f$.

Let us now check that

$u\in C_B^2[a,b]$. The left boundary condition is

$$B_{1}u={\alpha }_0(u_a{F}_b+u_b{F}_a)(a)+{\alpha }_1(u_a^{\prime }{F}_b+u_b^{\prime }{F}_a)(a)=F_b(B_1{u}_a)+F_a(B_1{u}_b).$$
The first term is zero since

$B_1{u}_a=0$ 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

$u\in C_B^2[a,b]$.

$◻$

Lemma 4 implies that if

${\varphi }_n\in C[a,b]$ satisfies

$$G{\varphi }_n={\lambda }_n{\varphi }_n$$
then it must also be an eigenvector of

$L$ with eigenvalue

$1/{\lambda }_n$ since

$${\varphi }_n=LG{\varphi }_n={\lambda }_{n}L{\varphi }_n.$$
Note that such a

${\varphi }_n$ is automatically in

$C_B^2[a,b]$ since it is in the range of

$G$.

Extension to selfadjoint

$\mathcal{G}$

Let

$\mathcal{G}:L^2[a,b]\to L^2[a,b]$ be defined by the same integral formula as

$G$; since

$\mathcal{G}$ agrees with

$G$ on

$C[a,b]$ it is an extension of

$G$.

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Lemma 5.

$\mathcal{G}$ is compact and selfadjoint. Moreover,

$\mathrm{ran}(\mathcal{G})\subset C[a,b]$ and

$\ker (\mathcal{G})=\{0\}$.
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

$ϵ>0$. Since

$K$ is continuous on a compact set it is uniformly continuous; choose

$\delta$ so that

$|K(y,\xi )-K(x,\xi )|<ϵ$ for all

$\xi \in [a,b]$ whenever

$|x-y|<\delta$. We now have

$$|(\mathcal{G}f)(x)-(\mathcal{G}f)(y)|\le {\int }_a^bϵ|f(x)|dx\le ϵ|b-a|\Vert f\Vert ,$$
establishing contiunity.

Assume

$\mathcal{G}f=0$. Then, for every continuous

$\varphi$,

$$0=(\varphi ,\mathcal{G}f)=(\mathcal{G}\varphi ,f)=(G\varphi f),$$
so

$f$ is orthogonal to every vector in

$\mathrm{ran}(G)=C_B^2[a,b]$. Since this set is dense in

$L^2[a,b]$ (homework), we conclude that

$f=0$.

$◻$

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

$\mathcal{G}$ has an ONB of eigenvectors

$\mathcal{G}{\varphi }_n={\lambda }_n{\varphi }_n$, with

${\lambda }_n\ne 0$. Since

$\mathrm{ran}(\mathcal{G})\subset C[a,b]$, we must have

${\varphi }_n\in C[a,b]$ whence

$$G{\varphi }_n={\lambda }_n{\varphi }_n.$$
By the argument in the previous section,

$1/{\lambda }_n$ are the desired eigenvalues of

$L$, as desired.

$◻$

See the class notes for a simple argument showing how to remove the Assumption by the estimate

$(\varphi ,L\varphi )>C\Vert \varphi {\Vert }^2$ 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.