An Application: Sturm-Liouville Systems

Consider the Sturm-Liouville operator $Lh = -h''$ with $a=0$, $b=1$, and for each of the following boundary conditions find the eigenvalues $\left\{ \lambda_{n}\right\}$, the eigenvectors $\left\{ e_{n}\right\}$, and the Green function $g(x,y)$:

In Theorem 6.12 show that $\sum_{n=1}^{\infty } \lambda_{n}^{-2} < \infty$ (see Exercise 5.3).

Theorem 6.12 (context): Assume (6.4): if $h \in \mathcal{D}$ and $Lh=0$, then $h=0$. Then there is a sequence $\left\{ \lambda_{1}, \lambda_{2}, \ldots \right\}$ of real numbers and a basis $\left\{ e_{1}, e_{2}, \ldots \right\}$ for $L^{2}[a,b]$ such that

1. $0 < \left|\lambda_{1}\right| < \left|\lambda_{2}\right| < \cdots$ and $\left|\lambda_{n}\right| \rightarrow \infty$.

2. $e_{n} \in \mathcal{D}$ and $Le_{n} = \lambda_{n} e_{n}$ for all $n$.

3. If $\lambda \neq \lambda_{n}$ for any $\lambda_{n}$ and $f \in L^{2}[a,b]$, then there is a unique $h$ in $\mathcal{D}$ with $Lh - \lambda h = f$.

4. If $\lambda = \lambda_{n}$ for some $n$ and $f \in L^{2}[a,b]$, then there is an $h$ in $\mathcal{D}$ with $Lh - \lambda h = f$ if and only if $\left\langle f, e_{n} \right\rangle = 0$. If $\left\langle f, e_{n} \right\rangle = 0$, any two solutions of $Lh - \lambda h = f$ differ by a multiple of $e_{n}$.

In Theorem 6.12 show that $h \in \mathcal{D}$ if and only if $h \in L^{2}[a,b]$ and $\sum_{n=1}^{\infty } \lambda_{n}^{2} \left| \left\langle h, e_{n} \right\rangle \right|^{2} < \infty$. If $h \in \mathcal{D}$, show that $h(x) = \sum_{n=1}^{\infty } \left\langle h, e_{n} \right\rangle e_{n}(x)$, where this series converges uniformly and absolutely on $[a,b]$.

Theorem 6.12 (context): Assume (6.4): if $h \in \mathcal{D}$ and $Lh=0$, then $h=0$. Then there is a sequence $\left\{ \lambda_{1}, \lambda_{2}, \ldots \right\}$ of real numbers and a basis $\left\{ e_{1}, e_{2}, \ldots \right\}$ for $L^{2}[a,b]$ such that

1. $0 < \left|\lambda_{1}\right| < \left|\lambda_{2}\right| < \cdots$ and $\left|\lambda_{n}\right| \rightarrow \infty$.

2. $e_{n} \in \mathcal{D}$ and $Le_{n} = \lambda_{n} e_{n}$ for all $n$.

3. If $\lambda \neq \lambda_{n}$ for any $\lambda_{n}$ and $f \in L^{2}[a,b]$, then there is a unique $h$ in $\mathcal{D}$ with $Lh - \lambda h = f$.

4. If $\lambda = \lambda_{n}$ for some $n$ and $f \in L^{2}[a,b]$, then there is an $h$ in $\mathcal{D}$ with $Lh - \lambda h = f$ if and only if $\left\langle f, e_{n} \right\rangle = 0$. If $\left\langle f, e_{n} \right\rangle = 0$, any two solutions of $Lh - \lambda h = f$ differ by a multiple of $e_{n}$.

In Theorem 6.12(c), show that $h(x) = \sum_{n=1}^{\infty } (\lambda_{n} - \lambda )^{-1} \left\langle f, e_{n} \right\rangle e_{n}(x)$ and this series converges uniformly and absolutely on $[a,b]$.

Theorem 6.12(c) (context): Assume (6.4). If $\lambda \neq \lambda_{n}$ for any $\lambda_{n}$ and $f \in L^{2}[a,b]$, then there is a unique $h$ in $\mathcal{D}$ with $Lh - \lambda h = f$, where $\left\{ \lambda_{n}\right\}$ and $\left\{ e_{n}\right\}$ are as in Theorem 6.12.

In Theorem 6.12(d), show that if $f \perp e_{n}$ and $Lh - \lambda_{n} h = f$, then $h(x) = \sum_{j \neq n} (\lambda_{j} - \lambda_{n})^{-1} \left\langle f, e_{j} \right\rangle e_{j}(x) + \alpha e_{n}(x)$ for some $\alpha$, where the series converges uniformly and absolutely on $[a,b]$.

Theorem 6.12(d) (context): Assume (6.4). If $\lambda = \lambda_{n}$ for some $n$ and $f \in L^{2}[a,b]$, then there is an $h$ in $\mathcal{D}$ with $Lh - \lambda h = f$ if and only if $\left\langle f, e_{n} \right\rangle = 0$. If $\left\langle f, e_{n} \right\rangle = 0$, any two solutions of $Lh - \lambda h = f$ differ by a multiple of $e_{n}$, where $\left\{ \lambda_{n}\right\}$ and $\left\{ e_{n}\right\}$ are as in Theorem 6.12.

This exercise demonstrates how to handle the case in which $\ker L \neq (0)$.