📂Lebesgue Spaces

Orthogonality of Solutions to the Regular Sturm-Liouville Problem

Theorem¹

Assume that $\lambda_{n}, \lambda_{m}$ are distinct eigenvalues of the regular S-L problem, and $u_{n}, u_{m}$ are the eigenfunctions corresponding to each eigenvalue with real values. Then, $u_{n}, u_{m}$ are orthogonal to each other in the $L_{w}^{2}(a,b)$ space. That is,

$$\int _{a} ^{b} u_{n}(x)u_{m}(x)w(x)dx=0$$

Explanation

Regular Sturm-Liouville Problem

The differential equation $(1)$ is defined on the interval $[a,b]$ and is called a regular Sturm-Liouville problem when it satisfies the following two conditions:

$(\text{i})$ For all $x \in [a,b]$, $p(x)>0$, $w(x)>0$

$(\text{ii})$ $(c_{1},c_{2})\ne (0,0)$ and for constants such that $(d_{1},d_{2})\ne (0,0)$, the following boundary conditions are fulfilled:

$$\begin{cases} c_{1}u(a) + c_{2}u^{\prime}(a) =0 \\ d_{1}u(b) + d_{2}u^{\prime}(b) =0 \end{cases}$$

Proof

Since $u_{n}, u_{m}$ are solutions of the S-L problem, the following equation holds:

$$\begin{align} \left[ p(x)u_{n}^{\prime}(x) \right]^{\prime}+\left[ q(x) +\lambda_{n} w(x) \right]u_{n}(x) =&\ 0 \\ \left[ p(x)u_{m}^{\prime}(x) \right]^{\prime}+\left[ q(x) +\lambda_{m} w(x) \right]u_{m}(x) =&\ 0 \end{align}$$

Calculating $(1)\times u_{m}-(2)\times u_{n}$ yields:

$$\begin{align*} && & \left[ p(x)u^{\prime}_{n}(x) \right]^{\prime}u_{m}(x)+\left[ q(x)+\lambda_{n} w(x) \right]u_{n}(x)u_{m}(x) \\ && &- \left[ p(x)u^{\prime}_{m}(x) \right]^{\prime}u_{n}(x)-\left[ q(x)+\lambda_{m} w(x) \right]u_{m}(x)u_{n}(x) = 0 \end{align*}$$

$$\begin{align*} \implies (\lambda_{n}-\lambda_{m})w(x)u_{n}(x)u_{m}(x)=&\ \left[ p(x)u^{\prime}_{m}(x) \right]^{\prime}u_{n}(x)-\left[ p(x)u^{\prime}_{n}(x) \right]^{\prime}u_{m}(x) \\ =&\ \left[ \left( p(x)u_{m}^{\prime}(x) \right)u_{n}(x)-\left( p(x)u_{n}^{\prime}(x) \right)u_{m}(x) \right] ^{\prime} \end{align*}$$

Now, integrating both sides of the above equation from $a$ to $b$ yields:

$$\begin{equation} \begin{align*} \left( \lambda_{n}-\lambda_{m} \right)\int _{a} ^{b}u_{n}(x)u_{m}(x)w(x)dx =&\ \left[ \left( p(x)u_{m}^{\prime}(x) \right)u_{n}(x)-\left( p(x)u_{n}^{\prime}(x) \right)u_{m}(x) \right] _{a}^{b} \\ =&\ p(b)\left[ u_{m}^{\prime}(b) u_{n}(b)- u_{n}^{\prime}(b) u_{m}(b) \right] \\ & -p(a)\left[ u_{m}^{\prime}(a)u_{n}(a)-u_{n}^{\prime}(a) u_{m}(a) \right] \end{align*} \end{equation}$$

Due to the boundary conditions of the regular S-L problem, we obtain the following equation for $(d_{1},d_{2})\ne (0,0)$:

$$\begin{align} d_{1}u_{n}(b)+d_{2}u_{n}^{\prime}(b) =&\ 0 \\ d_{1}u_{m}(b)+d_{2}u_{m}^{\prime}(b) =&\ 0 \end{align}$$

Without loss of generality, let’s assume $d_{1} \ne 0$. Calculating $(4)\times u_{m}^{\prime}(b)-(5)\times u_{n}^{\prime}(b)$ yields:

$$\begin{align*} &&\left( d_{1}u_{n}(b)+d_{2}u_{n}^{\prime}(b) \right)u_{m}^{\prime}(b)-\left( d_{1}u_{m}(b)+d_{2}u_{m}^{\prime}(b) \right)u_{n}^{\prime}(b) =&\ 0 \\ \implies && d_{1}\left( u_{n}(b)u_{m}^{\prime}(b)-u_{m}(b)u_{n}^{\prime}(b) \right) =&\ 0 \end{align*}$$

However, since we assumed $d_{1} \ne 0$, it follows that $\left( u_{n}(b)u_{m}^{\prime}(b)-u_{m}(b)u_{n}^{\prime}(b) \right)=0$. Therefore, we can see that the first term of the last line of $(3)$ becomes $0$, and by the same logic, the second term of the last line of $(3)$ also becomes $0$. Hence, we obtain the following equation:

$$\left( \lambda_{n}-\lambda_{m} \right)\int _{a} ^{b}u_{n}(x)u_{m}(x)w(x)dx=0$$

Since $\lambda_{n} \ne \lambda_{m}$,

$$\int _{a} ^{b}u_{n}(x)u_{m}(x)w(x)dx=0$$

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Corollary

Consider the Sturm-Liouville differential equation on a finite closed interval $[a,b]$.

$$\left[ p(x)u^{\prime}(x) \right]^{\prime}+\left[ q(x) +\lambda w(x) \right]u(x)=0$$

Let’s say for all $x\in(a,b)$, $p(x)>0$ and $w(x)>0$. Then,

$(\text{i})$ when $p(a)=p(b)=0$, equation $(0)$ holds.

$(\text{ii})$ if $p(a)=p(b)$, and $u(a)=u(b)$, $u^{\prime}(a)=u^{\prime}(b)$, then equation $(0)$ holds.

Proof

If $(\text{i})$ or $(\text{ii})$, since the last line of $(3)$ all becomes $0$, $(0)$ holds.

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