The eigenfunctions of the Sturm Liouville BVP

$$\int_0^a \sin \left( \frac{n \pi x}{a} \right) \sin \left( \frac{m \pi x}{a} \right) \ dx = 0$$ when $m \not= n$.

satisfy the integral relationship

$$\int^b_a r(x) \phi_n(x) \phi_m(x) \ dx = 0$$

If $m \not= n$

where $\phi_1, \phi_2, \phi_3, \dots$ are eigenfunctions, and $\phi_n$ corresponds to the eigenvalues $\lambda_n$.

Hence the set of functions for the Sturm-Liouville problem orthogonal on the interval of interest, with regards to the weight function $r(x)$.


Since $\phi_n$ and $\phi_m$ are eigenfunctions, they must satisfy the ODE

$$\frac{d}{dx} \left( p(x) \frac{d \phi_n}{dx} \right) + q(x) \phi_n = - \lambda r(x) \phi_n \ \ \ (1)$$

$$\frac{d}{dx} \left( p(x) \frac{d \phi_m}{dx} \right) + q(x) \phi_m = - \lambda r(x) \phi_m \ \ \ (2)$$

Multiply (1) by $\phi_m$ and (2) by $\phi_n$ and subtract:

$$\phi_m \frac{d}{dx} \left( p(x) \frac{d \phi_n}{dx} \right) - \phi_n \frac{d}{dx} (p(x) \frac{d \phi_m}{dx} = (\lambda_m - \lambda_n)r(x) \phi_n \phi_m$$

The left-hand side can be expressed as

$$\frac{d}{dx} \left( p(x) \left( \phi_m \frac{d \phi_n}{dx} - \phi_n \frac{d \phi_m}{dx} \right) \right)$$

Therefore, the equation becomes

$$\frac{d}{dx} \left( p(x) \left( \phi_m \frac{d \phi_n}{dx} - \phi_n \frac{d \phi_m}{dx} \right) \right) = (\lambda_m - \lambda_n)r(x) \phi_n \phi_m$$

Integrate both sides with respect to $x$ from $a$ to $b$:

$$\left[ p(x) \left( \phi_m \frac{d \phi_n}{dx} - \phi_n \frac{d \phi_m}{dx} \right) \right]^b_a = (\lambda_m - \lambda_n) \int^b_a r(x) \phi_n \phi_m \ dx$$

What I'm trying to figure out is how integrating both sides with respect to $x$ from $a$ to $b$ gets us

$\left[ p(x) \left( \phi_m \frac{d \phi_n}{dx} - \phi_n \frac{d \phi_m}{dx} \right) \right]^b_a = (\lambda_m - \lambda_n) \int^b_a r(x) \phi_n \phi_m \ dx$

As I understand it, this is an application of the fundamental theorem of calculus part 2 (called part 2 in my textbook):

If $f$ is continuous over $[a, b]$ and $F$ is any antiderivative of $f$ on $[a, b]$, then

$$\int_a^b f(x) \ dx = F(b) - F(a)$$

Can someone please demonstrate how this was done?


Note that: $$\int_{a}^{b}\frac{df}{dx}dx=f(b)-f(a)$$

It is the fundamental theorem of calculus. The $\lambda$'s are constant and therefore come out of the integral.


Your functions $f_n(x)=\sin(n\pi x/a)$ are solutions of $$ -\frac{d^2}{dx^2}f=\lambda f,\;\;\; 0 \le x \le a, \\ f(0)=f(a)=0. $$ So you have a weight function $r(x)\equiv 1$, coefficient $p\equiv 1$, and potential $q\equiv 0$. The eigenvalues are $\lambda_n=n^2\pi^2/a^2$. Therefore, the general theory gives $$ \int_{0}^{a}f_n(x)f_m(x)dx = 0,\;\; n \ne m,\;\; n,m=1,2,3,\cdots. $$


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