Theorem:
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 $r(x)$.
Proof:
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?
Best Answer
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.