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# 10.6: Sturm-Liouville theory

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In the end we shall want to write a solution to an equation as a series of Bessel functions. In order to do that we shall need to understand about orthogonality of Bessel function – just as sines and cosines were orthogonal. This is most easily done by developing a mathematical tool called Sturm-Liouville theory. It starts from an equation in the so-called self-adjoint form

$[r(x) y'(x)]' + [p(x)+\lambda s(x)] y(x) = 0 \label{eq:selfadj}$

where $$\lambda$$ is a number, and $$r(x)$$ and $$s(x)$$ are greater than 0 on $$[a,b]$$. We apply the boundary conditions

\begin{aligned} a_1 y(a)+ a_2 y'(a)&=0,\nonumber\\ b_1 y(b)+ b_2 y'(b)&=0,\end{aligned}

with $$a_1$$ and $$a_2$$ not both zero, and $$b_1$$ and $$b_2$$ similar.

Theorem $$\PageIndex{1}$$

If there is a solution to (\ref{eq:selfadj}) then $$\lambda$$ is real.

Proof

Assume that $$\lambda$$ is a complex number ($$\lambda = \alpha + i \beta$$) with solution $$\Phi$$. By complex conjugation we find that

\begin{aligned} [r(x) \Phi'(x)]' + [p(x)+\lambda s(x)] \Phi(x) &= 0\nonumber\\ {}[r(x) (\Phi^*)'(x)]' + [p(x)+\lambda^* s(x)] (\Phi^*)(x) &= 0\end{aligned}

where $$*$$ note complex conjugation. Multiply the first equation by $$\Phi^*(x)$$ and the second by $$\Phi(x)$$, and subtract the two equations: $(\lambda^*-\lambda)s(x) \Phi^*(x)\Phi(x)=\Phi(x)[r(x) (\Phi^*)'(x)]'-\Phi^*(x)[r(x) \Phi'(x)]'.$ Now integrate over $$x$$ from $$a$$ to $$b$$ and find

$(\lambda^*-\lambda)\int_a^bs(x) \Phi^*(x)\Phi(x)\,dx = \int_a^b\Phi(x)[r(x) (\Phi^*)'(x)]'-\Phi^*(x)[r(x) \Phi'(x)]'\,dx$

The second part can be integrated by parts, and we find

\begin{aligned} (\lambda^*-\lambda)\int_a^b s(x) \Phi^*(x)\Phi(x)\,dx &= \left[\Phi'(x) r(x) (\Phi^*)'(x)-\Phi^*(x)r(x) \Phi'(x)\right|^b_a \nonumber\\ &= r(b)\left[\Phi'(b) (\Phi^*)'(b)-\Phi^*(b)\Phi'(b)\right] \nonumber\\&& -r(a)\left[\Phi'(a) (\Phi^*)'(a)-\Phi^*(a)\Phi'(a)\right] \nonumber\\ &=0,\end{aligned}

where the last step can be done using the boundary conditions. Since both $$\Phi^*(x)\Phi(x)$$ and $$s(x)$$ are greater than zero we conclude that $$\int_a^bs(x) \Phi^*(x)\Phi(x)\,dx > 0$$, which can now be divided out of the equation to lead to $$\lambda=\lambda^*$$.

Theorem $$\PageIndex{2}$$

Let $$\Phi_n$$ and $$\Phi_m$$ be two solutions for different values of $$\lambda$$, $$\lambda_n\neq \lambda_m$$, then $\int_a^b s(x) \Phi_n(x) \Phi_m(x)\,dx = 0.$

Proof

The proof is to a large extent identical to the one above: multiply the equation for $$\Phi_n(x)$$ by $$\Phi_m(x)$$ and vice-versa. Subtract and find $(\lambda_n-\lambda_m)\int_a^b s(x) \Phi_m(x)\Phi_n(x)\,dx = 0$ which leads us to conclude that $\int_a^b s(x) \Phi_n(x) \Phi_m(x)\,dx = 0.$

Theorem $$\PageIndex{3}$$

Under the conditions set out above

1. There exists a real infinite set of eigenvalues $$\lambda_0,\ldots,\lambda_n, \ldots$$ with $$\lim_{n\rightarrow \infty} = \infty$$.
2. If $$\Phi_n$$ is the eigenfunction corresponding to $$\lambda_n$$, it has exactly $$n$$ zeroes in $$[a,b]$$. No proof shall be given.
Proof

No proof shall be given.

Clearly the Bessel equation is of self-adjoint form: rewrite $x^2 y'' + xy' + (x^2-\nu^2) y = 0$ as (divide by $$x$$) $[x y']' + (x-\frac{\nu^2}{x}) y = 0$ We cannot identify $$\nu$$ with $$\lambda$$, and we do not have positive weight functions. It can be proven from properties of the equation that the Bessel functions have an infinite number of zeroes on the interval $$[0,\infty)$$. A small list of these:

$\begin{array}{lclllll} J_0 &:& 2.42&5.52 &8.65&11.79&\ldots\\ J_{1/2} &:& \pi & 2 \pi & 3\pi & 4\pi & \ldots\\ J_{8} &:& 11.20 & 16.04 & 19.60 & 22.90 & \ldots \end{array}$