I'm having problems doing this BVP eigenvalue problem, and would appreicate if someone can help me through the next step
(Question was:)
By setting $y=u/x$, find all eigenvalues and eigenfunctions of the Sturm-Liouville problem:
$$xy''+2y'+\lambda xy=0$$
for $x \in [\pi, 2\pi]$ with BV's of $y(\pi)=y(2\pi)=0$
So I managed the first step, doing $y=u/x$ I changed the ODE into:
$$u''+\lambda u=0$$
So then i'm supposed to find the lambda eigenvalues, but i'm having issues.
The only examples i've ever seen involve $y''+\lambda y=0$ with BV's $y(0)=y(L)=0$ never with $y=u/x$ and $y(\pi)=y(2\pi)=0$
I understand that I need to try $\lambda > 0$, with general solution:
$$u= a\cos(\sqrt{\lambda}x)+b\sin(\sqrt{\lambda}x)$$
(therefore:)
$$y= a\cos(\sqrt{\lambda}x)+b\sin(\sqrt{\lambda}x)/x$$
But I don't know where to go from there, do I just sub the BV's $\pi$ and $2\pi$ in? I'm really confused. Any help would be greatly appriecated thanks.
Best Answer
Yes, the solutions of the original equation are of the form $$ y(x) = \frac{1}{x}\left[a\cos(\sqrt{\lambda}x)+b\sin(\sqrt{\lambda}x)\right], $$ where $a$ and $b$ must be chosen so that $y(\pi)=y(2\pi)=0$. The $1/x$ term doesn't affect the equations $y(\pi)=0$ and $y(2\pi)=0$. So you have a matrix system $$ \left[\begin{array}{cc}\cos(\sqrt{\lambda}\pi) & \sin(\sqrt{\lambda}\pi) \\ \cos(2\sqrt{\lambda}\pi) & \sin(2\sqrt{\lambda}\pi)\end{array}\right] \left[\begin{array}{c} a \\ b\end{array}\right] =\left[\begin{array}{c} 0\\ 0\end{array}\right]. $$ There is a non-trivial solution (i.e., an eigenfunction) iff the determinant of the 2x2 coefficient matrix is 0: $$ \sin(2\sqrt{\lambda}\pi-\sqrt{\lambda}\pi)=\sin(2\sqrt{\lambda}\pi)\cos(\sqrt{\lambda}\pi)-\cos(2\sqrt{\lambda}\pi)\sin(\sqrt{\lambda}\pi)=0. $$ So $\lambda=n^{2}$ where $n=0,1,2,3,\ldots$, and any scalar multiple $c$ yields a solution $$ \left[\begin{array}{c}a\\b\end{array}\right] =c\left[\begin{array}{c}0\\1\end{array}\right]. $$ So the eigenfunctions are, up to scalar multiples, $y_{n}(x)=\frac{1}{x}\sin(nx)$ for $n \ge 1$. These eigenfunctions are orthogonal with respect to the inner-product $$ (f,g) = \int_{\pi}^{2\pi}f(x)g(x)x^{2}dx. $$ The normalized eigenfunctions (unit length in above inner-product), are $$ \left\{\sqrt{\frac{2}{\pi}}\frac{1}{x}\sin(nx)\right\}_{n=1}^{\infty}. $$ This is because $$ \left(\frac{1}{x}\sin(nx),\frac{1}{x}\sin(nx)\right)=\int_{\pi}^{2\pi}\sin^{2}(nx)dx = \frac{\pi}{2}. $$ The weight function $x^{2}$ in the inner product comes from putting the equation into selfadjoint form $$ x^{2}y''+2xy + \lambda x^{2}y = 0, $$ $$ (x^{2}y')'+\lambda x^{2}y = 0. $$ The function multiplying $y$ is then the weight function, and the operator $$ Ly = \frac{1}{x^{2}}(x^{2}y')' $$ becomes selfadjoint with respect to this weighted inner-product.