Boundary value problem and eigenvalues (Sturm-Liouville)

Question

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.

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Write your solution as $$ A\sin\pars{\root{\lambda}\bracks{x - {3\pi \over 2}}} + B\cos\pars{\root{\lambda}\bracks{x - {3\pi \over 2}}} $$ We get $$\left\lbrace% \begin{array}{rcrcl} -\sin\pars{\root{\lambda}\,{\pi \over 2}}A & + & \cos\pars{\root{\lambda}\,{\pi \over 2}}B & = & 0 \\[1mm] \sin\pars{\root{\lambda}\,{\pi \over 2}}A & + & \cos\pars{\root{\lambda}\,{\pi \over 2}}B & = & 0 \end{array}\right. $$ Eigenvalues are given by $\sin\pars{\root{\lambda}\pi} = 0\quad\imp\quad \root{\lambda_{n}}\pi = n\pi$ where $n \in \mathbb{Z}$: $\lambda_{n} = n^{2}$. \begin{align} &\left\lbrace% \begin{array}{rcl} \mbox{Eigenfunction} & \propto & \sin\pars{2n\bracks{x - {3\pi \over 2}}} \propto \sin\pars{2nx} \\ \mbox{Eigenvalue}: && 4n^{2}\,,\quad n \in \mathbb{Z} \end{array}\right. \\[3mm] &\left\lbrace% \begin{array}{rcl} \mbox{Eigenfunction} & \propto & \cos\pars{\bracks{2n + 1}\bracks{x - {3\pi \over 2}}} \propto \sin\pars{\bracks{2n + 1}x} \\ \mbox{Eigenvalue}: && \pars{2n + 1}^{2}\,,\quad n \in \mathbb{Z} \end{array}\right. \end{align}