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I tried looking on the Mathematica forum for a similar solution but unfortunately haven't found one.

I was given the following Sturm-Liouville differential equation problem:

$x^2y''+5xy'+\lambda y=0$

$y(1)=y(e^{\pi}) = 0$

I displayed the equation in the traditional Sturm-Liouville manner and found that the weight function equals $x^3$, but I am having trouble finding the eigenvalues for this problem.

We were told in class that because the coefficients of the ODE are functions of $x$ and not constants, we should guess a solution for $y$ in the form of $x^m$ and continue from there finding the eigenvalues.

Doing that I got a simple quadratic equation involving $m$:

$m^2 -4m +\lambda = 0$

But because it is a quadratic equation where p and q are both non zero I can't seem to find a way to divide the eigenvalues into the common categories which we usually did in class (for $\lambda <0, >0, =0$), like in this case for instance:

$m^2 + \lambda = 0$

I believe the solution is very simple but just can't find it, help would be much appreciated!

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If you substitute $x^m$ into $x^2y''+5xy'+\lambda y=0$, you get $$ m(m-1)+5m+\lambda = 0 \\ m^2+4m+\lambda = 0 \\ m =\frac{-4\pm\sqrt{16-4\lambda}}{2}=-2\pm\sqrt{4-\lambda} $$ The corresponding solutions are $$ \frac{1}{x^2}x^{\sqrt{4-\lambda}}, \frac{1}{x^2}x^{-\sqrt{4-\lambda}}. $$ To put $x^2y''+5xy'+\lambda y=0$ into Sturm-Liouville form, multiply by $x^3$: $$ -(x^5y')'=\lambda x^3y \\ -\frac{1}{x^3}(x^5y')'=\lambda y $$ This problem is considered in $L^2_{x^3}(1,e^{\pi})$, where $x^3$ is the weight function for the space. The solutions of the above when $\lambda=0$ are both in $L^2_{x^3}(1,e^{\pi})$ because they are regular on $(1,e^{\pi})$. To solve the eigenvalue problem, search for $A,B$ such that $$ y(x)=A\frac{1}{x^2}x^{\sqrt{4-\lambda}}+B\frac{1}{x^2}x^{-\sqrt{4-\lambda}} $$ satisfies $$ y(1)=0,\;\; y(e^{\pi})=0. $$ The first condition is satisfied by setting $B=-A$, which gives \begin{align} y(x)&=\frac{A}{x^2}\left[x^{\sqrt{4-\lambda}}-x^{-\sqrt{4-\lambda}}\right] \\ &=\frac{A}{x^2}\left[e^{i\sqrt{\lambda-4}\ln x}-e^{-i\sqrt{\lambda-4}\ln x}\right] \\ &=\frac{A}{x^2}2i\sin(\sqrt{\lambda-4}\ln x) \end{align} The condition $y(e^{\pi})=0$ gives the eigenvalue equation $$ \sin(\sqrt{\lambda-4}\pi)=0. $$

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    $\begingroup$ Thank you very much!!! $\endgroup$
    – RandyMarsh
    Nov 19, 2021 at 5:03

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