Here is the problem:

$$ x^2y''-xy+\lambda y = 0,\quad y(1)=0,\quad y(L)=0,\quad L>0 $$

I am asked to find the Eigenvalues and Eigenfunction.

I can't figure out how to get a general equation for y. I tried integrating factors but that was a mess, I'm not sure if there's a better way to do this. It might be a Sturm Liouville equation, but I'm not sure to to solve those. Any help would be appreciated.

Here is the solution given:

$ \lambda_n=1+(n\pi/ln(L))^2,\quad y_n(x)=xsin(n\pi ln(x)/ln(L));\quad n=1,2,3... $

  • 2
    $\begingroup$ This is a second order Cauchy-Euler equation so it has two solutions of the form $y=x^m$ where $m$ satisfies the auxiliary equation $m(m-1)-m+\lambda=0$. In the event that the discriminant is negative $m$ will be complex with $m=a\pm bi$ for some $a,b$. In that case the general solution will be of the form $y=e^{ax}(c_1\sin(b\ln(x))+c_2\cos(b\ln(x)))$. You will have to use your boundary conditions to get the exact answer. $\endgroup$ – John Wayland Bales Mar 29 '16 at 1:50
  • 1
    $\begingroup$ Oops, my bad. It's not Cauchy-Euler. There is no $y^\prime$ term. $\endgroup$ – John Wayland Bales Mar 29 '16 at 2:01

There is likely a typo in the equation. The solution you give solves $$ x^2y''-xy'+\lambda_ny=0. $$ As John mentioned, this is an Euler equation with characteristic equation $m(m-1)-m+\lambda =0$. For $\lambda\leq1$, $m$ is real and there is no solution with $y(L)=0$. For $\lambda>1$ we get solutions $$ y(x)=c_1 x \cos(\sqrt{\lambda-1}\,\log x) + c_2 x \sin(\sqrt{\lambda-1}\,\log x). $$ The condition $y(0)=0$ forces $c_1=0$. The condition $y(L)=0$ forces $$ \sin(\sqrt{\lambda-1}\,\log L)=0, $$so $$\sqrt{\lambda-1}\log L=n\pi,\ \ \ n\in\mathbb N$$ This gives us eigenvalues $$ \lambda_n=1+\left(\frac{n\pi}{\log L}\right)^2 $$ and eigenvectors $$ y_n(x)=x\,\sin\left(\frac{n\pi}{\log L}\,\log x\right). $$

  • $\begingroup$ You da bomb dot com. May your limits never be undefined. $\endgroup$ – Spuds Mar 29 '16 at 18:37
  • $\begingroup$ Sorry I have one quick question about this - why are you considering $\lambda >1$? When looking for eigenvalues, I thought you just look at $\lambda <0,\quad \lambda = 0,\quad \lambda >0$. $\endgroup$ – Spuds Mar 29 '16 at 20:00
  • $\begingroup$ You look at what you have. Here the characteristic equation is $m^2-2m+\lambda$. This has real solution for $\lambda\geq1$ (which you can check cannot satisfy $y(0)=y(L)=0$), and non-real solutions for $\lambda>1$. $\endgroup$ – Martin Argerami Mar 29 '16 at 22:54
  • $\begingroup$ @Spuds: since you are new to the site, please consider upvoting the answer if you found it useful (not just mine, but any answer that you find useful; that's how this site works). $\endgroup$ – Martin Argerami Mar 29 '16 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.