Sturm-Liouville

**morganfor** · May 8th 2011, 04:48 PM

How does one find the eigenvalues and eigenfunctions of

-(x^2y')'=lambda *y

with 1<x<e, and y(1)=y(e)=0

I'm completely lost. Thank you!

**TheEmptySet** · May 8th 2011, 08:14 PM

Originally Posted by morganfor

How does one find the eigenvalues and eigenfunctions of

-(x^2y')'=lambda *y

with 1<x<e, and y(1)=y(e)=0

I'm completely lost. Thank you!

This equation is of type Cauchy-Euler

Using the anzant

$y=x^r$ gives the characteristic equation

$\frac{d}{dx}(x^2\cdot\frac{d}{dx}x^r) = \lambda x^r$

$x^r(r^2+r+\lambda)=0$

From the boundary conditions we can see we need periodic solutions so the roots of the characteristic equation must be complex. Solving the above for r gives

$r^2+r+\lambda=0 \iff \left(r +\frac{1}{2}\right)^2=\frac{1-4\lambda}{4} \iff r=\frac{-1 \pm \sqrt{1-4\lambda}}{2}$

This implies that lambda must be positive

This gives solutions

$y=x^{-\frac{1}{2}}\left( c_1\cos\left( \frac{\sqrt{4\lambda-1}}{2} ln(x) \right)+c_2\sin\left(\frac{\sqrt{4\lambda-1}}{2} \ln(x)\right)\right)$

Using the boundary conditions to solve for lambda gives

$y(1)=c_1=0$

This gives

$y=c_2x^{-\frac{1}{2}}\sin\left(\frac{\sqrt{4\lambda-1}}{2} \ln(x)\right)$

Now with

$y(e)=0=\frac{c_2}{e}\sin\left(\frac{\sqrt{4\lambda-1}}{2}\right)$

This gives

$\pi k =\frac{\sqrt{4\lambda-1}}{2} , k \in \mathbb{Z} \iff \lambda = \frac{4\pi^2 k^2+1}{4}, k=1,2,3...$

**morganfor** · May 8th 2011, 09:12 PM

thanks!!

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