Sturm Liouville problem

ux0 · #1 $Old$ November 4th, 2009, 09:51 PM

Show that $\pi^2$ is an eigenvalue of the Sturm- Liouville problem

$X'' + \lambda X=0$

$X(0) - X'(0)=0$

$\pi^2X(1/2)+X'(1/2)=0$

and find a corresponding eigenfunction.

To start i posed that the solution of this problem will be in the form

$X(x)= A \sin{x\sqrt{\lambda}}+B\cos{x\sqrt{\lambda}}$

$X'(x)=A \sqrt{\lambda} \cos{x\sqrt{\lambda}}-B \sqrt{\lambda}\sin{x \sqrt{\lambda}}$

using the end point conditions i get

$X(0)-X'(0)=A \sin{0\sqrt{\lambda}}+B\cos{0\sqrt{\lambda}}- A \sqrt{\lambda} \cos{0\sqrt{\lambda}}+B \sqrt{\lambda}\sin{0 \sqrt{\lambda}}=0$

this clearly gives us...

$B-A\lambda = 0$

clearly If A=B=0 this would give us a dummy answer so i toss it, and say that first this equation to be true,

$B=\sqrt{\lambda}$

$A = 1$

Where i am stuck is when i use the second endpoint condition i get...

$\pi^2A\sin{\frac{\sqrt{\lambda}}{2}}+\pi^2B\cos{\frac{\sqrt{\lambda}}{2}}+A\lambda\cos{\frac{\sqrt{\lambda}}{2}}-B\lambda\sin{\frac{\sqrt{\lambda}}{2}}=0$

with substitution + algebra putting the sin on one side and the cos on the other we get...

$(B-\pi^2A)\sin{\frac{\sqrt{\lambda}}{2}}=(\pi^2B+A)\cos{\frac{\sqrt{\lambda}}{2}}$

divide both sides by cos.....

$\frac{\sin{\frac{\sqrt{\lambda}}{2}}}{\cos{\frac{\sqrt{\lambda}}{2}}}=\frac{(\pi^2B+A)}{(B-\pi^2A)}$

substitute,

$b= \sqrt{\lambda}$

$A = 1$

implies

$\frac{\sin{\frac{\pi}{2}}}{\cos{\frac{\pi}{2}}}=\frac{\pi^3 +1}{\pi - \pi^2}$

SO how do i use this to prove $\pi^2$ is an eigenvalue, then how do i find my function.... please help

Opalg · #2 $Old$ November 5th, 2009, 02:18 AM

Quote:

Originally Posted by ux0 $View Post$

Show that $\pi^2$ is an eigenvalue of the Sturm- Liouville problem

$X'' + \lambda X=0$

$X(0) - X'(0)=0$

$\pi^2X(1/2)+X'(1/2)=0$

and find a corresponding eigenfunction.

To verify that $\pi^2$ is an eigenvalue, you just have to put $\lambda = \pi^2$ in the equation, and show that it then has a nonzero solution. So let $X(x) = A\sin\pi x + B\cos\pi x$ and see what the initial conditions $X(0) - X'(0)=0$ and $\pi^2X(1/2)+X'(1/2)=0$ then tell you about A and B.

ux0 · #3 $Old$ November 5th, 2009, 09:03 AM

once again, doing it with this method, it tells me that

$A=1$

$B=\sqrt{\lambda}=\pi$

But in either case

initial condition a)

$B-A\pi=0$

initial condition b)

$A\pi^2-B\pi=0$

they both equal zero, does this mean that $\pi^2$ is an eigenvalue of the Sturm-L-Problem?

if so... how do i find the eigen function?

Opalg · #4 $Old$ November 5th, 2009, 10:22 AM

Quote:

Originally Posted by ux0 $View Post$

... how do i find the eigen function?

You have already found it! It is the function $A\sin\pi x + B\cos\pi x$ with A=1 and B=π, namely $f(x) = \sin\pi x + \pi\cos\pi x$ .