Eigenvalues and eigenfunctions

[c299792458]

Homework Statement

How does one find all the permissible values of b for -{d\over dx}(-e^{ax}y')-ae^{ax}y=be^{ax}y with boundary conditions y(0)=y(1)=0?

Thanks.

Homework Equations

See above

The Attempt at a Solution

I assume we have a discrete set of \{b_n\} where they can be regarded as eigenvalues? After that how does one find the corresponding \{y_n\}? I am sure we substitute the \{b_n\} into the equation, but then I still don't know how this equation is solved. Please help! Perhaps it is easier to find the permissible b's if we write the equation in the form y''+ay'+(a+b)y=0?

 

[HallsofIvy]

Well, it would be y''+ ay'- (a+b)y= 0, but yes, that is the simplest way to approach this problem. What is the general solution to that equation? What must b equal in order that there be a solution to this boundary value problem? Remember that the characteristice equation may, depending upon b, have real or complex roots.

 

[c299792458]

@HallsofIvy:

Would the general solution be y(x)=A\exp[\frac{-a+\sqrt{a^2+4(a+b)}}{2}]+B\exp[\frac{-a-\sqrt{a^2+4(a+b)}}{2}]
And then the BC's mean that A+B=0 and \exp[\frac{-a+\sqrt{a^2+4(a+b)}}{2}]-\exp[\frac{-a-\sqrt{a^2+4(a+b)}}{2}]=0, therefore we need \sqrt{a^2+4(a+b)}=0 i.e. b={-1\over 4}(a^2+4a)? Is this the only permissible b?

Thanks.

 

[HallsofIvy]

No. You are assuming, incorrectly, that the solution must be of the form Ce^{r_1x}+ De^{r_2x}. That is true only if the characteristic equation has two real roots. In fact, what you give is NOT a "permissible value of b". With that value of b, your characteristic equation reduces to (r+ a/2)^2= 0 so that the only solution is -a/2. In that case, the general solution to the equation is y(x)= Ce^{-ax/2}+ Dte^{-ax/2}. The condition that y(0)= 0 gives C= 0 and then the condition that y(1)= 0 gives De^{-a/2}= 0 so that D= 0. That is not a non-trivial solution.

As I said before, look at complex roots to the characteristic equation.

 

[c299792458]

@HallsofIvy:

Thanks. So the general solution is y(x)=\exp({-ax\over 2})[A\sin({\sqrt{-a-4(a+b)}\over 2}x)+B\cos({\sqrt{-a-4(a+b)}\over 2}x)].

Then B=0 and we need A\sin({\sqrt{-a-4(a+b)}\over 2}x)=0 for A\neq 0 so {\sqrt{-a-4(a+b)}\over 2}=n\pi

Thanks again.