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.