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Eigenvalues and eigenfunctions

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    How does one find all the permissible values of [itex]b [/itex] for [itex]-{d\over dx}(-e^{ax}y')-ae^{ax}y=be^{ax}y[/itex] with boundary conditions [itex]y(0)=y(1)=0[/itex]?

    Thanks.



    2. Relevant equations
    See above


    3. The attempt at a solution

    I assume we have a discrete set of [itex]\{b_n\}[/itex] where they can be regarded as eigenvalues? After that how does one find the corresponding [itex]\{y_n\}[/itex]? I am sure we substitute the [itex]\{b_n\}[/itex] 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 [itex]b[/itex]'s if we write the equation in the form [itex]y''+ay'+(a+b)y=0[/itex]?
     
  2. jcsd
  3. Sep 26, 2011 #2

    HallsofIvy

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    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.
     
  4. Sep 26, 2011 #3
    @HallsofIvy:

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

    Thanks.
     
  5. Sep 26, 2011 #4

    HallsofIvy

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    No. You are assuming, incorrectly, that the solution must be of the form [itex]Ce^{r_1x}+ De^{r_2x}[/itex]. 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 [itex](r+ a/2)^2= 0[/itex] so that the only solution is [itex]-a/2[/itex]. In that case, the general solution to the equation is [itex]y(x)= Ce^{-ax/2}+ Dte^{-ax/2}[/itex]. The condition that y(0)= 0 gives C= 0 and then the condition that y(1)= 0 gives [itex]De^{-a/2}= 0[/itex] so that D= 0. That is not a non-trivial solution.

    As I said before, look at complex roots to the characteristic equation.
     
  6. Sep 26, 2011 #5
    @HallsofIvy:

    Thanks. So the general solution is [itex]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)].[/itex]

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

    Thanks again.
     
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