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Eigenvalue problem

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data

    find all eigen-values and eigen-functions for the initial boundary value problem:

    $$x^2y''+xy'-\lambda y=0$$
    Boundary Conditions:
    $$y(1)=y(e)=0$$

    2. Relevant equations


    3. The attempt at a solution

    i just wanted to know if my substitution in the Auxiliary equation is valid or not, i don't see why not but it's not what the book had.

    r4iEBrk.png
     
  2. jcsd
  3. Mar 23, 2015 #2

    HallsofIvy

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    You are confusing methods for two different types of differential equations.. [itex]l^2+ l- \lambda= 0[/itex] is the characteristic equation for the differential equation [itex]y''+ y'- \lambda= 0[/itex] with constant coefficients. This is an "equipotential equation" where each derivative is multiplied by x to a power equal to the degree of the derivative. It has a completely different characteristic equation.

    Looking for a solution of the form [itex]y= Cx^n[/itex], [itex]y'= Cnx^{n-1}[/itex] and [itex]y''= Cn(n- 1)x^{n-2}[/itex]. Putting that into the equation gives [itex]x^2(n(n-1)Cx^{n-2})+ x(nCx^{n-1})- \lambda (Cx^n)= n(n-1)Cx^n+ nCx^n- \lambda Cn= (n(n- 1)+ n- \lambda)x^n= 0[/itex]. In order for this to be 0 for all x, we must have [itex]n(n-1)+ n- \lambda= 0[/itex]. That is the characteristic equation for this problem.

    (Equivalently, the substitution [tex]t= ln(x)[/tex] converts an "equipotential equation" in x to an equation with constant coefficients in t.)
     
  4. Mar 23, 2015 #3

    Quantum Defect

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    Solution to quadratic equation is incorrect.
     
  5. Mar 23, 2015 #4

    Ray Vickson

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    As HallsofIvy has suggested: your auxiliary equation does not apply, because the LHS of your DE does not have constant coefficients. When they are functions of x, all bets are off. It is very important that you grasp this fact!
     
    Last edited: Mar 23, 2015
  6. Mar 23, 2015 #5

    vela

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    It might help you to understand how you can tell what you tried doesn't work. In your approach, you're assuming the solution has the form ##y=e^{mx}## where ##m## is a constant. You are free, of course, to try this, but it could lead to non-sensical results. If it does, you have to find a different method of solving the equation.

    If you had solved the resulting quadratic correctly, you should have gotten ##m x = -\frac 12 \pm \frac 12 \sqrt{1+4\lambda}##. Isolating ##m##, you find it depends on ##x## and is not a constant, but you had assumed it was a constant. You've reached a contradiction, which indicates that your initial assumption was not valid.
     
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