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Differential equation

  1. Mar 31, 2010 #1
    I have a problem with this equation

    [tex]\frac{d^2\Omega(\alpha)}{d\alpha^2}+\frac{(1+\Phi)+\Phi e^{-\alpha}}{(1+\Phi)-\Phi e^{-\alpha}}\frac{d\Omega(\alpha)}{d\alpha}-S(S+1)\Omega(\alpha)=0[/tex]

    Boundary conditions are



    How to to transform part before first derivative [tex]\frac{(1+\Phi)+\Phi e^{-\alpha}}{(1+\Phi)-\Phi e^{-\alpha}}[/tex]?
  2. jcsd
  3. Apr 5, 2010 #2
    [tex]\Phi[/tex] and [tex]S[/tex] are constants! Thanks for your answer!
  4. Apr 5, 2010 #3
    The general solution to the equation (maple obtained) is

    [tex]\Omega(x)= \frac{e^{-x/2}}{-\Phi-1+\Phi e^{-x}}(C_{1}e^{x(S+1/2)}+C_{2}e^{-x(S+1/2)})[/tex]

    The first boundary condition puts the constraint


    while the second one says

    [tex]\Pi_{p=-S}^{p=S} [p-[(1+2S)C_{2}+S-\Phi]] =0[/tex]

    which is a product with a finite number of terms. I assume S is an integer or half integer A solution is obtained by cancelling any factor of the product. You get a solution for any C2 satisfying

    [tex](1+2S)C_{2}+S-\Phi=p, -S\le p \le S [/tex]

    All this needs some checking
    Last edited: Apr 6, 2010
  5. Apr 5, 2010 #4
    Ok! But how to get solution without use of any computer programme? How to solve this with pencil and paper? What is the idea?
  6. Apr 6, 2010 #5
    Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

    [tex] u''(x)+A(x)u'(x)-S(S+1)u(x)=0[/tex]

    then it turns out that [tex]u(x)=exp(-\frac{1}{2}\int{A(x')dx'})v(x)[/tex] transforms the equation into

    [tex] v''+[-S(S+1)-A^2/4-A'/2]v=0[/tex]

    In this case, we have

    [tex] \int{A(x)dx}=-2ln[\frac{exp(-x/2)}{-1-\Phi+\Phi exp(-x)}][/tex]

    And the resulting equation for v(x) is


    resulting in the solution above. This equation has several lucky "coincidences", but the procedure to eliminate the first derivative term is standard, and very useful for non-homogeneous oscilators
  7. Apr 6, 2010 #6
    Are you sure about this?

    I get equation

  8. Apr 6, 2010 #7
    How you get this form [tex]v''(x)-(S+1/2)^{2}v(x)=0[/tex] from [tex] v''+[-S(S+1)-A^2/4-A'/2]v=0[/tex]?
  9. Apr 7, 2010 #8
    By substituting A
  10. Apr 8, 2010 #9
    [tex]A(x)=\frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}[/tex]

    [tex]A'(x)=[\frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}]'[/tex]

    [tex]A'(x)=\frac{-\Phi e^{-x}[(1+\Phi)-\Phi e^{-x}]-[(1+\Phi)+\Phi e^{-x}]\Phi e^{-x}}{[(1+\Phi)-\Phi e^{-x}]^2}=\frac{-2\Phi e^{-x}(1+\Phi)}{[(1+\Phi)-\Phi e^{-x}]^2}[/tex]

    [tex]\frac{-A'(x)}{2}=\frac{\Phi e^{-x}(1+\Phi)}{[(1+\Phi)-\Phi e^{-x}]^2}[/tex]

    [tex]\frac{-A^2}{4}=-\frac{\frac{(1+\Phi)^2}{4}+\frac{1}{2}(1+\Phi)\Phi e^{-x}+\frac{1}{4}\Phi^2 e^{-2x}}{[(1+\Phi)-\Phi e^{-x}]^2}[/tex]

    [tex]\frac{-A^2}{4}-\frac{A'(x)}{2}=-\frac{\frac{(1+\Phi)^2}{4}-\frac{1}{2}(1+\Phi)\Phi e^{-x}+\frac{1}{4}\Phi^2 e^{-2x}}{[(1+\Phi)-\Phi e^{-x}]^2}=-\frac{1}{4}[/tex]


  11. Apr 8, 2010 #10
    You are welcome
  12. Apr 10, 2010 #11
    I got

    [tex]\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=2ln[\frac{(1+\Phi)-\Phi e^{-x}}{\Phi e^{-x}}][/tex]

    I have done this integral like

    [tex]\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=\int\frac{1+\Phi}{(1+\Phi)-\Phi e^{-x}}dx+\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx[/tex]

    and got

    [tex]\int\frac{1+\Phi}{(1+\Phi)-\Phi e^{-x}}dx=ln[\frac{(1+\Phi)-\Phi e^{-x}}{\Phi e^{-x}}][/tex]


    [tex]\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=ln[{(1+\Phi)-\Phi e^{-x}}][/tex]
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