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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

    [tex]\Omega(0)=1[/tex]

    [tex][\Pi^S_{p=-S}(p-\frac{d}{d\alpha})]\Omega(\alpha)|_{\alpha=0}=0[/tex]

    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

    [tex]C_{1}+C_{2}=-1[/tex]

    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

    [tex]v''(x)-(S+1/2)^{2}v(x)=0[/tex]

    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

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

    [tex]\upsilon''(x)-(S+\frac{1}{2})^2\upsilon(x)=0[/tex]


    Thanks!
     
  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]

    and

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