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Second nonlinear DE for exciton diffusion

  1. Jun 4, 2010 #1
    Dear Everyone,

    I am working on a physic problem of exciton diffusion involved in organic optoelectronics.

    It is in the form of

    y''+a*y+b*y^2=0.

    Is there a general solution to this equation?

    Thanks!

    elfine
     
    Last edited: Jun 4, 2010
  2. jcsd
  3. Jun 5, 2010 #2
    I don't know how to handsolve the equations, but I sent it through Mathematica, and you don't want to see the answer. Although I posted it anyway.

    Code (Text):

    in = DSolve[y''[x] + a*y[x] + b*y[x]^2 == 0, y[x], x]
    out = Solve[(4 EllipticF[
          ArcSin[\[Sqrt]((Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3] -
                y[x])/(-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] +
                Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
                 3]))], (Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] -
             Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
              3])/(Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 1] -
             Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3])]^2 (Root[-3 C[1] +
              3 a #1^2 + 2 b #1^3 &, 2] -
           Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
            3]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 1] +
           y[x]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] +
           y[x]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3] +
           y[x]))/((-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 1] +
           Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &,
            3]) (-Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 2] +
           Root[-3 C[1] + 3 a #1^2 + 2 b #1^3 &, 3]) (C[1] - a y[x]^2 -
           2/3 b y[x]^3)) == (x + C[2])^2, y[x]]
     
    where Solve solves an equation for y[x] and Root finds the roots of an equation.
     
  4. Jun 5, 2010 #3
    Thank your reply.

    The attached picture is a given method for solving this kind equation, but it is uncompleted.

    For I am not good at math and I can't get a good solution from matlab, could anyone help me on this?

    Best wishes!

    elfine
     

    Attached Files:

  5. Jun 5, 2010 #4
    There is suppose to be a general solution to the general elliptic equation:

    [tex]y''=A+By+Cy^2+Dy^3[/tex]

    multiplying by y', integrating, and adjusting the constants we obtain the form:

    [tex]\left(y')^2=a+by+cy^2+dy^3+ey^4[/tex]

    and by a suitable change of variable [itex]z=z(y)[/itex] we can reduce it to the standard form:

    [tex]\left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2 z^2)[/tex]

    in which [itex]z(x)=\text{sn}(x,k)[/itex] where sn is the Jacobi elliptic sine function. We then invert the expression [itex]z=z(y)[/itex] to obtain the solution in y. However, the exact details of that procedure is a little unclear to me.
     
  6. Jun 6, 2010 #5
    Hi jackmell,
    Thanks a lot! Your suggestion is very valuable to me.
    I adjust my DE to
    [tex]\left(y')^2=a+cy^2+dy^3[/tex]
    with the boundary of
    [tex]\left y(\infty)=0[/tex]
    and the simplified equation is
    [tex]\left(y')^2=cy^2+dy^3[/tex]
    By subtitute
    [tex]\left t=\sqrt{c+dy}[/tex]
    I finally get
    [tex]\left \frac{adt}{t^2-b}=dx[/tex]
    And its primitive function is a Hyperbolic function.
    elfine
     
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