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Math Help - finding general solution (difficult differential equation)

  1. #1
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    finding general solution (difficult differential equation)

    Hi, all

    Can someone tell me what the general solution is for this differential equation, if any?


    u(x)''= A + b*u(x)^3 - u(x)


    To me there seems to be no general solution for it!


    Thanks in advance,

    Polo
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  2. #2
    MHF Contributor chisigma's Avatar
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    The DE is...

    u^{''} = a + b u^{3} - u = f(u) (1)

    ... so that it contains only u^{''} and u. The approach for solving such type of equation is the substitution...

    u^{''} = \frac{d u^{'}}{dx}= \frac{d u^{'}}{du} \frac{du}{dx} = u^{'} \frac{du{'}}{du} (2)

    ... so that (1) becomes...

    u^{'} du^{'} = (a + b u^{3} - u) du (3)

    The (3) is a separated variables first order ODE the solution of which is...

    u^{'2} = 2 (a u + \frac{b}{4} u^{4} - \frac{u^{2}}{2}) + c_{1} \rightarrow u^{'} = \pm \sqrt {\frac{b}{2} u^{4} - u^{2} + 2 a u + c_{1}} (4)

    The (4) is a separable variables first order ODE again the solution of which is...

    x = \pm \int \frac{du}{\sqrt {\frac{b}{2} u^{4} - u^{2} + 2 a u + c_{1}}} + c_{2} (5)

    Now the solution of the integral in (5) requires a little more effort ...

    Kind regards

    \chi \sigma
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  3. #3
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    Hi,


    Itís exactly with this part I have problems with.
    How do you solve such a beast (equation 5)?

    Is there any solution for it?


    Thanks in advance

    Polo

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  4. #4
    MHF Contributor chisigma's Avatar
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    Integrals of the type...

    \int f[x,\sqrt{p(x)}]\cdot dx

    ... where f(*) is a rational function and p(*) a polynomial od degree 3 or 4 are called elliptic integrals and their solution is even today not very confortable. See here...

    Elliptic Integral -- from Wolfram MathWorld

    Kind regards

    \chi \sigma
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  5. #5
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    Hey guys. That's really an interesting problem in transformation theory and involves the solution of the general elliptic equation y''=A+By+Cy^2+Dy^3. The following is adapted from "Introduction to Nonlinear Differential and Integral Equations" by H. Davis.

    We wish to get the DE into the form:

    \left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2 z^2)=\Delta^2(z)

    which has the solution z=\text{sn}(x,k) where \text{sn} is the Jacobi elliptic function. We can do that by writing y''=A-y+by^3 in it's elliptic form \left(y'\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)(y-\delta)=h^2\Delta_2^2(y) (multiply the general elliptic equation by y' and integrate) and then using the transformations:

    z^2=\frac{\beta-\delta}{\alpha-\delta}\frac{y-\alpha}{y-\beta}

    k^2=\frac{(\beta-\gamma)(\alpha-\delta)}{(\alpha-\gamma)(\beta-\delta)}

    M^2=\frac{(\beta-\delta)(\alpha-\gamma)}{4}

    and here is the challenging part. Show:

    \frac{1}{\Delta(z)}\frac{dz}{dx}=\frac{M}{\Delta_2  (y)}\frac{dy}{dx}

    That expression is a little easier to show if the factored elliptic equation is only a cubic and involving simpler transformation rules and is the version I went through. I assume the quartic is just more messy. Anyway, it then follows:

    z=\text{sn}(hMx,k)

    and by inverting the expression for z above, the expression for y follows. Seems to me though the roots in general are complex which then means the solution is complex-analytic. I say give a good 45 minute presentation on this Polo and you don't have to take the final exam. Some nice plots too.
    Last edited by shawsend; May 10th 2010 at 06:26 AM. Reason: correct formulas
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  6. #6
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    Dear Shawsend,


    Although that the factored elliptic equation in cubic form is for the moment more than I can handle. However, it seems doable.

    But I have to admit that higher factored elliptical equation is just for the moment above my head.

    Thank you very much for your help, I appreciate very much.

    N.B.: the thank you bottom does not seem to function
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