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Can it be solved: 2nd order non-linear diff. equation

  1. Mar 28, 2009 #1
    Hope I have posted this in the right section, this question is half differential equation and half finite difference method. The equation I have is a form of the Lucas Washburn equation, which is concerned with capillary rise:

    [tex]\rho\left[\left(z+\lambda\right)z''_{tt}+(z'_{t})^{2}\right]+Vzz'_{t}+\rho g z=F[/tex]

    [tex]\lambda,\rho[/tex], F & V are constants,initial conditions are z(0)=0, and z'(0)=0
    The Equation in another form:

    [tex]z\ddot{z}+\dot{z}^{2}=az\dot{z}+bz+c[/tex]

    Am I correct in thinking that this differential equation has no analytical solution? In light of that I want to try and solve for z(t) numerically using a finite difference method but am unsure about how to reform this equation into a from I can use. Any thoughts or suggestions would be greatly appreciated.

    Teller
     
    Last edited: Mar 28, 2009
  2. jcsd
  3. Mar 28, 2009 #2
    ...

    Maybe I'm misunderstanding you, but you could use any numerical integration technique - Euler's method, Runge Kutta, Leapfrog, etc. - on the "another form" after solving for the highest-order derivative in terms of the others.

    z'' = az' + b + c/z - z'*z'/z

    then

    z_(n+1) = z_(n) + z'_(n)*delta_t
    z'_(n+1) = z'_(n) + z''_(n)*delta_t
    z''_(n+1) = az'_(n+1) + b + c/z_(n+1) - z'_(n+1)*z'_(n+1)/z_(n+1)

    No?
     
  4. Mar 29, 2009 #3
    Surprisingly, the system is integrable as long as we have a = 0. The solution is not suitable for perturbation theory, however, since it must be given implicitly in terms of non-elementary functions.

    If a = 0 and c = 0, then the unknown function can be solved for in terms non-elementary functions.

    If a = 0 and b = 0 the solution is:

    [tex]\sqrt{c t^2+2 c c_2 t-\frac{e^{2 c_1}}{c}+c c_2^2}[/tex]

    I mention these special cases in case any of a,b,c are small perturbations, but it looks like in the general case the equation cannot be reduced to integration.
     
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