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Convergence of perturbative solutions to a non-linear diff eq

  1. Jul 24, 2014 #1
    Hi. First off, sorry for the not so descriptive title. If one of you finds a better tilte I will edit it.

    We have the equation
    \partial_{xx}\phi = -\phi + \phi^{3} + \epsilon \left(1- \phi^{2}\right)
    We will look for solutions satisfying
    \phi(\pm \infty) = \pm 1 \qquad \partial_{x}\phi\big|_{x=\pm \infty} = 0
    Integrating once we find
    \left(\partial_{x}\phi\right)^{2} = -\phi^{2} + \frac{\phi^{4}}{2} + 2\epsilon \left(\phi - \frac{\phi^{3}}{3}\right) + K
    Using the boundary conditions we find that no solution exists for [itex]\epsilon \neq 0[/itex]. For [itex]\epsilon=0[/itex] we find( assuming [itex]\phi(0) = 0[/itex])
    \phi_{0} = \tanh\left( \frac{x}{\sqrt{2}} \right)
    Now lets assume [itex]\epsilon << 1[/itex] and expand [itex]\phi = \sum \epsilon^{j}\phi_{j}[/itex] imposing the boundary conditions
    \phi_{0}(\pm \infty) = \pm 1 \qquad \phi_{j}(\pm \infty) = 0
    For j>0, and
    \partial_{x}\phi_{i}\big|_{x=\pm \infty} = 0
    For all i.
    Now what will the higher and higher order terms signify? And will this series ever converge towards anything( either stationary or periodic)? I worked out the expansion up to [itex]\phi_{5}[/itex] without getting any wiser( I can write down the solutions if anybody asks but it takes a bit of time.)
  2. jcsd
  3. Jul 28, 2014 #2
    It turns out I had miscalculated a solvabillity integral. The expansion breaks down already at first order. Sorry about that!

    On the odd chance that anybody would be interested in seeing the calculation, just let me know.
    Last edited: Jul 28, 2014
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