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First variation of convolution of two nonlinear functions

  1. May 14, 2014 #1
    A new variational principle is presented in this paper: http://arxiv.org/ftp/arxiv/papers/1112/1112.2286.pdf

    When trying to derive something like the equation of motion of a Duffing oscillator, I take the following approach:

    Set up the functional as such:
    $$
    \mathcal{I}\left(t,x(t),\dot{x}(t)\right) = \frac{1}{2}\left[\dot{x} * \dot{x} \right]+\frac{\beta}{2}\left[x * x \right]+\frac{\alpha}{4}\left[x^2 * x^2 \right]
    $$

    Where:
    $$
    \left[f * g\right]=\int^{t}_{0}f\left(t-\tau\right)g\left(\tau\right)\,\text{d}\tau
    $$

    Now, normally, I would take the first variation of the expression for $\mathcal{I}$ and then collect the similar variations $\delta x$ and equate their coefficient to zero, but it's not immediately obvious how one would do that with this expression:


    $$
    \mathcal{I}= \left[\delta\dot{x} * \dot{x} \right]+\beta\left[\delta x * x \right]+\alpha\left[x \delta x * x^2 \right]
    $$

    Using the integration by parts principle for convolutions, we get:

    $$
    \mathcal{I}= \left[\delta x * \left(\ddot{x}+\beta x\right) \right]+\left[x \delta x * x^2 \right]+\left.\delta x \dot{x}\right]^{t}_{0}
    $$

    Where it not for the nonlinear term in the last term of $\mathcal{I}$, I would have the full dynamics, namely:
    $$
    \ddot{x}+\beta x=0
    $$

    Although the dynamics should be:

    $$
    \ddot{x}+\beta x+\alpha x^3 =0
    $$

    But there is no way I can see of incorporating the convolution term with $x\delta x$ into the other terms convolved with $\delta x$.

    I've tried playing around with the properties of convolutions and so far have not found a solution.

    Is there a way to manipulate this convolutions to get the desired dynamics?
     
  2. jcsd
  3. May 14, 2014 #2
    Any method that would enable me to reexpress:
    $$
    \left[x \delta x * x^2 \right]
    $$

    As the variation convolved with something else would work.
     
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