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I am trying to solve this differential equation on which I've been stuck for several days now.

$$\frac{d X}{d t}=\frac{\int_{-\infty}^{\infty}\frac{\partial f}{\partial t}\frac{\partial f}{\partial x}dx}{\int_{-\infty}^{\infty}\left(\frac{\partial f}{\partial x}\right)^{2}dx} $$

where $f(x,t)$ is as smooth and integrable as you want it to be. Hey, I'm a physicist :)

I have also the normalisation $\int f(x,t)dx=\int f^{2}(x,t)dx=1 $

If for all $t$, $x_0(t)$ is a center of symmetry relative to x then $X(t)=x_0(t)$ is a solution of my equation. This lets me think that in the general, non-symmetric case, a solution of this equation might be related to a generalized mean of $f$. It does make sense when I do some simulations.

Rewriting this in the Fourier plane using Parseval identities leads to interesting formulas but I can't interpret them either. I also tried to think of it as $L^2$ inner products without any result.

That would be great if anyone had an idea on how to give some sense to this equation, particularily if a solution of it could be interpreted as a generalized mean of $f$. Or obviously to solve it if that proved to be possible.

The original problem is actually in dimension $N$ with gradients instead of derivatives relative to $x$ but any idea in dimension $1$ would be extremely welcome.

Thanks to whoever takes time considering this problem,

Olivier

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    $X$ is just a funcion of $t$. So, how does this unknown enters into the rhs? As you write it this can be straightforwardly integrated.2012-01-16
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    X(t) is a primitive of the rhs but I don't know how to integrate that...2012-01-16
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    $dX/dt$ is independent of $X$, so any solution plus a constant shift is also a solution. So it doesn't make sense to call this a generalized mean. Maybe you can call the right-hand side a generalized velocity, though.2012-09-22

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