# Solution of a differential equation that would be a generalized mean?

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. – Jon Jan 16 '12 at 14:02
X(t) is a primitive of the rhs but I don't know how to integrate that... – Oli Jan 16 '12 at 23:58
$dX/dt$ 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. – Rahul Sep 22 '12 at 2:41

A first attempt can be accomplished by iteration. So, consider

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

that can be rewritten as

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

and so

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

and finally

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

and this is amenable to an iterative technique. E.g. one can choose

$$X_0=0$$

and so

$$X_1=-\frac{\int_{-\infty}^{\infty}\frac{\partial f}{\partial t}\frac{\partial f}{\partial x}dx}{\int_{-\infty}^{\infty}\frac{\partial}{\partial t}\left(\frac{\partial f}{\partial x}\right)^{2}dx}$$

that is already useful for computations. But, of course, this will depend on the problem at hand and where all this started from. A better choice for $X_0$ could be found after some attempts.

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Thanks for your help. I'll see if I can work something from this – Oli Jan 18 '12 at 8:29

Suppose $f$ is a conserved quantity transported by a velocity field $u$, so that it obeys the advection equation, $$f_t + (fu)_x = 0.$$ Then it turns out that $$u = - \frac{f_t + fu_x}{f_x},$$ so $\tilde u := -f_t/f_x$ is an estimate of $u$ under the assumption that its divergence $u_x$ is negligible. Your right-hand side is the average of $\tilde u$ weighted by $f_x^2$. So it gives some kind of an estimated mean of the velocity $u$ that $f$ is being transported by.

The integral of the right-hand side over time is therefore some sort of a generalized displacement. But since it is only unique up to translation, it doesn't make sense to call it a mean; it depends on your choice of initial condition $X(0)$. Also, if $f$ undergoes compression or rarefaction (so $u_x$ is not negligible), the results may no longer be sensible. I can imagine that if $f$ evolves in a complicated asymmetrical fashion but eventually returns to its initial distribution $f(x,T) = f(x,0)$, you might not have $X(T) = X(0)$. But I don't have an explicit counterexample.

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