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I have been looking at a problem requires the solution of an equation of the form: $$\int_{x_0}^{x_1} f(x,y)^{-n} \left(\frac{\partial}{\partial y} f(x,y)\right)^m dx = 0$$ for integer values of $m$ and $n$ greater than 1.

What non-trivial functions satisfy this? How would you go about finding a single solution as an example?

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I guess, are there non-trivial solutions is the first question... – Lucas Dec 16 '12 at 18:35
up vote 1 down vote accepted

Here's a less trivial solution. Without loss of generality, translate your integration variable so the bounds of the integral are $[-a,a]$, where $a=(x_1-x_0)/2$.

$m$ and $n$ have opposite parity:

When $m$ and $n$ have opposite parity, a solution of the form $f(x,y) = \sum_{i} \alpha_i(x) \beta_i(y)$, where each of the $\alpha_i(x)$ is an odd function is a solution. This means that both $f(x,y)$ and $\frac{\partial f}{\partial y}$, are odd functions of $x$ and that consequently $f^{-n} (\frac{\partial f}{\partial y})^m$ is as well, meaning the integral will vanish for any value of $y$.

$m$ and $n$ are both even:

In this case $f^n$ and $(\frac{\partial f}{\partial y})^m$ are both positive, which requires that either $f(x,y)=0$ or $\frac{\partial f(x,y)}{\partial y}=0$ for each value of $(x, y)$. I think this pretty much means that the only solutions are trivial, ie $f(x,y)=g(x)$ for some function $g(x)$.

$m$ and $n$ are both odd:

Need to think about this a bit more.

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Write $f(x,y)=\alpha(x)\beta(y)$. Then for $n\ne m$, any $\alpha(x)$ that satisfies $\int_{x_0}^{x_1} \alpha(x)^{m-n} dx = 0$ should provide a solution for arbitrary $\beta(y)$. For example, $\alpha(x) = \sin(2\pi k\frac{x-x_0}{x_1-x_0})^{1/(m-n)}$, for integral k.

If $n=m$, then you can choose $\alpha(x)$ arbitrarily, as long as you choose $\beta(y)=c$, but maybe you would consider that to be a trivial solution.

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I would consider it a trivial solution, but that does not mean that your response is not appreciated :) – Lucas Dec 19 '12 at 6:11
Actually the $(m-n)$th root isn't guaranteed to be real when $m$ and $n$ have the same parity (for negative arguments). So this solution is really only good when $m$ and $n$ have the same parity. – GregP Dec 20 '12 at 17:17

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