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Find the function $f=f(x,u,u')$ for which the equality

$$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial u'} \right)=\frac{\partial^2}{\partial x^2}\left( u+u^2\right)$$ holds.

Here, $u=u(x)$ and $u'=\partial u/\partial x$. It is presumed that $u(x)\rightarrow0$ as $x \rightarrow \pm \infty$.

This is what I have tried so far: Integrate once to find $$\left(\frac{\partial f}{\partial u'} \right)=\frac{\partial}{\partial x}\left( u+u^2\right)$$ Can I now do this and somehow resolve through per partes? $$f=\int \frac{\partial}{\partial x}\left( u+u^2\right)\partial u'$$ Maybe the differential $\partial u'=\partial(du/dx)$ can be in some way simplified.

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drabus
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1 Answers1

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I have found an $f$ that satisfies your equation.

$$f=\left[\frac{1}{2}+u\right](u’)^2+g(x,u)$$

where $g(x,u)$ is some unknown function that would need to be found using boundary or some other data.

To make this as simple as possible, let’s do this the easy way and start with your final equation.

$$f=\int\frac{\partial}{\partial x}(u+u^2)\;\partial u’$$

It is best to interpret this integral representation as the anti-derivative. With that said, there isn’t much stopping you from taking the anti-derivative w.r.t. $u’$. Differentiate inside the expression and do so.

$$f=\int(u’+2uu’)\;\partial u’=\frac{1}{2}(u’)^2+u(u’)^2+g(x,u)$$

The extra term $g(x,u)$ is necessary as we took the anti derivative w.r.t. $u’$. Plugging the result back into your equation verifies that this is a solution.

Aside: Your PDE closely resembles that of the Euler-Lagrange equation.

$$\frac{d}{dx}\frac{\partial f}{\partial u’}=\frac{\partial f}{\partial u}$$

If $f$ we’re to satisfy this equation, it would be called a “Lagrangian”. However, I believe this would result in a different $f$ than your given problem. Just an extra tid-bit.

MasterYoda
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  • @LeeDavidChungLin Sorry, I did this early this morning when I had little time and forgot to update it with an explanation. Thanks for the reminder. – MasterYoda Feb 15 '20 at 06:13
  • It's a perfectly good answer now. Sorry to see you getting downvoted (wasn't me). – Lee David Chung Lin Feb 15 '20 at 06:33
  • To the three people downvoting this, can you at least explain why? Not sure what is wrong with my answer. – MasterYoda Feb 15 '20 at 19:14
  • @MasterYoda, thank you very much, you are indeed correct as far I can see. Interesting tid-bit about the Lagrangian. Do you have any clue about how the function would change if it satisfied the Lagrangian? I will also try to find it and add it to the answer. – drabus Feb 16 '20 at 18:22
  • @MasterYoda I had down voted for not providing an explanation and putting it as answer. Have up voted now. – mathisgood Feb 17 '20 at 20:59
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    @drabus For the Lagrangian, I so far have only found $f$’s that depend on $u’’$. I haven’t found any that satisfy $f=f(x,u,u’)$ for your equation. That being said I haven’t shown that it must depend on $u’’$, so maybe there is a Lagrangian that satisfies your PDE and I’m just not clever enough to find it. – MasterYoda Feb 19 '20 at 19:51
  • @MasterYoda I have also been looking for a solution with no luck. Even Mathematica is helpless here. – drabus Feb 20 '20 at 08:06