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Let $f_n \in C^2(\bar{\Omega})$ be a sequence satisfying

$\Delta f_n - f_n^3 \to 0 \ \ {\rm in} \ \ L^2(\Omega)$

where $\Omega \subset {\mathbb R}^2$ is bounded and open with a smooth boundary. Is it necessarily true that $\|f_n^3\|_{L^2(\Omega)}$ and $\|\Delta f_n\|_{L^2(\Omega)}$ are uniformly bounded? If not, can you give a counterexample?

If there is a counterexample, I imagine it would involve $f_n$ becoming unbounded on $\partial \Omega$. If so, is it possible that this statement is true for $f_n \in C^2_c(\bar{\Omega})$ (i.e. if $f_n$ all have compact support in $\Omega$.)

The reason I believe it may be true is that the condition $\Delta f_n - f_n^3 \to 0$ seems incompatible with the sequence $f_n$ having an increasing positive interior maximum or a decreasing negative interior minimium as $\Delta f_n$ and $-f_n^3$ would have the same signs and not cancel out. I have tried to come up with counterexamples in one dimension but have had no luck yet.

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Is "$\Delta$" equal $\partial^2/\partial x^2 + \partial^2/\partial y^2$ or the negative of that? I think both definitions are in use, and the answer to your question might depend on the sign. – Noam D. Elkies Oct 26 '11 at 18:23
Yes, I'm using $\Delta = \partial^2/\partial x^2 + \partial^2/\partial y^2$. I think you are right that the answer would be different with $\Delta$ replaced by $-\Delta$. – Jeff Oct 26 '11 at 19:27
In one dimension $f_n(x) = 2^{1/2} / (x-x_n)$ satisfies $\Delta f_n - f_n^3 = 0$, and if $x_n \rightarrow \partial \Omega$ from the outside then $\|f_n^3\| \rightarrow \infty$. Yes, such $f_n$ are unbounded on $\partial \Omega$. – Noam D. Elkies Oct 26 '11 at 21:26
up vote 3 down vote accepted

A counterexample in one dimension: take $\Omega:=(0,1)$ and $f_n(x):=\frac{\sqrt 2}{x+\frac{1}{n}}$. Then $f''_n(x)-f_n^3(x)=0$ while $\| f''_n \| _{2,\Omega}=\|f^3_n\|_{2,\Omega}=O(n^{5/2}) \, .$

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Thanks! Nice counterexample. This is kind of what I was expecting, $f_n$ becoming unbounded at the boundary. Any idea what happens if we restrict $f_n$ to have compact support? – Jeff Oct 26 '11 at 21:27
Then if you multiply $\Delta f_n - f_n^3 := h_n\to 0$ in $L^2(\Omega)$ by $f_n$ and integrate on the domain you get $\|\nabla f_n\|^2_2 + \|f_n\|^ 4_ 4 = o(\|f_n\|_2)$, which implies that the LHS goes to zero. – Pietro Majer Oct 26 '11 at 22:01
Thanks, don't know why I didn't see that before, it's not so hard. – Jeff Oct 26 '11 at 23:00

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