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.



$L^2$ boundeness of a sequence
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 $\| f''\_n \| \_{2,\Omega}=\|f^3\_n\|\_{2,\Omega}=O(n^{5/2}) \, .$


