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I have a question that is clearly not research level, but it's confusing me so I will ask anyway.
There must be some little logic flaw I am missing. Take $\Omega$ a bounded smooth domain in $\mathbb R^N$ and assume $ \lambda_k$ is the $k$ eigenvalue of $ -\Delta$ in $H^1_0(\Omega)$.

Let $v$ denote a smooth solution of $$-\Delta v - \lambda^2 v = \lambda^2 \mbox{ in } \Omega$$ with $ v=0$ on $ \partial \Omega$ and we assume $ \lambda^2 \neq \lambda_k$ for any $k$ but with $ \lambda^2> \lambda_1$. Then we know that $v$ must be negative somewhere.

Now consider $u$ given by $v= e^{\lambda u}-1$ and note that $ v \ge 0$ in $\Omega$ exactly when $ u \ge 0$ in $ \Omega$. So we expect that $u$ must be negative somewhere. Also note that $u$ must be smooth since $v$ is smooth. Also note that $u$ satisfies

$$-\Delta u = \lambda ( \lvert \nabla u\rvert^2+1) \mbox{ in } \Omega$$ with $u=0$ on $ \partial \Omega$ and hence we can apply the maximum principle to see that $u \ge 0$ in $ \Omega$.

Clearly I am missing something.

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  • $\begingroup$ I purposely didn't write $u$ as a function of $v$ since it involves the ln function and I was scared of this. I thought the way I wrote it above gets around this but I guess it doesn't. If $v$ gets close to -1 then $ u \rightarrow -\infty$ and hence i guess my claim that since $v$ is smooth hence $u$ must be smooth is the (or at least 'a') problem. $\endgroup$
    – Math604
    Nov 18, 2021 at 18:45

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If you apply the maximum principle, at a point $p$ where the function $v$ reaches its minimum, you get $-\lambda^2 v(p) \geq \lambda^2$ so $v(p) \leq -1$. In particular, the function $u$ is not globally defined as it has to go to $-\infty$ at least at $p$.

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  • $\begingroup$ Yes, I agree. The flaw is that $u$ does not reach (all of) $\partial\Omega$ : it is defined on a smaller domain, as you said, where $v$ is positive. $\endgroup$
    – username
    Nov 18, 2021 at 22:20
  • $\begingroup$ Another way is to use the same argument proving that $v$ must be negative somewhere to show that must be $-1$ somewhere. Write $v=-1+u$ with $-\Delta u-\lambda^2 u=0$ and $u=1$ at the boundary. Then $u$ must be negative somewhere. $\endgroup$
    – Giorgio Metafune
    Nov 18, 2021 at 22:59
  • $\begingroup$ To answer @username : $u$ is defined in a neighborhood of $\partial \Omega$ as $v$ is smooth and $v \equiv 0$ on the boundary. The problem is what happens on the inside (away from the boundary) of $\Omega$. $\endgroup$
    – Romain Gicquaud
    Nov 19, 2021 at 9:06
  • $\begingroup$ thanks for the comments. $\endgroup$
    – Math604
    Nov 19, 2021 at 10:00

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