# Simple elliptic pde problem

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.

• 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. Commented Nov 18, 2021 at 18:45

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 $$p$$.
• 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. Commented Nov 18, 2021 at 22:20
• 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. Commented Nov 18, 2021 at 22:59
• 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 $\Omega$. Commented Nov 19, 2021 at 9:06