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.
Best Answer
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 $p$.