Positive first eigenvalue; operator satisfies maximum principle

Question

I am attempting to understand a paper. They have $u$ is a stable smooth solution of $ -\Delta u = f(u)$ in $ \Omega$ with $ u=0$ on $ \partial \Omega$ where $\Omega$ is a bounded domain in Euclidean space. By stable they mean the first eigenvalue of the linearized operator is strictly positive; ie. there is some $ \mu>0$ and $ \phi>0$ which solves $$-\Delta \phi - f'(u) \phi=\mu \phi \; \; \mbox{ in } \; \; \Omega$$ with $ \phi=0$ on $ \partial \Omega$. They then go on to say that $ \sup_\Omega | \nabla u| = \sup_{\partial \Omega} | \nabla u|$ by the maximum principle since $ \mu$ is positive and since $ u_{x_i}$ solves $-\Delta (u_{x_i}) - f'(u) u_{x_i}=0$ in $ \Omega$.

The part I don't understand is this final claim about the maximum of the gradient being obtained on the boundary. I assume this must come directly off the standard maximum principle but I don't see it.

For instance I assume they are saying if $ C(x)>0$ is some smooth function and there is some $ \phi>0$ in $ \Omega$ and $ \mu>0$ which satisfies $ -\Delta \phi - C(x) \phi = \mu \phi$ in $ \Omega$ with $ \phi=0$ on $ \partial \Omega$ and $ -\Delta \psi - C(x) \psi=0$ in $ \Omega$ that we must have $ \sup_{\Omega} | \psi| = \sup_{\partial \Omega} | \psi| $; but I am convinced I can come up with trivial counter examples to this final statement. I assume clearly I must be missing something obvious or. Any comments would be helpful.

Answer 1

I don't know if that's helpful, but here is a comment (not an answer). While it is possibly not true that $u_i$ has an extremum on $\partial\Omega$, it is true that $$ v_i:=\frac{u_i}{\phi} $$ does. Note the algebraic identity $$ b\Delta \left(ab\right)=\textrm{div}\left(b^2Da\right)+ab\Delta b, $$ and apply it to $a=v_i$ and $b=\phi$ to obtain \begin{eqnarray*} 0&=&-\phi \Delta u_i - f^\prime(u_i)\phi u_i \\ &=& -\textrm{div}\left(\phi^2Dv_i\right)+v_i\phi\left( -\Delta \phi - f^\prime(u_i)\phi \right) \\ &=&-\textrm{div}\left(\phi^2Dv_i\right) +\mu\phi^2 v_i, \end{eqnarray*} and now the right hand-side is an operator with positive coefficients. Normally this is used with a Neumann or Periodic eigensolution, which does not vanish on the boundary, and therefore it gives meaningful bounds for the gradients. Here since $\phi$ vanishes, what it says is less clear.