Is this a consequence of the maximum principle for superharmonic functions?

Question

Let $\Omega\subset\mathbb{R}^n$ be bounded and suppose we have $$ \Delta u_{x_1}+f'(u)u_{x_1}=0\textrm{ in }\Omega, $$ and, moreoever, know that $u_{x_1}\geq 0, u_{x_1}\neq 0$ in $\Omega$.

Here $u_{x_1}$ denotes the partial derivative in the first coordinate and $f$ is assumed to be Lipschitz continuous and differentiable.

Can we conclude that $u_{x_1}>0$ in $\Omega$?

This seems to be some direct consequence of maximum principle for superharmomic functions which says that a superharmonic functions does not attain its minium on the interior (and if it does, it is a constant function).

My idea is that due to $u_{x_1}\geq 0$ in $\Omega$, we know that, in $\Omega$, $$ 0=\Delta u_{x_1}+f'(u)u_{x_1}\geq \Delta u_{x_1} $$ which means that $u_{x_1}$ is super-harmonic.

Since $u_{x_1}\neq 0$, which means that it is not constants, the minimum uf $u_{x_1}$ is not attained in the interior but on $\partial\Omega$.

Which means that $u_{x_1}>0$ in $\Omega$.

Answer 1

Thanks for linking to your source in the comments. For the version of the maximum principle that's relevant here, it actually refers us on to another paper [GNN1] "Symmetry and Related Properties via the Maximum Principle" by Gidas, Wei-Ming and Nirenberg. Paraphrasing from that paper:

Assume a nonnegative $C^{2}$ solution $w$ for $Lw \le 0$ where \begin{equation*} Lw=\sum_{ij}a_{ij}w_{x_{i}x_{j}} + \sum_{i} b_{i} w_{x_{i}} + c w \end{equation*} uniformly elliptic with coefficients uniformly bounded in absolute value. If $w$ vanishes at some point in $\Omega$ then $w \equiv 0$.

For the proof we are referred on again, this time to a book, "Maximum principles in differential equations" by Protter and Weinberger. But handily we are also told that the subsequent formulation of the Hopf Lemma "follows by the same argument used to prove the maximum principle above ... and for the convenience of the reader we include the derivation". I don't have the book so I tried to adapt that derivation. \begin{eqnarray*} \text{Let}\qquad\quad\; v &=& e^{-\alpha x_{1}}w. \\ \text{Then}\qquad v_{x_{1}}&=&-\alpha e^{-\alpha x_{1}}w+e^{-\alpha x_{1}}w_{x_{1}} \\ v_{x_{1}x_{1}}&=&\alpha^{2}e^{-\alpha x_{1}}w-2\alpha e^{-\alpha x_{1}}w_{x_{1}}+e^{-\alpha x_{1}}w_{x_{1}x_{1}} \\ v_{x_{i}} &=& e^{-\alpha x_{1}}w_{x_{i}}\qquad\quad (i \neq 1) \\ v_{x_{i}x_{i}} &=& e^{-\alpha x_{1}} w_{x_{i}x_{i}} \\ e^{\alpha x_{1}}\Delta v &=& \Delta w-2\alpha w_{x_{1}}+\alpha^{2}w. \end{eqnarray*} Let's make life easier for ourselves by treating only our special case $\Delta w+ c w = 0$. \begin{eqnarray*} \text{Let}\qquad\qquad L'v &=& \Delta v + 2 \alpha v_{x_{1}}. \\ \text{Then}\qquad e^{\alpha x_{1}} L' v &=& \Delta w - \alpha^{2}w \\ \Delta w + c w &=& e^{\alpha x_{1}}L' v + \left(c+\alpha^{2}\right)w. \end{eqnarray*} Now choose $\alpha$ large enough to make $c+\alpha^{2}$ always nonnegative. We know that $0 \le w$ so \begin{equation*} e^{\alpha x_{1}}L'v \le \Delta w + c w = 0. \end{equation*} Now we can apply the usual strong maximum (or rather minimum) principle for supersolution $v$ because $L'v \le 0$ with $L'$ containing no zeroth order term. I.e. if $v$ attains its minimum at an interior point then it is constant. And since $0 \le v$ we can say that if $v = 0$ at any interior point then $v \equiv 0$ throughout $\Omega$.

The same holds for $w$; if $w=0$ at any interior point then $w \equiv 0$ throughout $\Omega$. But we are told it is not the case that $w \equiv 0$ and can therefore conclude that $0 < w$ in $\Omega$.

For $\Delta u_{x_{1}} + f'(u) u_{x_{1}}=0$ the situation is only slightly complicated by the additional dependence of coefficient $c$ on $u$. Because $f$ is Lipschitz, $f'$ is bounded so we can still find some appropriate $\alpha$ to make the proof above carry through.