I want to show that the following PDE has at most one solution

\begin{equation} -\Delta u +c(u)=f \text{ in } \Omega \end{equation} \begin{equation} u=0 \text{ on } \partial\Omega \end{equation}

where $\Omega\subset\mathbb{R}^n$ is open and bounded with smooth boundary $\partial\Omega$, $f$ and $c$ are smooth, and $c$ is non-decreasing, i.e. it satisfies $c(a)\geq c(b)$ if $a\geq b$.

Usually I would assume $\omega=u_1-u_2$ for two solutions $u_1$ and $u_2$, and proceed by the energy method. But this time I don't even know if the maximum/minimum principles apply, and even if they did I don't know what this leaves me for the PDE in $\omega$.


1 Answer 1


You can prove this by either energy methods or by maximum principles, but these methods need to be adapted to handle the fact your PDE is no longer linear. Here is how you can do it from each perspective:

  1. Via Maximum Principle/Comparison Principle

The trick here is to apply the Mean Value Theorem to $c$. Let $u,v \in C^2(\Omega)\cap C^0(\overline{\Omega})$ be solutions of the PDE. By the Mean Value Theorem, $$ c(u) - c(v) =\tilde{c}(x)(u-v) $$ for some smooth, nonegative function $\tilde{c}$ (smoothness follows from smoothness of $c$, and nonegativity from $c$ nondecreasing). Hence, if $w := u-v$ then $$ - \Delta w +\tilde{c}(x)w=0 \text{ in } \Omega \\ w =0 \text{ on } \partial \Omega, $$ and so it follows by the weak maximum principle that $w = 0$ in $\Omega$ i.e $u= v$.

As an aside, you can use this technique (i.e. using the Mean Value Theorem) to prove comparison theorems for a large class of quasilinear PDE or even fully nonlinear PDE, see for example Theorem 10.1 in Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger.

  1. Via Energy Methods

Solutions to your PDE are precisely the minimisers of the functional $$ J[w] = \int_\Omega \frac 1 2 \vert Dw\vert^2 + F(w) d x$$ over the set of functions $\mathcal{A} = \{ w \in C^2(\Omega) \, \vert \, w =0 \text{ on } \partial \Omega \} $ (normally $w$ is in a Sobolev space, but for simplicity suppose it is $C^2$) where $$ F(z) := \int^z_0 c(t) d t. $$ Since $c$ is nondecreasing the map $z \mapsto F(z)$ is convex ($F''(z) = c'(z) \geqslant 0$) and $p \mapsto \vert p \vert^2 $ is also convex (this follows from Cauchy-Schwarz), so the functional $J$ is convex. Hence, if $u,v \in \mathcal{A}$ are minimisers of $J$ i.e. solve your PDE and $w = \frac 1 2 (u +v)$ then $$J[w] \leqslant \frac 1 2 J[u] + \frac 1 2 J[v]. $$ If $c$ is strictly increasing then $z \mapsto F(z)$ is strictly convex, so the above inequality is an equality if and only if $u=v$ (I believe this still holds even if $c$ is not strictly increasing). Since $u,v$ are minimisers $J[u]=J[v]\leqslant J[w]$, so $$J[w] \leqslant \frac 1 2 J[u] + \frac 1 2 J[v] \leqslant J[w]. $$ Hence, $$J[w] = \frac 1 2 J[u] + \frac 1 2 J[v], $$ so $u=v$. If you haven't seen this before it may seem a little like black magic, but it a standard trick. If you want to read further, see Chapter 8 of Partial Differential Equation by Lawrence Evans, particularly Theorem 3 of 8.2.2.

  • $\begingroup$ Really well explained answer (+1). $\endgroup$ Commented Apr 30, 2021 at 13:58

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