Section 5.3 A semilinear comparison principle
Because (5.1) is non-linear, we will want to have maximum principles on hand which apply to non-linear problems. Such maximum principles are much harder to come by than for linear equations, but one can often make progress by rewriting a non-linear inequality so that appears linear. A prototypical example, which we will need in the sequel, is the following.
Lemma 5.4.
Let \(\Omega\) be bounded and connected and \(f \in C^1(\R)\text{,}\) and suppose \(u,v \in C^2(\Omega)
\cap C^1(\overline \Omega)\) satisfy \(\Delta u + f(u) \le \Delta v + f(v)\) in \(\Omega\) and \(u \ge v\) in \(\Omega\text{.}\) Then
- If \(u=v\) at a point in \(\Omega\text{,}\) then \(u \equiv v\text{.}\)
- If \(u=v\) at a point \(p \in \partial \Omega\) where there is an interior ball, then either \(\partial u/\partial n \lt \partial v/\partial n\) at \(p\) or else \(u \equiv v\text{.}\)
Proof.
Letting \(w=v-u\) as usual, we find
\begin{equation}
\Delta w + f(v)-f(u) \ge 0\text{.}\tag{5.3}
\end{equation}
In order to write this as a linear differential inequality for \(w\) alone, we use the fundamental theorem of calculus. For any \(x \in \overline
\Omega\) we have
\begin{align*}
f(v(x))-f(u(x))
\amp = \int_0^1 \frac d{dt} f\big(tv(x)+(1-t)u(x)\big)\, dt\\
\amp = \int_0^1 f'\big(tv(x)+(1-t)u(x)\big)(v(x)-u(x))\, dt\\
\amp = \bigg(\int_0^1 f'\big(tv(x)+(1-t)u(x)\big)\, dt\bigg)(v(x)-u(x))\\
\amp = c(x)w(x)\text{,}
\end{align*}
where here we have defined
\begin{equation*}
c(x) = \int_0^1 f'\big(tv(x)+(1-t)u(x)\big)\, dt\text{.}
\end{equation*}
Thus (5.3) can be rewritten as
\begin{equation}
\Delta w + cw \ge 0\text{.}\tag{5.4}
\end{equation}
Moreover, the assumptions on \(u\text{,}\) \(v\text{,}\) and \(f\) guarantee that \(c \in
C^0(\overline\Omega)\text{.}\) In particular, \(c\) is bounded, which we can also check by directly estimating the integral.
While we know absolutely nothing about the sign of \(c\text{,}\) we do know that \(w = v-u \le 0\) on \(\Omega\text{.}\) Thus, if \(w=0\) at some point \(p \in
\overline\Omega\text{,}\) then we automatically have \(\sup_\Omega w = 0\text{.}\) The result now follows by applying (the third cases of) Theorem 3.8 and Theorem 3.9 to (5.4).
Exercises Exercises
1. (PS9) Necessity of an assumption in Lemma 5.4.
Give counterexamples showing that either part of Lemma 5.4 can fail if the assumption that \(u \ge v\) in \(\Omega\) is weakened to \(u
\ge v\) on \(\partial\Omega\text{.}\)
Hint 1.
Take \(N=1\) and think back to Exercise 1.1.1.
Hint 2.
Consider solutions of \(u''+u=v''+v=0\text{.}\)
Solution.
Take \(N=1\text{,}\) \(\Omega=(0,2\pi)\text{,}\) and \(f(s)=s\text{.}\) Then \(u(x) = \sin x\) and \(v \equiv 0\) satisfy
\begin{equation*}
\Delta u + f(u) = 0 = \Delta v + f(v)
\end{equation*}
in \(\Omega\text{,}\) and \(u=v\) on \(\partial\Omega\) as well as at \(0 \in \Omega\text{.}\) But clearly we do not have \(u \ge v\) in \(\Omega\text{,}\) let alone \(u \equiv v\text{.}\) Moreover, at \(2\pi \in \partial\Omega\) we have
\begin{equation*}
\frac{\partial u}{\partial n}(2\pi) = u'(2\pi) = 1 \gt 0 = \frac{\partial v}{\partial n}(2\pi)\text{.}
\end{equation*}
