Section 3.3 The Hopf lemma and the strong maximum principle
The results that we have proved so far in this chapter are weaker than Theorem 1.4 in that they do not rule out the possibility that \(u\) achieves its maximum value at an interior point. In many applications of the theory this distinction is not particularly important, but in others, including in Chapter 5, it will be essential to have ‘strong’ maximum principles which rule out interior maxima.
Since we already have weak maximum principles at our disposal, our approach will be to first investigate the boundary behaviour of solutions to elliptic inequalities. The basic lemma in this direction is as follows.
Lemma 3.7. Hopf lemma for balls.
Let \(\Omega=B\) be a ball, and suppose that \(L\) is uniformly elliptic, \(u \in C^2(B) \cap C^1(\overline B)\) satisfies \(Lu \ge 0\) in \(B\text{,}\) and any of the conditions
- \(\displaystyle c \equiv 0\)
- \(c \le 0\) and \(\displaystyle\max_{\overline B} u \ge 0\)
- \(\displaystyle \max_{\overline B} u = 0\)
hold. If there is a point \(p \in \partial B\) such that \(u(x) \lt
u(p)\) for all \(x \in B\text{,}\) then the normal derivative
\begin{equation*}
\frac{\partial u}{\partial n}(p) \gt 0\text{.}
\end{equation*}
Note that the basic inequality \(\partial u/\partial n(p) \ge 0\) follows immediately from Lemma 2.48; the content of the lemma is that this inequality is strict.
Proof.
Changing variables (as in Exercise 3.1.2), we can assume without loss of generality that \(B = B_1(0)\text{.}\) Perhaps unsurprisingly at this point, our proof will involve two auxiliary functions. The first is
\begin{equation*}
z(x) = e^{-k\abs x^2} - e^{-k}\text{,}
\end{equation*}
where here \(k \gt 0\) is a (large) parameter to be determined. Note that \(0 \lt z \lt 1\) in \(B\) and \(z = 0\) on \(\partial B\text{.}\) We then consider
\begin{equation*}
v = u + \varepsilon z
\end{equation*}
where \(\varepsilon \gt 0\) is a (small) parameter to be determined.
Let \(M = u(p) = \sup_B u\text{.}\) We will apply the weak maximum principle to \(v\text{,}\) not on the entire ball \(B\text{,}\) but on an annular subset
\begin{equation*}
A = B_1(0) \without \overline{B_{1/2}(0)}\text{.}
\end{equation*}
Step 1.
First we choose \(\varepsilon \gt 0\) so that \(v \le M\) on the boundary \(\partial A = \partial
B_1(0) \cup \partial B_{1/2}(0)\text{.}\) Since \(v(p)=u(p)=M\text{,}\) this will imply that \(\max_{\partial A} v = v(p) = M\text{.}\)
On \(\partial B_1(0)\) we have \(u \le M\) and \(z = 0\text{,}\) and so certainly \(v = u+\varepsilon z \le M\text{.}\) On \(\partial B_{1/2}(0)\) we have
\begin{equation*}
v = u+\varepsilon z \le \max_{\partial B_{1/2}(0)} u + \varepsilon\text{.}
\end{equation*}
Since \(u \lt M\) on \(\partial B_{1/2}(0)\) by hypothesis, we can ensure \(v \le
M\) by picking
\begin{gather*}
\varepsilon = M - \max_{\partial B_{1/2}(0)} u \gt 0.
\end{gather*}
Step 2.
Next, we choose \(k\) such that \(Lz \ge 0\) in \(A\text{.}\) Calculating
\begin{align*}
\partial_j e^{-k\abs x^2} \amp= -2ke^{-k\abs x^2}x_j,\\
\partial_{ij} e^{-k\abs x^2} \amp= e^{-k\abs x^2}
\big( 4k^2 x_i x_j - 2k\delta_{ij} \big),
\end{align*}
we find that
\begin{align*}
e^{k\abs x^2} Lz \amp= 4k^2
a_{ij}x_i x_j
-2k
a_{ii}- 2k b_{i} x_i+ c( 1- e^{-k(1-\abs x^2)} ).
\end{align*}
By uniform ellipticity, we have
\begin{gather*}
a_{ij} x_i x_j \ge \lambda_0 \abs x^2 \ge \tfrac 14 \lambda_0
\end{gather*}
on \(\overline A\text{,}\) and hence
\begin{gather*}
e^{k\abs x^2} Lz \ge k^2\lambda_0-2k
\sup_A \sum_{i=1}^N(|{a_{ii}}| + 2|b_i|) - \sup_A |c|.
\end{gather*}
Choosing \(k \gt 0\) sufficiently large, the above estimate implies \(Lz
\ge 0\) and hence \(Lv \ge 0\) in \(A\text{.}\) Note that it is crucial here that we are working on \(A\) and not the entire ball \(B\text{.}\)
Step 3.
By Step 2, we have \(Lv = Lu + \varepsilon Lz
\ge 0\) in \(A\text{,}\) and by Step 1 we have \(\max_{\partial A} v = v(p) = M\text{.}\) Applying Theorem 3.2, we conclude that \(\max_{\overline A} v = v(p)\text{,}\) and so Lemma 2.48 implies \(\partial v/\partial n \ge 0\) at \(p\text{.}\) Rearranging this inequality, we find
\begin{align*}
\frac{\partial u}{\partial n}(p)
\amp =
\frac{\partial v}{\partial n}(p) - \varepsilon \frac{\partial z}{\partial n}(p)\\
\amp \ge
- \varepsilon \frac{\partial z}{\partial n}(p)\\
\amp = 2k\varepsilon e^{-k} \gt 0
\end{align*}
as desired.
Theorem 3.8. Strong maximum principle.
Suppose \(\Omega\) is connected, \(L\) is uniformly elliptic, \(u \in C^2(\Omega)
\cap C^0(\overline \Omega)\) satisfies \(Lu \ge 0\text{,}\) and any of the conditions
- \(\displaystyle c \equiv 0\)
- \(c \le 0\) and \(\displaystyle\sup_\Omega u \ge 0\)
- \(\displaystyle \sup_\Omega u = 0\)
hold. If \(\sup_\Omega u\) is attained at a point in \(\Omega\text{,}\) then \(u\) is constant.
Proof.
Assume that \(M = \sup_\Omega u\) is attained at some point in \(\Omega\text{,}\) and that \(u \not \equiv M\text{.}\) We first claim that there exists a ball \(B\) with \(\overline B \subseteq \Omega\text{,}\) \(u \lt M\) on \(B\text{,}\) and \(u(p) = M\) for some \(p \in \partial B \subseteq \Omega\text{.}\)
Let \(\Omega^-\) be the set of points in \(\Omega\) where \(u \lt M\text{,}\) and let \(\Omega^M = \Omega \without \Omega^-\) be the set of points in \(\Omega\) where \(u =
M\text{.}\) We will use connectedness to produce a point \(q \in \partial\Omega^- \cap
\Omega^M\text{.}\) Note that by assumption both \(\Omega^-\) and \(\Omega^M\) are nonempty. Moreover, since \(u\) is continuous, \(\Omega^-\) is open. Since \(\Omega\) is connected, this implies that \(\Omega^M\) is not open. Thus there exists a point \(q \in \Omega^M\) such that, for all \(r \gt 0\text{,}\) \(B_r(q) \not \subseteq \Omega^M\text{.}\) Taking \(r \gt 0\) small enough that \(B_r(q) \subset
\Omega\text{,}\) this implies that \(B_r(q) \cap \Omega^- \ne \varnothing\text{.}\) Since \(q \in \Omega^M = \Omega \without
\Omega^-\text{,}\) we deduce that \(q \in \partial \Omega^-\) and hence \(q \in \partial\Omega^- \cap \Omega^M\) as desired.
Now let \(x_0\) be any point in \(\Omega^-\) which is closer to \(q\) than it is to \(\partial\Omega\text{.}\) Then \(B=B_r(x_0)\) with \(r = \dist(x_0,\partial\Omega^-)\) has the desired properties and the claim is proved.
\begin{equation*}
\frac{\partial u}{\partial n} (p) = n \cdot \nabla u (p) \gt 0 \text{.}
\end{equation*}
In particular, \(\nabla u(p) \ne 0\text{.}\) But \(p\) is an interior point where \(u\) achieves its maximum, and so \(\nabla u(p) = 0\) by Proposition 2.34, a contradiction.
Using Theorem 3.8, we can now strengthen Lemma 3.7 and generalise it to a wider class of domains \(\Omega\text{.}\) We will call this the ‘Hopf lemma’; some authors instead call it the ‘boundary point lemma’ or the ‘Hopf boundary point lemma’.
Theorem 3.9. Hopf lemma.
Suppose that \(\Omega\) is connected, \(L\) is uniformly elliptic, and \(u \in C^2(\Omega) \cap C^1(\overline\Omega)\) satisfies \(Lu\ge 0\) in \(\Omega\text{,}\) and one of the conditions
- \(\displaystyle c \equiv 0\)
- \(c \le 0\) and \(\displaystyle\sup_\Omega u \ge 0\)
- \(\displaystyle \sup_\Omega u = 0\)
holds. If \(\sup_\Omega u = u(p)\) for some \(p \in \partial\Omega\) where \(\Omega\) satisfies the interior ball property, then either
- \(\partial u/\partial n \gt 0\) at \(p\text{;}\) or else
- \(u\) is constant,
where here \(n\) is the outward unit normal to an interior ball \(B\) at \(p\text{.}\)
Proof.
Let \(B\) be an interior ball at \(p\text{,}\) and note that \(u \in C^2(B) \cap
C^1(\overline B)\text{.}\) If \(u(x) \lt u(p)\) for all \(x \in B\) then Lemma 3.7 implies \(\partial u/\partial n(p) \gt 0\text{.}\) If \(u(y) = u(p) =
\sup_\Omega u\) for some \(y \in B\text{,}\) the Strong maximum principle implies that \(u\) is constant.
Exercises Exercises
1. (PS5) Normal derivatives of radial functions.
Let \(B=B_r(x_0)\) be a ball, and suppose that \(u \in C^1(\overline
B)\) is given by
\begin{equation*}
u(x) = U(\abs{x-x_0})
\end{equation*}
for some function \(U \in C^1(\R)\text{.}\) For any \(p \in \partial B\text{,}\) show that
\begin{equation*}
\frac{\partial u}{\partial n}(p) = U'(r) \text{.}
\end{equation*}
Hint.
Think back to the first part of Exercise 2.3.2.
Solution.
Writing \(u = U \circ \rho\) where \(\rho(x)=\abs{x-x_0}\text{,}\) the chain rule gives (for \(x \in B_r(x_0) \without \{x_0\}\))
\begin{equation*}
\nabla u(x) = U'(\rho(x)) \nabla \rho(x)
= U'(\abs{x-x_0}) \frac{x-x_0}{\abs{x-x_0}}.
\end{equation*}
In particular, for \(p \in \partial B_r(x_0)\) we have
\begin{equation*}
\nabla u(p) = U'(r) n,
\end{equation*}
where \(n\) is the unit normal vector from Definition 2.47. Thus
\begin{equation*}
\frac{\partial u}{\partial n}(p)
= n \cdot \nabla u(p)
= U'(r) \abs n^2
= U'(r),
\end{equation*}
where in the last step we have used that \(n\) is a unit vector.
2. Minimum principles.
State and prove minimum principle analogues of
(a)
Theorem 3.8 and
Hint.
Consider the function \(-u\text{,}\) and use the fact that \(\inf_\Omega u =
-\sup_\Omega(-u)\text{.}\)
(b)
Hint.
Same as for the previous part.
3. A strong comparison principle.
Let \(\Omega\) be bounded and connected and let \(L\) be uniformly elliptic with \(c \le 0\text{.}\) If \(u,v \in C^2(\Omega) \cap C^0(\overline \Omega)\) satisfy \(Lu \le
Lv\) in \(\Omega\) and \(u \ge v\) on \(\partial \Omega\text{,}\) show that \(u \gt v\) in \(\Omega\) unless \(u \equiv v\text{.}\)
Hint.
First apply Proposition 3.4. Then, apply Theorem 3.8 to the difference \(w = v-u\text{.}\) If \(c \equiv 0\text{,}\) then the argument is simpler; you may want to try that case first.
Solution.
Applying Proposition 3.4, we have immediately that \(u \ge v\) in \(\Omega\text{.}\) Thus the difference \(w = v-u \le 0\) in \(\Omega\text{,}\) and if \(u=v\) at some point in \(\Omega\) then \(w\) achieves \(\max_{\overline\Omega} w = 0\) at this point. But \(Lw = Lv-Lu \ge 0\text{,}\) and so by the Strong maximum principle this is only possible if \(w
\equiv 0\text{,}\) i.e. \(u \equiv v\text{.}\)
4. (PS5) Uniqueness for other boundary conditions.
Let \(\Omega\) be bounded and connected with the interior ball property and let \(L\) be uniformly elliptic with \(c \le 0\text{.}\) Suppose that \(u,v \in
C^2(\Omega) \cap C^1(\overline \Omega)\) satisfy
\begin{equation}
\left\{
\begin{alignedat}{2}
L u \amp= Lv \amp\quad\amp \ina \Omega, \\
\frac{\partial u}{\partial n} \amp= \frac{\partial v}{\partial n} \amp\quad\amp \ona \partial \Omega.
\end{alignedat}
\right.\tag{3.4}
\end{equation}
(a)
Show that \(u\) and \(v\) differ by a constant.
Hint.
Consider the difference \(w=u-v\text{,}\) and apply the weak or strong maximum principle followed by Theorem 3.9. If \(c \equiv 0\text{,}\) then the argument is simpler; you may want to try that case first.
Solution.
Consider the difference \(w=u-v\text{.}\) Since \(w\) is continuous and \(\Omega\) is bounded, either \(\max_{\overline\Omega} w \ge 0\) or \(\min_{\overline\Omega} w \le 0\) or both. Replacing \(w\) with \(-w\) if necessary, we can assume that \(M =
\max_{\overline\Omega} w \ge 0\text{.}\) Suppose that \(w\) is not constant. Then, since \(Lw = Lu-Lv = 0\) in \(\Omega\text{,}\) Theorem 3.8 implies that \(w \lt M\) in \(\Omega\text{,}\) and hence that \(w(p)=M\) for some \(p
\in \partial \Omega\text{.}\) But then Theorem 3.9 implies that \(\partial w/\partial n \gt 0\) at \(p\text{,}\) which is a contradiction.
(b)
If \(c \not \equiv 0\text{,}\) conclude that \(u \equiv v\text{.}\)
Solution.
As \(w=u-v\) is constant, we have \(Lw = cw = 0\) in \(\Omega\text{.}\) If \(c \not \equiv 0\text{,}\) then the only way for this to happen is if \(w \equiv
0\text{.}\)
Comment.
In this and other problems students sometimes claim, implicitly or explicitly, that \(c \not \equiv 0\text{,}\) plus the overall assumption that \(c
\le 0\text{,}\) implies \(c \lt 0\text{.}\) While it ends up being a very minor detail in this problem, it is important to understand that this is not true. To understand why, let’s unpack what the inequalities for the coefficient function \(c\) mean, precisely, in the context of this unit:
- \(c \equiv 0\) is shorthand for “\(c(x) = 0\) for all \(x \in \Omega\)”. The negation of this, which we write as \(c \not \equiv 0\text{,}\) is therefore “there exists \(x \in \Omega\) such that \(c(x) \ne 0\)”.
- \(c \le 0\) is by default interpreted “\(c \le 0\) in \(\Omega\)”, which in turn is shorthand for “\(c(x) \le 0\) for all \(x \in \Omega\)”. Similarly for \(c \lt 0\text{.}\)
Thus \(c \not \equiv 0\) combined with \(c \le 0\) in \(\Omega\) implies that there exists \(x \in \Omega\) such that \(c(x) \lt 0\text{.}\) It does not imply that \(c(x) \lt 0\) for all \(x \in
\Omega\text{.}\) For an explicit example, take for instance \(\Omega = (-1,1) \subset \R\) and \(c(x)=-x^2\text{.}\) Then \(c \le 0\) in \(\Omega\) and \(c \not \equiv 0\text{,}\) but \(c(0)=0\) and so it is not true that \(c \lt 0\) in \(\Omega\text{.}\)
5. Connectedness and the strong maximum principle.
Give a counterexample showing that the conclusion of Theorem 3.8 can fail (albeit in a rather uninteresting way) if \(\Omega\) is not connected.
6. (PS5) Maximum principles in the reverse order.
Suppose for simplicity that \(\Omega\) is both bounded and connected. We proved our maximum principles for elliptic equations in roughly the following order:
- Weak maximum principle for \(c \equiv 0\)
- Weak maximum principle for \(\max_{\overline \Omega} u = 0\)
- Strong maximum principle for \(c \equiv 0\)
- Strong maximum principle for \(\sup_\Omega u = 0\)
It is interesting to note that one can also work in the reverse order, in the sense that (6) implies (1)–(5).
(a)
Use (6) to prove (4).
Hint.
The function \(\tilde u = u - \sup_\Omega u\) satisfies \(\sup_\Omega \tilde
u = 0\text{.}\)
Solution.
Here and in what follows, let \(L\) be uniformly elliptic and \(u \in
C^2(\Omega) \cap C^0(\overline \Omega)\) satisfy \(L u \ge 0\) in \(\Omega\text{.}\) Note that we never have to get our hands dirty by, e.g., introducing a complicated comparison function.
Assuming that \(c \equiv 0\) and that \(\sup_\Omega u\) is achieved at some point \(x \in \Omega\text{,}\) we need to show that \(u\) is constant. Consider the function \(\tilde u = u - \sup_\Omega u\text{.}\) Then \(\tilde u\) achieves \(\sup_\Omega \tilde u = 0\) at \(x\text{.}\) Moreover, since \(c \equiv 0\) we have \(L\tilde u = Lu \ge 0\text{.}\) Thus (6) implies that \(\tilde u\) is constant, and hence that \(u\) is constant as desired.
(b)
Use (6) to prove (5).
Solution.
Assuming now that \(c \le 0\) and that \(\sup_\Omega u \ge 0\) is achieved at some point \(x \in \Omega\text{,}\) we need to show that \(u\) is constant. Again consider the function \(\tilde u = u - \sup_\Omega u\text{.}\) Then \(\tilde u\) achieves \(\sup_\Omega \tilde u = 0\) at \(x\text{.}\) Moreover, since \(c \le
0\) we have
\begin{gather*}
L\tilde u = Lu - c \sup_\Omega u \ge Lu \ge 0.
\end{gather*}
Thus (6) implies that \(\tilde u\) is constant, and hence that \(u\) is constant as desired.
(c)
Use (4)–(6) to prove (1)–(3).
Solution.
Since \(\Omega\) is bounded, we know that \(\sup_\Omega = \max_{\overline
\Omega} u\) is achieved at some point \(x^* \in \overline \Omega\text{.}\) If \(x^* \in \partial\Omega\) then
\begin{gather}
\max_{\overline \Omega} u = \max_{\partial\Omega} u\tag{✶}
\end{gather}
obviously holds, so it remains to consider the possibility that \(x^* \in
\Omega\text{.}\) Whichever case (1)–(3) of the weak maximum principle we are are in, the parallel case (4)–(6) of the strong maximum principle now implies that \(u\) is constant, and so (✶) certainly holds. (Indeed, since \(x^* \in \partial\Omega \subseteq \overline\Omega\) we have \(\max_{\partial\Omega} u \le \max_{\overline \Omega} u = u(x^*) \le \max_{\partial\Omega} u\text{.}\))
