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MA40203: Theory of Partial Differential Equations

Miles Wheeler

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.
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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.
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Note that the basic inequality ∂u/∂n(p)≥0 follows immediately from Lemma 2.48; the content of the lemma is that this inequality is strict.
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Proof.

Changing variables (as in Exercise 3.1.2), we can assume without loss of generality that B=B1(0). Perhaps unsurprisingly at this point, our proof will involve two auxiliary functions. The first is
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z(x)=e−k|x|2−e−k,
where here k>0 is a (large) parameter to be determined. Note that 0<z<1 in B and z=0 on ∂B. We then consider
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v=u+εz
where ε>0 is a (small) parameter to be determined.
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Figure 3.1. The annular set A in the proof of Lemma 3.7.
Let M=u(p)=supBu. We will apply the weak maximum principle to v, not on the entire ball B, but on an annular subset
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A=B1(0)∖B1/2(0)―.
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Step 1.

First we choose ε>0 so that v≤M on the boundary ∂A=∂B1(0)∪∂B1/2(0). Since v(p)=u(p)=M, this will imply that max∂Av=v(p)=M.
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On ∂B1(0) we have u≤M and z=0, and so certainly v=u+εz≤M. On ∂B1/2(0) we have
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v=u+εz≤max∂B1/2(0)u+ε.
Since u<M on ∂B1/2(0) by hypothesis, we can ensure v≤M by picking
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ε=M−max∂B1/2(0)u>0.
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Step 2.

Next, we choose k such that Lz≥0 in A. Calculating
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∂je−k|x|2=−2ke−k|x|2xj,∂ije−k|x|2=e−k|x|2(4k2xixj−2kδij),
we find that
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ek|x|2Lz=4k2aijxixj−2kaii−2kbixi+c(1−e−k(1−|x|2)).
By uniform ellipticity, we have
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aijxixj≥λ0|x|2≥14λ0
on A―, and hence
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ek|x|2Lz≥k2λ0−2ksupA∑i=1N(|aii|+2|bi|)−supA|c|.
Choosing k>0 sufficiently large, the above estimate implies Lz≥0 and hence Lv≥0 in A. Note that it is crucial here that we are working on A and not the entire ball B.
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Step 3.

By Step 2, we have Lv=Lu+εLz≥0 in A, and by Step 1 we have max∂Av=v(p)=M. Applying Theorem 3.2, we conclude that maxA―v=v(p), and so Lemma 2.48 implies ∂v/∂n≥0 at p. Rearranging this inequality, we find
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∂u∂n(p)=∂v∂n(p)−ε∂z∂n(p)≥−ε∂z∂n(p)=2kεe−k>0
as desired.
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Proof.

Assume that M=supΩu is attained at some point in Ω, and that u≢M. We first claim that there exists a ball B with B―⊆Ω, u<M on B, and u(p)=M for some p∈∂B⊆Ω.
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Figure 3.2. The ball B in the proof of Theorem 3.8.
Let Ω− be the set of points in Ω where u<M, and let ΩM=Ω∖Ω− be the set of points in Ω where u=M. We will use connectedness to produce a point q∈∂Ω−∩ΩM. Note that by assumption both Ω− and ΩM are nonempty. Moreover, since u is continuous, Ω− is open. Since Ω is connected, this implies that ΩM is not open. Thus there exists a point q∈ΩM such that, for all r>0, Br(q)⊈ΩM. Taking r>0 small enough that Br(q)⊂Ω, this implies that Br(q)∩Ω−≠∅. Since q∈ΩM=Ω∖Ω−, we deduce that q∈∂Ω− and hence q∈∂Ω−∩ΩM as desired.
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Now let x0 be any point in Ω− which is closer to q than it is to ∂Ω. Then B=Br(x0) with r=dist(x0,∂Ω−) has the desired properties and the claim is proved.
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Applying Lemma 3.7 to u on B, we find that the outer normal derivative
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∂u∂n(p)=n⋅∇u(p)>0.
In particular, ∇u(p)≠0. But p is an interior point where u achieves its maximum, and so ∇u(p)=0 by Proposition 2.34, a contradiction.
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Using Theorem 3.8, we can now strengthen Lemma 3.7 and generalise it to a wider class of domains Ω. We will call this the ‘Hopf lemma’; some authors instead call it the ‘boundary point lemma’ or the ‘Hopf boundary point lemma’.
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Figure 3.3. The interior ball B in the Hopf lemma.
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Proof.

Let B be an interior ball at p, and note that u∈C2(B)∩C1(B―). If u(x)<u(p) for all x∈B then Lemma 3.7 implies ∂u/∂n(p)>0. If u(y)=u(p)=supΩu for some y∈B, the Strong maximum principle implies that u is constant.
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Note that neither Theorem 3.8 nor Theorem 3.9 require Ω to be bounded.
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Exercises Exercises

1. (PS5) Normal derivatives of radial functions.

Let B=Br(x0) be a ball, and suppose that u∈C1(B―) is given by
u(x)=U(|x−x0|)
for some function U∈C1(R). For any p∈∂B, show that
∂u∂n(p)=U′(r).
Hint.
Think back to the first part of Exercise 2.3.2.
Solution.
Writing u=U∘ρ where ρ(x)=|x−x0|, the chain rule gives (for x∈Br(x0)∖{x0})
∇u(x)=U′(ρ(x))∇ρ(x)=U′(|x−x0|)x−x0|x−x0|.
In particular, for p∈∂Br(x0) we have
∇u(p)=U′(r)n,
where n is the unit normal vector from Definition 2.47. Thus
∂u∂n(p)=n⋅∇u(p)=U′(r)|n|2=U′(r),
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)
Hint.
Consider the function −u, and use the fact that infΩu=−supΩ(−u).

3. A strong comparison principle.

Let Ω be bounded and connected and let L be uniformly elliptic with c≤0. If u,v∈C2(Ω)∩C0(Ω―) satisfy Lu≤Lv in Ω and u≥v on ∂Ω, show that u>v in Ω unless u≡v.
Hint.
First apply Proposition 3.4. Then, apply Theorem 3.8 to the difference w=v−u. If c≡0, then the argument is simpler; you may want to try that case first.
Solution.
Applying Proposition 3.4, we have immediately that u≥v in Ω. Thus the difference w=v−u≤0 in Ω, and if u=v at some point in Ω then w achieves maxΩ―w=0 at this point. But Lw=Lv−Lu≥0, and so by the Strong maximum principle this is only possible if w≡0, i.e. u≡v.

4. (PS5) Uniqueness for other boundary conditions.

Let Ω be bounded and connected with the interior ball property and let L be uniformly elliptic with c≤0. Suppose that u,v∈C2(Ω)∩C1(Ω―) satisfy
(3.4){Lu=Lv in Ω,∂u∂n=∂v∂n on ∂Ω.
(a)
Show that u and v differ by a constant.
Hint.
Consider the difference w=u−v, and apply the weak or strong maximum principle followed by Theorem 3.9. If c≡0, then the argument is simpler; you may want to try that case first.
Solution.
Consider the difference w=u−v. Since w is continuous and Ω is bounded, either maxΩ―w≥0 or minΩ―w≤0 or both. Replacing w with −w if necessary, we can assume that M=maxΩ―w≥0. Suppose that w is not constant. Then, since Lw=Lu−Lv=0 in Ω, Theorem 3.8 implies that w<M in Ω, and hence that w(p)=M for some p∈∂Ω. But then Theorem 3.9 implies that ∂w/∂n>0 at p, which is a contradiction.
(b)
If c≢0, conclude that u≡v.
Solution.
As w=u−v is constant, we have Lw=cw=0 in Ω. If c≢0, then the only way for this to happen is if w≡0.
Comment.
In this and other problems students sometimes claim, implicitly or explicitly, that c≢0, plus the overall assumption that c≤0, implies c<0. 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≡0 is shorthand for “c(x)=0 for all x∈Ω”. The negation of this, which we write as c≢0, is therefore “there exists x∈Ω such that c(x)≠0”.
  • c≤0 is by default interpreted “c≤0 in Ω”, which in turn is shorthand for “c(x)≤0 for all x∈Ω”. Similarly for c<0.
Thus c≢0 combined with c≤0 in Ω implies that there exists x∈Ω such that c(x)<0. It does not imply that c(x)<0 for all x∈Ω. For an explicit example, take for instance Ω=(−1,1)⊂R and c(x)=−x2. Then c≤0 in Ω and c≢0, but c(0)=0 and so it is not true that c<0 in Ω.

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 Ω is not connected.

6. (PS5) Maximum principles in the reverse order.

Suppose for simplicity that Ω is both bounded and connected. We proved our maximum principles for elliptic equations in roughly the following order:
  2. Weak maximum principle for c≤0 and maxΩ―u≥0
  3. Weak maximum principle for maxΩ―u=0
  5. Strong maximum principle for c≤0 and supΩu≥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 u~=u−supΩu satisfies supΩu~=0.
Solution.
Here and in what follows, let L be uniformly elliptic and u∈C2(Ω)∩C0(Ω―) satisfy Lu≥0 in Ω. Note that we never have to get our hands dirty by, e.g., introducing a complicated comparison function.
Assuming that c≡0 and that supΩu is achieved at some point x∈Ω, we need to show that u is constant. Consider the function u~=u−supΩu. Then u~ achieves supΩu~=0 at x. Moreover, since c≡0 we have Lu~=Lu≥0. Thus (6) implies that u~ is constant, and hence that u is constant as desired.
(b)
Use (6) to prove (5).
Solution.
Assuming now that c≤0 and that supΩu≥0 is achieved at some point x∈Ω, we need to show that u is constant. Again consider the function u~=u−supΩu. Then u~ achieves supΩu~=0 at x. Moreover, since c≤0 we have
Lu~=Lu−csupΩu≥Lu≥0.
Thus (6) implies that u~ is constant, and hence that u is constant as desired.
(c)
Use (4)–(6) to prove (1)–(3).
Solution.
Since Ω is bounded, we know that supΩ=maxΩ―u is achieved at some point x∗∈Ω―. If x∗∈∂Ω then
(✶)maxΩ―u=max∂Ωu
obviously holds, so it remains to consider the possibility that x∗∈Ω. 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∗∈∂Ω⊆Ω― we have max∂Ωu≤maxΩ―u=u(x∗)≤max∂Ωu.)
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