Strong Maximum Principle

2. Maximum Principles

2.1 Guide

In this chapter we discuss maximum principles and their applications. Two kinds of

maximum principles are discussed. One is due to Hopf and the other to Alexandro. The

former gives the estimates of solutions in terms of the L

∞

-norm of the nonhomogeneous

terms, while the latter gives the estimates in terms of the L

-norm. Applications include

various a priori estimates and the moving plane method.

Most of the statements in Section 2.2 are rather simple. One probably needs to go

over Theorem 2.11 and Proposition 2.13. Section 2.3 is often the starting point of the

a priori estimates. Section 2.4 can be omitted in the rst reading, as we will look at it

again in Section 5.2. The moving plane method explained in Section 2.6 has many recent

applications. We choose a very simple example to illustrate such a method. The result

goes back to Gidas-Ni-Nirenberg, but the proof contains some recent observations in the

paper [1]. The classical paper of Gilbarg-Serrin [7] may be a very good supplement to

this chapter. It may also be a good idea to assume the Harnack inequality of Krylov-

Safanov in Section 5.3 and to ask students to improve some of the results in the paper

[7].

2.2 Strong Maximum Principle

Suppose Ω is a bounded and connected domain in R

. Consider the operator L in

Ω,

Lu ≡ a

(x)D

u + b

(x)D

u + c(x)u

for u ∈ C

(Ω) ∩ C(Ω). We always assume that a

, b

, and c are continuous and hence

bounded in Ω and that L is uniformly elliptic in Ω in the following sense:

(x)ξ

≥ λ|ξ|

for any x ∈ Ω and any ξ ∈ R

for some positive constant λ.

Chapter 2: Maximum Principles

Lemma 2.1. Suppose u ∈ C

(Ω) ∩C(Ω) satises Lu > 0 in Ω with c(x) ≤ 0 in Ω. If u

has a nonnegative maximum in Ω, then u cannot attain this maximum in Ω.

Proof. Suppose u attains its nonnegative maximum of Ω in x

∈ Ω. Then D

u(x

) = 0

and the matrix B = (D

)) is seminegative denite. By the ellipticity condition the

matrix A = (a

)) is positive denite. Hence the matrix AB is seminegative denite

with a nonnegative trace, that is, a

u(x

) ≤ 0. This implies Lu(x

) ≤ 0, which

is a contradiction.

Remark 2.2. If c(x) ≡ 0, then the requirement for nonnegativeness can be removed.

This remark also holds for some results in the rest of this section.

Theorem 2.3 (Weak Maximum Principle). Suppose u ∈ C

(Ω)∩C(Ω) satises Lu ≥ 0

in Ω with c(x) ≤ 0 in Ω. Then u attains on ∂Ω its nonnegative maximum in Ω.

Proof. For any ε > 0, consider w(x) = u(x) + εe

αx

with α to be determined. Then we

have

Lw = Lu + εe

αx

+ b

α + c).

Since b

and c are bounded and a

(x) ≥ λ > 0 for any x ∈ Ω, by choosing α > 0 large

enough we get

(x)α

+ b

(x)α + c( x) > 0 for any x ∈ Ω.

This implies Lw > 0 in Ω. By Lemma 2.1, w attains its nonnegative maximum only on

∂Ω, that is,

sup

Ω

w ≤ sup

∂Ω

Then we obtain

sup

Ω

u ≤ sup

Ω

w ≤ sup

∂Ω

≤ sup

∂Ω

+ ε sup

x∈∂Ω

αx

We nish the proof by letting ε → 0.

As an application we have the uniqueness of solution u ∈ C

(Ω) ∩ C(Ω) to the

following Dirichlet boundary value problem for f ∈ C(Ω) and φ ∈ C(∂Ω)







Lu = f in Ω,

u = φ on ∂Ω,

if c(x) ≤ 0 in Ω.

Remark 2.4. The boundedness of domain Ω is essential, since it guarantees the exis-

tence of a maximum and a minimum of u in Ω. The uniqueness does not hold if the

domain is unbounded. Some examples are given in Remark 1.9. Equally important is

the nonpositiveness of the coecient c.

Section 2.2: Strong Maximum Principle

Example 1. Set Ω = {(x, y) ∈ R

: 0 < x < π, 0 < y < π}. Then u = sin x sin y is a

nontrivial solution for the problem







∆u + 2u = 0 in Ω,

u = 0 on ∂Ω.

Theorem 2.5 (Hopf Lemma). Let B be an open ball in R

with x

∈ ∂B. Suppose

u ∈ C

(B) ∩C(B ∪{x

}) satises Lu ≥ 0 in B with c(x) ≤ 0 in B. Assume in addition

that

u(x) < u(x

) for any x ∈ B and u(x

) ≥ 0.

Then for each outward direction ν at x

with ν · n(x

) > 0 there holds

lim inf

t→0

[u(x

) − u(x

− tν)] > 0.

Remark 2.6. If in addition u ∈ C

(B ∪ {x

}), then we have

∂u

∂ν

) > 0.

Proof. We may assume that the center of B is at the origin with radius r. We assume

further that u ∈ C(B) and u(x) < u(x

) for any x ∈ B \{x

} (since we can construct a

tangent ball B

to B at x

and B

⊂ B).

Consider v(x) = u(x) + εh(x) for some nonnegative function h. We will choose

ε > 0 appropriately such that v attains its nonnegative maximum only at x

. Denote

Σ = B ∩ B

1/2r

). Dene h(x) = e

−α|x|

− e

−αr

with α to be determined. We check

in the following that

Lh > 0 in Σ.

Direct calculation yields

Lh = e

−α|x|







4α

i,j=1

(x)x

− 2α

i=1

(x) − 2α

i=1

(x)x

+ c







− ce

−αr

≥ e

−α|x|







4α

i,j=1

(x)x

− 2α

i=1

(x) + b

(x)x

] + c







By the ellipticity assumption, we have

i,j=1

(x)x

≥ λ|x|

≥ λ





So for α large enough, we conclude Lh > 0 in Σ. With such h, we have Lv = Lu+εLh > 0

Chapter 2: Maximum Principles

in Σ for any ε > 0. By Lemma 2.1, v cannot attain its nonnegative maximum inside Σ.

Next we prove that for some small ε > 0 v attains at x

its nonnegative maximum.

Consider v on the boundary ∂Σ.

• For x ∈ ∂Σ ∩ B, since u(x) < u( x

), we have u(x) < u(x

) − δ for some δ > 0.

Take ε small such that εh < δ on ∂Σ ∩ B. Hence for such ε we have v(x) < u(x

)

for x ∈ ∂Σ ∩B.

• On Σ ∩ ∂B, h(x) = 0 and u(x) < u(x

) for x 6= x

. Hence v(x) < u(x

) on

Σ ∩ ∂B \ {x

} and v(x

) = u(x

Therefore we conclude

v(x

) − v(x

− tν)

≥ 0 for any small t > 0.

Hence we obtain by letting t → 0

lim inf

t→0

[u(x

) − u(x

− tν)] ≥ −ε

∂h

∂ν

By denition of h, we have

∂h

∂ν

) < 0.

This nishes the proof.

Theorem 2.7 (Strong Maximum Principle). Let u ∈ C

(Ω)∩C(Ω) satisfy Lu ≥ 0 with

c(x) ≤ 0 in Ω. Then the nonnegative maximum of u in Ω can be assumed only on ∂Ω

unless u is a constant.

Proof. Let M be the nonnegative maximum of u in Ω. Set Σ = {x ∈ Ω : u(x) = M }. It

is relatively closed in Ω. We need to show Σ = Ω.

We prove by contradiction. If Σ is a proper subset of Ω , then we may nd an open

ball B ⊂ Ω \ Σ with a point on its boundary belonging to Σ. (In fact, we may choose a

point p ∈ Ω \ Σ such that d(p, Σ) < d(p, ∂Ω) rst and then extend the ball centered at

p. It hits Σ before hitting ∂Ω.) Suppose x

∈ ∂B ∩ Σ. Obviously we have Lu ≥ 0 in B

and

u(x) < u(x

) for any x ∈ B and u(x

) = M ≥ 0.

Theorem 2.5 implies

∂u

∂n

) > 0 where n is the outward normal direction at x

to the

ball B. Since x

is the interior maximal point of Ω, Du(x

) = 0. This leads to a

contradiction.

Corollary 2.8 (Comparison Principle). Suppose u ∈ C

(Ω) ∩C(Ω) satises Lu ≥ 0 in

Ω with c(x) ≤ 0 in Ω. If u ≤ 0 on ∂Ω, then u ≤ 0 in Ω. In fact, either u < 0 in Ω or

u ≡ 0 in Ω.

Section 2.2: Strong Maximum Principle

In order to discuss the boundary value problem with general boundary condition, we

need the following result, which is just a corollary of Theorems 2.5 and 2.7.

Corollary 2.9. Suppose Ω has the interior sphere property and that u ∈ C

(Ω)∩C

(Ω)

satises Lu ≥ 0 in Ω with c(x) ≤ 0. Assume u attains its nonnegative maximum at

∈ Ω. Then x

∈ ∂Ω and for any outward direction ν at x

to ∂Ω

∂u

∂ν

) > 0

unless u is constant in Ω.

Application. Suppose Ω is bounded in R

and satises the interior sphere property.

Consider the following boundary value problem







Lu = f in Ω

∂u

∂n

+ α(x)u = φ on ∂Ω

(*)

for some f ∈ C(Ω) and φ ∈ C(∂Ω). Assume in addition that c(x) ≤ 0 in Ω and α(x) ≥ 0

on ∂Ω. Then problem (*) has a unique solution u ∈ C

(Ω) ∩C

(Ω) if c 6≡ 0 or α 6≡ 0. If

c ≡ 0 and α ≡ 0, problem (*) has a unique solution u ∈ C

(Ω) ∩ C

(Ω) up to additive

constants.

Proof. Suppose u is a solution to the following homogeneous equation:







Lu = 0 in Ω,

∂u

∂n

+ α(x)u = 0 on ∂Ω.

CASE 1. c 6≡ 0 or α 6≡ 0. We want to show u ≡ 0.

Suppose that u has a positive maximum at x

∈ Ω. If u ≡ const > 0, this contradicts

the condition c 6≡ 0 in Ω or α 6≡ 0 on ∂Ω. Otherwise x

∈ ∂Ω and

∂u

∂n

) > 0 by

Corollary 2.9, which contradicts the boundary value. Therefore u ≡ 0.

CASE 2. c ≡ 0 and α ≡ 0. We want to show u ≡ const.

Suppose u is a nonconstant solution. Then its maximum in Ω is assumed only on

∂Ω by Theorem 2.7, say at x

∈ ∂Ω. Again Corollary 2.9 implies

∂u

∂n

) > 0. This is a

contradiction.

The following theorem, due to Serrin, generalizes the comparison principle with no

restriction on c(x).

Theorem 2.10. Suppose u ∈ C

(Ω) ∩C(Ω) satises Lu ≥ 0. If u ≤ 0 in Ω, then either

u < 0 in Ω or u ≡ 0 in Ω.

Chapter 2: Maximum Principles

Proof. We present two methods.

METHOD 1. Suppose u(x

) = 0 for some x

∈ Ω. We will prove that u ≡ 0 in Ω.

Write c(x) = c

(x)−c

−

(x) where c

(x) and c

−

(x) are the positive part and negative

part of c(x), respectively. Hence u satises

u + b

u − c

−

u ≥ −c

u ≥ 0.

So we have u ≡ 0 by Theorem 2.7.

METHOD 2. Set v = ue

−αx

for some α > 0 to be determined. By Lu ≥ 0, we

have

v + [α(a

+ a

) + b

v + (a

+ b

α + c)v ≥ 0.

Choose α large enough such that a

+ b

α + c > 0. Therefore v satises

v + [α(a

+ a

) + b

v ≥ 0.

Hence we apply Theorem 2.7 to v to conclude that either v < 0 in Ω or v ≡ 0 in Ω.

The next result is the general maximum principle for the operator L with no restric-

tion on c(x).

Theorem 2.11. Suppose there exists a w ∈ C

(Ω) ∩ C

(Ω) satisfying w > 0 in Ω and

Lw ≤ 0 in Ω. If u ∈ C

(Ω) ∩ C(Ω) satises Lu ≥ 0 in Ω, then

cannot assume in

Ω its nonnegative maximum unless

≡ const. If, in addition, assumes its nonnegative

maximum at x

∈ ∂Ω and

6≡ const, then for any outward direction ν at x

to ∂Ω

there holds

∂

∂ν





) > 0

if ∂Ω has the interior sphere property at x

Proof. Set v =

. Then v satises

v + B

v +





v ≥ 0

where B

= b

w. We may apply Theorem 2.7 and Corollary 2.9 to v.

Remark 2.12. If the operator L in Ω satises the condition of Theorem 2.11, then the

comparison principle applies to L. In particular, the Dirichlet boundary value problem







Lu = f in Ω,

u = φ on ∂Ω,

has at most one solution.

Section 2.2: Strong Maximum Principle

The next result is the so-called maximum principle for a narrow domain.

Proposition 2.13. Let d be a positive number and ϵ be a unit vector such that |(y −

x) · ϵ| < d for any x, y ∈ Ω. Then there exists a d

> 0, depending only on λ and

the sup-norm of b

and c

, such that the assumptions of Theorem 2.11 are satised if

d ≤ d

Proof. By choosing ϵ = (1, 0, . . . , 0) we suppose Ω lies in {0 < x

< d}. Assume in

addition |b

|, c

≤ N for some positive constant N. We construct w as follows. Set

w = e

αd

− e

αx

> 0 in Ω. By direct calculation we have

Lw = −(a

+ b

α)e

αx

+ c(e

αd

− e

αx

) ≤ −(a

+ b

α) + N e

αd

Choose α so large that

+ b

α ≥ λα

− N α ≥ 2N.

Hence Lw ≤ −2N + Ne

αd

= N (e

αd

− 2) ≤ 0 if d is small such that e

αd

≤ 2.

Remark 2.14. Some results in this section, including Proposition 2.13, hold for un-

bounded domain. Compare Proposition 2.13 with Theorem 2.32.