570 CHAPTER 29. CALCULUS OF VECTOR FIELDS

Lemma 29.1.4 Suppose h and ∂h∂x are continuous on the rectangle R = [c,d]× [a,b].

Then 29.2 holds.

Proof: Let ∆x be such that x,x+∆x are both in [c,d]. By Theorem 15.12.4 on Page 323there exists δ > 0 such that if |(x, t)− (x1, t1)|< δ , then∣∣∣∣∂h

∂x(x, t)− ∂h

∂x(x1, t1)

∣∣∣∣< ε

b−a.

Let |∆x|< δ . Then ∣∣∣∣∫ b

a

h(x+∆x, t)−h(x, t)∆x

dt −∫ b

a

∂h∂x

(x, t) dt∣∣∣∣

≤∫ b

a

∣∣∣∣h(x+∆x, t)−h(x, t)∆x

− ∂h∂x

(x, t)∣∣∣∣dt

=∫ b

a

∣∣∣∣∂h(x+θ t∆x)∂x

− ∂h∂x

(x, t)∣∣∣∣dt <

∫ b

a

ε

b−adt = ε.

Here θ t is a number between 0 and 1 and going from the second to the third line is anapplication of the mean value theorem.

The second formula of Theorem 29.1.3 states ∇×∇φ = 0. This suggests the followingquestion: Suppose ∇×f = 0, does it follow there exists φ , a scalar field such that ∇φ = f?The answer to this is often yes and a theorem will be given and proved after the presentationof Stokes’ theorem. This scalar field φ , is called a scalar potential for f .

29.1.3 The Weak Maximum Principle

There is also a fundamental result having great significance which involves ∇2 called the

maximum principle. This principle says that if ∇2u ≥ 0 on a bounded open set U , then u

achieves its maximum value on the boundary of U .

Theorem 29.1.5 Let U be a bounded open set in Rp and suppose

u ∈C2 (U)∩C(U)

such that ∇2u ≥ 0 in U. Then letting ∂U =U \U, it follows that

max{

u(x) : x ∈U}= max{u(x) : x ∈ ∂U} .

Proof: If this is not so, there exists x0 ∈U such that

u(x0)> max{u(x) : x ∈ ∂U} ≡ M

Since U is bounded, there exists ε > 0 such that

u(x0)> max{

u(x)+ ε |x|2 : x ∈ ∂U}.

Therefore, u(x)+ ε |x|2 also has its maximum in U because for ε small enough,

u(x0)+ ε |x0|2 > u(x0)> max{

u(x)+ ε |x|2 : x ∈ ∂U}