Calculus of Real and Complex Variables

4.3 Conservative Vector Fields

Recall the gradient of a scalar function x → F

,x ∈ ℝ^p. It was the vector

( ) | Fx1. | |( .. |) Fx p

the transpose of the matrix of the derivative of F. (It is best not to worry too much about this distinction between the gradient and the derivative at this point.)

Proof: By Theorem 4.2.3 there exists η ∈ C¹

such that γ = η, and γ = η such that

||∫ ∫ || || f ⋅dγ − f ⋅dη|| < ε. γ η

Then from Theorem 4.2.5, since η is in C¹

, it follows from the chain rule and the fundamental theorem of calculus that

Therefore,

|| ∫ || ||(F (γ(b))− F (γ (a)))− f ⋅dγ|| < ε γ

and since ε > 0 is arbitrary, This proves the theorem. ■

You can prove this theorem another way without using that approximation result.

Proof: Define the following function

{ F-(γ(t))−F(γ(s|γ))(−t)f−(γγ((ss)))|⋅(γ(t)−γ(s))if γ (t)− γ (s) ⁄= 0 h(s,t) ≡ 0 if γ(t)− γ(s) = 0

Then h is continuous at points

. To see this note that the above reduces to

o(γ(t)− γ(s)) -|γ-(t)−-γ-(s)|-

thus if

→, the continuity of γ requires this to also converge to 0.

Claim: Let ε > 0 be given. There exists δ > 0 such that if

< δ, then < ε.

Proof of claim: If not, then for some ε > 0, there exists

where < 1∕n but ≥ ε. Then by compactness, there is a subsequence, still called such that s_n → s. It follows that t_n → s also. Therefore,

0 = lim |h(sn,tn)| ≥ ε n→∞

a contradiction.

Thus whenever

< δ,

F-(γ(t))-−-F (γ-(s))−-f (γ-(s))⋅(γ(t)−-γ(s))< ε |γ (t) − γ (s)|

and so

|F (γ(t))− F (γ (s))− f (γ (s))⋅(γ (t)− γ (s))| ≤ ε|γ(t) − γ(s)|

So let

< δ and also let δ be small enough that

|∫ | || n∑ || ||γ f⋅dγ − f (γ(ti−1))⋅(γ(ti) − γ (ti−1))|| < ε k=1

Therefore,

| | ||∫ ∑n || ||γ f⋅dγ − F (γ (ti))− F (γ (ti−1))|| ≤ k=1

||∫ ∑n || || f⋅dγ − f (γ (ti−1)) ⋅(γ (ti)− γ (ti−1))||+ | γ k=1 |

| | ||∑n || || f (γ (ti−1))⋅(γ (ti)− γ (ti−1)) − (F (γ(ti))− F (γ(ti− 1)))|| k=1

n∑ ≤ ε + ε|γ (ti)− γ (ti−1)| ≤ ε (1 + V (γ,[a,b])) k=1

It follows that

Therefore, since ε is arbitrary,

∫ f⋅dγ = F (γ (b))− F (γ(a)) ■ γ

Proof: The first part was proved in Theorem 4.3.1. It remains to verify the existence of a potential in the situation of path independence.

Let x₀ ∈ Ω be fixed. Let S be the points x of Ω which have the property there is a bounded variation curve joining x₀ to x. Let γ_x0x denote such a curve. Note first that S is nonempty. To see this, B

⊆ Ω for r small enough. Every x ∈ B is in S. Then S is open because if x ∈ S, then B ⊆ Ω for small enough r and if y ∈ B, you could go take γ_x0x and from x follow the straight line segment joining x to y. In addition to this, Ω ∖S must also be open because if x ∈ Ω ∖S, then choosing B ⊆ Ω, no point of B can be in S because then you could take the straight line segment from that point to x and conclude that x ∈ S after all. Therefore, since Ω is connected, it follows Ω ∖ S = ∅. Thus for every x ∈ S, there exists γ_x0x, a bounded variation curve from x₀ to x.

Define

∫ F (x) ≡ f ⋅dγx0x γx0x

F is well defined by assumption. Now let l_x

denote the linear segment from x to x + te_k. Thus to get to x + te_k you could first follow γ_x0x to x and from there follow lx

(x+tek)

to x + te_k. Hence

∫ F-(x+tek)−-F-(x)= 1 f ⋅dl t t lx(x+te ) x(x+tek) k

1∫ t = t 0 f (x + sek) ⋅ekds → fk (x)

by continuity of f. Thus ∇F = f. ■

Proof: Using Lemma 4.2.9, this condition about closed curves is equivalent to the condition that the line integrals of the above theorem are path independent. This proves the corollary. ■

Such a vector valued function is called conservative. Summarizing the above we have the following major theorem which is called the fundamental theorem of line integrals.