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.)
Theorem 4.3.1Letγ :
[a,b]
→ ℝpbe continuous and of bounded variation. Also suppose ∇F = f on Ω,an open set containingγ∗and f is continuous on Ω. Then
||∫ ||
|| f⋅dγ − (F (γ (b))− F (γ(a)))||
| γ |
||∫ ∑n ||
= || f⋅dγ − F (γ(ti))− F (γ(ti−1))|| ≤ ε(1+ V (γ))
γ k=1
Therefore, since ε is arbitrary,
∫
f⋅dγ = F (γ (b))− F (γ(a)) ■
γ
Corollary 4.3.3Ifγ :
[a,b]
→ ℝpis continuous, has bounded variation, is a closed curve,γ
(a)
= γ
(b)
,andγ∗⊆ Ω where Ω is an open set on which ∇F = f, then
∫
f ⋅dγ = 0.
γ
Theorem 4.3.4Let Ω be a connected open set and let f : Ω → ℝpbe continuous. Then f has a potential Fif and only if
∫
f ⋅dγ
γ
is path independent for allγa bounded variation curve such thatγ∗is contained in Ω. This means theabove line integral depends only onγ
(a)
andγ
(b)
.
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 γx0x denote such a curve. Note first that S is nonempty. To see this,
B
(x0,r)
⊆ Ω for r small enough. Every x ∈ B
(x0,r)
is in S. Then S is open because if x ∈ S, then
B
(x,r)
⊆ Ω for small enough r and if y ∈ B
(x,r)
, you could go take γx0x 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
(x,r)
⊆ Ω, no point of B
(x,r)
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 γx0x, a bounded variation curve from x0 to
x.
Define
∫
F (x) ≡ f ⋅dγx0x
γx0x
F is well defined by assumption. Now let lx
(x+tek)
denote the linear segment from x to x + tek. Thus
to get to x + tek you could first follow γx0x to x and from there follow lx
(x+tek)
to x + tek.
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. ■
Corollary 4.3.5Let Ω be a connected open set and f : Ω → ℝp. Then f has a potential if andonly if every closed,γ
(a)
= γ
(b)
, bounded variation curve contained in Ω has the propertythat
∫
f ⋅dγ = 0
γ
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.
Theorem 4.3.6Let f : Ω → ℝpbe a C1vector field, meaning the partial derivatives exist and arecontinuous. Also let Ω be open and connected. Then forγa continuous bounded variation curve, thefollowing are equivalent.
f is conservative meaning f = ∇F.
∫γf⋅dγis path independent wheneverγ∗⊆ Ω.
∫γf⋅dγ = 0wheneverγis a closed curve, meaning thatγ :