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The gradient theorem: the line integral of a gradient is the endpoint increment

Statement

Let $U \subseteq R^{n}$ be open, let $ϕ : U \to R$ be $C^{1}$ , and let $γ : [a, b] \to U$ be piecewise- $C^{1}$ . Then

$\int_{γ} \nabla ϕ \cdot d r = ϕ (γ (b)) - ϕ (γ (a)) .$

Facts & Assumptions

Given: The open set, potential, and path in the Statement, with an admissible partition $a = t_{0} < \dots < t_{m} = b$ when $a < b$ .

[L1]

The vector line integral is the sum of the integrals of $⟨ \nabla ϕ (γ (t)), v_{i} (t) ⟩$ over the smooth pieces (Scalar line integrals with respect to arc length and vector-field line integrals).

[L2]

For a scalar function, the gradient lists its partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L3]

The total-derivative chain rule is $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L4]

If a continuous function on $[u, v]$ has an integrable interior derivative $q$ , then $\int_{u}^{v} q$ is its endpoint increment (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative).

Proof

technique · direct

1.1

If $a = b$ , [L1] makes the line integral zero and the two endpoint values agree. Assume henceforth that $a < b$ . On the interior of the $i$ th smooth piece, [L2] and [L3] give $\frac{d}{d t} ϕ (γ (t)) = ⟨ \nabla ϕ (γ (t)), γ^{'} (t) ⟩ .$

givenL1L2L3algebra

2.1

The continuous derivative extension on that piece is the integrand in [L1]. Applying [L4] gives $\int_{t_{i}}^{t_{i + 1}} ⟨ \nabla ϕ (γ (t)), v_{i} (t) ⟩ d t = ϕ (γ (t_{i + 1})) - ϕ (γ (t_{i})) .$

step 1.1L1L4

3.1

Sum step 2.1 over the finite partition. All interior endpoint values cancel, leaving $ϕ (γ (b)) - ϕ (γ (a))$ , and [L1] identifies the left side with the line integral.

step 2.1L1algebra

4.1

For a constant path the integrand is zero and the endpoints coincide, so both sides are zero. The same conclusion holds whenever merely $γ (a) = γ (b)$ .

L1step 3.1algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources

J. Lebl, Basic Analysis II, Theorem 9.3.1 (standard reference, not scraped)