KindiakMath

Multivariate Line Integrals

In this exercise, we establish several results in vector calculus that we will need to further our understanding in complex analysis. These results are analogous to the results involving contour integrals.

Definition 1. Call \boldsymbol{ \gamma } a curve in \mathbb R^n if there exists a continuous map \mathbf r : [a, b] \to \mathbb R^n such that \boldsymbol{ \gamma } = \mathbf r([a, b]). We say that \boldsymbol{ \gamma } is smooth if \boldsymbol{ \gamma }' is continuous. If, additionally, that \| \boldsymbol{ \gamma }'(t) \| \neq 0 for any t \in [a, b], then we define the line integral of \mathbf F : \mathbb R^n \to \mathbb R^n over \boldsymbol{ \gamma } by

\displaystyle \int_{\boldsymbol{ \gamma }} \mathbf F \cdot \mathrm d\mathbf r := \int_a^b \langle \mathbf F(\mathbf r(t)), \mathbf r'(t)\rangle\, \mathrm dt \equiv \int_a^b \langle \mathbf F \circ \mathbf r, \mathbf r'\rangle.

Abbreviating \mathbf F = (F_1,\dots, F_n)^{\mathrm T}, \mathbf r = (x_1,\dots, x_n)^{\mathrm T} and \mathrm d x_k = x_k'(t)\, \mathrm dt, define

\displaystyle \int_{\boldsymbol{ \gamma }} \sum_{k=1}^n F_k\, \mathrm dx_k:= \int_{\boldsymbol{ \gamma }} \mathbf F \cdot \mathrm d\mathbf r.

Problem 1. Let \boldsymbol{\gamma} = \mathbf r([a, b]) be a smooth curve and f : D \to \mathbb R, where D \subseteq \mathbb R^n is a domain. Prove that if \nabla f is continuous, then

\displaystyle \int_{\boldsymbol{\gamma}} \nabla f \cdot \mathrm d\mathbf r = f( \mathbf r(b) ) - f( \mathbf r(a) ).

(Click for Solution)

Solution. By the chain rule,

(f \circ \mathbf r)'(t_0) = (f'(\mathbf r(t_0)))(\mathbf r'(t_0)) = \langle (\nabla f)(\mathbf r(t_0)), \mathbf r'(t_0) \rangle.

By Definition 1 and the fundamental theorem of calculus,

\begin{aligned} \int_{\boldsymbol{\gamma}} \nabla f \cdot \mathrm d\mathbf r &= \int_a^b \langle (\nabla f) \circ \mathbf r, \mathbf r' \rangle = \int_a^b (f \circ \mathbf r)' = (f \circ \mathbf r)(b) - (f \circ \mathbf r)(a). \end{aligned}

Definition 2. Call \mathbf F conservative on a domain D if there exists a function f : D \to \mathbb R such that \nabla f = \mathbf F.

Definition 3. Call a curve \boldsymbol{\gamma} = \mathbf r([a, b]) closed if \mathbf r(a) = \mathbf r(b). We say that \boldsymbol{\gamma} has starting point \mathbf r(a) and ending point \mathbf r(b).

Problem 2. Let D \subseteq \mathbb R^n be a domain and \mathbf F : D \subseteq \mathbb R^n \to \mathbb R^n be continuous. Prove that the following are equivalent:

  • \mathbf F is conservative on D.
  • \int_{\boldsymbol{\gamma}} \mathbf F \cdot \mathrm d\mathbf r = 0 whenever \boldsymbol{\gamma} is closed.
  • \int_{\boldsymbol{\gamma}_1} \mathbf F \cdot \mathrm d\mathbf r = \int_{\boldsymbol{\gamma}_2} \mathbf F \cdot \mathrm d\mathbf r for any two curves \boldsymbol{\gamma}_1, \boldsymbol{\gamma}_2 that share the same start and end points.
(Click for Solution)

Solution. We prove one step at a time. If \mathbf F is conservative on D, then there exists a function f : D \to \mathbb R^n such that \nabla f = \mathbf F. By Problem 1, if \boldsymbol{\gamma} is closed, then \mathbf r(a) = \mathbf r(b) so that

\displaystyle \int_{\boldsymbol{\gamma}} \mathbf F \cdot \mathrm d\mathbf r = f(\mathbf r(b)) - f(\mathbf r(a)) = 0.

Given two smooth curves \boldsymbol{\gamma}_1, \boldsymbol{\gamma}_2, define the closed smooth curve \boldsymbol{\gamma}_1 + (-\boldsymbol{\gamma}_2):

\begin{aligned} \int_{\boldsymbol{\gamma}_1} \mathbf F \cdot \mathrm d\mathbf r - \int_{\boldsymbol{\gamma}_2} \mathbf F \cdot \mathrm d\mathbf r &= \int_{\boldsymbol{\gamma}_1} \mathbf F \cdot \mathrm d\mathbf r + \int_{-\boldsymbol{\gamma}_2} \mathbf F \cdot \mathrm d\mathbf r \\ &= \int_{\boldsymbol{\gamma}_1+ (-\boldsymbol{\gamma}_2)} \mathbf F \cdot \mathrm d\mathbf r = 0. \end{aligned}

Finally, suppose the path-independence of the line integral. Fix \mathbf x_0 \in D. Define the map f : D \to \mathbb R^n as follows: for any \mathbf x \in D and curve \boldsymbol{\gamma}_{\mathbf x} = \mathbf r([0, 1]) denote the straight-line path such that \mathbf r(0) = \mathbf x_0 and \mathbf r(1) = \mathbf x, declare

\displaystyle f(\mathbf x) := \int_{\boldsymbol{\gamma}_{\mathbf x}} \mathbf F \cdot \mathrm d\mathbf r.

By construction f(\mathbf x_0) = 0. Furthermore,

\begin{aligned} \int_{\boldsymbol{\gamma}_{\mathbf x}} \mathbf F(\mathbf x_0) \cdot \mathrm d\mathbf r &= \int_0^1 \langle \mathbf F(\mathbf x_0) , \mathbf r' \rangle = \int_0^1 \langle \mathbf F(\mathbf x_0) , \mathbf x \rangle = \langle \mathbf F(\mathbf x_0) , \mathbf x \rangle. \end{aligned}

Since \mathbf F is continuous, for any k > 0, there exists \delta > 0 such that

\| \mathbf v \| < \delta \quad \Rightarrow \quad \| \mathbf F(\mathbf x_0 + \mathbf v) - \mathbf F(\mathbf x_0) \| < k \cdot \epsilon.

By adapting the ML-inequality for multivariable functions,

\begin{aligned} | f(\mathbf x_0 + \mathbf v) - f(\mathbf x_0) - \langle \mathbf F(\mathbf x_0) , \mathbf v \rangle | &= \left| \int_{ \boldsymbol{\gamma}_{\mathbf v} } \mathbf F(\mathbf x_0+\mathbf v) \cdot \mathrm d\mathbf r - \int_{\boldsymbol{\gamma}_{\mathbf v}} \mathbf F(\mathbf x_0) \cdot \mathrm d\mathbf r\right| \\ &= \left| \int_{\boldsymbol{\gamma}_{\mathbf v}} (\mathbf F(\mathbf x_0+\mathbf v) - \mathbf F(\mathbf x_0)) \cdot \mathrm d\mathbf r\right| \\ &\leq \| \mathbf F(\mathbf x_0+\mathbf v) - \mathbf F(\mathbf x_0) \| \cdot \| \boldsymbol{\gamma}_{\mathbf v} \| \\ &\leq k \cdot \epsilon \cdot \|\mathbf v \|. \end{aligned}

Setting k = 1/2 yields

\langle \mathbf F(\mathbf x_0) , \cdot \rangle = f'(\mathbf x_0) = \langle (\nabla f)(\mathbf x_0), \cdot \rangle \quad \Rightarrow \quad (\nabla f)(\mathbf x_0) = \mathbf F(\mathbf x_0).

Since \mathbf x_0 \in D is arbitrary, \nabla f = \mathbf F, as required.

—Joel Kindiak, 15 Aug 25, 1700H

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