Application: Conservative Vector Fields

Section 10.4 Application: Conservative Vector Fields

We saw in Section 8.2 that every conservative vector field is closed. We also mentioned that the converse is true if \(D\) is simply connected. In this section we want to give a proof of this fact for plane vector fields. Recall from (8.5) that closed for a plane vector field, \(\vect f=(f_1,f_2)\text{,}\) means that

\begin{equation} \frac{\partial}{\partial x_1}f_2 -\frac{\partial}{\partial x_2}f_1=0\text{.}\tag{10.3} \end{equation}

This is exactly the term appearing in (10.1). Simply connected means that every closed curve, \(C\text{,}\) in \(D\) can be continuously shrunk to a point in \(D\) without ever leaving \(D\) (see Figure 8.9). In particular this means that the domain, \(D_C\text{,}\) enclosed by \(C\) is a subset of \(D\text{.}\) Hence it follows from (10.3) and Green's Theorem that

\begin{equation*} 0=\iint_{D_C}\frac{\partial}{\partial x_1}f_2 -\frac{\partial}{\partial x_2}f_1\,dx_1\,dx_2 =\oint_C\vect f\cdot\,d\vect x\text{.} \end{equation*}

This shows that if \(D\) is simply connected and \(\vect f\) is closed then the integral over every closed piecewise continuous curve is zero. By Proposition 8.2 the field \(\vect f\) is conservative.

Example 10.11. Conservative vector field.

Show that \(\vect f(x,y)=(2xy,x^2-y^3)\) is a conservative vector field on \(\mathbb R^2\text{.}\)

Solution.

\(\mathbb R^2\) is clearly simply connected, so we have to check whether \(\vect f\) is closed. We compute the relevant derivatives:

\begin{gather*} \frac{\partial}{\partial x}(x^2-y^3)=2x\\ \frac{\partial}{\partial y}2xy=2x\text{.} \end{gather*}

As the two derivatives are the same \(\vect f\) is closed and thus conservative.

Vector Calculus: An Introduction

Search Results:

Section 10.4 Application: Conservative Vector Fields

Example 10.11. Conservative vector field.