Conservative vector fields

It is important to devote a section to this as conservative fields. We start off with a definition.

Definition 2.6 A vector field $\boldsymbol{F}$ is conservative if there exists a scalar function $V(x, y)$ such that:

$V$ is then called a potential function. Note that the above apply to three-dimensional scalar functions as well, i.e. $V=V(x, y, z)$.

We then introduce a useful theorem for conservative vector fields:

Fundamental theorem for conservative fields

Let $\boldsymbol{F}=\nabla V$ be a conservative field ( $\mathrm{V}$ is the potential of $\boldsymbol{F}$ ) on a domain $D$. If $\mathcal{C}$ is a path from points $A$ and $B$ in $D$, then:

The proof of Theorem 2.3 is as follows: we consider a conservative field, $\boldsymbol{F}=\nabla V$, where $V=V(x, y)$ and a given parametrisation $\boldsymbol{r}(t)$ of curve $\mathcal{C}$ (where $a \leq t \leq b$ ) such that $V=V(x(t), y(t)$ ). Using Eq. (2.56) the line integral of $\boldsymbol{F}$ is:

By the chain rule, the integrand in Eq. (2.65) is expressed as:

Plugging Eq. (2.66) in Eq. (2.65) yields:

Example 2.8 Evaluate $\int_{\mathcal{C}} \boldsymbol{F} \cdot d \boldsymbol{r}$ where $\boldsymbol{F}$ is conservative, i.e. $\boldsymbol{F}=\nabla V$ and $V(x, y, z)=\cos (\pi x)+\sin (\pi y)-x y z$ where $\mathcal{C}$ is any curve connecting point $A$ to $B$ where $A=(1,1 / 2,2)$ and $B=(2,1,-1)$.

Solution

Using Eq. (2.62), we have:

Observe that we are not given a specific path for $\mathcal{C}$ and we do not require to have the gradient vector, $\nabla V$. Using Theorem 2.3 , we only need the endpoints to evaluate the line integral of the conservative field $\boldsymbol{F}$.

Suppose we have two paths, $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ that have the same initial and final points (say, $A$ and $B$, respectively). Now, if a vector field $\boldsymbol{F}$ is conservative (i.e. $\boldsymbol{F}=\nabla V$ ) then, according to Theorem 2.3, we have the following results for the line integrals of $\boldsymbol{F}$ along $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ :

Therefore, if $\boldsymbol{F}$ is conservative,

for any pair of curves $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ with the same endpoints. The line integral of the conservative vector $\boldsymbol{F}$ from $A$ to $B$ is independent of the path since the result only depends on the initial and final points.

Suppose now that we have a closed curve; the latter is a curve or path whose initial and final points are the same (e.g. a circle is a closed curve). We refer to the line integral of $\boldsymbol{F}$ around a closed path as the circulation of $\boldsymbol{F}$ and denote the integral sign with a circle in the middle, i.e. $\oint$. The following theorem applies to the line integral of a conservative field around a closed path.

Conservative fields around a closed curve

Let $\boldsymbol{F}=\nabla V$ be a conservative field on a domain $\mathcal{D}$ and assume $\mathcal{C}$ is a closed path from points $A$ and $B$ (closed implies $A=B$ ), then:

i.e. Eq. (2.70) states that the circulation of $\boldsymbol{F}$ around a closed path is zero.

The proof of Theorem 2.4 is a straightforward result from Theorem 2.3. Since $A=B$, then Eq. (2.62) gives $V(B)-V(A)=0$.

Finding potential functions

So far, we know how to show that a vector field is conservative given a potential function, $V$. However, this is not an efficient way of telling whether a given vector field, $\boldsymbol{F}=\left(F_{1}, F_{2}, F_{3}\right)$, where $F_{1}, F_{2}$ and $F_{3}$ are the components of the vector field, is conservative. We start off with the following theorem about the quality of cross-partials of a conservative field.

Cross-partial property of a conservative vector field

If the vector field $\boldsymbol{F}=\left(F_{1}, F_{2}, F_{3}\right)$ is conservative, then $\underline{\nabla \times \mathbf{F}=0}$. From Definition 2.4 of the curl of a vector, we see that

So, according to Theorem 2.5 , equality of the cross partials indicates that $\boldsymbol{F}$ is conservative. This theorem is valid provided the domain $D$ on which $\boldsymbol{F}$ is defined is simply-connected, i.e. one that has no holes, as shown in Fig. 2.12, leading to Theorem 2.6:

(a)

(b)

Figure 2.5: (a) Simply-connected regions and (b) nonsimply-connected regions.

Existence of a potential function

Let $\boldsymbol{F}$ be a vector field on a simply-connected region, $D$. Then, if $\boldsymbol{F}$ satisfies the cross-partial condition given by Eq. (2.71), $\boldsymbol{F}$ is conservative.

Next, we will look at an example of finding a potential function $V$, given a two-dimensional vector field, $\boldsymbol{F}=\left(F_{1}, F_{2}\right)$ using the equality of the partials conditions given by Eq. (2.71). Example 2.9 Suppose we have a vector field, $\boldsymbol{F}=\left(2 x y+y^{3}, x^{2}+\right.$ $\left.3 x y^{2}+2 y\right)$. Show that $\boldsymbol{F}$ is conservative and find a potential function, $V(x, y)$.

Solution According to Theorem 2.6, $\boldsymbol{F}$ is conservative if defined on a simply-connected domain, $D$ and if there exists equality of the partials. In the case of two dimensions, the partial $\frac{\partial}{\partial z}$ is zero and hence we only need to show that $\frac{\partial F_{1}}{\partial y}=\frac{\partial F_{2}}{\partial x}$.

$\frac{\partial F_{1}}{\partial y}=\frac{\partial}{\partial y}\left(2 x y+y^{3}\right)=2 x+3 y^{2}, \quad \frac{\partial F_{2}}{\partial x}=\frac{\partial}{\partial x}\left(x^{2}+3 x y^{2}+2 y\right)=2 x+3 y^{2}$

Further, the entire $x y$-plane is a simply-connected region and therefore $\boldsymbol{F}$ is conservative. Now, from Eq. (2.61), we know that:

The equations above tell us that $V(x, y)$ is an antiderivative of $F_{1}$ regarded as a function of $x$ alone and of $F_{2}$ regarded as a function of $y$ alone. To determine $V(x, y)$ therefore, we integrate $F_{1}$ with respect to $x$ and $F_{2}$ with respect to $y$ :

note that in order to obtain the general antiderivative of $F_{1}$ with respect to $x$, we the usual 'constant' of integration is now a function of the variable being kept constant, i.e. $y$. Similarly, we have:

where, agan, we have added a function of the variable being kept constant, i.e. $x$. Of course Eqs. (2.72) and (2.73) must be equal. Hence, comparing the two equations, we have:

The potential function is given by $V(x, y)=x^{2} y+x y^{3}+y^{2}$.