Checking path independence for a given vector field by evaluating the integral over every closed curve is an impossible task! Hence, in practice, we need other criteria to check whether a given vector field is conservative. In this section we want to introduce such a criterion.
Assume for the moment that \(\vect f=(f_1,\dots,f_N)\) is a continuously differentiable gradient vector field with potential \(V\) on some open subset \(D\) of \(\mathbb R^N\text{.}\) By the symmetry of the second partial derivatives (see Proposition 4.38) and since \(\vect f=\grad V\) we have
for all \(\vect x\in D\) and \(1\leq i,j\leq N\text{.}\) We have seen in the previous section that gradient and conservative fields are the same. Hence we have the following proposition.
Proposition8.7.
Suppose that \(\vect f=(f_1,\dots,f_N)\) is a continuously differentiable conservative vector field on \(D\subset\mathbb R^N\text{.}\) Then
for all \(\vect x\in D\) and \(1\leq i,j\leq N\text{.}\)
Hence for a vector field \(\vect f\) to be conservative, (8.4) must be satisfied. The question arising is whether (8.4)guarantees that \(\vect f\) is conservative. Unfortunately, the answer is NO in general (see Example 8.12 below)! However, under additional assumptions on \(D\) we will be able to show that (8.4) is sufficient to ensure that \(\vect f\) is a gradient field. Because of their significance in any `field theory,' vector fields satisfying condition (8.4) have a special name.
Definition8.8.Closed vector field.
A vector field, \(\vect f=(f_1,\dots,f_N)\) is said to be closed on \(\vect D\subset\mathbb R^N\) if (8.4) holds for all \(\vect x\in D\) and \(1\leq i,j\leq N\text{.}\)
We mentioned above that it depends on the domain \(D\) whether every closed vector field on \(D\) is conservative or not. The condition on \(D\) is a topological one. We need to make sure that every closed curve in \(D\) can be continuously shrunk to a point in \(D\) without ever leaving \(D\text{.}\)Figure 8.9 shows a simply connected domain in the plane. If a plane domain has a hole like the one in Figure 8.10, a curve around the hole cannot be shrunk to a point in \(D\text{.}\)
Figure8.9.A simply connected domain.
Figure8.10.A domain not simply connected.
One can then prove the following theorem.
Theorem8.11.
Every closed vector field on a simply connected domain in \(\mathbb R^N\) is conservative.
A proof of the above theorem in full generality is out of the scope of these notes. Some special cases can easily be obtained as applications of the theorems of Green and Stokes in Part IV. For \(N=2\) and \(N=3\) we will give a complete proof in Section 10.4 and Section 13.3.In the following sections we look at the special cases of \(N=2\) and \(N=3\text{.}\)
