Understanding the conditions for path-independence

The function

$$\left(\vphantom{ \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} }\right. {\frac{{-y}}{{x^2+y^2}}} {\frac{{x}}{{x^2+y^2}}} \left.\vphantom{ \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} }\right)$$

is a favorite example function for driving home the conditions for path-independence.

Recall that one condition for path independence is the following. For a simply connected domain, a continuously differentiable vector field F is path-independent if and only if its curl is zero.

For a vector field in two dimensions, F(x, y), the analogue of curl is

$${\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial y}}}$$

It turns out that for our example F(x, y), the curl is zero.

$${\frac{{\partial F_2}}{{\partial x}}} {\frac{{1}}{{x^2+y^2}}} {\frac{{x(2x)}}{{(x^2+y^2)^2}}} {\frac{{y^2-x^2}}{{(x^2+y^2)^2}}}$$

$${\frac{{\partial F_1}}{{\partial y}}} {\frac{{1}}{{x^2+y^2}}} {\frac{{y(2y)}}{{(x^2+y^2)^2}}} {\frac{{y^2-x^2}}{{(x^2+y^2)^2}}}$$

So, therefore, $${\frac{{\partial F_2}}{{\partial x}}}$$ - $${\frac{{\partial F_1}}{{\partial y}}}$$ = 0.

Can we conclude F is path-independent? No, the test does not apply because F is a two-dimensional vector field and its domains has a hole in it. F is defined everywhere except at the point (0, 0).

We still don't know if F is path-independent or path-dependent. Let's try another test.

Since the point of showing this example is to drive home the conditions for path-independence, it is likely the example was chosen to illustrate some of the subtlety involved. If the example has zero curl and is path-independent, the example won't be particularly instructive. Most likely, the example is path-dependent. So, let's try a test for path-dependence.

If we can find a closed path along which the integral of F is nonzero, then we can conclude F is path-dependent. The fact that F is not defined at the origin is key. So, let's try a path that goes around the origin. The simplest such path is the unit circle.

A counterclockwise parametrization of the unit circle is g(t) = (cos t, sin t), for 0$\le$t$\le$2$\pi$. If C is the counterclockwise unit circle, then we can calculate

$$\left(\vphantom{\frac{-\sin t}{\cos^2 t + \sin^2 t}, \frac{\cos t}{\cos^2 t+ \sin^2 t}}\right. {\frac{{-\sin t}}{{\cos^2 t + \sin^2 t}}} {\frac{{\cos t}}{{\cos^2 t+ \sin^2 t}}} \left.\vphantom{\frac{-\sin t}{\cos^2 t + \sin^2 t}, \frac{\cos t}{\cos^2 t+ \sin^2 t}}\right)$$ F(cos t, sin t) = (- sin t, cos t).

Therefore,

$$\int_{C}^{} \int_{0}^{{2\pi}}$$

$$\int_{0}^{{2\pi}}$$

$$\int_{0}^{{2\pi}} \int_{0}^{{2\pi}} \pi$$

Ah ha! Just as we suspected, the circulation around C is not zero. F is path-dependent.

The circulation can be clearly seen by plotting the vector field F. It's difficult to plot, because the vector field blows up at the origin. But this figure gives you the idea.

What about the circulation around a closed path B

$$\int_{B}^{}$$

that doesn't encircle the origin? If B doesn't encircle the origin, then F is defined everywhere inside B. The part of the domain of F inside B has no holes in it. Since F is defined everywhere in the region D inside B, we can use Green's theorem to conlude that

$$\int_{B}^{} \int_{D}^{} \left(\vphantom{\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}}\right. {\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial y}}} \left.\vphantom{\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}}\right) \int_{D}^{}$$

Indeed, the circulation of F around B is zero. In fact, the circulation around any closed curve that does not contain the origin is zero.