Show that $F=\left(x, xy\right)$ is not the gradient of a scalar function

Question

Suppose $F=\left(x, xy\right)$. For $F$ to be a gradient of a function $f$, $x=\frac{\partial }{\partial x}\left(f\right)$ and $xy=\frac{\partial }{\partial y}\left(f\right)$. My notes state that this can be verified with Clairaut theorem, that is, by checking if $\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}\left(f\right)\right)=\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}\left(f\right)\right)$, which, in this case, it isn't true.

But why does it make sense to use the theorem?

Answer 1

If you have studied the differential operator "curl", you must know that the curl of a gradient is zero (https://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_of_gradient_is_zero)

Here, using notations $$F=\left(f_1(x,y)=x, f_2(x,y)=xy\right)$$

$$curl(F)=\partial f_2/\partial x - \partial f_1/\partial y=y - 0 =y$$

Should $F$ be a gradient, i.e., $F=grad(G)$ for some function $G$, we would have :

$$curl(grad(G))=0$$

which is not the case.

Answer 2

From the first equation $x = f_x$, we deduce $$ 0 = \frac{\partial^2 f}{\partial x\partial y} $$ by differentiation w.r.t $y$. From the second equation $xy = f_y$, we deduce $$ y = \frac{\partial^2 f}{\partial y\partial x} $$ by differentiation w.r.t $x$. Using the theorem to conclude does only make sense if it's assumptions apply. Here, the equality of mixed derivatives must be true. Thus, there is no such function $f$ that is sufficiently smooth.

Answer 3

Suppose $(\phi,\psi)$ is the gradient of some function $f,$ then we must have $$\mathrm df=f_x\mathrm dx+f_y\mathrm dy=\phi\mathrm dx+\psi\mathrm dy.$$

Thus we must have $f_x=\phi.$ Integrating with respect to $x$ then gives $f=\int\phi\mathrm dx.$ Similarly, we must also have $f_y=\psi,$ so that $f=\int\psi\mathrm dy.$ It then must be the case that $$\int\phi\mathrm dx=\int\psi\mathrm dy.$$ That is, differentiating with respect to $x,$ we must have $\phi=\int\psi_x\mathrm dy,$ and now differentiating with respect to $y$ we must have $$\phi_y=\psi_x.$$

So this condition is necessary for the given vector to be some gradient. It is easy to see that it is also sufficient provided some continuity conditions are satisfied, which have been assumed anyway for the necessity case to be proved.