Is it possible for $\nabla f=y\mathbf{i}-x\mathbf{j}$

Question

is it possible for a function $f(x,y)$ satisfying $\nabla f=y\mathbf{i}-x\mathbf{j}$ ?

Since $\frac{\partial^2 f}{\partial x\partial y}=-1$ and $\frac{\partial^2 f}{\partial y\partial x}=1$, thus $f$ should be singular.

I find a similar answer in wikipedia "Symmetry of second derivatives",

$f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ for $(x,y)\ne(0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$, however, this function just satisfies at $(x,y)=(0,0)$.

Does anybody know a function satisfies the condition in the whole variables domain?

Answer 1

You're requiring:

$f_x=y$ and $f_y=-x$

This requires

$f=xy+A(y)$ and $f=-xy+B(x)$

which implies $2xy=B(x)-A(y)$. Try and show this is impossible (remember, no y can appear in B and no x can appear in y). Try and consider some arbitrary function of x while holding y constant, then changing the constant.

Now no one says $f$ has to be twice continuously differentiable, so the symmetry of the second derivative doesn't have to hold, but the above already rules a function like this out.