How to check there exists a function on given domain whose derivative is given function?

Question

I am going through below multiple choice Question,

Question: let $F_1, F_2: \mathbb{R^2} →\mathbb{R}$ such that, $$F_1(x_1, x_2)=\frac{-x_2}{x_1^2+x_2^2} \text{ and } F_2(x_1, x_2)=\frac{x_1}{x_1^2+x_2^2}.$$ Then which of the following is/are correct?

1. $\frac{∂F_1}{∂x_2} = \frac{∂F_2}{∂x_1}$

2. there exists function $f: \mathbb{R^2}-\{(0,0)\} →\mathbb{R}$ such that, $\frac{∂f}{∂x_1} =F_1$ and $\frac{∂f}{∂x_2} =F_2$

3. there exists No function $f: \mathbb{R^2}-\{(0,0)\} →\mathbb{R}$ such that, $\frac{∂f}{∂x_1} =F_1$ and $\frac{∂f}{∂x_2} =F_2$

4. there exists a function $f:D→\mathbb{R}$ where $D$ is open disc of radius 1 centered at $(2,0)$, which satisfies, $\frac{∂f}{∂x_1} =F_1$ and $\frac{∂f}{∂x_2} =F_2$ on $D$.

My attempt: clearly

$$\frac{∂F_1}{∂x_2} =\frac{x_2^2 -x_1^2}{\left(x_1^2+x_2^2\right)^2} =\frac{∂F_2}{∂x_1}$$

So that (a) is ✓ (correct). But I have no idea about other options. Kindly please help me, facing trouble from hours!! :-(

Answer 1

Let's assume such function $f$ exists, so $\frac{\partial f}{\partial x_1} = F_1$. Then, $$ f(x_1, x_2) = \int \frac{-x_2 dx_1}{x_1^2 + x_2^2} = -\arctan(x_1/x_2) + C(x_2) $$ Thus, $$ \frac{\partial f}{\partial x_2} = \ldots $$ Can you compute the partial derivative, set it equal to $F_2$ and see if it matches, and then finish the problem?

UPDATE

Differentiating, we get $$ \frac{x_1}{x_1^2+x_2^2} = F_2 = \frac{\partial f}{\partial x_2} = \frac{x_1}{x_1^2+x_2^2} + C'(x_2) $$ which implies $C'(x_2) = 0$, so $C$ is a constant, not depending on $x_2$. Thus, $$ f(x_1, x_2) = -\arctan(x_1/x_2) + C $$ seems to satisfy the algebraic conditions of the problem. But notice this function is not defined anywhere at $x_2=0$. Can you finish the problem now?

Answer 2

The vector field ${\bf F}=(F_1,F_2)$ is nothing else but $\nabla{\arg}$, where ${\rm arg}(x,y)$ denotes the polar angle of $(x,y)$, modulo $2\pi$. With this in mind it becomes clear that only 2) is false. But we need a proof of this not making use of the "background information" given here.

In order to prove that the field ${\bf F}$ has no potential $f$, even though its curl is $\equiv0$, we need a stronger tool: Integrate ${\bf F}$ along the unit circle $$\gamma:\quad t\mapsto (\cos t,\sin t)\qquad(0\leq t\leq2\pi)\ .$$ You obtain $$\int_\gamma {\bf F}\cdot d{\bf z}=\int_0^{2\pi}\bigl((-\sin t)(-\cos t)+\cos t\cos t\bigr)\>dt=2\pi\ .$$ Since this integral is $\ne0$ the field ${\bf F}$ cannot have a potential.