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!! :-(

  • 1
    $\begingroup$ Are you allowed to use Complex Analysis? $F_1$ and $F_2$ are nothing but real and imaginary parts of $i/z$ and basic facts from CA show that 2) is false and the others are true. $\endgroup$ Feb 13 '18 at 6:19
  • $\begingroup$ @kavi sir, yes we are allowed to use complex analysis. But what "Basic fact are you telling about"? $\endgroup$ Feb 13 '18 at 8:43
  • $\begingroup$ If 2. is true then $i/z$ becomes a derivative. The integral of a derivative overf any closed path is always 0. However the integral of $i/z$ over a a circle centered at 0 is easy to calculate from the definition and it is $-2\pi$. Hence 2. is false. This also shows that 3. is true. For 4. just note that f has a power series expansion in a neighborhood of $(2,0)$ and we can intergate the series term by term to get an analytic function g such that $g'=f$. From this it follows that 4. is true. $\endgroup$ Feb 14 '18 at 10:25

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?


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?

  • $\begingroup$ I don't think I can do it. But from my teacher, they given me hint that, limit of $F_1$ and $F_2$ doesn't exists as $(x,y)→(0,0)$ , so that such function $f$ do not exists! But I could not get it!! Please help me with more clarification. $\endgroup$ Feb 13 '18 at 4:45
  • $\begingroup$ @AkashPatalwanshi see update $\endgroup$
    – gt6989b
    Feb 13 '18 at 4:53
  • $\begingroup$ Sir you said in below lines that, function is not defined anywhere at $x_2= 0$. So it not defined on $\{(0,a): a∈\mathbb{R} \}$. Hence such a function $f$ do not exists on given domain ? Is am I right? But then what about option d) ? $\endgroup$ Feb 13 '18 at 4:58
  • $\begingroup$ It is possible to extend this solution to the line $x_2 = 0$ using $\arctan(x_1 / x_2) = \frac{\pi}{2} - \arctan(x_2 / x_1)$. However, what actually happens is that you can't get a function on all of the punctured plane without violating the condition that $f$ is single-valued. $\endgroup$ Feb 13 '18 at 5:34
  • 1
    $\begingroup$ Another way to get a contradiction is: if $\nabla f = (F_1, F_2)$, then $\int_C (F_1 \, dx_1 + F_2 \, dx_2) = f(C_e) - f(C_s)$ where $C_e$ is the ending point of $C$ and $C_s$ is the starting point. On the other hand, if you let $C$ be the unit circle centered at 0, it is straightforward to calculate $\int_C (F_1 \, dx_1 + F_2 \, dx_2) \ne 0$ so you get a contradiction. $\endgroup$ Feb 13 '18 at 5:37

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.

  • $\begingroup$ Thank you, @christian sir, for clarifying the concepts behind this topic. $\endgroup$ Feb 14 '18 at 4:57

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