2

Show that $\textbf F$ is a gradient field by giving a scalar function $f$ on $\Omega^+$ such that $\nabla f=\textbf F$.

$\textbf F = (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0)$, $\Omega^+ = \{(x,y)|x,y>0\}$.

My attempt:

We must have that $\frac{\partial f}{\partial x} = \frac{-y}{x^2+y^2}$, so I obtained that $f = -arctan(x/y)+g(y,z)$.

Also $\frac{\partial f}{\partial y} = \frac{x}{x^2+y^2}$, so I got that $f = arctan(y/x)+h(x,z)$.

And $\frac{\partial f}{\partial z} = 0$ tells us that $f$ is in terms of $x$ and $y$.

But now I am stuck. How is it possible that $-arctan(x/y)+g(y,z)=arctan(y/x)+h(x,z)$?

Bobby Lee
  • 1,030
  • 1
    I would like to add that for your last bit you should have $g(y,z)=g(y)$ and $h(x,z)=h(x)$ – Ellya Mar 30 '14 at 21:26
  • 1
    Notice that, for $x,y>0$, $\frac{\pi}{2} - \arctan (x/y) \equiv \arctan(y/x)$. – FH93 Mar 30 '14 at 22:12

1 Answers1

1

To add to the comments from @ellya and @FH93, another way of tackling the problem is to move to polar coordinates.

Using $$ r^2 = x^2 + y^2 \quad \text{and} \quad \theta = \tan^{-1}(y/x) $$ we have $$ \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial\theta}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial\theta}{\partial y} \end{bmatrix} \begin{bmatrix} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial\theta} \end{bmatrix} $$ If we invert the Jacobian matrix and multiply it with the left hand side, we get $$ \begin{bmatrix} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial\theta} \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$ Therefore, $f = f(\theta) = \theta = \tan^{-1}(y/x)$.

Biswajit Banerjee
  • 600