Elementary Second partial derivative

Question

If $V=\frac{xy} {(x^2+y^2)^2}$ and $x=r\cos\theta$, $y=r\sin\theta$ show that $\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=0$

My attempt : $\frac{\partial {V}} {\partial{x}} =y\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}\\$

$\frac{\partial {V}} {\partial{y}} =x\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}$

$\frac{\partial{V} } {\partial{r}}= \frac{\partial {V}} {\partial{x}}\frac{\partial {x}} {\partial{r}}+\frac{\partial {V}} {\partial{y}}\cdot\frac{\partial {y}} {\partial{r}}\\\frac{\partial {V}} {\partial{r}}=\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\}$

$\frac{\partial^2{V} } {\partial{r}^2} =\frac{\partial} {\partial{r}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\}\\=$ $\frac{\partial} {\partial{\theta}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\} \frac{\partial{\theta} }{\partial{r}}$

At this point it is getting clumsy. Any hint on this.

Answer 1

Hint: Write $$V=\frac{\sin2\theta}{2r^2}.$$Now take partial derivative with respect to $r$ and $\theta$.

Answer 2

We have, by substituting our expressions for $x$ and $y$ $$V=\frac{r^2\cos(\theta)\sin(\theta)}{(r^2\cos^2(\theta)+r^2\sin^2(\theta))^2}=\frac{r^2\cos(\theta)\sin(\theta)}{(r^2(\cos^2(\theta)+\sin^2(\theta)))^2}=\frac{r^2\cos(\theta)\sin(\theta)}{r^4}=\frac{\cos(\theta)\sin(\theta)}{r^2}.$$ We can even utilise a double angle formula $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ to arrive to $$V=\frac{\sin(2\theta)}{2r^2},$$ which will make our derivatives easier.

So now, we compute the partial derivatives $$\frac{\partial V}{\partial r}=-\frac{\sin(2\theta)}{r^3},\quad\frac{\partial^2V}{\partial r^2}=\frac{3\sin(2\theta)}{r^4},$$ $$\frac{\partial V}{\partial\theta}=\frac{\cos(2\theta)}{r^2},\quad\frac{\partial^2V}{\partial\theta^2}=-\frac{2\sin(2\theta)}{r^2}.$$

Then, substituting into the expression $$\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=\frac{3\sin(2\theta)}{r^4}-\frac{\sin(2\theta)}{r^4}-\frac{2\sin(2\theta)}{r^4}=0,$$ as desired.