If $r=\sqrt{x^2+y^2}$ and $u=f(r)$ then prove $ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} =f''(r)+\frac{1}{r}f'(r)$

Question

We have $r=\sqrt{x^2+y^2}$ and $u=f(r)$ then prove , from chain rule I know that $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}.\frac{\partial r}{\partial x}$$ and $$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}.\frac{\partial r}{\partial y}$$ Hence , $$\frac{\partial r}{\partial x}=\frac{x}{r}$$ similarly $$\frac{\partial r}{\partial y}=\frac{y}{r}$$ therefore $$\frac{\partial u}{\partial x}=f'(r).\frac{x}{r} $$ and $$\frac{\partial u}{\partial y}=f'(r).\frac{y}{r} $$ finally $$ \frac{\partial^2 u}{\partial x^2}=f''(x).\frac{x}{r}+f'(r).\frac{r^2-x^2}{r^3}$$ finally $$ \frac{\partial^2 u}{\partial y^2}=f''(x).\frac{y}{r}+f'(r).\frac{r^2-y^2}{r^3}$$ Now summation of both double partial derivative clearly is't equal to one given in question , where did I go wrong ?

Answer 1

You're making an error in taking the second derivative.

\begin{align} \frac{\partial}{\partial x} \left(f'(r) \frac{x}{r}\right) &= \frac{x}{r} \frac{\partial}{\partial x} f'(r) + f'(r)\frac{\partial}{\partial x}\frac{x}{r} \\ &= \frac{x}{r}\left( f''(r) \frac{x}{r}\right) + f'(r) \left(\frac{1}{r} - \frac{x}{r^2} \frac{\partial}{\partial x}r \right) \\ &= f''(r)\frac{x^2}{r^2} + f'(r)\left(\frac{1}{r} - \frac{x^2}{r^3}\right) \end{align}

doing the same thing for $y$, we'll end up with \begin{align} u_{xx} + u_{yy} &= f''(r)\left(\frac{x^2+ y^2}{r^2} \right) + f'(r)\left(\frac{2}{r} - \frac{x^2 + y^2}{r^3}\right) \\ &= f''(r) + \frac{f'(r)}{r} \end{align}