If $v = v(r)$ and $r^2 = \sum_{i=1}^{n} x_i^2$, then prove $$\sum v_{x_i x_i} = v_r ((n-1)/r) + v_{rr}$$
I start by:
$$\frac{\partial r}{\partial x_i} = \frac{x_i}{r}$$
$$\frac{\partial v}{\partial x_i} = \frac{\partial v}{\partial r} \frac{\partial r}{\partial x_i} = \frac{\partial v}{\partial r} \frac{x_i}{r}$$
Now, $$\frac{\partial^2 v}{\partial x^2_i} = \frac{\partial}{\partial x_i }\left(\frac{\partial v}{\partial r} \frac{x_i}{r}\right)$$
From here, I did bifurcation into two methods:
Method 1:
$$\frac{\partial^2 v}{\partial x^2_i} = \frac{\partial}{\partial x_i }\left(\frac{\partial v}{\partial r} \frac{x_i}{r}\right)\\ = \frac{\partial^2 v}{\partial x_i\partial r}\frac{x_i}{r} + \frac{\partial v}{\partial r}\frac{1}{r} - x_i\frac{\partial v}{\partial r}\frac{1}{r^2}\frac{\partial r}{\partial x_i}\\ = \frac{\partial^2 v}{\partial x_i\partial r}\frac{x_i}{r} + \frac{\partial v}{\partial r}\frac{1}{r} - \frac{\partial v}{\partial r}\frac{x_i^2}{r^3}$$
Apply sigma
$$\sum\frac{\partial^2 v}{\partial x^2_i} =\frac{n-1}{r} \frac{\partial v}{\partial r} + \color{red}{\sum \frac{\partial^2 v}{\partial x_i\partial r}\frac{x_i}{r}}$$
Method 1(A):
Wrote the red term as $$\frac{\partial}{\partial r}\left(\frac{\partial v}{\partial r}\right) \frac{\partial r}{\partial x_i} \frac{x_i}{r} = \frac{\partial^2 v}{\partial r^2} \frac{x_i^2}{r^2}$$ which on summing gives required result!
Method 1(B):
Wrote red part as:
$$\frac{\partial }{\partial r}\left(\frac{\partial v}{\partial x_i}\right)\frac{\partial r}{ \partial x_i} = \frac{\partial ^2 v}{\partial x_i^2}$$
which is obviously not correct, but doesnt appear where error occurs.
Method 2:
Using chain rule:
$$\frac{\partial^2 v}{\partial x^2_i}= \frac{\partial}{\partial x_i }\left(\frac{\partial v}{\partial r} \frac{x_i}{r}\right) = \frac{\partial}{\partial r }\left(\frac{\partial v}{\partial r} \frac{x_i}{r}\right) \frac{\partial r}{\partial x_i}\\ = \frac{x_i}{r} \left( \frac{x_i}{r} \frac{\partial^2 v}{\partial r^2} + \color{red}{ \frac{1}{r} \frac{\partial v}{\partial r}\frac{\partial x_i}{\partial r}}-\frac{\partial v}{\partial r}\frac{x_i^2}{r^2}\right)$$
On applying sigma and noting $\sum x_i \frac{\partial x_i}{\partial r} = r$ we get wrong answer:
$$\sum\frac{\partial^2 v}{\partial x_i^2} = \frac{\partial ^2 v}{\partial r^2} + 0$$
Here what went wrong?
I am really confused here, as to when we use chain rule without encountering error. Here I know $v$ is function of $r$ which is function of all $x_i$. So it is known that $\frac{\partial v}{\partial x_i} = \frac{\partial v}{\partial r} \frac{\partial r}{\partial x_i}$. Where do I go wrong everytime?
Thank you a lot!
Your troubles stem from dubious notation: The same letter $v$ is used for a certain function $r\mapsto v(r)$ of one variable $r$, and then also for the nested function $$x\mapsto f(x):=v\bigl(r(x)\bigr)$$ of the vector variable $x$. – We are told to compute the Laplacian $$\Delta f(x):=\sum_{i=1}^n{\partial^2 f\over\partial x_i^2}\ .$$ You started correctly with $${\partial f\over\partial x_i}=v'(r){\partial r\over\partial x_i}={v'(r)\over r}\,x_i\ .$$ Now the next step: $${\partial^2 f\over\partial x_i^2}={d\over dr}{v'(r)\over r} \cdot {\partial r\over \partial x_i}\cdot x_i+{v'(r)\over r}={r v''(r)-v'(r)\over r^2}\>{x_i^2\over r}+{v'(r)\over r}\ .$$ Summing over $i$ then gives the result you were asked to prove.