Answer to Question #275800 in Calculus for Hafsa Tahsin

Question

[SADT9] The Laplacian of a function f of n variables x 1 ,x 2 ,*** x n denoted nabla^ 2 f is defined by

nabla^ 2 f(x 1 ,x 2 ,***,x n ):= partial^ 2 f partial x 1 ^ 2 + partial^ 2 f partial x 2 ^ 2 +***+ partial^ 2 f partial x n ^ 2

Now assume that f depends only on r where r=(x 1 ^ 2 +x 2 ^ 2 +***+x n ^ 2 )^ 1 2 i.e. f(x 1 ,x 2 ,***,x n )=g(r) for some function g. Show that, for x 1 ,x 2 ,***,x n ne0 ,

nabla^ 2 f= n-1 r g^ prime (r)+g^ prime prime (r)

Answer 1

r=(x 1 ^ 2 +x 2 ^ 2 +***+x n ^ 2 )^ 1 2 i.e. f(x 1 ,x 2 ,***,x n )=g(r) for some function g. Show that, for x 1 ,x 2 ,***,x n ne0 ,nabla^ 2 f= n-1 r g^ prime (r)+g^ prime prime (r)

$r=\sqrt{x_1^2+x_2^2+...+x_n^2}$

$f(x_ 1 ,x_ 2 ,...,x _n )=g(r)$

$\frac{\partial r}{\partial x_n}=\frac{x_n}{\sqrt{x_1^2+x_2^2+...+x_n^2}}$

$\frac{\partial^2 r}{\partial x^2_n}=\frac{\sqrt{x_1^2+x_2^2+...+x_n^2}-x^2_n/\sqrt{x_1^2+x_2^2+...+x_n^2}}{x_1^2+x_2^2+...+x_n^2}$

$\nabla^2f=n/r-\frac{x_1^2+x_2^2+...+x_n^2}{\sqrt{x_1^2+x_2^2+...+x_n^2}(x_1^2+x_2^2+...+x_n^2)}=\frac{n-1}{r}$