Answer to Question #278341 in Calculus for marie

Question

The Laplacian of a function f of n variables x₁, x₂,⋯x_n, denoted ∇²f is defined by ∇²f(x1, x2,⋯xn) := (∂²f/∂x₁²)+(∂²f/∂x₂²)+...+(∂²f/∂x_n²)

Now assume that f depends only on r where r= (x₁²₊ x²₂₊⋯+x²_n)^1/2

i.e. f(x₁,x₂,⋯,x_n)=g(r), for some function g

. Show that, for x₁,x₂,⋯,x_n≠0, ∇²f=[(n-1)/r]g′(r)+g′′(r)

Answer 1

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

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

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

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

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