Answer to Question #275782 in Calculus for Nawrin A

Question

The Laplacian of a function f

 of n

 variables x

1

,x

2

,⋯x

n

, denoted ∇

2

f

 is defined by

∇

2

f(x

1

,x

2

,⋯,x

n

):=∂

2

f

∂x

2

1

+∂

2

f

∂x

2

2

+⋯+∂

2

f

∂x

2

n

Now assume that f

 depends only on r

 where r=(x

2

1

+x

2

2

+⋯+x

2

n

)

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

≠0

,

∇

2

f=n−1

r

g

′

(r)+g

′′

(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)

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

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

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

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

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