Question #278341

The Laplacian of a function f of n variables x1, x2,⋯xn, denoted ∇2f is defined by ∇2f(x1, x2,⋯xn) := (∂2f/∂x12)+(∂2f/∂x22)+...+(∂2f/∂xn2)

Now assume that f depends only on r where r= (x12+ x22+⋯+x2n)1/2

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

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



1
Expert's answer
2021-12-13T16:57:36-0500

r=x12+x22+...+xn2r=\sqrt{x_1^2+x_2^2+...+x_n^2}

f(x1,x2,...,xn)=g(r)f(x_ 1 ,x_ 2 ,...,x _n )=g(r)


rxn=xnx12+x22+...+xn2\frac{\partial r}{\partial x_n}=\frac{x_n}{\sqrt{x_1^2+x_2^2+...+x_n^2}}


2rxn2=x12+x22+...+xn2xn2/x12+x22+...+xn2x12+x22+...+xn2\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}


2f=n/rx12+x22+...+xn2x12+x22+...+xn2(x12+x22+...+xn2)=n1r\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}

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