Question

check my work

Answer 1

Given a scalar point function
\[f(r)\]
where
\[r=\sqrt{x^2+y^2+z^2}\]
Prove that
\[\nabla^{2}f(r)=f''(r)+\frac{2}{r}f'(r)\]
\[\nabla^{2}f(r)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\frac{\partial^2f}{\partial z^2}\]
\[\therefore \frac{\partial}{\partial x}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x})+\frac{\partial}{\partial y}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial y})+\frac{\partial}{\partial z}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial z})\]
Now
\[r^2=x^2+y^2+z^2\]
\[\implies 2r\frac{\partial r}{\partial x}=2x\]\[\implies \frac{\partial r}{\partial x}=\frac{x}{r}\]
\[\implies \frac{\partial r}{\partial y}=\frac{y}{r}\]\[\implies \frac{\partial y}{\partial z}=\frac{z}{r}\]
Now LHS again,
\[\frac{\partial}{\partial x}(f'(r)\frac{x}{r})+\frac{\partial}{\partial y}(f'(r)\frac{y}{r})+\frac{\partial }{\partial z}(f'(r)\frac{z}{r})\]
\[\frac{x}{r}.\frac{\partial f'}{\partial x}+\frac{1}{r}.f'(r)+xf'(r).\frac{\partial}{\partial x}(r^{-1})\]
\[+\frac{y}{r}.\frac{\partial f'}{\partial y}+\frac{1}{r}.f'(r)+yf'(r)\frac{\partial}{\partial y}(r^{-1})\]
\[+\frac{z}{r}\frac{\partial f'}{\partial z}+\frac{1}{r}.f'(r)+yf'(r).\frac{\partial }{\partial z}(r^{-1})\]
Applying chain rule
\[\frac{x}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial x}+xf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial x}\]
\[+\frac{y}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial y}+yf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial y}\]
\[+\frac{z}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial z}+zf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial z}\]
\[+\frac{3}{r}.f'(r)\]
Now we have
\[\frac{x^2}{r^2}.f''(r)-\frac{x^2}{r^3}.f'(r)+\frac{y^2}{r^2}.f''(r)-\frac{y^2}{r^3}.f'(r)+\frac{z^2}{r^2}.f''(r)-\frac{z^2}{r^3}.f'(r)+\frac{3}{r}.f'(r)\]
\[\frac{3}{r}.f'(r)+f''(r)(\frac{x^2+y^2+z^2}{r^2})-\frac{1}{r}.f'(r)(\frac{x^2+y^2+z^2}{r^2})\]
\[\implies \frac{3}{r}.f'(r)+f''(r)-\frac{1}{r}.f'(r)=f''(r)+\frac{2}{r}.f'(r)\]

Answer 2

are you allowed spherical?

https://gyazo.com/abacaf4b950f6e57251b8fdd887162b8

if so, it comes down to expanding out the first term on the RHS of this as everything else = 0

Answer 3

idk, it's just a proof ques, am I right though??

Answer 4

I haven't spherical coordinates before, I did with rectangular coordinate system

Answer 5

i'd need a cold towel to wade through that!!!

i'll print it out and see if i can help....

Answer 6

haven't used*

Answer 7

@ganeshie8 gave me a medal so it must be good, no need to print it out :)

Answer 8

this looks shorter to me

and my head now hurts

Answer 9

Oh you used the quotient rule, I used triple product rule
\[\frac{d}{dx}(u(x).v(x).w(x))=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}\]

Answer 10

wow your method is indeed quite fast @IrishBoy123