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Let $f(x,y,z)\in C^2 (\mathbb{R}^3 ) $ and assume that there exist a function $g:\mathbb{R}\to \mathbb{R}$ for which $g(u)=f(x,y,z)$ where $u=x^2 + y^2 + z^2 $ .

Prove that $f_{xx} + f_{yy} + f_{zz} $ depends only on $u$ .

I have no idea how to prove it... I guess I should differentiate something but have no idea what.

Will someone please help ?

Thanks !

homogenity
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2 Answers2

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Hint: Explicitly calculate the $f_i$ using the chain rule. E.g. $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} g(u(x,y,z))=?$

vuur
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  • Well, I guess that it equals: $ g'(u) \cdot \frac{\partial u}{\partial x} $ , and thus: $f_{xx} = g''(u)\cdot u_x ^2 + g'(u)\cdot u_{xx} $ . . so we have: $ f_{xx} + f_{yy} +f_{zz} = 4u\cdot g''(u) + 6g'(u) $ which obviously depends only on $u$ . Thanks a lot ! – homogenity Apr 17 '14 at 11:25
  • Changing your question in this way is bad style. At least leave the first question there. To answer your modified question: Simply derive $\frac{\partial f}{\partial x}$ again by $x$, this time using product and chain rule. Edit: yes, that looks about right! – vuur Apr 17 '14 at 11:27
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Hint: $f(x,y,z)=g(x^2+y^2+z^2)$ is homogeneous of degree $2$, thus it satisfies the corresponding Euler differential equation: $$f_x x+f_y y+f_zz=2f.$$

user1337
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