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If $U=f(r)$ where $r^2 = x^2 + y^2 +z^2$ , what is $U_{xx} + U_{yy} + U_{zz}=$
The result must be in terms of $r, f(r)$ and its derivatives only

I have obtained $f_x(r)=f^{'}(r)\frac{\partial r}{\partial x}= f^{'}(r) \frac{x}{r}$
I'm confused with $f_{xx}(r)$
mahes
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1 Answers1

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We have, as you write correctly, $$ \def\p#1#2#3{\frac{\partial^{#3} #2}{\partial #1^{#3}}}\p x{}{} f(r) = f'(r)\p x r{} $$ Taking another derivative, we have, using the product rule first $$ \p x{}2 f(r) = \p x{}{}\left(f'(r) \p xr{}\right) = \p x{}{} f'(r) \p xr{} + f'(r) \p xr2$$ Using the chain rule in the first term as for the first derivative, gives $$ \p x{}2 f(r) = f''(r) \left(\p xr{}\right)^2 + f'(r)\p xr2 $$ Now, $$ \p xr{} = \frac xr, \quad \p xr2 = \frac{r - \frac{x^2}r}{r^2} = \frac{r^2 - x^2}{r^3}$$ Hence $$ \p x{}2 f(r) = f''(r) \frac{x^2}{r^2} + f'(r) \frac{r^2 - x^2}{r^3} $$ Due to symmetry, the $y$- and $z$-derivatives are analogous.

martini
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