In a question asking to find a function $f$ such that for $\Vert y\Vert <1$, satisfies $\displaystyle f(y)=\int_{S^3}{1\over |x-y|^2}|dx|$, an included hint suggested finding $\nabla \cdot \nabla f$ and I don't understand how to differentiate under an volume integral.

Actually its $d\sigma_3(x)$, where $d\sigma(x)$ which was said to be a notation of a surface area, if I am correct (I am looking for the it in the notes)

It seems to have to do with average of a function, but a brief solution to this question first says that by computing the $\Delta f$ "one can see that it is zero", and then it immediately discusses the average value and orientations and all, but only after presumably calculating it. is it really that obvious that it includes no leads on the computation itself?

I've been spending hours on it, trying to find other ranges, but all it does is confuse me even more. Could use some help here.

Laplace Equation: $\displaystyle\Delta g=\nabla \cdot \nabla g=\sum_{i=1}^{d}{{\partial^2 g}\over \partial x_i^2}$

  • $\begingroup$ What does it mean $|\mathrm{d}x|$? It's $\mathrm{d}\Vert x\Vert$? (I'm used to differentiate absolute value ($|x|$) and vector norm ($\Vert x\Vert$) in general) $\endgroup$ Aug 28, 2023 at 23:00
  • $\begingroup$ Hi @MathAttack I addressed it, please check my modified question, although it's possible I was mistaken in the exact definition, hoping you'll recognize the notation $\endgroup$
    – Meitar
    Aug 28, 2023 at 23:01
  • 1
    $\begingroup$ Chain rule says $$\nabla\frac{1}{|x-y|^2} = \frac{2(x-y)}{|x-y|^4}$$ I'm sure you can compute the divergence from here on your own by using product rule: $\nabla\cdot(f\:\mathbf{F}) = \nabla f \cdot \mathbf{F} + f\:\nabla\cdot\mathbf{F}$ $\endgroup$ Aug 29, 2023 at 0:04

1 Answer 1


Chain rule allows us to deduce a higher dimensional power rule. We know that the gradient of a scalar should be a vector and the scaling should follow power rule to make sense. Therefore a good guess would be

$$\nabla|x|^n = n|x|^{n-2}x$$

and you can verify this simplifies to normal power rule in $1$ dimension. You can prove this by chain rule

$$\nabla (|x|^2) = \nabla (x\cdot x) = 2x \implies \nabla|x|^n = \nabla(|x|^2)^{\frac{n}{2}} = \frac{n}{2}(|x|^2)^{\frac{n-2}{2}}\:2x = n|x|^{n-2}x$$

We can immediately compute the gradient then divergence of the integrand via chain rule and product rule

$$\nabla\frac{1}{|x-y|^2} = -\frac{1}{\left(|x-y|^2\right)^2}\cdot 2(y-x) = \frac{2(x-y)}{|x-y|^4}$$

$$\nabla\cdot\frac{2(x-y)}{|x-y|^4} = \frac{2(-4)}{|x-y|^4} - \frac{4(x-y)}{\left(|x-y|^2\right)^3}\cdot2(y-x) = \frac{-8}{|x-y|^4}+\frac{8|x-y|^2}{|x-y|^6} = 0$$

which also integrates to $0$, hence the integrand and the integral are harmonic on the unit ball in $\Bbb{R}^4$.


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