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Let $f$ be a $C^{2}$ function on an open set which contains $\overline{B(x, r)} .$ We can use first Green's formula and then differentiation under the integral sign to obtain $$ \int_{B(x, t)} \Delta f d \lambda=\int_{S(x, t)} \frac{\partial f}{\partial n_{e}} d \sigma \quad(0<t \leq r) $$ where $\partial / \partial n_{e}$ denotes the exterior normal derivative, and then $$ \int_{B(x, t)} \Delta f d \lambda=t^{N-1} \int_{S} \frac{\partial}{\partial t} f(x+t y) d \sigma(y) $$ Source : Classical Potential Theory, Armitage and Gardiner.

I didnt understand what they mean by exterior normal derivative and how they get the second formula ? Any help is really appreciated !

BRH
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  • Please define all involved data. What is $B(x,t)$? (And what is $x$, what is $t$?) Where is $f$ defined? What is $S(x,t)$ and what is $S$ (without $x,t$ as arguments)? The exterior normal derivative of a function is in such cases the derivation of a function defined on some boundary of a manifold with boundary included in some $\Bbb R^N$ in the direction of the normal to the boundary pointing outside (into the exterior). I suppose $S(x,t)$ is the sphere centered in $x$ with radius $t$ and by abuse, $S$ (without parameters) is $S(0,1)$. Then parametrize $S(x,t)$ as $S\ni y\to x+ty\in S(x,t)$. – dan_fulea Oct 28 '21 at 09:35
  • I hope now is everything clear. That radial derivative brings the factor $t^{N-1}$ when the ball $B(x,t)$ is in $N$ dimensions. So the sphere is in $(N-1)$ dimensions, and the Jacobian involved brings the factor $t^{N-1}$ into play. – dan_fulea Oct 28 '21 at 09:38
  • Yes it is! Thank you so much ! $B(x,t)={x\in \mathbb R^N : |x|<1}$ and $S$ is not defined there but I think it's the unit sphere. – BRH Oct 28 '21 at 09:58
  • So, the normal derivative of a function $f$ is just $\nabla f.n$ with $n$ the normal derivative ! – BRH Oct 28 '21 at 09:59
  • The derivative of a function $g$ (defined on $\Bbb R^N$ or at least in some neighbourhood of $S(x,t)$) in the normal direction, applied in a point $x+ty$ is the derivative of the function of one variable $\epsilon$:$$\epsilon\to g(x+(t+\epsilon)y)\ .$$ – dan_fulea Oct 28 '21 at 10:27
  • I do now ! Thank you for helping me! – BRH Oct 28 '21 at 10:40

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