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This is Question 2.18 from Gilbarg and Trudinger, chapter 2.

We are given that $\Omega$ is open bounded smooth boundary. Now fix $x_0\in \Omega$ and a constant $c>0$ such that $B(x_0,c)\subset\subset \Omega$. Next, given $u$ such that $\Delta u=0$ in $\Omega$ and another $2$ positive constants $a$ and $b$ such that $a<b<c$ and $b^2=ac$. Then, the question asks us to prove the following simple and beautiful equation.

$$ \int_{\partial B(0,1)}u(x_0+aw)u(x_0+cw)\,dSw=\int_{\partial B(0,1)}u^2(x_0+bw)\,dSw $$

My try:

Define $$ v(w):=u(x_0+aw)u(x_0+cw)- u^2(x_0+bw)$$ If I can prove $v$ is harmonic, then by Mean Value Theorem I would be done, since $$ 0=v(0)=\int_{\partial B(0,1)} v(w)\,dSw$$ Hence, by $v\in C^2$, I just compute $\Delta v$ and obtain that, after simplification, $$ \Delta v(w)=2b^2(\nabla u(x_0+aw)\cdot\nabla u(x_0+cw)-\nabla u(x_0+bw)\cdot\nabla u(x_0+bw)) $$ I feel I am close but I can not go future from here. Any hint would be very welcome!

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  • $\begingroup$ Generally, you should not expect nonlinear transformations to preserve harmonicity. A counterexample: $u(x,y)=x+xy$ is harmonic but $v(x,y)=b^2(a+c-2b)x^2y$ is not. $\endgroup$ – user147263 Nov 25 '14 at 6:24
  • $\begingroup$ Yes you were right... I shouldn't expect my try will work. Thx for the counter example ! $\endgroup$ – spatially Nov 25 '14 at 14:03
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Let $B$ be the unit ball. Applying Green's second identity to $\phi(w)=u(aw)$ and $\psi(w)=u(cw)$ yields $$ \int_B \psi \frac{\partial \phi}{\partial n}= \int_B \phi \frac{\partial \psi}{\partial n}\tag{1} $$ hence (by the Chain Rule) $$ a \int_B u(cw) u_r(aw)= c \int_B u(aw)u_r(cw) \tag{2} $$ where $u_r$ is the derivative of $u$ in the radial direction.

On the other hand, differentiating the function $$ I(a) = \int_{\partial B}u(aw)u((b^2/a)w) \tag{3} $$ we get $$ I'(a) = \int_{\partial B}\left(u_r(aw)u((b^2/a)w)-\frac{b^2}{a^2}u(aw)u_r((b^2/a)w)\right) =0 \tag{4} $$ according to (2). Thus, $I$ is a constant function, which implies the claim.

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  • $\begingroup$ $a$ is a constant. How you define a function $I$ of $a$? Also, if we suppose we can do that, then (2) doesn't imply (4) because $$ a \int_{\partial B} u(cw) u_r(aw) \neq a \int_{\partial B} u(cw) u_r(aw) w$$ $\endgroup$ – Ruzayqat Aug 1 '15 at 4:14
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1- $a<b<c \text{ and }b^2 = ac\implies a= b\cdot\dfrac{b}{c}:=b\alpha$ also $c= \dfrac{b}{\alpha}$. Consider $f: (0,1)\mapsto \mathbb{R}$ with $$f(\alpha) =\int_{|\omega|=1}u(x_0+b\alpha\omega)u(x_0+\frac{b}{\alpha}\omega)d\omega $$

Then $f$ is well defines and smooth with vanishing derivative thanks to Harmonicity of $u$ and the Green identity as follows:

\begin{eqnarray} \frac{d}{dr}\left(\int_{|\omega|=1}u(x_0+r\omega)d\omega.\right)&= & \int_{|\omega|=1} \nabla u(x_0 +r\omega)\cdot w d\omega\\ &=& \frac{1}{r^{n-1}}\int_{|y-x_0|= r} \nabla u(y)\cdot \frac{y}{r}d\sigma(y)\\ &=&\frac{1}{r^{n-1}}\int_{|y-x_0|<r} \Delta u(y) dy =0 \end{eqnarray} where $0<r<c.$ Therefore, $f$ is constant whence $$f(1) = \lim_{\alpha \to 1^-} f(\alpha)= \int_{|\omega|=1}u^2(x_0+b\omega)d\omega.$$

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