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Let $U_2,U_3$ be the open unit balls in $\mathbb{R}^2,\mathbb{R}^3$ respectively.

Fact 1: (From Rudin's real complex analysis) Let $u:\partial U_2\rightarrow \mathbb{R}$ be continuous, then there exists a unique continuous function $u_{*}:\overline{U_2}\rightarrow \mathbb{R}$ such that $u_{*}|U_2$ is harmonic. The function $u_{*}$ is given by: $$ u_{*}(r\cos\theta,r\sin\theta)=\begin{cases} \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1-r^2}{1-2r\cos(\theta-t)+r^2}u(\cos(t),\sin(t)) dt & \textrm{if } 0\leq r<1 \textrm{ } \\ u(r\cos\theta,r\sin\theta) & \textrm{if } \,r=1{ } \end{cases} $$

Question 1: Let $v:\partial U_3\rightarrow \mathbb{R}$ be continuous, Is there a continuous function $u_{*}:\overline{U_3}\rightarrow \mathbb{R}$ such that $u_{*}|U_3$ is harmonic ?

Question 2: Is there a known way to construct the function $v_{*}$ in question 1 like how the function $u_{*}$ was constructed in fact 1 ?

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4 Answers 4

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Yes, the result and the method generalises to all dimensions. Let $\omega_n$ be the $n-1$-dimensional volume of the unit sphere in $\mathbb{R}^n$, then

$$P(x,y) = \frac{1}{\omega_n} \cdot \frac{\lVert y\rVert^2-\lVert x\rVert^2}{\lVert y-x\rVert^n}$$

is the Poisson kernel, and the Dirichlet problem on the unit sphere is solved by the Poisson integral

$$v_\ast (x) = \int_{\lVert y\rVert = 1} P(x,y)\cdot v(y)\,dS(y),$$

where $dS(y)$ is the surface measure on the unit sphere.

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  • $\begingroup$ Very interesting. My initial guess was that this result is for dimension 2 only, since I proved it in the book using tools from complex analysis and holomorphic functions $f:\mathbb{C}\rightarrow \mathbb{C}$ $\endgroup$
    – Amr
    Jan 22, 2014 at 19:27
  • $\begingroup$ The connection between harmonic and holomorphic functions in dimension $2$ makes for a particularly easy proof of Poisson's integral formula in dimension $2$, but the result is general. $\endgroup$ Jan 22, 2014 at 19:31
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In fact this works in any dimension. What you are asking is a special case of the following Cauchy problem: to find a function $u\in C^2(\Omega)$ for $\Omega\subset\mathbb{R}^n$ open such that $$\left\{\begin{array}{ll}\Delta u = 0 & \text{in}\,\,U\\ u = u_0 & \text{on}\,\,\partial U\end{array}\right. $$ where $u_0$ is a given continuous function on the boundary. The first equation is also called Laplace equation. The solution $u$ is called harmonic and is in fact not only twice differentiable, but even analytic.

Let us assume $\Omega$ to be $B(0,1)$ the unit ball. Then the solution can be represented explicitly by first constructing a fundamental solution using Green's function for the unit ball.

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  • $\begingroup$ the solution is infinitely differentiable for all dimensions !! I already knew that the solution was had partial derivates for all orders in dimesnion 2, because every 2D harmonic function is locally the real part of a holomorphic function $\endgroup$
    – Amr
    Jan 22, 2014 at 19:30
  • $\begingroup$ Yes, analytic means infinitely differentiable and representable as a power series. $\endgroup$
    – J.R.
    Jan 22, 2014 at 19:32
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Yes this can be done using the method of Green's functions. If you specify the values of a continuous function on the surface of the unit sphere (This is called a potential function) then you can extend that function into the interior of the sphere as a harmonic function.

$$ \Phi(\vec{x}) = \frac{1}{4\pi} \int_S \Phi(a,\theta',\phi') \frac{a(a^2-x^2)}{(x^2+a^2-2ax \cos(\gamma))^{3/2} } d \Omega' $$

Where $a$ is the radius of the sphere.

$$ \vec{x} = x\left[\cos(\phi)\sin(\theta) \vec{e}_1+\sin(\phi)\sin(\theta) \vec{e}_2+\cos(\theta) \vec{e}_3 \right] $$

$$ d\Omega' = \sin(\theta')d\theta' d\phi'$$

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There's a Poisson Integral Representation for every dimension. There's a good Wikipedia page for this: http://en.wikipedia.org/wiki/Poisson_kernel . And, you'll find corresponding ones for half-spaces on the same page, too. You can read more of the details on the Wikipedia page.

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