Consider the equation in 2 dimension case ,$\Delta u(x_1,x_2)=0$ ,$u=1$ for $|x|=1$, and $u \rightarrow 0$ as $|x|$ tends to $\infty$.
Do we have a solution?
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Let's asume that a solution exists and consider the conformal transformation $$(y_1,y_2) = \frac{(x_1,-x_2)}{x_1^2 + x_2^2},$$ that maps the exterior of the disk to the interior, and consider the new function $$v(x_1,x_2) = u(y_1,y_2).$$ Because the transformation is conformal, and $u$ is harmonic, then $v$ is harmonic as well. So $$ \Delta v = 0 \quad \forall|x|\leq 1,$$ $$ v = 1 \quad \forall|x|= 1,$$ and the condition at infinity translate into $v(0,0) = 0$. But using the mean value property for $v$ $$v(0,0) = \frac{1}{2\pi}\int_{|x| = 1} v(x) dx = 1,$$ which is a contradiction.
themaker
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Write the equation in polar coordinates. Since the boundary value at $|x|=1$ has circular symmetry, look for a solution depending only on $r$.
Julián Aguirre
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Yes, but if we consider the equation $\Delta u_R(x_1,x_2)=0$ ,$u_R=1$ for $|x|=1$, and $u_R = 0$ if $|x|=R$, we have $u=C\log(|x|/R)$, what happen if we let $R \rightarrow \infty$? – mnmn1993 Nov 21 '16 at 13:39
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1The general solution is $u=A\log r+B$. There are no bounded solutions. – Julián Aguirre Nov 21 '16 at 13:42
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So,there is no solution for my question? – mnmn1993 Nov 21 '16 at 13:48
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1No such solution exists. – Julián Aguirre Nov 21 '16 at 13:48
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@JuliánAguirre This only proof that there are not bounded solution with radial symmetry. – themaker Nov 21 '16 at 14:04