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Here's problem.

Let $U$ be a bounded domain in $\mathbb{R}^{n}$ and $\vec{b} : \mathbb{R}^{n} \to \mathbb{R}^{n}$ and $g: \mathbb{R}^{n} \to \mathbb{R}$ be continuous. Show that there can be at most one solution $u \in C^{2}(U) \cap C(\bar{U})$ of the equation;

$-\bigtriangleup u + \bar{b}\cdot Du = 0 $ in $U$

$u = g $ in $\partial U$.

First of all, I think using maximum principle to this; If $x_0 \in U$ is maximum point, then $\Delta U(x_0) = 0$,so $\bigtriangleup u (x_0) = 0$. However I don't know how to expand this approach. Anyone have idea?

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  • $\begingroup$ May be $\bigtriangleup u(x_0)<0$ and $D u(x_0)=0$? $\endgroup$ – gaoxinge Jan 29 '14 at 12:18
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    $\begingroup$ @gaoxinge Is $\bigtriangleup u(x_0)$ negative only? I think 0 is also possible value. If it is negative, then $u$ has no maximal point, and similarly no minimal point on $U$. Then it is contradictory since it violates extreme value theorem. $\endgroup$ – user124697 Jan 29 '14 at 12:21
  • $\begingroup$ Oh, it can take $0$ and $\bigtriangleup u(x_0)\leq{0}$@user124697 $\endgroup$ – gaoxinge Jan 29 '14 at 12:25
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    $\begingroup$ @gaoxinge Yes. By given condition, $\bigtriangleup u(x_0) = 0$. But I don't know what do I have to do solve this problem... Thank you! $\endgroup$ – user124697 Jan 29 '14 at 12:28
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It is not clear to me what $\bar{b}$, $D$ and $\bar{U}$ are. And is $U$ a single open set or a disjoint number of sets, $\bar{U}$ being its closure? Assuming that $\bar{b}=\mathbf{b}\in \mathbb{R}^{n}$ and $% D=\partial _{\mathbf{x}}$ then $$ (-\partial _{\mathbf{x}}^{2}+\mathbf{b\cdot }\partial _{\mathbf{x}})u=0,\;% \mathbf{x}\in U,\;u(\mathbf{x})=g(\mathbf{x}),\;\mathbf{x}\in \partial U. $$ Writing $$ v(\mathbf{x})=\exp [-\frac{1}{2}\mathbf{x\cdot b}]u(\mathbf{x}), $$ we obtain $$ (-\partial _{\mathbf{x}}^{2}+\frac{1}{4}b^{2})v(\mathbf{x})=0,\;\mathbf{x}% \in U,\;v(\mathbf{x})=\exp [-\frac{1}{2}\mathbf{x\cdot b}]g(\mathbf{x})=h(% \mathbf{x}),\;\mathbf{x}\in \partial U. $$ In a few simple cases it is possible to obtain $v(\mathbf{x})$ and hence $u(% \mathbf{x})$ explicitly and has the required properties.

In one dimension let $U=[-d,+d]$. Then, for $x\in U$ the general solution is a linear combination of the eigenfunctions of $-\partial _{\mathbf{x}}^{2}$ at the eigenvalue $-\frac{1}{4}b^{2}$, $$ v(x)=\alpha \exp [ikx]+\beta \exp [-ikx],\;x\in (-d,+d),\;k^{2}=-\frac{1}{4}% b^{2}, $$ so $$ v(x)=\alpha \exp [\frac{bx}{2}]+\beta \exp [-\frac{bx}{2}] $$ Now $$ v(-d)=\alpha \exp [-\frac{bd}{2}]+\beta \exp [+\frac{bd}{2}% ]=h(-d),\\v(+d)=\alpha \exp [+\frac{bd}{2}]+\beta \exp [-\frac{bd}{2}]=h(+d), $$ which gives unique $\alpha $ and $\beta $, so there is only one solution, which has the required properties.

In general $v(\mathbf{x})$ is again a linear combination of the eigenfunctions of $-\partial _{\mathbf{x}}^{2}$ at the eigenvalue $-\frac{1} {4}b^{2}$ for $\mathbf{x}$ in $U$.

In three dimensions they satisfy $$ (-\partial _{\mathbf{x}}^{2}+\frac{1}{4}b^{2})\varphi =(-\frac{1}{x}\partial _{x}^{2}x+\frac{\mathbf{l}^{2}}{x^{2}}+\frac{1}{4}b^{2})\varphi =0. $$ Let ($\mathbf{x}=x\mathbf{e}_{\mathbf{x}}$) $$\begin{eqnarray*} \varphi _{l}^{m}(\mathbf{x}) &=&\rho _{l}(x)Y_{l}^{m}(\mathbf{e}_{\mathbf{x}% }),\;l=0,1,2,\ldots \\ (-\frac{1}{x}\partial _{x}^{2}x+\frac{l(l+1)}{x^{2}}+\frac{1}{4}b^{2})\rho _{l}(x) &=&0. \end{eqnarray*}$$ The general expression for $v(\mathbf{x})$ is then (see, for instance, A. Messiah, Quantum Mechanics I) $$ v(\mathbf{x)=}\sum_{l,m}c_{lm}\varphi _{l}^{m}(\mathbf{x})=\sum_{l,m}c_{lm}% \rho _{l}(x)Y_{l}^{m}(\mathbf{e}_{\mathbf{x}}),\;\mathbf{x}\in U. $$ Now consider the special case that $U$ is a sphere with radius $r$ centered in the origin and suppose further that $h(\mathbf{x})=h$, fixed, for all $% \mathbf{x}\in \partial U$. Then $$ v(r\mathbf{e}_{\mathbf{x}})=\sum_{l,m}c_{lm}\rho _{l}(r)Y_{l}^{m}(\mathbf{e}% _{\mathbf{x}})=h, $$ which requires $l=m=0$, leaving $$ v(r\mathbf{e}_{\mathbf{x}})=c_{00}\rho _{0}(r)=h $$ Here $$ (-\frac{1}{x}\partial _{x}^{2}x+\frac{1}{4}b^{2})\rho _{0}(x)=0\;\Rightarrow \;(-\partial _{x}^{2}+\frac{1}{4}b^{2})x\rho _{0}(x)=0, $$ so $$\begin{eqnarray*} x\rho _{0}(x) &=&\alpha \exp [\frac{bx}{2}]+\beta \exp [-\frac{bx}{2}], \\ \rho _{0}(x) &=&\frac{1}{x}\{\alpha \exp [\frac{bx}{2}]+\beta \exp [-\frac{bx% }{2}]\} \end{eqnarray*}$$ Regularity in the origin requires $\beta =-\alpha $, so $$\begin{eqnarray*} \rho _{0}(x) &=&\alpha \frac{1}{x}\{\exp [\frac{bx}{2}]-\exp [-\frac{bx}{2}% ]\}, \\ v(\mathbf{x}) &=&c_{00}\alpha \frac{1}{x}\{\exp [\frac{bx}{2}]-\exp [-\frac{% bx}{2}]\}, \\ v(r\mathbf{e}_{\mathbf{x}}) &=&c_{00}\alpha \frac{1}{r}\{\exp [\frac{br}{2}% ]-\exp [-\frac{br}{2}]\}=h, \end{eqnarray*}$$ which fixes $c_{00}\alpha $.

For a more general $h=h(r\mathbf{e}_{\mathbf{x} })$ it can be expanded in spherical harmonics, giving $c_{lm}.$ Now $\rho _{l}(x)$ is associated with a spherical Bessel function.

In general, finding a solution for $v(\mathbf{x})$ in $U$ and then using the boundary conditions on $\partial U$ should work but may be complicated for more general $U$ and $h(\mathbf{x}).$

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