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I have trouble solving the following Evans' PDE problem. I would appreciate it if someone could help me solving it. Thank you very much in advance.

Let $U\subset \mathbb{R}^n$ be an bounded domain with smooth boundary. Assume $u$ is a smooth solution of $$ Lu=-\sum_{i,j=1}^na^{i,j}u_{x_ix_j}=f \ \text{in} \ U, \ \ u=0 \ \text{on} \ \partial U, $$ where $f$ is bounded. Fix $x^0 \in \partial U$. A $barrier$ at $x^0$ is a $C^2$ function $w$ such that $$ Lw\ge 1, \ \ w(x^0)=0, \ \ w|_{\partial U}\ge 0. $$ Show that if $w$ is a barrier at $x^0$, there exists a constant $C$ such that $$ |Du(x^0)|\le C|\frac{\partial w}{\partial \nu}(x^0)|. $$ Note that we assume $a^{i,j}$ are smooth and satisfy uniform ellipcity.

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    Do you mean $Lw \ge 1$ in the definition of a barrier?2012-12-13
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    Fixed the typo in the definition.2012-12-14

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Give two functions $v_{1}= u + \Vert f \Vert_{\infty} w$, $v_{2} = -u + \Vert f \Vert_{\infty} w$. Now, calculate $Lv_1, Lv_2$, and then use maximum principle to obtain the answer .....

Good Luck

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Note that $\nabla u(x_0)$ and $\nabla w(x_0)$ are parallel to each other and to $\nu(x_0)$.

Suppose you have somehow found two $C^2$ functions $w_1$ and $w_2$ with the properties $w_1(x_0) = 0 = w_2(x_0)$ , $w_1 \ge 0 \ge w_2$ on $\partial \Omega$, and $Lw_1 \ge Lu \ge Lw_2$ in $\Omega$.

What can you say about $w_1, w_2, u$? What can you say about their gradients at $x_0$? And can you perhaps construct such functions from what you have?

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    I don't really understand the first sentence. Why are those vectors parallel to each other?2012-12-14
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    Use the assumptions that $u$ and $w$ both attain local minima at $x_0$, when restricted to $\partial \Omega$. This is not pde, it's just calculus of several variables. If $f$ is smooth and defined in a neighborhood of a smooth surface $S$ and $f \ge 0$ on $S$, $f(x_0) = 0$ for $x_0 \in S$, then $\nabla f(x_0)$ is parallel to $\nu$, a unit normal of $S$ at $x_0$. This is also known as Lagrange's multiplier rule.2012-12-14
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    Sorry for annoying question, but I still don't see why $u$ and $w$ attains local minima at $x_0$. Assuming this fact, your argument makes a perfect sense.2012-12-15
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    First convince yourself that $\nabla u(x_0)$ is parallel to $\nu(x_0)$. Do you see why?2012-12-15
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    Yes, $\partial U$ is a contour of $u$, so $\nabla u(x_0)$ is parallel to $\nu(x_0)$.2012-12-15
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    I now see your argument. Weakly minimum principle implies $x_0$ is the minimum of $w$ on $\overline{U}$.2012-12-15