2012.03.11 偏微分方程花括号（大括号）的使用方法

完成效果：

http://s10/middle/a3b863dagc00a507438c9&690
代码：

More definitely, the problem is
\begin{equation}\label{pb1}
\left \{
\begin{array}{rl}
    -\lambda^2 \Delta u=g(x)-e^{-u}, \quad & in \Omega \\
    u=h(x),\quad & on \partial\Omega \\
\end{array}
\right.
\end{equation}

注意：

1.两个方程行都加换行符\\

2.对齐的位置&

3.\right后面不要忘记带“.”,否则无法执行

2012.04.21 公式对齐的命令

http://s2/middle/84fde258gbe272f460741&690

That $E$ is lower semi-continuous, hold since $E$ is both convex and Gateaux-differentiable.
\[\begin{aligned}
E(u)=&\int_\Omega \frac{1}{2} \lambda^2 |Du|^2+\frac{1}{2}ku^2-f(x)udx\\
\geqslant &\int_\Omega \frac{1}{2} \lambda^2 |Du|^2+\frac{1}{2}ku^2dx-\frac{1}{2}\int_\Omega f(x)^2dx-\frac{1}{2}\int_\Omega u^2dx\\
\geqslant &\int_\Omega \frac{1}{2} \lambda^2 |Du|^2+\frac{1}{2}(k-1)u^2dx-\frac{1}{2}\int_\Omega f(x)^2dx\\
\geqslant &C_0 \|u\|_{1,2}^2 -C_1,~\forall u \in M
\end{aligned}\]
where $C_0=\min {\{\frac{1}{2} \lambda^2, \frac{1}{2}(k-1)\}}, C_1=\frac{1}{2}\int_\Omega f(x)^2dx$.