Example 12.2.3 (Linear Example).label

1. (1)

   Consider the linear PDE $x_{1}\pd_{1}u+2x_{2}\pd_{2}u+\pd_{3}u=3u$ and $u=\p(x_{1},x_{2})$ on $S$ given by $x_{3}=0$ then

   \begin{align*}\der{x_1}{t}=x_{1}&&\der{x_2}{t}=2x_{2}&&\der{x_3}{t}=1&&\der{u}{t}=3u\end{align*}

   
   

   then

   \begin{align*}(x_{1},x_{2},x_{3},u)|_{t=0}=(s_{1},s_{2},0,\p(x_{1},x_{2}))\end{align*}

   
   

   so

   \begin{align*}x_{1}=s_{1}e^{t}&&x_{2}=s_{2}e^{2}t&&x_{3}=t+s_{3}&&u=s_{4}e^{3t}\end{align*}

   
   

   By initial condition $s_{3}=0,s_{4}=\p(x_{1},x_{2})$. Express $t,s$ as funtions of $x$ we have $t=x_{3},s_{1}=e^{-x_3}x_{1},s_{2}=e^{-2x_3}x_{2}$ thus

   \begin{align*}u=\p(e^{-x_3}x_{1},e^{-2x_3}x_{2})e^{3x_3}\end{align*}

   
   

   We check the Cauchy condition

   \begin{align*}u|_{x_3=0}=\p(x_{1},x_{2})\cd 1\end{align*}

   
   

   and check the PDE is satisfied

   \begin{align*}x_{1}[\p_{,1}e^{-x_3}]+2x_{2}[\p_{,2}e^{-2x_3}]+[-x_{1}\p_{,1}e^{-x_3}+\p_{,2}(-2x_{2})e^{-2x_3}]e^{3x_3}+\p 3e^{3x_3}=3u\end{align*}

   
   
2. (2)

   $u\pd_{1}u+\pd_{2}u=1$ with $S: x_{1}=x_{2}=s$ for $0<s<1$ and $u|_{S}=\frac{1}{2}s$ so

   \begin{align*}\der{x_1}{t}=u&&\der{x_2}{t}=1&&\der{u}{t}=1\end{align*}

   
   

   then

   \begin{align*}(x_{1},x_{2},y)|_{t=0}=(s,s,\frac{1}{2}s)\end{align*}

   
   

   and

   \begin{align*}\det\begin{pmatrix}\pder{x_1}{s}&a_{1}(s,s,\frac{1}{2}s)\\ \pder{x_2}{s}&a_{2}(s,s,\frac{1}{2}s)\end{pmatrix}=\det\begin{pmatrix}1&\frac{1}{2}s\\ 1&1\end{pmatrix}=1-\frac{1}{2}s\neq 0\end{align*}

   
   

   Solving for $x_{1},x_{2},u$ we have

   \begin{align*}x_{2}(s,t)=t+s&&u(s,t)=t+\frac{1}{2}s&&x_{1}(s,t)=\frac{1}{2}t^{2}+\frac{1}{2}st+s\end{align*}

   
   

   Collecting terms we have solve for $s$ and $t$

   \begin{align*}x_{2}-x_{1}=-\frac{1}{2}t(-2+\UBC{t+s}{x_2})\implies t=\frac{2(x_{2}-x_{1})}{2-x_{2}}\end{align*}

   
   

   and

   \begin{align*}x_{2}=t+s\implies s=x_{2}-t=x_{2}-\frac{2(x_{2}-x_{1})}{2-x_{2}}=\frac{-x_{2}^{2}+2x_{1}}{2-x_{2}}\end{align*}

   
   

   Lastly

   \begin{align*}u(x)=t(x)+\frac{1}{2}s(x)=\frac{4x_{2}-2x_{1}-x_{2}^{2}}{2(2-x_{2})}\end{align*}

   
   

   We check that the Cauchy condition are satisfied on $x_{1}=x_{2}=s$

   \begin{align*}u=\frac{4s-2s-s^{2}}{2(2-s)}=\frac{2s-s^{2}}{2(2-s)}=\frac{1}{2}s\end{align*}

   
   

   Check that $u$ solves the PDE

   \begin{align*}\pd_{1}u=\frac{-2}{2(2-x_{2})}=\frac{1}{x_{2}-2}&&\pd_{2}u=\frac{1}{2}\frac{(4-2x_{2})(2-x_{2})-(4x_{2}-2x_{1}-x_{2}^{2})(-1)}{(2-x_{2})^{2}}=\frac{1}{2}\frac{2(2-x_{2})^{2}+4x_{2}-2x_{1}-x_{2}^{2}}{(2-x_{2})^{2}}\end{align*}

   
   

   so

   \begin{align*}u\pd_{1}u+\pd_{2}u=-\frac{4x_{2}-2_{x}1-x_{2}^{2}}{2(2-x_{2})^{2}}+\frac{2(2-x_{2})^{2}+4x_{2}-x_{2}^{2}-2x_{1}}{2(2-x_{2}^{2})}=1\end{align*}

   
   

   We note that at $s=2$ then the solution to pde we wrote isn’t defined since the IFT step breaks down. Indeed $s=2$ the $\det$ step gives $0$ hence not invertible.