Example 12.3.2 (What can go wrong?).label

1. (1)

   Lack of Uniqueness

   \begin{align*}\begin{cases}\pd_{t}\pd_{x} u=0&u(x,0)=\p_{0}(x)\\ (x,t)\in\R^{2}&\pd_{t}u(x,0)=\p_{1}(x)\end{cases}\end{align*}

   
   

   $t=0$ in characteristic. Observe

   \begin{align*}(\pd_{t}\pd_{x}u)(x,0)=0\implies \pd_{x}\p_{1}(x)=0\implies \p_{1}\text{ const}\end{align*}

   
   

   So the Cauchy problem has no solutions unless $\p_{1}$ is constant. Suppose that $\p_{1}$ is constant then

   \begin{align*}u(x,t)=\p_{0}(x)+f(t)\end{align*}

   
   

   where $f$ is any function such that $f'(0)=\p_{1}$ then we have no uniqueness of solution.

2. (2)

   Dependence of Cauchy Data $\pd_{x}^{2}u-\pd_{t}u=0$, $t=0$ in characteristic. Suppose that $u$ is a solution with $u(x,0)=\p_{0}(x)$. Then $(\pd_{t}u)(x,0)=\pd_{xx}u(x,0)=\p_{0}''(x)$. Then $\p_{1}\define (\pd_{t}u)(x,0)$ is determined by $\p_{0}$ i.e. choosing $\p_{0}$ forces the choices of $\p_{1}$. Looking at characteristic tell us this also since

   \begin{align*}\chi=\xi_{x}^{2}=0\iff\xi_{x}=0\end{align*}

   
   

   so $(\xi_{x},\xi_{t})$ characteristic normal direction is given by $(0,const)$ in particular $t=const$ are characteristic curves.