Remark 12.3.11 (J. Hadamard).label Does the unique $C^{\om}$ solution of a CK equation depend “continuously” on the Cauchy data in any reasonable sense? No. Consider
\begin{align*}\begin{cases}\pd_{1}^{2}u+\pd_{2}^{2}u=0\\ u(x_{1},0)=0&&(\pd_{2}u)(x_{1},0)=ke^{-\sqrt{k}}\sin(kx_{1})\end{cases}\end{align*}
$x_{2}=0$ we have $S$ is non characteristic
\begin{align*}\pd_{2}^{2}u=-\pd_{1}^{2}u\end{align*}
There exists unique $C^{\om}$ solution
\begin{align*}u(x_{1},x_{2})=e^{-\sqrt{k}}\sin(kx_{1})\sinh(kx_{2})\end{align*}
As $k\to\infty$, Cauchy data and all derivatives $\to 0$. For $x_{2}>0$, $u(x_{1},x_{2})\to\infty$. OTOH $\pd_{1}^{2}u+\pd_{2}^{2}u=0$, $u(x_{1},0)=0,\pd_{2}u(x_{1},0)$ has unique $C^{\om}$ solution $u(x_{1},x_{2})=0$ for all $(x_{1},x_{2})\in\R^{2}$.
Post a Comment