Show that the function \(u=u(x,y)\) given by \[u(x,y)=\frac{e^{ny}-e^{-ny}}{2n^2}\sin nx\quad \text{for}\ n\in {\mathbf N}\] is the solution on \(D=\{(x,y)\ | x^2+y^2<1\}\) of the Cauchy problem for the Laplace equation \[\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=0,\quad u(x,0)=0,\quad \frac{\partial u}{\partial y}(x,0)=\frac{\sin nx}{n}.\]
Show that there exist points \((x,y)\in D\) such that \(\displaystyle{\limsup_{n\to\infty} |u(x,y)|=\infty}\).
Solution
Write \[ u_n(x,y)=\frac{\sinh(ny)}{n^2}\sin(nx). \] Then \[ (u_n)_{xx}=-\sinh(ny)\sin(nx), \qquad (u_n)_{yy}=\sinh(ny)\sin(nx), \] so \(\Delta u_n=0\). Also \[ u_n(x,0)=0, \qquad (u_n)_y(x,0)=\frac{\cosh(0)}n\sin(nx)=\frac{\sin(nx)}n. \] Thus \(u_n\) solves the stated Cauchy problem.
Take the fixed point \[ (x,y)=\left(\frac\pi6,\frac12\right). \] It lies in \(D\), because \[ \frac{\pi^2}{36}+\frac14<1. \] For \(n=6k+3\) we have \(|\sin(n\pi/6)|=1\), and hence \[ \left|u_n\left(\frac\pi6,\frac12\right)\right| =\frac{\sinh(n/2)}{n^2}. \] Since \(\sinh(n/2)/n^2\to\infty\), it follows that \[ \limsup_{n\to\infty} \left|u_n\left(\frac\pi6,\frac12\right)\right|=\infty. \] This is Hadamard’s instability phenomenon: the Cauchy data \(\sin(nx)/n\) tend uniformly to \(0\), while the corresponding harmonic solutions are unbounded at a fixed interior point along a subsequence.