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
Goal: (1) Verify that \(u_n(x,y) = \frac{e^{ny} - e^{-ny}}{2n^2}\sin(nx)\) solves, on \(D = \{x^2 + y^2 < 1\}\), the Cauchy problem \(\Delta u = 0\), \(u(x,0) = 0\), \(u_y(x,0) = \frac{\sin(nx)}{n}\); (2) show there exist points \((x,y) \in D\) with \(\limsup_{n \to \infty} \abs{u_n(x,y)} = \infty\).
1. Verify the boundary conditions at \(y = 0\).
\(u_n(x,0) = \frac{e^0 - e^0}{2n^2}\sin(nx) = 0\). And \(\dd{u_n}{y} = \frac{n e^{ny} + n e^{-ny}}{2n^2}\sin(nx) = \frac{e^{ny} + e^{-ny}}{2n}\sin(nx)\), so \(\dd{u_n}{y}(x,0) = \frac{2}{2n}\sin(nx) = \frac{\sin(nx)}{n}\).
2. Verify \(\Delta u_n = 0\) on \(D\).
\({\partial^2 u_n \over \partial x^2} = -n^2 u_n\) (since \({d^2 \over dx^2}\sin(nx) = -n^2\sin(nx)\)), and \({\partial^2 u_n \over \partial y^2} = \frac{n^2 e^{ny} - n^2 e^{-ny}}{2n^2}\sin(nx) = \frac{e^{ny} - e^{-ny}}{2}\sin(nx) = n^2 u_n\). Hence \(u_{xx} + u_{yy} = -n^2 u_n + n^2 u_n = 0\).
3. For (2), choose \(x \in (0,1)\) with \(x/\pi\) irrational.
Any such \(x\) has \(\abs{x} < 1\), so \((x,y) \in D\) whenever \(y\) is small enough that \(x^2 + y^2 < 1\); e.g. \(x = 1/\pi\) and \(y = 1/2\) works since \(1/\pi^2 + 1/4 < 1\).
4. Along a subsequence of \(n\)’s, \(\abs{\sin(nx)} \ge 1/2\).
The sequence \(nx \bmod 2\pi\) is equidistributed (Weyl’s criterion), because \(x/2\pi = (x/\pi)/2\) is irrational; hence \(\abs{\sin(nx)} \ge 1/2\) for infinitely many \(n\).
5. For fixed \(y > 0\), \(\abs{u_n(x,y)} \to \infty\) along that subsequence.
Along the subsequence of 4, \(\abs{u_n(x,y)} \ge \frac{1}{2}\frac{e^{ny} - e^{-ny}}{2n^2} \ge \frac{e^{ny}}{4n^2}\qty(1 - e^{-2ny})\), and for fixed \(y > 0\), \(e^{ny}/n^2 \to \infty\) and \(1 - e^{-2ny} \to 1\). (For \(y < 0\) the same argument applies with \(e^{-ny}\) in place of \(e^{ny}\).)
6. Q.E.D.
<1>1–<1>2 verify (1); <1>3–<1>5 verify (2) at any point \((x,y) \in D\) with \(y \neq 0\) and \(x/\pi\) irrational, e.g. \((1/\pi, 1/2)\).