Consider $\Delta u=0$ in $\mathbb{R}^2,$ with the Cauchy data: $$u=0, \frac{\partial u}{\partial x_2} = \frac{1}{n} \sin(n x_1),~~\textrm{on}~\{ X_2 = 0 \}.$$
My approach: WLOG let $X_1(x_1)=X(x)$ and $X_2(x_2)=Y(y).$ Now separation of variables, $u(x , y)=X(x) Y(y)$ yields: $$\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)}=-\lambda~~\textrm{for}~\lambda >0.$$ Now solving the two ODEs yield: $$X(x)=A \cos \sqrt{\lambda}x + B \sin \sqrt{\lambda}x~~,~~Y(y)=C \cosh\sqrt{\lambda}y + B \sinh \sqrt{\lambda}y.$$
Can someone please explain me how to use the Cauchy data to complete the given problem. Thank you for your time.
HINT : $$u(x,y)=\left(A \cos \sqrt{\lambda}x + B \sin \sqrt{\lambda}x\right)\left(C \cosh\sqrt{\lambda}y + D \sinh \sqrt{\lambda}y\right).$$ $$\begin{cases} \left(\frac{\partial u}{\partial y}\right)_{y=0} =\left(A \cos \sqrt{\lambda}x + B \sin \sqrt{\lambda}x\right)D\sqrt{\lambda} =\frac{1}{n} \sin(n x) \quad\to\quad \begin{cases} \sqrt{\lambda}=? \\ A=? \\ BD=? \end{cases} \\ u(x,0)=\left(A \cos \sqrt{\lambda}x + B \sin \sqrt{\lambda}x\right)C=0 \quad\to\quad C=? \end{cases}$$