Consider the Laplace equation in two dimensions: $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0$$. Construct a sequence of real analytic functions $(u_k)_{k=1}^\infty$ with $u_k:\mathbb{R}^2\to\mathbb{R}$ solving the equation such that for any $l\in\mathbb{N}{:}$ $$\begin{aligned}\sup_x\left|(\partial_x)^lu_k(x,0)\right|+\sup_x\left|(\partial_x)^l\partial_yu_k(x,0)\right|\to0\end{aligned}$$ as $k \to \infty$, but such that for any $e>0$, $\sup_x|u_k(x,\epsilon)|\to\infty.$ What does this mean for the well-posedness of the Cauchy problem for Laplace’s equation?
Attempt: I was given a hint to first solve the Cauchy problem for the initial surface ${y = 0}$ with data $u(x, 0)$ = $\cos kx$, $u_y(x, 0) = 0$ by seeking a solution of the form $u(x, y) = X(x)Y (y).$
After substituting in the last equation I got $-X(x)Y^{\prime\prime}(y)+X^{\prime\prime}(x)Y(y)=0$, but I don't have any ideas how to move forward in this problem
Following the hint, let's seek a solution to Laplace's equation of the form $u(x,y)=Y(y)\cos kx$. Substituting this ansatz in the PDE, we obtain the equation $(-k^2Y(y)+Y''(y))\cos kx=0$, which has as general solution $Y(y)=C_1\cosh ky +C_2\sinh ky.$ Imposing the conditions $u(x,0)=\cos kx, u_y(x,0)=0$, we find $C_1=1$ and $C_2=0$, so $u(x,y)=\cos kx \cosh ky$. Since Laplace's equation is linear, we have the freedom to multiply $u$ by a constant. Using this freedom, we define $u_k$ as $$ u_k(x,y)=C_k\cos kx \cosh ky. \tag{1} $$ Now we must show that it is possible to choose $C_k$ so that $(1)$ satisfies the conditions $$ \lim_{k\to\infty}\left(\sup_{x\in{\mathbb{R}}}\left|(\partial_x)^{\ell}u_k(x,0)\right|+\sup_{x\in{\mathbb{R}}}\left|(\partial_x)^{\ell}\partial_yu_k(x,0)\right|\right)=0\qquad \forall\ell\in\mathbb{N}, \tag{2} $$ $$ \lim_{k\to\infty}\sup_{x\in{\mathbb{R}}}|u_k(x,\epsilon)|=\infty\qquad \forall\epsilon>0, \tag{3} $$ that is, $$ \lim_{k\to\infty}C_k\,k^{\ell}=0\qquad \forall\ell\in\mathbb{N}, \tag{$2'$} $$ $$ \lim_{k\to\infty}C_k\cosh k\epsilon=\infty\qquad \forall\epsilon>0. \tag{$3'$} $$ It's easy to verify that $C_k=e^{-\sqrt{k}}$ satisfies both $(2')$ and $(3')$.