= Solution
Choose
$$
\boxed{u_n(x,y)=\frac1n\cos(nx)\cosh(ny),\qquad
v_n(x,y)=-\frac1n\sin(nx)\sinh(ny)}.
$$
Direct differentiation gives $u_{ny}=v_{nx}$ and $v_{ny}=-u_{nx}$. Both functions are smooth and $2\pi$-periodic in $x$, with initial data $f_n(x)=n^{-1}\cos nx$, $v_n(x,0)=0$. Their squared norms are
$$
\int_0^{2\pi}|f_n|^2dx=\frac\pi{n^2}\to0,\qquad
\int_0^{2\pi}|u_n(x,y)|^2dx=\frac\pi{n^2}\cosh^2(ny)\to\infty
$$
for every fixed $y\ne0$. Thus \b[continuous dependence on the initial data fails in this $L^2$ topology], even on smooth analytic data. The <Cauchy problem for a partial differential equation> is ill-posed in the sense of <Hadamard well-posedness>; existence of analytic solutions from the preceding theorem does not supply stability for an elliptic Cauchy problem.
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