Points off for the rude response to Otis' sincere attempt to help you!

Deanne is correct, but you can see this simply even with the operator $\partial_{x_1}$ in $\mathbb R^n$. If $u(0, x_2, \ldots, x_n)$ is nowhere analytic, then the
solution of $\partial u/ \partial_{x_1} = 0$ is constant in $x_1$ but never analytic in the orthogonal directions.

On the other hand, there is a class of operators known as analytic hypoelliptic (which includes both the Laplacian for any analytic metric and the associated heat operator) for which such a statement is true: $L u = f$ and $f$ analytic implies $u$ analytic.

However as the paper you reference indicates, all sorts of wild things can happen in general; that paper gives an example of a constant coefficient system and analytic ``initial data'' where the solution is not analytic. Actually, that is a rather nonstandard sort of initial value problem where Cauchy data for one component is posed on one hypersurface and Cauchy data for the second component is posed on a transverse hypersurface.

systemof linear PDE's with analytic coefficients and initial data, but non-analytic solutions. This should be contrasted with the Cauchy-Kowalewski theorem, which says that for a linear PDE (for a single function) analytic data and coefficeints imply analyticity of solutions and ( in light of @Bazin's answer) suprisingly existence. – Otis Chodosh Aug 15 '12 at 23:50