Laplace's equation - trial I was asked to solve this problem
$$
\begin{cases}
u_{xx}+u_{yy}=0, \quad x\in \mathbb{R}, y>0\\
u(x,0)=f(x)\\
u_y(x,0)=g(x)
\end{cases}
$$
I tried to solve by using the wave-equation strategy.
If the general solution (computed by characteristic polynomial) is
$$u(x,y)=\Phi_1(x-iy)+\Phi_2(x+iy)$$
we have just to apply the conditions. I'll show the passages below
$$
\begin{cases}
u(x,0)=f(x)\\
u_y(x,0)=g(x)
\end{cases}
$$
$$
\begin{cases}
\Phi_1(x)+\Phi_2(x)=f(x)\\
-i\Phi'_1(x)+i\Phi'_2(x)=g(x)
\end{cases}
$$
$$
\begin{cases}
\Phi_2(x)+\Phi_1(x)=f(x)\\
\displaystyle \Phi'_2(x)-\Phi'_1(x)=\frac{g(x)}{i}
\end{cases}
$$
$$
\begin{cases}
\Phi_2(x)+\Phi_1(x)=f(x)\\
\Phi'_2(x)-\Phi'_1(x)=-i g(x)
\end{cases}
$$
Solving, it yields:
$$
\begin{cases}
\displaystyle \Phi_1(x)=\frac{1}{2} f(x) + \frac{i}{2}\int_0^{x} g(s)ds - c\\
\displaystyle \Phi_2(x)=\frac{1}{2} f(x) - \frac{i}{2}\int_0^{x} g(s)ds + c
\end{cases}
$$
So, if
$$u(x,y)=\Phi_1(x-iy)+\Phi_2(x+iy)$$
then 
$$u(x,y)=\frac{1}{2} f(x-iy) + \frac{i}{2}\int_0^{x-iy} g(s)ds - c + \frac{1}{2} f(x+iy) - \frac{i}{2}\int_0^{x+iy} g(s)ds + c$$
and putting in order, I obtain
$$u(x,y)=\frac{1}{2} [f(x-iy) + f(x+iy)] - \frac{j}{2}\int_{x-iy}^{x+iy} g(s)ds$$
Was I right in doing these computations? Am I allowed to use the same strategy I used for wave equation?
Can I leave function $f$ indicated like that or am I obliged to convert them by using the properties of olographic functions? 
Thanks for your attention.
 A: Your solution works if $f$ and $g$ are entire (analytic) functions. In the wave equation, the corresponding method works for arbitrary differentiable or even worse functions, so there is quite a bit of difference.
Also, your solutions are very sensitive to tiny perturbations of the data. As they say "the Cauchy problem for the Laplace equation is ill-posed in the sense of Hadamard". 
To see the ill-posedness, choose $f_k(x)=\frac1k \cos(kx)$ and $g_k(x)=0$. Then the  solution of the problem found by your method is given by $u_k(x,y)=\frac1k \cos(kx)\cosh(ky)$. If you let $k\to\infty$, $f_k$ tends to $0$ uniformly, but $u_k(0,1)=\frac1k \cosh(k)$ is not close to the solution $u=0$ of the problem with $f=0$ and $g=0$. So you have initial values that are arbitrarily close ($f_k$ is close to $f$, and $g_k$ is even equal to $g$) where the solutions are far apart from each other.
You can't even fix this by asking for also some derivatives of $f_k$ being close to those of $f$: the example $f_k(x)=\frac{1}{k^n}\cos(kx)$ has its first $(n-1)$ derivatives tend to zero as $k\to\infty$, but $u_k(x,y)$ does not tend to the zero function.
