PDE IVP : $\Delta u(x,y)=e^{-x^4-y^4}, \; \; u_y(x,0) = e^{-|x|}\sin(x)$ Exercise :

Give the solution of the following IVP in integral form :
  $$\begin{cases} \Delta u(x,y)=e^{-x^4-y^4}, &  (x,y) \in (-\infty,\infty)\times(0,\infty) \\ u_y(x,0)=e^{-|x|}\sin(x), & x\in(-\infty, \infty)\end{cases}$$

Question :
A general way in such cases is forming the characteristics problem and finding two integral curves in order to form the integral solution of the IVP. 
But in this case, there are $2$ things I have never encountered again :
(1) The function $e^{-x^4-y^4}$ does not have a standard integral, thus the characteristic's problem is a mess (can't be solved with what I know until now anyway)
(2) The initial value is described in the form of a derivative of the solution and also has an absolute value.
Any tips or thorough explanation/solution will be much appreciated !
 A: The question asks to give the solution in integral form, so Green's function is the way to go here.
For the general problem
\begin{align}
\nabla^2u &= \phi(x,y), & (x,y) &\in D \\
\nabla u\cdot \textbf{n} &= f(x), & (x,y) &\in \partial D &
\end{align}
where $D =\{(x,y): y > 0\}$ is the upper-half plane, the corresponding Green's function satisfies
\begin{align} 
\nabla^2G &= \delta(\xi-x,\eta-y), & (\xi,\eta) &\in D  \\ 
\nabla G \cdot \textbf{n} &= 0, & (\xi,\eta) &\in \partial D
\end{align}
We can use the method of images to derive
$$ G(\xi,\eta,x,y) = \frac{1}{4\pi}\Big[\ln\big[(\xi-x)^2+(\eta-y)^2\big] + \ln\big[(\xi-x)^2+(\eta+y)^2\big] \Big] $$
Green's second identity gives
$$ \iint_D (u\nabla^2G - G\nabla^2u)dA = \int_{\partial D} (u\nabla G - G\nabla u)\cdot\textbf{n}\ dS $$
Plugging in appropriate values gives
\begin{align}
u(x,y) &= \iint_D G\phi\ dA - \int_{\partial D} G f\ dS \\
&= \int_{0}^\infty \int_{-\infty}^\infty G\phi(\xi,\eta)\ d\xi d\eta - \int_{-\infty}^\infty G\big|_{\eta=0}f(\xi)\ d\xi
\end{align}
