Laplace equation with initial conditions So I have Laplace equation:
$$ u_{xx}+u_{yy}= 0 $$
and initial conditions
$$ u(0,y)=0, \;\: u_x(0,y)=y  $$
And I have to solve it.
My solution:
If we assume that the solution is of the form:  $$ u(x,y)=X(x)Y(y) $$
Then follows: $$ u(0,y)=X(0)Y(y)=0; \:\; u_x(0,y)=X'(0)Y(y)=y.$$
Furthermore: $$ u_{xx}(x,y)=X''(x)Y(y); \:\; u_{yy}(x,y)=X(x)Y''(y) $$
From that, we get: $$ \begin{aligned}&\;u_{xx}+u_{yy}=X''(x)Y(y)+X(x)Y''(y)=0 \\ \\\Leftrightarrow& \; X''(x)Y(y)=-X(x)Y''(y)  \\\\
 \Leftrightarrow& \; -\frac{X(x)}{X''(x)}=\frac{Y(y)}{Y''(y)}  \end{aligned}$$
My problem:
Every time I try to go further I get:
$X(0)=0$, which then implies, that $X'(0)=0$, which furthermore is in contradiction with $X'(0)Y(y)=y$
This is a part where I didn't know where to go further, everything I try to insert doesn't end well and I am not sure, whether my approach was good at all. I would appreciate any kind of help!
 A: From your calculation, we have to solve
$$(1)\left\{\begin{aligned}&X''+\lambda X=0\\
&X(0)=0\end{aligned}\right.\qquad\text{and}\qquad 
(2)\left\{\begin{aligned}&Y''-\lambda Y=0\\
&Y(y)=ky
\end{aligned}\right.$$
where $\lambda$ and $k=(X'(0))^{-1}$ are constants.
The nonzero solutions of $(1)$ are
$$X(x)=\left\{\begin{aligned}
&c_1\sin(\sqrt{\lambda} x),\quad&&\text{if } \lambda>0\\
&c_1e^{\sqrt{-\lambda} x}-c_1e^{-\sqrt{-\lambda} x},\quad&&\text{if } \lambda<0\\
& c_1 x, \quad&&\text{if } \lambda=0
\end{aligned}\right.\tag{3}$$
with $c_1\neq 0$. From system $(2)$, we have
$$Y(y)=ky\quad \Rightarrow \quad Y'(y)=k\quad \Rightarrow \quad Y''(y)=0\quad \Rightarrow \quad \lambda=0.$$
Thus, the only possibility is the third case of $(3)$:
$$X(x)=c_1 x,\qquad  Y(y)=ky=(X'(0))^{-1}y=\frac{1}{c_1} y.$$
and the desired solution is
$$u(x,y)=X(x)Y(y)=(c_1 x)\left(\frac{1}{c_1}y\right)=xy.$$
A: Of course, only inspection is sufficient to find the solution $u(x,y)=xy$ ( through trial and error). I suppose that the question is how analytically ? They are several methods. For example :
It is well known that the general solution of
$$u_{xx}+u_{yy}=0\tag 1$$
can be expressed on the form :
$$u(x,y)=F(y+ix)+G(y-ix)\tag 2$$
in which $F$ and $G$ are arbitrary complex functions in the most general case.
The conditions considered here are :
$$u(0,y)=F(y)+G(y)=0\quad\implies\quad F'(y)+G'(y)=0$$
$$u_x(0,y)=iF'(y)-iG'(y))=y$$
Solving for $F'$ leads to :
$$F'(y)=-\frac{i}{2}y$$
Integrating wrt $y$ leads to :
$$F(y)=-\frac{i}{4}y^2+c$$
$$G(y)=-F(y)=\frac{i}{4}y^2-c$$
Now the functions $F$ and $G$ are known. We put them into Eq.$(2)$ where the variables are not $y$ but are $(y+ix)$ and $(y-ix)$ respectively.
$$u=\left(-\frac{i}{4}(y+ix)^2+c\right)+\left(\frac{i}{4}(y-ix)^2-c\right)$$
Expend and simplify. The result is :
$$u(x,y)=xy$$
A: After a little tinkering with multilinear functions, I found a solution to be $u(x,y)=xy$.
