let $R:= (0,1) \times (0,1) \subset \mathbb{R^2}$ be the unit square and let $u(x,y)$ be the answer of $\Delta u(x,y)=0 \space in \space R,$ and $u(x,y)=f(x,y) \space on \space \partial R$
Find a solution $u$ that satisfies the following boundary conditions:
$\mathrm{f}(x, y) = \begin{cases} x & \text{if}\space y = 0 \\ x^2 & \text{if}\space y=1\\ 0 & \text{if}\space x=0\\ 1 & \text{if}\space x=1 \end{cases}$
My attempt:
rewriting the conditions:
$\begin{cases} u(x,0)=x & \text{for}\space x\in [0,1] \\ u(x,1)=x^2 & \text{for}\space x\in [0,1]\\ u(0,y)=0 & \text{for}\space y\in [0,1]\\ u(1,y)=1 & \text{for}\space y\in [0,1] \end{cases}$
With separation of variables method:
$u(x,y)=X(x)Y(y)$
$\Rightarrow \Delta u=X''Y+Y''X=0 $$\quad \Rightarrow \frac{X''}{X}=-\frac{Y''}{Y}=-k$
$X''+kX=0$
$Y''-kY=0$
1:
$X''+kX=0$ takes the solution:
$X(x)=A\cos(\sqrt{k}x)+B\sin(\sqrt{k}x)$ $\quad \Rightarrow X(0)=0 \space \Rightarrow A=0$
Now I don't know what to do with $X(1)=1$. Can someone help please.
Notice that $w(x,y):=x+y(x^2-x)$ satisfies the equation $\Delta w = 2y$ and the boundary conditions $w(x,0)=x$, $w(x,1)=x^2$, $w(0,y)=0$, and $w(1,y)=1$. Therefore, if $u$ is a solution to the original problem, then $v:=u-w$ is a solution to the Poisson equation $$ \Delta v=-2y \tag{1} $$ with homogeneous boundary conditions \begin{align} v(x,0)&=0, \\ v(x,1)&=0, \\ v(0,y)&=0, \\ v(1,y)&=0. \tag{2} \end{align} This problem can be solved using Fourier series. Write $v(x,y)$ as $$ v(x,y)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{mn}\sin(m\pi x)\sin(n\pi y) \tag{3} $$ and plug into $(1)$: $$ -\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{mn}(m^2+n^2)\pi^2\sin(m\pi x)\sin(n\pi y)=-2y. \tag{4} $$ Using the orthogonality of the Fourier basis, one obtains the following expression for the coefficients $A_{mn}$: $$ A_{mn}=\frac{8}{(m^2+n^2)\pi^2}\int_0^1\int_0^1y\sin(m\pi x)\sin(n\pi y)\,dxdy. \tag{5} $$ The evaluation of the double integral is straightforward.
Addendum
For completeness, let me show another method of solution (the one suggested by @Artem): write $u$ as $u=u_1+u_2$, where $u_1$ and $u_2$ are solutions to the Laplace equation satisfying the following boundary conditions: $$ \begin{cases} u_1(x,0)=x, \\ u_1(x,1)=x^2, \\ u_1(0,y)=0, \\ u_1(1,y)=0, \end{cases} \qquad \text{and} \qquad \begin{cases} u_2(x,0)=0, \\ u_2(x,1)=0, \\ u_2(0,y)=0, \\ u_2(1,y)=1. \end{cases} \tag{A.1} $$ Using the method of separation of variables, and respecting the homogeneous boundary conditions in each case, we find the following expressions for $u_1$ and $u_2$: \begin{align} u_1(x,y)&=\sum_{n=1}^{\infty}\sin(n\pi x)[A_n\cosh(n\pi y)+B_n\sinh(n\pi y)], \tag{A.2} \\ u_2(x,y)&=\sum_{n=1}^{\infty}C_n\sinh(n\pi x)\sin(n\pi y). \tag{A.3} \end{align} The coefficients $A_n$, $B_n$ and $C_n$ are determined by the remaining boundary conditions: \begin{align} u_1(x,0)&=\sum_{n=1}^{\infty}A_n\sin(n\pi x)=x \implies A_n=2\int_0^1 x\sin(n\pi x)\,dx, \tag{A.4} \\ u_1(x,1)&=\sum_{n=1}^{\infty}[A_n\cosh(n\pi)+B_n\sinh(n\pi)]\sin(n\pi x)=x^2 \\ &\implies A_n\cosh(n\pi)+B_n\sinh(n\pi)=2\int_0^1 x^2\sin(n\pi x)\,dx, \tag{A.5} \\ u_2(1,y)&=\sum_{n=1}^{\infty}C_n\sinh(n\pi)\sin(n\pi y)=1 \\ &\implies C_n=\frac{2}{\sinh(n\pi)}\int_0^1\sin(n\pi y)\,dy. \tag{A.6} \end{align}