solve laplace equation $ U_{xx} + U_{yy} = 0 $ with given boundary condition : $ (x,y) \in (0,a) \times (0,b) $ $\begin{cases} U_x(0,y) = U(x,0) = U_x(a,y) = 0 \\ U(x,b)= c= constant \hspace{3mm} ,0 < x < a_1 < a \\ U_y(x,b) = 0 \hspace{3mm} ,a_1 < x < a \end{cases}$
my approach: I decided to introduce a linking condition $U(a_1,y) = f(x)$ , therefore we will continue solving two parts separately and connect them at the end .
first problem:
$ U_{xx} + U_{yy} = 0 , \hspace{4mm} (x,y) \in (0,a_1) \times (0,b)$
boundary conditions: $\begin{cases} U_x(0,y) = U(x,0) = 0\\U(a_1,y) = f(x) \\ U(x,b)= c= constant \end{cases}$
solution of first problem: $$U(x,y)=v(x,y)+w(x,y)$$ dirichlet BC $\implies w(x,y)=0+\frac{y}{b}\times c=\frac{yc}{b}$ $$\implies u(x,y)=\frac{yc}{b}+v(x,y)$$ $$\implies \begin{cases}U_x(0,y)=v_x(0,y)=0\\U(a_1,y)=\frac{yc}{b}+v(a_1,y) \implies v(a_1,y)=f(y)-\frac{yc}{b} \end{cases}$$
so all BCs are now homogenous except one side and we can solve problem for $v(x,y)$:
$$\implies v(x,y)= \sum_{n=1}^{\infty} A_n\sin(\frac{n\pi}{b}y)\cosh(\frac{2n-1}{2b}\pi x)$$
$$\implies U(x,y)= \frac{yc}{b}+\sum_{n=1}^{\infty} A_n\sin(\frac{n\pi}{b}y)\cosh(\frac{2n-1}{2b}\pi x)$$
$$U(a_1,y)=f(y)= \frac{yc}{b}+\sum_{n=1}^{\infty} A_n\sin(\frac{n\pi}{b}y)\cosh(\frac{2n-1}{2b}\pi a_1)$$ $A_n$ could be found using fourier series of $f(x)$, therefore the first problem is solved
second problem:
$ U_{xx} + U_{yy} = 0 , \hspace{4mm} (x,y) \in (a_1,a) \times (0,b)$
boundary conditions: $\begin{cases} U(x,0) = U_x(a,y) = U_y(x,b)=0\\U(a_1,y) = f(x) \end{cases}$
solution of second problem: BCs are all homogenous except the introduced linking condition , if the rectangular was attached to y-axis I knew how to simply solve it but now I don't know what should be placed in $......$ below $$ U(x,y)= \sum_{n=1}^{\infty} B_n\sin(\frac{2n-1}{2b}\pi y)\cosh(.......x)$$
I think it should be solved in this way but I'm not sure :
$$ U(x,y)= \sum_{n=1}^{\infty} B_n\sin(\frac{2n-1}{2b}\pi y)\cosh(\frac{2n-1}{2b}\pi(x-a))$$
and the other difficulty is if both above problems are solved how can I sum up them to achieve a single solution for the main problem ?
thanks for your help!