I have a rectangular region where $-1<x<1$ and $-1<y<1$. The boundary value problem for this region is given by $$ \kappa^2\frac{\partial^2u_1}{\partial x^2}+\delta^2\frac{\partial^2u_1}{\partial y^2}+\delta (\kappa^2-1)\frac{\partial^2u_2}{\partial x \partial y}=A,\tag{1} $$ $$ \delta^2\kappa^2\frac{\partial^2u_2}{\partial y^2}+\frac{\partial^2u_2}{\partial x^2}+\delta (\kappa^2-1)\frac{\partial^2u_1}{\partial x \partial y}=0,\tag{2} $$
$$ \left.\kappa^2\frac{\partial u_1}{\partial x}+\delta(\kappa^2-2)\frac{\partial u_2}{\partial y}\right|_{x=\pm 1}=B^{\pm },\quad \left.\delta\frac{\partial u_1}{\partial y}+(\kappa^2-2)\frac{\partial u_2}{\partial x}\right|_{y=\pm 1}=0,\quad u_2(x,\pm 1)=0. \tag{3} $$ where $u_1$ and $u_2$ are displacements and $\kappa$, $\delta$, $A$ and $B^{\pm}$ are known constants.
Eqs (1) and (2) can be transformed to a fourth-order PDE in respect to one of sought for displacements. For example, in terms of $u_2$, we obtain $$ u_{2,1111}+ a\, u_{2,1122} + b\, u_{2,2222}=0\tag{4} $$ where $a=2\delta^2$ and $b=\delta^4$. Therefore, it can be presented in operator form as $$ (\partial_1^2+\lambda_1^2\partial_2^2)(\partial_1^2+\lambda_2^2\partial_2^2)=0\tag{5} $$ $\lambda_1^2+\lambda_2^2=a$ and $\lambda_1^2\lambda_2^2=b$. Hence, the solution of (5) can be expressed $$ u_2=\sum_{i=1}^{2}\phi_i(\lambda_i x,y) $$ where $\phi_i$ are arbitrary plane-harmonic function. Similarly, the other displacement $u_1$ may be found again the sum of two plane-harmonic functions which are harmonic conjugate functions of $\phi_i$. Unfortunately, after applying the boundary conditions (3) I couldn't find any proper functions for the solutions. Does anyone have an idea about the solution to the problem or suggest any book I can use?
(I kept the previous clarification now below, for the sake of the comments)
So motivated by the third set of boundary conditions one can attempt the following Ansatz:
$$u_2=\sum_{k=1}^{\infty}\sin(\tfrac{k\pi}{2}(x+1))\cdot r_k(y)$$ Since everything is linear we solve for one fixed $k$ first and then take the sum in the end. To this end we plug $u_2=\sin(\tfrac{k\pi}{2}(x+1))r_k(y)$ into the forth order equation and get $$ 0=\frac{\kappa^2}{16}\sin(\tfrac{k\pi}{2}(x+1))\left(k^4\pi^4r_k(y)-8\delta^2k^2\pi^2r_k''(y)+16\delta^4r_k^{(4)}(y)\right)\\ \Rightarrow r_k(y) = (C_2y+C_1)e^{\frac{k\pi}{2\delta}y} + (C_4y+C_3)e^{-\frac{k\pi}{2\delta}y} $$
Next, we use equation $(2)$ to calculate $u_1$. This is achieved by simply integrating once over $x$ and once over $y$.
$$ u_1(x,y)=\cos(\tfrac{k\pi}{2}(x+1))\left((C_4y+C_3+C_4\tfrac{2\delta(\kappa^2+1)}{k\pi(\kappa^2-1)})e^{\frac{k\pi}{2\delta}y}-(C_2y+C_1-C_2\tfrac{2\delta(\kappa^2+1)}{k\pi(\kappa^2-1)})e^{-\frac{k\pi}{2\delta}y}\right)+F_1(y)+F_2(x) $$
In the next step, we use the first equation and arrive at
$$ F_2''(x)\kappa^2+F_1''(y)\delta^2=A\\ \Rightarrow F_2(x)=F_{22}x^2+F_{21}x+F_{20}\\ \Rightarrow F_1(y)=\frac{A-2\kappa^2F_{22}}{2\delta^2}y^2+F_{11}y+F_{10} $$ Which altogether yields:
$$ u_1(x,y)=\cos(\tfrac{k\pi}{2}(x+1))\left((C_4y+C_3+C_4\tfrac{2\delta(\kappa^2+1)}{k\pi(\kappa^2-1)})e^{\frac{k\pi}{2\delta}y}-(C_2y+C_1-C_2\tfrac{2\delta(\kappa^2+1)}{k\pi(\kappa^2-1)})e^{-\frac{k\pi}{2\delta}y}\right)+\frac{A-2\kappa^2F_{22}}{2\delta^2}y^2+F_{11}y+F_{10}+F_{22}x^2+F_{21}x+F_{20}\\ u_2(x,y)=\sin(\tfrac{k\pi}{2}(x+1))\left((C_2y+C_1)e^{\frac{k\pi}{2\delta}y} + (C_4y+C_3)e^{-\frac{k\pi}{2\delta}y}\right) $$
At this point we would normally take the sum over $k$, and then try to fit the parameters (all dependend on $k$ of course) to solve the remaining boundary conditions.
At this point however (and that is exactly why i wanted to know the exact problem!) we run into two problems:
However, for $2A=B^+-B^-$ due to the first point we can easily deduct many solutions which do not even require the sum for example:
\begin{align} u_1(x,y)&=\frac{Ax^2+2(A+B^-)x}{2\kappa^2}\\ u_2(x,y)&=0 \end{align}
This is not an answer but its too long for a comment:
I don't see how you obatin the stated coefficients of the forth order equation:
$$ \kappa^2\frac{\partial^2}{\partial x^2} (2)+\delta^2\frac{\partial^2}{\partial y^2} (2) -\delta(\kappa^2-1)\frac{\partial^2}{\partial x\partial y} (1)\\ \Leftrightarrow \delta^2\left(\kappa^2u_{2,xxxx}+(\delta^2+2\kappa^2-1)u_{2,xxyy}+\delta^2\kappa^2u_{2,yyyy}\right)\\ \Leftrightarrow u_{2,xxxx}+\underbrace{\frac{\delta^2+2\kappa^2-1}{\kappa^2}}_{=a}u_{2,xxyy}+\underbrace{\delta^2}_{=b}u_{2,yyyy} $$