I am struggling with a problem of finding a solution $\phi(x,y)$ for biharmonic equation in a rectangular region $(x,y)\in [0,a]\times [0,b]$ i.e.
$$\frac{\partial^4 \phi}{\partial x^4}+2\frac{\partial^4 \phi}{\partial x^2\partial y^2}+\frac{\partial^4 \phi}{\partial y^4}=0$$
with boundary conditions
\begin{align} \frac{\partial^2 \phi}{\partial y^2}(0,y)&=0\\ \frac{\partial^2 \phi}{\partial x\partial y}(0,y)&=0\\ \frac{\partial^2 \phi}{\partial y^2}(a,y)&=0\\ \frac{\partial^2 \phi}{\partial x\partial y}(a,y)&=0\\ \frac{\partial^2 \phi}{\partial x^2}(x,b)&=a\cos\alpha+\frac{3\sin\alpha}{a}(2x-a)\\ \frac{\partial^2 \phi}{\partial x\partial y}(x,b)&=\frac{6x\sin\alpha}{a^2}(x-a)\\ \frac{\partial^2 \phi}{\partial x\partial y}(x,0)&=C\frac{\partial^2 \phi}{\partial x^2}(x,0) \end{align} where $\alpha$ and $C$ are some constants.
I have tried to express the solution as
$$\phi(x,y)=\sum_{m=0}^{\infty} (x-a)\sin\left(\frac{2m\pi x}{a}\right)\left(A_m\cosh\left(\frac{2m\pi y}{a}\right)+B_m\sinh\left(\frac{2m\pi y}{a}\right)\right)$$
which satisfies the first four of the boundary conditions, one is even able to satisfy by expanding the functions on the right hand sides of the fifth and sixth boundary conditions into Fourier series. However, I was unable to satisfy the last boundary condition
I would be very glad if you have any suggestions for a solution.
Thank you.