I would like to solve the fourth order PDE
\begin{align*} \dfrac{\partial^{4}u}{\partial y^{4}} & =\dfrac{\partial^{2}u}{\partial x^{2}}\text{ in }{\Omega=\left\{ (x,y):x\in[0,1],y\in[0,\infty)\right\} }\\ \end{align*}
subject to
\begin{align*} u & =-1\text{ on }y=0,\\ \dfrac{\partial^{3}u}{\partial y^{3}}-c\dfrac{\partial^{2}u}{\partial y^{2}} & =0\text{ on }y=0,\\ u & \sim0\text{ as }y\to\infty,\\ \dfrac{\partial u}{\partial x} & =0\text{ on }x=0,\\ -\dfrac{\partial u}{\partial x}+d\dfrac{\partial^{2}u}{\partial y^{2}} & =0\text{ on }x=1, \end{align*} where $c,d$ are positive constants.
This looks like a problem that can be tackled using separation of variables. I have deduced a solution of the form
\begin{align*} u(x,y)={\rm Re}\left(\sum_{n=0}^{\infty}{\cal P}_{n}{\rm e}^{-\left(\sqrt{R_{n}+\phi}+{\rm i}\sqrt{R_{n}-\phi}\right)y}\cosh\left(\phi+{\rm i}\left(n+\tfrac{1}{2}\right)\pi\right)x\right) \end{align*}
where $\mathcal{P}_n\in\mathbb C$,
$$ \phi={\rm arctanh}(1/d)$$
and
$$ R_{n}=\sqrt{\phi^{2}+(n+\tfrac{1}{2})^{2}\pi^{2}}.$$
The problem now is that I cannot seem to figure out how to use the boundary conditions involving the second derivatives of $u$ to determine the coefficients $\mathcal{P}_n$. I am aware that in certain limiting cases (such as $\phi \to 0$) that it is possible to write the solution in terms of eigenfunctions and then use orthogonality, but how would I do it for this problem?
Is there an alternative way in which I could have written down the solution to make it more amenable to an alternative method? Is there anything to be gained by formulating an adjoint problem?