How can I solve this PDE?
$$\dfrac{\partial^4 u}{\partial x^4} + 2\dfrac{\partial^4 u}{\partial x^2\partial y^2} + \dfrac{\partial^4 u}{\partial y^4} = 0$$
On a square $\Omega = \left[0, \ 1\right]\times \left[0, \ 1\right]$ and boundary conditions $$u(x, \ y) = 0 \ \ \text{on} \ \partial \Omega$$ $$\begin{align*}\left[\dfrac{\partial u}{\partial x}\right]_{x=0} & = 0 \\ \left[\dfrac{\partial u}{\partial x}\right]_{x=1} & = 0 \\ \left[\dfrac{\partial u}{\partial y}\right]_{y=0} & = 0 \\ \left[\dfrac{\partial u}{\partial y}\right]_{y=1} & = \sin ^2\left(\pi x\right)\end{align*}$$
I tried using separation of variables, like
$$u(x, \ y) = X(x) \cdot Y(y)$$
But I got stuck with
$$X^{(4)} Y + 2X^{(2)}Y^{(2)} + XY^{(4)} = 0$$
I also tried to write $X(x)$ and $Y(y)$ as
$$X(x) = \sum_{i=1}^{\infty}a_{i} \cdot \sin \left(i\pi x\right)+b_{i}\cdot \cos \left(i\pi x\right)$$ $$Y(y) = \sum_{j=1}^{\infty}c_{j} \cdot \sinh \left(j\pi y\right)+d_{j}\cdot \cosh \left(j\pi y\right)$$
And finding the coefficients, but it feels like 'guessing' using my previous experience of solving Laplace's equation.
EDIT: As commented, the previous initial conditions were insufficient to get a unique solution.
EDIT 2: I tried to solve using this MSE question but I got stuck.
I found the eigenfunction in $x$ and $y$ direction, but now I don't know how to proceed and mix them. In the link, the eigenfunctions in $x$ and $y$ were the same.
My attempt:
Step 1: Find the eigenfunctions on $x$-direction
$$\dfrac{d^4 u}{dx^4} - \lambda^4 u = 0$$
For some $\lambda \in \mathbb{R}^{+}$ and boundary conditions $u(0) = u'(0) = u(1) = u'(1) = 0$
For $\lambda = 0$ $$u(x) = k_0 + k_1 x + k_2 x^2 + k_3 x^3$$ $$u'(x) = k_1 + 2k_2 x + 3k_3 x^2$$ Applying the boundary conditions we see that $$k_0 = k_1 = k_2 = k_3 = 0$$ That means $u(x) = 0$, which is impossible
For $\lambda > 0$
$$u(x) = k_0 \underbrace{\cos \lambda x}_{a(x)} + k_1 \underbrace{\sin \lambda x}_{b(x)} + k_2 \underbrace{\cosh \lambda x}_{c(x)} + k_3 \underbrace{\sinh \lambda x}_{d(x)}$$ $$u'(x) = -k_0 \sin \lambda x + k_1 \cos\lambda x + k_2 \sinh \lambda x + k_3 \cosh \lambda x$$ Applying boundary conditions
$$\underbrace{\begin{bmatrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ \cos \lambda & \sin \lambda & \cosh \lambda & \sinh \lambda \\ -\sin \lambda & \cos \lambda & \sinh \lambda & \cosh \lambda \end{bmatrix}\begin{bmatrix}k_0 \\ k_ 1 \\ k_2 \\ k_3\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}}_{\mathbf{A}}$$ The trivial solution is when $c_i = 0$, but we search for the non-trivial solution. That happens when the determinant is zero $$\det \mathbf{A} = 0 \Leftrightarrow \cos \lambda \cdot \cosh \lambda = 1$$ The three first non-zero solutions are $$\begin{align*}\lambda_1 & \approx 1.5056\pi \\ \lambda_2 & \approx 2.4998 \pi \\ \lambda_2 & \approx 3.5000 \pi & \end{align*}$$ Then, we rewrite $u(x)$ as
$$u(x) = \sum_i k_{0i} \left(a_i(x)-c_i(x)\right) + k_{1i} \cdot \left(b_i(x)-d_i(x)\right)$$Then $$\begin{align*}\dfrac{d^4a_i}{dx^4} & = \lambda^4 a_i & \ \ \ \ \ \ \ \ & \dfrac{d^2 a_i}{dx^2} = -\lambda^2 a_{i} \\ \dfrac{d^4b_i}{dx^4} & = \lambda^4 b_i & & \dfrac{d^2 b_i}{dx^2} = -\lambda^2 b_{i} \\ \dfrac{d^4c_i}{dx^4} & = \lambda^4 c_i & & \dfrac{d^2 c_i}{dx^2} = \lambda^2 c_{i} \\ \dfrac{d^4d_i}{dx^4} & = \lambda^4 d_i & & \dfrac{d^2 d_i}{dx^2} = \lambda^2 d_{i} \end{align*}$$
Step 2: Find the eigenfunctions on $y$-direction:
Doing the same as Step 1 $$\dfrac{\partial^4u}{\partial y^4} - \mu^4 u = 0$$
- For $\mu = 0$ $$u(y) = k_0 + k_1 y + k_2 y^2 + k_3 y^3$$ Putting the boundary conditions $u(0) = u'(0) = u(1) = 0$ and $u'(1) = 1$, then $$u(y) = -y^2 (1-y)$$
- For $\mu > 0$ $$u(y) = k_0 \cos \mu y + k_1 \sin \mu y + k_2 \cosh \mu y + k_3 \sinh \mu y$$ Applying the boundary condition $u(0) = u'(0) = 0$ $$u(y) = k_0 \left(\cos \mu y - \cosh \mu y\right) + k_1 \left(\sin \mu y - \sinh \mu y\right)$$ Applying the boundary condition $u(1) = 0$ and $u'(1) = 1$ $$\underbrace{\begin{bmatrix}\cos \mu - \cosh \mu & \sin \mu - \sinh \mu \\ -\sin \mu - \sinh \mu & \cos \mu - \cosh \mu\end{bmatrix}}_{\mathbf{A}}\begin{bmatrix}k_0 \\ k_1 \end{bmatrix} = \begin{bmatrix}0 \\ 1 \end{bmatrix}$$ The matrix has determinant $$\det \mathbf{A} = 0 \Leftrightarrow \cos \mu \cosh \mu = 1$$ So, must have $\cos \mu \cosh \mu \ne 1$ to have a unique solution
Step 3: Suppose $u(x, \ y)$ is a linear combination of eigenfunctions:
$$u(x, \ y) = \sum_{i=1}^{\infty} f_{i}(y) \cdot \left(a_i(x)-c_i(x)\right) + g_i(y) \cdot \left(b_i(x) - d_i(x)\right)$$
$$\begin{align*}\nabla^4 u = \sum_{i=1}^{\infty} & a_i\left(\lambda_{i}^{4}-2\lambda_i^2 \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^4}{\partial y^4}\right)f_i \\ + & b_i \left(\lambda_{i}^{4}-2\lambda_i^2 \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^4}{\partial y^4}\right)g_i \\ - & c_i\left(\lambda_{i}^{4}+2\lambda_i^2 \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^4}{\partial y^4}\right)f_i \\ - & d_i\left(\lambda_{i}^{4}+2\lambda_i^2 \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^4}{\partial y^4}\right)g_i\end{align*}$$ What should I use for $f$ and $g$?
