I have the following system of PDE's
$$ B\big( -\partial_x^2 W^{00}+\partial_x \partial_t W^{01}\big) -2\beta_0 W^{00}=-\frac{\alpha_0}{2}\\ B\big(\partial_t^2 W^{00}-\partial_x \partial_t W^{01}\big)-2\beta_0 W^{00}=\frac{\alpha_0}{2}\\ B\big(- \partial_x^2W^{01}+\partial_x \partial_t W^{00}\big)-2\beta_0W^{01}=0\\ B\big(-\partial_t^2 W^{01}-\partial_x \partial_t W^{00}\big) -2\beta_0 W^{01}=0 $$
which, according to some Lagrangian, should determine the functions $W^{01}$ and $W^{00}$. On first sight this is a system of four differential equations determining two function and is probably overconstrained and unsolvable. I'm not too familiar with situations like this and am wondering if there any methods for determining whether this really has no solutions.
Any additional comments/thoughts would be greatly appreciated.
Adding the the second to the first, and subtracting the fourth from the third yields, after simplification, \begin{align*} B(\partial_t^2-\partial_x^2)W^{00}-4\beta_0W^{00}&=0 \\ B((\partial_t^2-\partial_x^2)W^{01}+2\partial_x\partial_tW^{00})&=0. \end{align*} What's most important about this step is that the first equation is now de-coupled. It looks like a hyperbolic pde in $W^{00}$. Assuming you have non-zero boundary conditions, use the appropriate methods for hyperbolics. Once you've solved that one, plug the result into the second equation and solve that hyperbolic equation for $W^{01}$.