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.

  • 1
    $\begingroup$ Two comments: you could easily combine the first and second equations, and the third and fourth equations, to get it down to two. So it doesn't strike me as over-constrained. Also, everything is linear. Are $B$ and $\beta_0$ and $\alpha_0$ constants? If so, you have linear with constant coefficients, which is the most straight-forward type of pde to solve. $\endgroup$ Sep 24, 2018 at 19:19
  • $\begingroup$ So you're saying that I should solve eq 1 for $W^{00}$, substitute into eq 2, and similarly for eq 3 and 4 in order to boil it down to two equations? This would give $B\partial_t^2W^{00}+B \partial_1^2W^{00}-2B \partial_x \partial_tW^{01}=\alpha_0$ and $-B\partial_t^2W^{01}+\partial_x^2 W^{01}-2\partial_x \partial_t W^{00}=0$. Yes $B$, $\beta_0$, and $\alpha_0$ are constants. $\endgroup$
    – Kenny H
    Sep 24, 2018 at 19:36
  • $\begingroup$ I would go at it like this: set the first equation LHS equal to the negative of the second LHS (because both are equal to $-\alpha_0/2$), and the third equation LHS equal to the fourth equation LHS, because both vanish. $\endgroup$ Sep 24, 2018 at 19:42

1 Answer 1


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 $W^{01}$.


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