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I came across this question

Set of coupled partial differential equations

And the answer is clear. However, what if the solution of the second equation does not appear in the first equation for example $$\frac{\partial f}{\partial x} + \frac{\partial g}{\partial x} + \frac{\partial f}{\partial y}=a$$ $$\frac{\partial g}{\partial x} + \frac{\partial g}{\partial y}=b$$ Would this still be considered a coupled system?

Also, in the theory of PDEs, we usually derive weak formulations. Say for example we have a system of two equations posed on a domain $\Omega \subset \mathbb{R}^n$ \begin{equation} \dfrac{\partial u}{\partial t} + L(u,v) = f, \end{equation} \begin{equation} \dfrac{\partial v}{\partial t} + P(v) = g, \end{equation} \begin{align} \text{Plus Boundary and Initial conditions.} \end{align}

Where $u$ and $v$ are the unknowns, $f$ and $g$ are given data, and $L$ and $P$ are two differential operators.

How would you tackle such a problem to obtain a weak formulation and the existence and uniqueness of a solution?

My thought is to treat the second equation separately and prove the existence and uniqueness of $v$. Then, treat the first equation to find $u$ assuming $v$ is given. Does this make sense?

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Yes, your thoughts are correct, as the solution $v$ of the second PDE can be derived without any knowledge on $u$. When this is done, you can solve the first PDE to get the solution $u$, which then of course depends on $v$. But note that this only works, if there is no extra dependence in the boundary conditions.

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  • $\begingroup$ Thank you! This is helpful. $\endgroup$ Jun 22, 2023 at 21:49

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