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If you solve a given PDE (Navier stoke's, Euler, heat eqn, advection eqn, etc...) using FVM, is this PDE supposed to be valid at every cell in the discretized domain, or only in the global domain as a whole?

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  • $\begingroup$ It is very unclear what you mean by the "pde being valid" The pde can only be applied to a proposed solution that is continuous and differentiable. The FVM (and FEM) are based on integral formulations of the governing equations. They are imposed on every cell, and if the method is conservative, so that every flux describes transfer between cells without loss of the conserved quantities, then they hold globally as well. $\endgroup$ – Philip Roe Apr 13 '18 at 15:49
  • $\begingroup$ The pde cannot be expected to apply if the solution is discontinuous because such a solution does not have derivatives, but any solution can always be integrated, hence the integral viewpoint. Obviously, a discrete solution with finitely many degrees of freedom cannot reproduce all the properties of a solution with infinite degrees of freedom. Any numerical solution must begin by deciding which properties to insist on and which properties to let go of. $\endgroup$ – Philip Roe Apr 13 '18 at 15:50
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As with many questions that are more about philosophy than science, there are at least three different ways of answering your question, each with a fairly solid argument:

  1. Given that the finite volume method requires discretising the problem, there is nowhere where the exact continuum PDE holds.

  2. However, we do create one equation representing a discretised version of the original PDE for each cell, so in this sense there is an equation "valid" in each cell.

  3. Unless your original system is very simple, or you make some very unusual choices, then the final system will couple the degrees of freedom representing each cell. It's in this sense, when attempting to solve the problem, that there are global solutions only.

Lets work through producing a finite difference method and see when each bit kicks in. Since we're choosing to use the FVM, I'm going to assume that the original PDE looks like $$\frac{\partial \tau }{\partial t} +\nabla\cdot \mathbf{F}(\tau) = 0.$$ Fortunately all the equations you ask about can be written in this conservative form. Now for each cell we write an integral equation, $$\int_{e_i}\frac{\partial \tau }{\partial t} +\nabla\cdot \mathbf{F}(\tau)d\mathbf{x} = 0.$$ Integrating by parts, and pulling the differentiation outside the integral gives $$\frac{d \tau }{dt} \int_{e_i}\tau d\mathbf{x} =\int_{\delta e_i} \mathbf{F}(\tau)\cdot d\mathbf{S}.$$ This is the last time we have an (integral) equation morally the same as the original PDE,(q.v. answer [1]).

From this point onwards we choose to express the integral of the flux, $$\int_{f(e_i,e_j)} \mathbf{F}(\tau)\cdot d\mathbf{S},$$ directed through a facet, $f(e_i,e_j)$, bordering cells $e_i$ and $e_j$, in terms of our cell integrals, $\int_{e_i}\tau d\mathbf{x}$ and $\int_{e_j}\tau d\mathbf{x}$. Now the equation has been discretised.

If we choose (weirdly) to have terms in $f(e_i,e_j)$ only given as functions of $\int_{e_i}\tau d\mathbf{x}$, then the individual degrees of freedom are uncoupled, and we just have one independently soluble equation for each of the $\int_{e_i}\tau d\mathbf{x}$ (strongly suggesting answer [2] for your original question).

However, this means giving up local conservation of $\tau$, one of the desirable properties of the FVM, which needs terms in $f(e_i,e_j)$ and $f(e_j,e_i)$ to match. Once we do this, the variables are coupled, and we can only find a solution by attacking the global solution. At this point we could argue for answer [3] instead.

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