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I have solved numerically the following system: \begin{cases} \partial_t{u} + \partial_x{v} = 0 \\ \partial_t{v} + \frac{1}{\varepsilon^2}\partial_x{u} = -\frac{1}{\varepsilon^2}(v-f(u)) \end{cases} with periodic boundary conditions by finite volume method (IMEX-Runge Kutta).

I was wondering if, as it happens in a lot of other systems, in this case, I should expect to have a mass conservation property.

In the scalar case that kind of property can be verified by checking if $\forall n\in\mathbb{N}$ the following holds:

$$ \sum_{i\in\mathbb{N}} u_i^n = \sum_{i\in\mathbb{N}} u_i^{n+1}$$

but in my vector case I don't know how to eventually proceed. Should I consider any particular norm?

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  • $\begingroup$ The first equation is still in divergence form and if you use a finite volume method, you still have conservation both at pde and numerical level. $\endgroup$ – cfdlab Mar 7 at 2:23
  • $\begingroup$ @cfdlab I think that the OP is also interested in how to see numerically the conservation, since in the scalar case is trivial, but in this case the solution is a vector of two components, so what quantity should one check for conservation? $\endgroup$ – VoB Mar 7 at 9:11
  • $\begingroup$ The $v$ equation is not a conservation law, so it will not be a conserved quantity. $u$ is clearly conserved if you use a finite volume method. We dont know what is the scheme for $v$, so it is not possible to say. $\endgroup$ – cfdlab Mar 9 at 11:53
  • $\begingroup$ I have used Rusanov numerical flux, which in this linear case is just Conservative Lax-Friedrichs and the $u$ variable is in fact conserved. Thanks for the answer for the $v$ variable, which in fact is not conserved by my scheme and hence now I understand why it is correct. $\endgroup$ – Dadeslam Mar 9 at 12:14
  • $\begingroup$ @Dadeslam can you close this question by accepting one of the answers that is best in your view. $\endgroup$ – cfdlab Mar 15 at 4:58
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Yes, mass conservation holds, because in fact you are solving some sort of a relativistic diffusion equation. Why:

$$\partial_{t} u + \partial_{x} v = 0$$

and:

$$\partial_{t} v + \frac{1}{\epsilon^{2}} \partial_{x} u = -\frac{1}{\epsilon^{2}} (v - f(u))$$

But from first equation:

$$\partial_{tt} u + \partial_{xt} v = 0$$

and from the second one:

$$\partial_{tx} v + \frac{1}{\epsilon^{2}} \partial_{xx} u = -\frac{1}{\epsilon^{2}} (\partial_{x} v - f^{'}(u) \partial_{x} u)$$

But $\partial_{xt} v = \partial_{tx} v$, so:

$$-\epsilon^{2} \partial_{tt} u + \partial_{xx} u = \partial_{t} u + f^{'}(u) \partial_{x} u$$

or by taking $\epsilon^{2} = \frac{1}{c_{s}^{2}}$, where $c_{s}$ is some sort of speed of sound:

$$\frac{1}{c_{s}^{2}} \frac{\partial^{2} u}{\partial t^{2}} + \frac{\partial u}{\partial t} - \frac{\partial^{2} u}{\partial x^{2}} + f^{'}(u) \frac{\partial u}{\partial x} = 0$$

If I take the convective velocity as $w = f^{'}(u)$:

$$\frac{1}{c_{s}^{2}} \frac{\partial^{2} u}{\partial t^{2}} + \frac{\partial u}{\partial t} - \frac{\partial ^{2} u}{\partial x^{2}} + w \frac{\partial u}{\partial x} = 0$$

Now define: $$\frac{D}{Dt} = \frac{\partial}{\partial t} + w \frac{\partial }{\partial x}$$

and d'Alembert (wave) operator:

$$\Box^{2} = -\frac{1}{c_{s}^{2}} \frac{\partial^{2}}{\partial t^{2}} + \frac{\partial^{2}}{\partial x^{2}}$$

So finally:

$$\frac{D u}{D t} = \Box^{2} u$$

Taking integration from both side ($\Omega$ is your computational domain, in 1D case it is just a line):

$$\frac{D}{D t} \int_{\Omega} u(x,t) dx = \int_{\Omega} \Box^{2} u dx = \int_{\partial \Omega} \Box u da = 0$$

If you have Neumann boundary condition of $\Box u = 0$ at the wall. So:

$$\frac{D}{D t} \int_{\Omega} u dx = 0$$

or:

$$\int_{\Omega} u dx = C$$

Where $C$ is just a constant equal to total mass of your system.

Update: Note that same thing is true for $v$, because you have this as well for $v$ variable:

$$\frac{D v}{D t} = \Box^{2} v$$

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    $\begingroup$ It looks clear to me now that $u$ needs to be preserved, but since we are dealing with a balance law in the second equation of the system and I am getting some dissipative effect in the numerical solution, I am not understanding now why we should have a conservation even in $v$. Can you please explain it to me? $\endgroup$ – Dadeslam Mar 8 at 12:37
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It's clear that the mass of $v$ is not conserved in general. Just take $u(x,t=0) = 0$ and choose $f$ so that $f(0)=0$. Then

$$\left. \frac{\partial v}{\partial t}\right|_{t=0} = -\frac{v}{\epsilon^2}.$$

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