# jump conditions for Poisson/Darcy equation in primal form versus mixed form

Consider the Darcy equation, $$\mathbf{v} + \dfrac{k}{\mu_0}\nabla p = \mathbf{f} \\ \mathrm{div}\; \mathbf{v} = 0$$

If the coefficient $k$ is piecewise constant across an interface $\Gamma$ in the domain, we have that

(1) $$\mathrm{jump}(k \cdot p\mathbf{n}) = 0 \rightarrow \mathrm{jump}(k\cdot p) =0$$ over the interface $\Gamma$

On the other hand, the system can also be written as $$\mathrm{div}(k \nabla p) = \mu_0f_1 = \mathrm{div}\mathbf{f}$$ in which case the jump condition now is

(2) $$\mathrm{jump}(k \nabla p\cdot\mathbf{n}) = 0$$ over $\Gamma$

I am confused as to how to reconcile the two.

Secondly, does this dictate the choice of finite element spaces used in the solution ? For example, using a piecewise continuous polynomial for $p$ in 1) would be wrong ?

Your jump condition is wrong. To see this, let's assume for a moment that $\mu=1$ because it plays no real role in your formulation. Then, if you want to integrate the $\nabla p$ term and still want to get a symmetric formulation, you need to start with the first equation in the form $$k^{-1} \mathbf v + \nabla p = k^{-1} \mathbf f,$$ which ultimately leads to the jump condition $[p]=0$ -- in other words, the pressure must be continuous.
How do you see that this is the right jump condition? Multiply the equation with a test function $\phi$ and integrate over a (arbitrary) part of the domain $\Omega_1$, for example one of the subdomains where $k$ is constant. After integrating by parts, you have $$(\phi,k^{-1} \mathbf v)_{\Omega_1} - (\nabla\cdot\phi, p)_{\Omega_1} + (\phi,p\mathbf n)_{\partial\Omega_1}= (\phi,k^{-1} \mathbf f)_{\Omega_1}.$$ Now do the same on $\Omega_2=\Omega\backslash\Omega_1$ and you get $$(\phi,k^{-1} \mathbf v)_{\Omega_2} - (\nabla\cdot\phi, p)_{\Omega_2} + (\phi,p\mathbf n)_{\partial\Omega_2}= (\phi,k^{-1} \mathbf f)_{\Omega_2}.$$ In the last equation, the sign of the normal vector is of course outward from $\Omega_2$, whereas in the first equation it is outward from $\Omega_1$. Now add these two equations together and you get $$(\phi,k^{-1} \mathbf v)_{\Omega} - (\nabla\cdot\phi, p)_{\Omega} + (\phi,p\mathbf n)_{\partial\Omega} + (\phi,[p]\mathbf n)_{\Gamma} = (\phi,k^{-1} \mathbf f)_{\Omega},$$ where $\Gamma$ is the interface between $\Omega_1$ and $\Omega_2$ and $[p]$ is the jump of $p$ on the interface. $\mathbf n$ is the normal from $\Omega_1$ info $\Omega_2$.
We could, on the other hand, also have multiplied the original equation by $\phi$ and instead integrated over the entire domain right away. This would have shown that $\mathbf v$ must satisfy the equation $$(\phi,k^{-1} \mathbf v)_{\Omega} - (\nabla\cdot\phi, p)_{\Omega} + (\phi,p\mathbf n)_{\partial\Omega} = (\phi,k^{-1} \mathbf f)_{\Omega}.$$ Comparing with the equation immediately above, it is clear that we have the condition $$(\phi \cdot \mathbf n,[p])_{\Gamma} = 0.$$ Because the normal traces $\phi \cdot \mathbf n$ of functions in $H(div)$ are in $L^2(\Gamma)$, this implies that $[p]=0$ in $L^2(\Gamma)$.
• Oh! I think my understanding of the jump condition itself is an issue. I thought, like the case of discontinuous thermal conductivity in the diffusion equation, the $k$ must appear in the jump condition and must be associated with some kind of flux term. If you could make an explanatory remark, that would be very helpful. Sep 2, 2014 at 1:46