# 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. – me10240 Sep 2 '14 at 1:46