Weak Formulation of the Conductivity Equation

Start from the Neumann Problem

$$\nabla \cdot (\gamma \nabla u)=0\text{in }\Omega ,\gamma {\partial }_{\nu }u=g\text{on }\partial \Omega .$$

Step 1: test. Multiply by a test function $v\in H^1(\Omega )$ and integrate. This is the $L^2(\Omega )$ inner product with $v$:

$${\int }_{\Omega }v\nabla \cdot (\gamma \nabla u)\mathrm{d}x=0.$$

Step 2: integrate by parts. Green’s first identity with $\mathbf{F}=\gamma \nabla u$ gives

$${\int }_{\Omega }v\nabla \cdot \mathbf{F}\mathrm{d}x={\int }_{\partial \Omega }v\mathbf{F}\cdot \nu \mathrm{d}s-{\int }_{\Omega }\nabla v\cdot \mathbf{F}\mathrm{d}x.$$

Since $\gamma \nabla u\cdot \nu =\gamma {\partial }_{\nu }u$:

$${\int }_{\partial \Omega }v\gamma {\partial }_{\nu }u\mathrm{d}s-{\int }_{\Omega }\gamma \nabla u\cdot \nabla v\mathrm{d}x=0.$$

Step 3: insert the boundary condition. The Neumann condition is natural: it enters through the boundary integral.

$$\boxed{{\int }_{\Omega }\gamma \nabla u\cdot \nabla v\mathrm{d}x={\int }_{\partial \Omega }gv\mathrm{d}s\forall v\in H^1(\Omega ).}$$

Only first derivatives appear, so $u\in H^1(\Omega )$ and $\gamma \in L^{\infty }(\Omega )$ suffice. Testing with $v\equiv 1$ gives the compatibility condition ${\int }_{\partial \Omega }g=0$.

Dirichlet case. For $u=f$ on $\partial \Omega$ the condition is essential: it is built into the trial space ($\mathrm{tr}u=f$). The test functions vanish on the boundary ($v\in H_0^1$), so the boundary integral disappears:

$${\int }_{\Omega }\gamma \nabla u\cdot \nabla v\mathrm{d}x=0\forall v\in H_0^1(\Omega ).$$

With interior sources $h$: add ${\int }_{\Omega }hv$ to the right-hand side.

Existence and uniqueness follow from the Lax-Milgram Theorem. Discretisation leads to the Galerkin Method.

References

1. L. C. Evans (2010). Partial Differential Equations, 2nd ed. AMS GSM 19. doi:10.1090/gsm/019
2. S. C. Brenner, L. R. Scott (2008). The Mathematical Theory of Finite Element Methods, 3rd ed. Springer. doi:10.1007/978-0-387-75934-0