Weak formulation and well-posedness of the advection-reaction-diffusion problem

For now we assume homogenous Dirichlet boundary conditions. After multiplying (44) with a test function \(v\) and integrating by parts the term \(-(\epsilon \Delta u , v)_{\Omega}\), we arrive at the following weak formulation of the ADR problem:

Find \(u \in H^1_0(\Omega) = V\) s.t. for all \(v \in V\)

(47)\[a(u,v) := \epsilon(\nabla u, \nabla v)_{\Omega} + (\boldsymbol{b}\cdot\nabla u, v)_{\Omega} + (c u , v)_{\Omega} = (f, v)_{\Omega} =: l(v).\]

Here, we used the notation \((\alpha, \beta)_{U} = \int_{U} \alpha \beta \dU\). Now we want to apply the Lax-Milgram Theorem to prove the weak formulation (47) is well-posed. As \(l(\cdot)\) is defined exactly as in the Poisson problem discussed in Weak formulation of partial differential equations, the boundedness of \(l\) is clear. But in contrast to the previous cases, the bilinear form \(a(\cdot, \cdot)\) is not symmetric, and therefore it it less trivial to show that the bilinear form \(a(\cdot, \cdot)\) indeed is both coercive and bounded.

Coercivity

First, we want to show that there exists a constant \(C_a > 0\) such that

\[ a(u, u) \geqslant \alpha \|u\|_{V}^2 \quad \foralls u \in V \]

for a suitably chosen norm \(\| \cdot \|_{V}\). Since we \(V = H^1_0(\Omega)\), we pick \(\| \cdot \|_V = \| \nabla (\cdot) \|_{\Omega }\), which is equivalent to the \(H^1\)-norm, thanks to the Poincare inequality. Now, we set \(u=v\) in (47) and obtain

\[ a(u,u) = \epsilon \|\nabla u\|_{\Omega}^2 + (\boldsymbol{b}\cdot\nabla u, u)_{\Omega} + (c u , u)_{\Omega}. \]

(eq-coerc-adr-form-step1) Now the troublesome term is the nonsymmetric part \( (b\cdot\nabla u, u)_{\Omega} \) and to handle it, we first observe that

\[ \nabla\cdot (\boldsymbol{b} uv) = \boldsymbol{b}\cdot\nabla u v + \boldsymbol{b}\cdot\nabla v u + \nabla\cdot \boldsymbol{b} u v. \]

and then use the famous the Gauß/Divergence theorem to see that

\[(\boldsymbol{b}\cdot \bfn u, v)_{\partial \Omega} = \int_{\Omega} \nabla\cdot (\boldsymbol{b} uv) dx = (\boldsymbol{b}\cdot\nabla u, v)_{\Omega} + (\boldsymbol{b}\cdot\nabla v, u)_{\Omega} + (\nabla\cdot\boldsymbol{b} v, u)_{\Omega}.\]

So the term \( (\boldsymbol{b}\cdot\nabla u, v)_{\Omega} \) can be rewritten as

\[(\boldsymbol{b}\cdot\nabla u, v)_{\Omega} = - (u, \boldsymbol{b}\cdot\nabla v)_{\Omega} - (\nabla\cdot\boldsymbol{b} v, u)_{\Omega} + (\boldsymbol{b}\cdot \bfn u, v)_{\partial \Omega}.\]

For \(u,v\in H^1_0(\Omega)\), the boundary term vanishes. Moreover, in the typical case \(\nabla \cdot \boldsymbol{b} = 0\) case, we observe that \((\boldsymbol{b}\cdot\nabla u, v)_{\Omega}\), i.e.

\[(\boldsymbol{b}\cdot\nabla u, v)_{\Omega} -(u, \boldsymbol{b}\cdot\nabla v)_{\Omega}\]

and thus \((\boldsymbol{b}\cdot\nabla u, u)_{\Omega} = 0\)!

In the more general case where \(\nabla \cdot \boldsymbol{b} \neq 0\), we can cleverly rewrite \((\boldsymbol{b}\cdot\nabla u, v)_{\Omega}\) as

\[\begin{split}(\boldsymbol{b}\cdot\nabla u, v)_{\Omega} &= \dfrac{1}{2}(\boldsymbol{b}\cdot\nabla u, v)_{\Omega} + \dfrac{1}{2}(\boldsymbol{b}\cdot\nabla u, v)_{\Omega} \\ &= \dfrac{1}{2}(\boldsymbol{b}\cdot\nabla u, v)_{\Omega} - \dfrac{1}{2}(u, \boldsymbol{b}\cdot\nabla v)_{\Omega} - \dfrac{1}{2}(\nabla\cdot\boldsymbol{b} u, v)_{\Omega} + \dfrac{1}{2}\underbrace{(\boldsymbol{b}\cdot \bfn u, v)_{\partial \Omega}}_{= 0}.\end{split}\]

Combining this expression with \((c u, v)_{\Omega}\) yields

\[(\boldsymbol{b}\cdot\nabla u, u)_{\Omega} + (c u, u)_{\Omega} = ((c - \tfrac{1}{2}\nabla\cdot\boldsymbol{b}) u, u)_{\Omega},\]

and after recalling our general assumption that \(c(x) - \tfrac{1}{2}\nabla\cdot\boldsymbol{b}(x) \geqslant c_0 > 0\) for all \(x\in \Omega\), we finally conclude that for all \(u\in V\)

\[\begin{split}a(u,u) &= \epsilon \|\nabla u\|_{\Omega}^2 + ((c - \tfrac{1}{2}\nabla\cdot\boldsymbol{b}) u, u)_{\Omega} \\ &\geqslant \epsilon \|\nabla u\|_{\Omega}^2 + c_0 \|u\|_{\Omega}^2 \geqslant \epsilon \|\nabla u\|_{\Omega}^2.\end{split}\]

If one prefers to work with the \(H^1\)-norm, we could have replaced the last inequality by \(\geqslant \min\{\epsilon, c_0\} \|u\|_{H^1(\Omega)}\) or simply used the Poincare inequality.

Boundedness

The proof of boundedness is rather straightforward and involves mainly the use of the Cauchy-Schwarz inequality and applications of the to translate \(L^2\)-norms into \(\|\cdot\|_V\)-norms. To this end we observe that for all \(u,v\in V\)

\[\begin{split}a(u,v) &= \epsilon(\nabla u, \nabla v)_{\Omega} + (\boldsymbol{b}\cdot\nabla u, v)_{\Omega} + (c u , v)_{\Omega} \\ &\leqslant \epsilon \|\nabla u\|_{\Omega} \|\nabla v\|_{\Omega} + \|\boldsymbol{b}\|_{L^{\infty}(\Omega)}\|\nabla u\|_{\Omega} \|v\|_{\Omega} + \|c\|_{L^{\infty}(\Omega)} \|u\|_{\Omega}\|v\|_{\Omega} \\ &\leqslant \epsilon \|\nabla u\|_{\Omega} \|\nabla v\|_{\Omega} + \|\boldsymbol{b}\|_{L^{\infty}(\Omega)} \|\nabla u\|_{\Omega} C_P \|\nabla v\|_{\Omega} + \|c\|_{L^{\infty}(\Omega)} C_P \|\nabla u\|_{\Omega} C_P \|\nabla v\|_{\Omega} \\ &\leqslant \underbrace{ ( \epsilon + \|\boldsymbol{b}\|_{L^{\infty}(\Omega)} C_P + \|c\|_{L^{\infty}(\Omega) C_P^2} ) }_{C_a} \|\nabla u\|_{\Omega} \|\nabla v\|_{\Omega}.\end{split}\]