Bounding Weak Coercitivity with Robin Boundary Conditions

Question

Question

I am trying to bound the weak coercitivity of the bilinear form in the context of a parabolic boundary value problem with Robin Boundary Conditions. I am following the approach mentioned in Numerical Approximation of Partial Differential Equations by Quarteroni and Valli (pp. 365-366):

These are the unclear steps from the book:

"By applying the trace theorem and an interpolation result (see Theorems 1.3.1 and 1.3.7), for each $\epsilon > 0$ one has: $$ \|v\|_{0, \partial \Omega_2}^2 \leq \epsilon\|Dv\|_{0}^2 + C_\epsilon\|v\|_{0}^2" $$

So far I have been able to use the trace theorem, thus proving that: $$ \|v\|_{0,\partial \Omega} \leq C\|v\|_{1,\Omega} $$

However, I am unsure how to choose the parameters $s_1$ and $s_2$ to apply the mentioned interpolation theorem.

Theorem 1.3.7 (Interpolation theorem)

Assume that $\Omega$ is an open set of $\mathbb{R}^d$ with a Lipschitz continuous boundary. Let $s_1 < s_2$ be two real numbers, and define $ r = (1-\theta)s_1 + \theta s_2$, where $0 \leq \theta \leq 1$. There exists a constant $ C > 0 $ such that: $$ \|v\|_r \leq C \|v\|_{s_1}^{1-\theta} \|v\|_{s_2}^\theta \quad \forall v \in H^{s_2}(\Omega). $$

I also suspect that the following inequality must be used: $$ \|\nabla v\|_0 \|v\|_0 \leq \epsilon \|\nabla v\|_0^2 + \frac{1}{4\epsilon}\|v\|_0^2 $$

Any insights on how to choose $s_1$ and $s_2$, corrections or suggestions on how to proceed would be greatly appreciated.

Notation Clarification

$\| v \|_{0,\Omega} = \|v\|_{L^2(\Omega)} $and $\| v \|_{1,\Omega} = \|v\|_{H^1(\Omega)}$.

Answer 1

Consider the following parabolic partial differential equation (PDE) which incorporates time-dependency:

\begin{cases} \partial_t u -\nabla \cdot (\mu \nabla u) + \mathbf{b} \cdot \nabla u + \nabla \cdot (\mathbf{c}u)+ \sigma u = f & \text{in } \Omega \times (0,T], \\ u = g_D & \text{on } \Gamma_D \times (0,T], \\ \mu \nabla u \cdot \mathbf{n} + \gamma u = g_N & \text{on } \Gamma_R \times (0,T], \\ u(\cdot,0) = u_0 & \text{in } \Omega. \end{cases}

Where $ u_0 $ is the initial condition and $ T $ is the final time.

In this instance, the bilinear form originating from the weak formulation has the following form:

$$ a(\tilde{u}, v)=\int_{\Omega} \mu \nabla \tilde{u} \cdot \nabla v \, d \Omega - \int_{\Omega} \tilde{u} \mathbf{b} \cdot \nabla v \, d \Omega + \int_{\Omega} v\mathbf{c} \cdot \nabla \tilde{u} \, d\Omega + \int_{\Omega} \sigma \tilde{u} v \, d \Omega + \int_{\Gamma_R} \gamma \tilde{u} v \, d\Gamma $$

Now I will prove the weak coercitivity constant

$$\alpha(v,v) \geq \mu_0 \|\nabla v\|_0^2 - \|\mathbf{b}-\mathbf{c}\|_{L^\infty(\Omega)}\|\nabla v\|_0\|v\|_0- \| \sigma \|_{L^\infty(\Omega)}^2 \|v\|_0^2 -\|\gamma\|_{L^\infty(\Omega)} \|v\|_{0,\partial \Omega}^2 $$ $$ \alpha(v,v) + \|\mathbf{b}-\mathbf{c}\|_{L^\infty(\Omega)}\|\nabla v\|_0\|v\|_0+ \| \sigma\|_{L^\infty(\Omega)}^2 \|v\|_0^2 +\|\gamma\|_{L^\infty(\Omega)} \|v\|_{0,\partial \Omega}^2\geq \mu_0 \|\nabla v\|_0^2 $$ Using the following inequalities: $$ \|\nabla v\|_0\|v\|_0 \leq \epsilon \|\nabla v \|_0^2+ \frac{1}{4\epsilon}\|v\|_0^2 $$ $$ \|v\|_{L^2(\partial \Omega)}^2\leq C'\|\nabla v \|_0\|v\|_0 \leq C' \biggl( \epsilon' \|\nabla v \|_0^2+ \frac{1}{4\epsilon'}\|v\|_0^2\biggl) $$ \begin{aligned} \alpha(v,v) + \|\mathbf{b}-\mathbf{c}\|_{L^\infty(\Omega)} \left( \epsilon \|\nabla v \|_0^2+ \frac{1}{4\epsilon}\|v\|_0^2\right) +\|a_0\|_{L^\infty(\Omega)}^2 \|v\|_0^2 + \\ +C\|\gamma\|_{L^\infty(\Omega)} (\|v\|_0^2+\|\nabla v\|_0^2)+ \|\gamma\|_{L^\infty(\Omega)} C' \biggl( \epsilon' \|\nabla v \|_0^2+ \frac{1}{4\epsilon'}\|v\|_0^2\biggl) \|v\|_0^2 \geq \mu_0 \|\nabla v\|_0^2 \end{aligned}

Next steps, don't get discouraged:

\begin{aligned} \alpha(v,v) + \left( \|\mathbf{b}-\mathbf{c}\|_{L^\infty(\Omega)}\frac{1}{4\epsilon} + \|a_0\|_{L^\infty(\Omega)} + C' \frac{1}{4\epsilon} \| \gamma \|_{L^\infty(\Omega)}\right) \|v\|_0^2 & \geq \\ \left(\mu_0 -\epsilon \|\mathbf{b}-\mathbf{c}\|_{L^\infty(\Omega)}-C'\epsilon' \|\gamma\|_{L^\infty(\Omega)}\right)\|\nabla v\|_0^2, \\ \\ a(v,v) + \lambda \|v\|_0^2 \geq \frac{\left(\mu_0 -\epsilon \|\mathbf{b}-\mathbf{c}\|_{L^\infty(\Omega)} -C'\epsilon' \|\gamma\|_{L^\infty(\Omega)}\right)}{1+ C_\Omega^2}\|v\|_1^2. \end{aligned}

Where we used in the last step the Poincaré inequality. Obviously the $\epsilon,\epsilon'$ can be chose arbitrarily small, so the constant on the righ hand can always be forced to be positive.