Entropy method for chemical reaction-diffusion systems II: A linear system

In the last post, I’ve shown you how does the entropy method work in the case of a single equation: heat equation with homogeneous Neumann boundary condition. The application of entropy method in systems is more involved and needs more efforts to bring out what we want.

In this post, we first try to apply the method to a linear system. The application to nonlinear systems is more complicated and will be revealed in next posts.

Modelling the reaction-diffusion system

Assuming that a reaction

\boxed{A \leftrightharpoons B}

takes place in a bounded domain \Omega\subset \mathbb R^n with the forward- and backward-reaction constant rates are normalised to 1. The corresponding linear RD system for two unknown functions a(x,t) and b(x,t) reads as

\begin{cases} a_t - d_a\Delta a = -a + b, \quad x\in \Omega\\ b_t - d_b\Delta b = a - b, \quad x\in\Omega\\ \nabla a \cdot \nu = \nabla b \cdot \nu = 0, \quad x\in \partial\Omega\end{cases}

and initial data a(x,0) = a_0(x) and b(x,0) = b_0(x).

Normalised volume and average value

Throughout this post, we will assume that \Omega has a normalised volume, that is |\Omega| = 1. For a function f, we denote by

\overline{f} = \int_{\Omega}f(x)dx

the average value of f.

Conservation of mass

The solution to this solution admits one conservation of mass 

\overline{a}(t) + \overline{b}(t) = \overline{a_0} + \overline{b_0} =: M for all t>0

Constant equilibrium

The unique equilibrium is (a_{\infty}, b_{\infty}) = \left(\frac{M}{2}, \frac{M}{2}\right).

Convergence to equilibrium?

To prove the convergence to equilibrium (a,b) \longrightarrow (a_{\infty}, b_{\infty}) as t\rightarrow +\infty we consider the quadratic entropy functional 

\boxed{E[a,b](t) = \|a\|^2 + \|b\|^2},

where \|\cdot\| is the usual L^2 norm, and its entropy dissipation

\boxed{D[a,b](t) = -\frac{d}{dt}E[a,b](t) = 2d_a\|\nabla a\|^2 + 2d_b\|\nabla b\|^2 + 2\|a - b\|^2}.

We now aim to prove an entropy-entropy dissipation estimate of the form

\boxed{D[a,b] \geq \lambda (E[a,b] - E[a_{\infty}, b_{\infty}])}\qquad \qquad (*)

for some \lambda >0.

To prove this inequality (*) we follow the following scheme, which is later turned out to be very effective to apply to nonlinear systems,

Step 0: (Decompose the difference of entropy)

Using the mass conservation \overline{a} + \overline{b} = M and values of the equilibrium we can show that

E[a,b] - E[a_{\infty}, b_{\infty}] =: E_1 + E_2


E_1 = E[a,b] - E[\overline{a}, \overline{b}] = \|a - \overline{a}\|^2 + \|b - \overline{b}\|^2


E_2 = E[\overline{a}, \overline{b}] - E[a_{\infty}, b_{\infty}] = (\overline{a} - a_{\infty})^2 + (\overline{b} - b_{\infty})^2.

In the next steps, we will try to control E_1 and E_2 separately.

Step 1: (Role of diffusion)

To control E_1, we use the Poincaré inequaltiy to have

\|\nabla a\|^2 \geq \lambda_1\|a - \overline{a}\|^2 and \|\nabla b\|^2 \geq \lambda_1\|b - \overline{b}\|^2.

Hence, we have

\frac{1}{2}D[a,b] \geq \min\{d_a, d_b\}\lambda_1 E_1.

Step 2: (Reaction of averages)

Applying the Jensen’s inequality \int_{\Omega}f^2dx \geq \left(\int_{\Omega}fdx\right)^2 (see here for more general inequalities) we have

\|a - b\|^2 \geq (\overline{a} - \overline{b})^2 = [(\overline{a} - a_{\infty}) - (\overline{b} - b_{\infty})]^2 = (\overline{a} - a_{\infty})^2 + (\overline{b} - b_{\infty})^2 = E_2

where we used the mass conservation \overline{a} + \overline{b} = M at the last step. Therefore, we have

\frac{1}{2}D[a,b] \geq \|a - b\|^2 \geq E_2.

Step 3: (Combining steps 1 and 2)

From steps 1 and 2, we easily see that

\boxed{D[a,b] \geq \mu(E[a,b] - E[a_{\infty}, b_{\infty}])}

with \mu = 2\min\{\min\{d_a, d_b\}\lambda_1,1\}, thus, by the Gronwall inequality

E[a,b](t) - E[a_{\infty}, b_{\infty}] \leq e^{-\mu t}(E[a_0,b_0] - E[a_{\infty},b_{\infty}]).

On the other hand, by the mass conservation we easily see that

E[a,b](t) - E[a_{\infty}, b_{\infty}] = \|a(t) - a_{\infty}\|^2 + \|b(t) - b_{\infty}\|^2.

In conclusion, we have proved

\boxed{\|a(t) - a_{\infty}\|^2 + \|b(t) - b_{\infty}\|^2 \leq e^{-\mu t}(E[a_0,b_0] - E[a_{\infty},b_{\infty}])}


About baotangquoc

Lecturer School of Applied Mathematics and Informatics Hanoi University of Science and Technology No 1, Dai Co Viet Street, Hanoi
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