## Exponential convergence to equilibrium for a class of chemical reactions

In this post, we will see how a Lyapunov functional will help us to prove the explicit convergence to equilibrium for some chemical reactions.

We consider a single reversible reaction

$2U \leftrightharpoons V$

with the forward and backward reaction rates are assumed to be $1$. Denote by $u(t), v(t)$ the concentrations of $U, V$ at time $t$ respectively. Using the law of mass action, we arrive at the following nonlinear system of ODEs:

$\begin{cases}u' = -2(u^2 - v),&t>0,\\ v' = u^2 - v, &t>0,\\ u(0) = u_0\geq 0, \quad v(0) = v_0\geq 0 \end{cases} (1)$

We note that the system (1) possesses the mass conservation

$u(t) + 2v(t) = u_0 + 2v_0 =: M \quad \text{ for all } t>0$

where we call $M > 0$ the initial mass.

The system (1) has a unique equilibrium $(u_\infty, v_\infty)$ which balances the reaction and satisfies the mass conservation:

$\begin{cases}u_{\infty}^2 = v_{\infty},\\ u_{\infty} + 2v_{\infty} = M\end{cases}$

We remark that, without satisfying the mass conservation, we would have many infinite stationary states to the system (1).

The main question is: do the concentrations tend to the equilibrium as $t\rightarrow +\infty$.

Theorem. There exists an explicit constant $\lambda>0$ depending on the initial mass $M$ such that,

$|u(t) - u_{\infty}| + |v(t) - v_{\infty}| \leq e^{-\lambda t}(|u_0-u_\infty| + |v_0 - v_\infty|)$.

The Theorem tells us that, we are not only able to show the convergence to equilibrium but also able to compute explicitly the rate of convergence. The idea of the proof is to use the free energy functional (or Boltzmann-type entropy functional).

Proof.

Multiplying the equation of $u$ with $log(u)$ and the equation of $v$ with $log(v)$ we have

$(ulogu - u)_t = u_tlog(u) = -2log(u)(u^2-v)$

and

$(vlogv - v)_t = v_tlog(v) = log(v)(u^2-v)$.

Summing up the two equations above we have

$\frac{d}{dt}(ulogu - u + vlogu - v) = -(u^2-v)log(u^2/v) \leq 0$

or equivalently

$\frac{d}{dt}E[u,v](t) = -D[u,v](t) \leq 0$

with $E[u,v] = ulogu - u + vlogv - v$ and $D[u,v] = (u^2 - v)log(u^2/v)$.

Remark. The functional $E$ is called free energy or entropy or Boltzmann-type entropy. In the latter case, we call $D$ entropy dissipation.

We note that $D[u,v] = 0$ iff $u^2 = v$, which combining with the mass conservation implies that,

$D[u,v] = 0 \quad \text{ if and only if } (u,v) = (u_\infty,v_\infty)$.

This gives us a hope that the convergence actually happens!

To do that, we need two useful estimates relating to the relative entropy:

(i) $E[u,v] - E[u_\infty,v_\infty] = ulog(u/u_\infty) - u + u_{\infty} + vlog(v/v_\infty) - v + v_\infty$; and

(ii) $E[u,v] - E[u_\infty,v_\infty] \geq C(|u-u_\infty|^2 + |v - v_\infty|^2)$.

Hints: The proof of (i) is easy with direct computations by using the mass conservation. The proof of (ii) is a little more tricky and is left as an exercise (for you, the reader, to check!)

The property (ii) tells us that, if we can prove the convergence of the entropy $E[u,v](t)$ to $E[u_\infty,v_\infty]$ as $t\rightarrow +\infty$, the we will get the convergence of solutions.

To do that, we first observe

$\frac{d}{dt}(E[u,v] - E[u_\infty,v_\infty]) = -D[u,v]$.

If we can prove that

$D[u,v] \geq \alpha(E[u,v] - E[u_\infty,v_\infty])$ (*)

then we have

$\frac{d}{dt}(E[u,v] - E[u_\infty,v_\infty]) \leq -\alpha(E[u,v] - E[u_\infty,v_\infty])$

which, by the help of Gronwall’s lemma, gives us $E[u,v](t) \longrightarrow E[u_\infty,v_\infty]$ exponentially as $t\rightarrow +\infty$ with the rate $\alpha$.

Therefore, our aim now is to prove the inequality (*). This inequality is called entropy-entropy dissipation estimate.

In order to do this, we use the idea of Bakry-Emery criterion: to investigate the second derivative of the relative entropy $E[u,v]-E[u_\infty,v_\infty]$ or equivalently the first derivative of the entropy dissipation $D[u,v]$.

By direct computations, we have

$\frac{d}{dt}D[u,v] = -(4u+1)(u^2 - v)log(u^2/v) - (u^2 - v)^2(4/u + 1/v) \leq -D[u,v]$

thanks to the positivity of solution (why do we have this? It would be nice if you could get the answer yourself). We thus get

$D[u,v](t) \leq e^{-t}D[u_0,v_0] \longrightarrow 0\quad \text{ as } t\rightarrow +\infty$.

Now, we integrate

$\frac{d}{dt}D[u,v] \leq -D[u,v]$

from $t$ to $+\infty$ and use $D[u,v](+\infty) = 0$ and $E[u,v](+\infty) = E[u_\infty,v_\infty]$ (one more why!!! Hints: $E[u,v]$ is a Lyapunov functional), we have

$-D[u,v](t) \leq -(E[u,v](t) - E[u_\infty,v_\infty])$

or equivalently

$D[u,v] \geq (E[u,v] - E[u_\infty,v_\infty])$.

We have proved (*) and thus complete the proof of Theorem.

Remark.

(i) One could think of extending the method to the case of spatially imhomogeneous case with diffusion, that is the concentration does not only depend on time but also on spatial varibles. The system hence writes as

$\begin{cases}u_t - d_u\Delta u = -2(u^2 - v), &x\in\Omega, t>0,\\ v_t - d_v\Delta v = u^2 - v, &x\in\Omega, t>0\end{cases}$

with homogeneous Neumann boundary condition. However, due to the appearance of spatial variables, the Bakry-Emery method leads to “nasty terms” which are (almost) impossible to control. There is a way to prove a corresponding version of (*), it was first given in a work of Desvillettes and Fellner.

(ii) By patient computations, one easily sees that the method would work for a more general reaction, which writes as

$\alpha_1U_1 + \alpha_2U_2 + \ldots + \alpha_NU_N \leftrightharpoons \beta_1V_1 + \beta_2V_2 + \ldots + \beta_MV_M$

for any number $N$ and $M$ of substances.

Lecturer School of Applied Mathematics and Informatics Hanoi University of Science and Technology No 1, Dai Co Viet Street, Hanoi
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### 3 Responses to Exponential convergence to equilibrium for a class of chemical reactions

1. Phi Ha says:

Hi bro, do u still remember me. I found that this post is quite interesting. However, if the chemical reaction is not too fast, then instead of a system of ODE you have a system of DDE (delay). Can you prove the same result in this case?

2. baotangquoc says:

How would your delay system look like? I haven’t known any results for delay systems using this entropy method. But I am happy to take a look.
P/s: Sure I remember you 😀

3. Phi Ha says:

The system would look like

$u' = -2 [u^2(t-\tau_1) - v(t-\tau_2)]$,
$v' = u^2(t-\tau_1) - v(t-\tau_2)$,

where $\tau_1 \geq 0, \tau_2 \geq 0$ are two time delays.