Spatiotemporal Evolution of Coinfection Dynamics: A Reaction–Diffusion Model

Abstract

This paper investigates the impact of spatial heterogeneity on the interaction between similar strains in a dynamical system of coinfecting strains with spatial diffusion. The SIS model studied is a reaction–diffusion system with spatially heterogeneous coefficients. The study considers two limiting cases: asymptotically slow and fast diffusion coefficients. When the diffusion coefficient is small, the slow system is shown to be a semilinear system of “replicator equations,” describing the spatiotemporal evolution of the strains’ frequencies. This system is of the reaction–advection–diffusion type, with an additional advection term that explicitly involves the heterogeneity of the associated neutral system. In the case of fast diffusion, traditional methods of aggregating variables are used to reduce the spatialized SIS problem to a homogenized SIS system, on which the results of the non-spatial model can be applied directly.

Appendices

Appendix A Proof for Theorem 7

The idea of our proof bases on the technique mentioned in [36].

Proof

Firstly, we make a convention for the norm using in this proof. For each $t \in \mathbb {R}^+$, for every $f_1, f_2 \in L^2\left( \Omega \times \mathbb {R},\mathbb {R}^n\right)$ we denote

$$\begin{aligned} \langle f_1, f_2 \rangle = \int _{\Omega }f_1\left( x,t\right) \cdot f_2\left( x,t\right) dx, \end{aligned}$$

where the $f_1\cdot f_2$ representing for the usual scalar product $\sum _{i=1}^{n}f^i_1 f^i_2$ in $\mathbb {R}^n$. This scalar product $\langle \cdot ,\cdot \rangle$ induces the norm

$$\begin{aligned} \left\| f\left( \cdot ,t\right) \right\| _2 = \left( \int _{\Omega }f\left( x,t\right) \cdot f\left( x,t\right) dx\right) ^{1/2} \end{aligned}$$

For the sake of convenience in this proof, we only write $\left\| \cdot \right\|$ instead of $\left\| \cdot \right\| _2$.

We do the same convention for $\langle g_1,g_2 \rangle$ and $\left\| g\left( \cdot ,t\right) \right\|$ for all $g_1,g_2,g \in C^1\left( \Omega \times \mathbb {R},\mathbb {R}^m\right)$.

Because in the finite dimensional space, all norms are equivalent, we then denote $|\cdot |$ to be the usual 2-Euclidean norm. Moreover, we recall the notation $A \prec 0$ for a symmetric matrix A if A is definitely negative, and $A \succ 0$ for definitely positive symmetric matrix.

First, let us show that the interval $[t_0,t_1]$ can be subdivided into subinterval $\Delta _k = [\tau _{k-1},\tau _k]$, where $k \in \{1,2,\dots ,N\}$ and $t_0 = \tau _0< \tau _1< \dots <\tau _N = t_1$ in such a way that for every k, there exists a symmetric matrix $P_k = P^T_k \succ 0$ for which

$$\begin{aligned} P_k A(x,t) + A^T(x,t)P_k \prec -I. \end{aligned}$$

(A1)

Indeed, since A(x, t) is a Hurwitz matrix for every $t \in [t_0,t_1]$, according to [37], there exists $P(x,t) = P^T(x,t) \succ 0$ such that

$$\begin{aligned} P(x,t) A(x,t) + A^T(x,t)P(x,t) \prec -I. \end{aligned}$$

Since A depends continuously on t, there exists an open interval $\Delta (t)$ such that $t \in \Delta (t)$ and

$$\begin{aligned} P(x,t)A(x,\tau ) + A^T(x,\tau )P(x,t) \prec -I, \quad \forall \tau \in \Delta (t). \end{aligned}$$

Now the open intervals $\Delta (t)$ with $t \in \left[ t_0,t_1\right]$ cover the whole closed bounded interval $\left[ t_0,t_1\right]$ and taking a finite number of $\tau _k$, $k=1,\dots ,N$ such that $\left[ t_0,t_1\right]$ is completely covered by $\Delta (\tau _k)$ yields the desired partition subdivision.

We can note that a strictly negative upper bound is not required on the real parts’ eigenvalues uniformly in space because the spatial domain is supposed to be compact.

Note that, from (A1), for all $y \in \mathbb {R}^m$ we have that

$$\begin{aligned} y^T\left( P_kA + A^TP_k\right) y \prec -y^T y. \end{aligned}$$

(A2)

Second, because F, G are continuously differential in x and t, then for every $\mu >0$ there exists $C,r >0$ such that

$$\begin{aligned}{} & {} \left\| F\left( f_0(x,t) + \bar{\delta }_f\left( x,t\right) ,g_0(x,t) + \bar{\delta }_g\left( x,t\right) ,x,t\right) - F\left( f_0,g_0,x,t\right) \right\| \nonumber \\{} & {} \quad \le C\left( \left\| \bar{\delta }_f\left( x,t\right) \right\| + \left\| \bar{\delta }_g\left( x,t\right) \right\| \right) \end{aligned}$$

(A3)

for all $t \in \mathbb {R}$, $\bar{\delta }_f\left( x,t\right) \in \mathbb {R}^n$, $\bar{\delta }_g\left( x,t\right) \in \mathbb {R}^m$ satisfying

$$\begin{aligned} \forall t \in [t_0,t_1], \forall x\in \Omega , \qquad |\bar{\delta }_f(x,t)| \le r, \qquad |\bar{\delta }_g (x,t)| \le r. \end{aligned}$$

For the sake of simplicity, we write $\bar{\delta }_f$ and $\bar{\delta }_g$ instead of $\bar{\delta }_f\left( x,t\right)$ and $\bar{\delta }_g\left( x,t\right)$. We now have the Taylor expansion as follows, noting that $G\left( f_0(x,t),g_0(x,t),x,t\right) = 0$,

$$\begin{aligned} G\left( f_0(x,t) + \bar{\delta }_f,g_0(x,t) + \bar{\delta _g},x,t\right) = A\left( x,t\right) \bar{\delta _g} + B(x,t)\bar{\delta }_f + o\left( |\bar{\delta _g}|\right) + o\left( |\bar{\delta _g}|\right) , \end{aligned}$$

(A4)

with B(x, t) is the Jacobian matrix of $G\left( \cdot ,\cdot ,t\right)$ with respect to the first variable.

For each $k = 1,\dots , N$, and $u \in \mathbb {R}^m$, set $|u|_k = \left( u^T P_k u\right) ^{1/2}$, then $|\cdot |_k$ is a norm in $\mathbb {R}^m$. Indeed, because $P_k \succ 0$ then $|\cdot |_k$ is well-defined, it suffices to check the condition $|u + v|_k \le |u|_k + |v|_k$, which is equivalent to

$$\begin{aligned} \left( u^TP_kv\right) ^2 \le \left( u^TP_ku\right) \left( v^TP_kv\right) . \end{aligned}$$

It now becomes

$$\begin{aligned} \left( \left( L^Tu\right) ^T\left( L^Tv\right) \right) ^2 \le \left( \left( L^Tu\right) ^T\left( L^Tu\right) \right) \left( \left( L^Tv\right) ^T\left( L^Tv\right) \right) , \end{aligned}$$

(A5)

thanks to the Cholesky’s factorization, which states that, if $P_k \succ 0$, there exist a square matrix such that $P_k = L^T_kL_k$. Note that, (A5) holds because of the inequality Cauchy-Schwarz. Hence, $|\cdot |_k$ is a norm in $\mathbb {R}^m$ and it is equivalent to an arbitrary norm in $\mathbb {R}^m$.

Then, for $\delta _f(x,t) = f(x,t) - f_0(x,t)$, $\delta _g(x,t) = g(x,t) - g_0(x,t)$, we have that

$$\begin{aligned} \left\{ \begin{array}{ll} \dfrac{d}{d t}\left\| \delta _f\right\| ^2 &{}\le C_1 \left( \left\| \delta _f\right\| + \left\| \delta _g\right\| \right) \left\| \delta _f\right\| ,\\ \epsilon \dfrac{d}{d t} \left\| \delta _g\right\| ^2_k &{}\le -q\left\| \delta _g\right\| ^2_kdt + C_1\left( \left\| \delta _f\right\| ^2 + \epsilon \right) dt \end{array} \right. \end{aligned}$$

(A6)

as long as $\delta _f$, $\delta _g$ are sufficiently small, where $C_1$, q are positive constants which do not depend on k.

Initially, for the sake of simplicity, in the following arguments, we write f, g instead of $f\left( x,t\right)$ and $g\left( x,t\right)$, respectively. Then, we have the equation for $\delta _f\left( x,t\right)$ as follows

$$\begin{aligned} \begin{aligned} \dfrac{\partial }{\partial t}\delta _f&= F\left( f_0 + \delta _f,g_0+\delta _g,x,t\right) - F\left( f_0 ,g_0,x,t\right) + K\delta _f \end{aligned}. \end{aligned}$$

By the convention of $\left\| \cdot \right\|$, we have that

$$\begin{aligned} \begin{aligned} \dfrac{d}{d t}\left\| \delta _f\right\| ^2&= \dfrac{d}{d t}\langle \delta _f,\delta _f\rangle = 2\langle \dfrac{\partial }{\partial t}\delta _f,\delta _f\rangle =\langle F\left( f_0 + \delta _f,g_0+\delta _g,x,t\right) \\&\quad - F\left( f_0 ,g_0,x,t\right) + K\delta _f,\delta _f\rangle \\&= \langle F\left( f_0 + \delta _f,g_0+\delta _g,x,t\right) - F\left( f_0 ,g_0,x,t\right) ,\delta _f\rangle + \langle K\delta _f,\delta _f\rangle \\&\le \left\| F\left( f_0 + \delta _f,g_0+\delta _g,x,t\right) - F\left( f_0 ,g_0,x,t\right) \right\| \left\| \delta _f\right\| \\&\quad + \langle K\delta _f,\delta _f\rangle \le C\left( \left\| \delta _f\right\| + \left\| \delta _g\right\| \right) \left\| \delta _f\right\| + \langle K\delta _f,\delta _f\rangle . \end{aligned} \end{aligned}$$

On the other hand, recalling that $K_f = a_f\left( x\right) \nabla + \Delta$ implies

$$\begin{aligned} \begin{aligned} \langle K_f \delta _f,\delta _f\rangle = \langle \Delta \delta _f,\delta _f\rangle + \langle a_f\left( x\right) \nabla \delta _f,\delta _f\rangle = -\int _{\Omega }\left| \nabla \delta _f\right| ^2dx + \int _{\Omega }a_f\left( x\right) \nabla \delta _f\cdot \delta _f dx \end{aligned} \end{aligned}$$

which leads to, when we apply the Young inequality for the term $\int _{\Omega }a\left( x\right) \nabla \delta _f\cdot \delta _f dx$,

$$\begin{aligned} \begin{aligned} \langle K_f \delta _f,\delta _f\rangle \le&-\int _{\Omega }\left| \nabla \delta _f\right| ^2dx + \max \limits _{x\in \Omega } \left( \left| a_f\left( x\right) \right| \right) \\&\left[ \dfrac{1}{\max \limits _{x\in \Omega } \left( \left| a_f\left( x\right) \right| \right) }\int _{\Omega }\left| \nabla \delta _f\right| ^2dx + C\left( \max \limits _{x\in \Omega }\left( \left| a_f\left( x\right) \right| \right) \right) \int _{\Omega }\left| \delta _f\right| ^2dx\right] , \end{aligned} \end{aligned}$$

where $\left| a_f\left( x\right) \right|$ is the matrix in which entries are absolute values of corresponding coordinates of $a_f\left( x\right)$.

Accordingly, we have the estimation for $\dfrac{d}{d t}\left\| \delta _f\right\| ^2$ as follows

$$\begin{aligned} \dfrac{d}{d t}\left\| \delta _f\right\| ^2 \le C_1\left( \left\| \delta _f\right\| + \left\| \delta _g\right\| \right) \left\| \delta _f\right\| \end{aligned}$$

(A7)

Next, we come to control the growth of $\left\| \delta _g\right\| _k$. We first observe that

$$\begin{aligned} \begin{aligned} \epsilon \dfrac{\partial }{\partial t}\delta _g =&G\left( f_0\left( x,t\right) +\delta _f, g_0\left( x,t\right) +\delta _g,x,t\right) \\&+ \epsilon K_g\delta _g + \epsilon \left[ K_g g_0\left( x,t\right) + \dfrac{\partial }{\partial t}g_0\left( x,t\right) + G_1(x)\cdot \nabla f_0\right] + \epsilon G_1(x)\cdot \nabla \delta _f. \end{aligned} \end{aligned}$$

We denote $\epsilon \left[ K_g g_0\left( x,t\right) + \dfrac{\partial }{\partial t}g_0\left( x,t\right) + G_1(x)\cdot \nabla f_0\right]$ as $O\left( \epsilon \right)$, then

$$\begin{aligned} \epsilon \dfrac{\partial }{\partial t}\delta _g = G\left( f_0\left( x,t\right) +\delta _f, g_0\left( x,t\right) +\delta _g,x,t\right) + \epsilon K_g\delta _g + O\left( \epsilon \right) + \epsilon G_1(x)\cdot \nabla \delta _f. \end{aligned}$$

(A8)

Using the Taylor expansion for G in (A4) and the equation (A8), we obtain the following computations

$$\begin{aligned} \begin{aligned} \epsilon \dfrac{d}{d t}\left\| \delta _g\right\| ^2_k&= \epsilon \dfrac{d}{d t}\langle \delta _g, P_k\delta _g\rangle = \epsilon \langle \dfrac{\partial }{\partial t}\delta _g,P_k\delta _g\rangle + \epsilon \langle \delta _g,P_k\dfrac{\partial }{\partial t}\delta _g\rangle \\&= \left( \langle A\delta _g, P_k\delta _g\rangle + \langle \delta _g,P_kA\delta _g\rangle \right) + 2B(x,t)\langle \delta _f,\delta _g\rangle \\&\quad + \langle o\left( \left| \delta _g\right| \right) + o\left( \left| \delta _g\right| \right) + O\left( \epsilon \right) ,P_k\delta _g + \delta _g\rangle \\&\quad + 2\epsilon \langle G_1(x)\cdot \nabla \delta _f,P_k\delta _g + \delta _g\rangle + \epsilon \langle K_g\delta _g, P_k\delta _g + \delta _g\rangle \\&= \langle \delta _g,\left( A^TP_k + P_kA\right) \delta _g\rangle + 2B(x,t)\langle \delta _f,\delta _g\rangle + \langle o\left( \left| \delta _f\right| \right) \\&\quad +o\left( \left| \delta _g\right| \right) + O\left( \epsilon \right) ,P_k\delta _g + \delta _g\rangle \\&\quad + 2\epsilon \langle G_1(x)\cdot \nabla \delta _f,P_k\delta _g + \delta _g\rangle + \epsilon \langle K_g\delta _g, P_k\delta _g + \delta _g\rangle \\&\le -\left\| \delta _g\right\| ^2 + 2C_1\left\| \delta _f\right\| \left\| \delta _g\right\| + \langle o\left( \left| \delta _f\right| \right) + o\left( \left| \delta _g\right| \right) + O\left( \epsilon \right) ,P\delta _g + \delta _g\rangle \\&\quad + 2\epsilon \langle G_1(x)\cdot \nabla ,P_k\delta _g + \delta _g\rangle + \epsilon \langle K_g\delta _g, P\delta _g + \delta _g\rangle . \end{aligned} \end{aligned}$$

(A9)

Using the Young inequality, we have the estimation for $\langle o\left( \left| \delta _f\right| \right) + o\left( \left| \delta _g\right| \right) + O\left( \epsilon \right) ,P\delta _g + \delta _g\rangle$ as follows

$$\begin{aligned} \langle o\left( \left| \delta _f\right| \right) + o\left( \left| \delta _g\right| \right) + O\left( \epsilon \right) ,P\delta _g + \delta _g\rangle \le O\left( \epsilon \right) + \tilde{C}\left\| \delta _g\right\| ^2, \qquad \text {with} \quad \tilde{C} \ll 1. \end{aligned}$$

(A10)

Alternatively, applying the Young inequality, we have that

$$\begin{aligned} \langle G_1(x)\cdot \nabla \delta _f,P_k\delta _g + \delta _g\rangle \le C\left( G_1\right) \left\| \nabla \delta _f\right\| ^2 + \left\| \delta _g\right\| ^2 \le C\left( G_1\right) + \left\| \delta _g\right\| ^2 \end{aligned}$$

(A11)

since $\nabla \delta _f$ is bounded in $\Omega$.

For the term $\langle K_g\delta _g, P\delta _g + \delta _g\rangle$, we get that

$$\begin{aligned} \begin{aligned} \langle K_g\delta _g, P\delta _g + \delta _g\rangle&=\langle \Delta \delta _g,P\delta _g \rangle + \langle \Delta \delta _g, \delta _g \rangle + \langle a_g\left( x\right) \nabla \delta _g,P\delta _g\rangle + \langle a_g\left( x\right) \nabla \delta _g,\delta _g\rangle \\&= \int _{\Omega }\Delta \delta _g\cdot P\delta _gdx + \int _{\Omega }\Delta \delta _g\cdot \delta _gdx + \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot P\delta _gdx \\&\quad + \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot \delta _gdx\\&= -\int _{\Omega }\nabla \delta _g\cdot \nabla \left( P\delta _g\right) dx - \int _{\Omega }\left| \nabla \delta _g\right| ^2dx + \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot P\delta _gdx \\&\quad + \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot \delta _gdx \\&= -\int _{\Omega }\nabla \delta _gP \nabla \delta _gdx - \int _{\Omega }\nabla \delta _g\cdot \left( \nabla P\right) \delta _gdx - \int _{\Omega }\left| \nabla \delta _g\right| ^2dx \\&\quad + \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot P\delta _gdx + \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot \delta _gdx. \end{aligned} \end{aligned}$$

Note that $P \succ 0$ then $\int _{\Omega }\nabla \delta _gP \nabla \delta _gdx \ge \lambda \left\| \nabla \delta _g\right\| ^2$. Applying the Young inequality once more for the terms

$$\begin{aligned} \int _{\Omega }\nabla \delta _g\cdot \left( \nabla P\right) \delta _gdx, \qquad \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot P\delta _gdx, \qquad \int _{\Omega }a_g\left( x\right) \nabla \delta _g\cdot \delta _gdx, \end{aligned}$$

we have that

$$\begin{aligned} \begin{aligned} \langle K_g\delta _g, P\delta _g + \delta _g\rangle&\le - \left( 1 + \lambda \right) \left\| \nabla \delta _g\right\| ^2 + \left( 1 + \lambda \right) \left\| \nabla \delta _g\right\| ^2 + C\left( 1 + \lambda \right) \left\| \delta _g\right\| ^2 \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \langle K_g\delta _g, P\delta _g + \delta _g\rangle \le C\left( 1 + \lambda \right) \left\| \delta _g\right\| ^2, \end{aligned}$$

(A12)

with $C\left( 1 + \lambda \right)$ denoting a constant depending on $1 + \lambda$.

Combining these equations (A9), (A10), (A11), and (A12), and noting that two norms $\left\| \cdot \right\| _k$ and $\left\| \cdot \right\| _k$ are equivalent, we observe that

$$\begin{aligned} \epsilon \dfrac{\partial }{\partial t}\left\| \delta _g\right\| ^2_k \le \left( 2\epsilon + \epsilon C\left( 1+\lambda \right) - 1\right) \left\| \delta _g\right\| ^2_k + C_1\left\| \delta _f\right\| \left\| \delta _g\right\| + C_1\epsilon \end{aligned}$$

which implies when $\epsilon$ small enough

$$\begin{aligned} \epsilon \dfrac{\partial }{\partial t}\left\| \delta _g\right\| ^2_k \le -q\left\| \delta _g\right\| ^2_k + C_1\left\| \delta _f\right\| \left\| \delta _g\right\| + C_1\epsilon \end{aligned}$$

(A13)

Thus, combine (A7) and (A13) and we obtain that

$$\begin{aligned} \dfrac{d}{dt}\left( \left\| \delta _f\right\| ^2 + \epsilon \dfrac{C_1}{q}\left\| \delta _g\right\| ^2\right) \le C_1\left\| \delta _f\right\| ^2 - \left\| \delta _g\right\| ^2_k + C_1\epsilon . \end{aligned}$$

(A14)

for some constant $C_1$ independent of k.

By the Gronwall’s inequality for $\left( \left\| \delta _f\right\| ^2 + \frac{\epsilon C_1}{q}\left\| \delta _g\right\| _k^2\right)$, for each $k \ge 1$, we can regard $\tau _{k-1}$ as the initial value, and then deduce that

$$\begin{aligned} \left\| \delta _f\left( \tau _{k-1} + \tau \right) \right\| ^2 \le e^{C_3\tau }\left( \left\| \delta _f\left( x,\tau _{k-1}\right) \right\| ^2 + \epsilon \dfrac{C_1}{q}\left\| \delta _g\left( x,\tau _{k-1}\right) \right\| _k^2\right) dx + C_1\epsilon \end{aligned}$$

for $\tau \in [0,\tau _k - \tau _{k-1}]$. With the aid of this bound for the growth of $|\delta _f|$, the second inequality of (A6) implies a bound for $\left\| \delta _g\right\| _k$ as following

$$\begin{aligned} \begin{aligned}&\left\| \delta _g\left( \tau _{k-1} + \tau \right) \right\| ^2_kdx \le e^{-q\tau /\epsilon }\left\| \delta _g\left( \tau _{k-1}\right) \right\| ^2_k\\&\quad + C_4\left( \left\| \delta _f\left( x,\tau _{k-1}\right) \right\| ^2dx + \epsilon \dfrac{C_1}{q}\left\| \delta _g\left( x,\tau _{k-1}\right) \right\| ^2_k\right) \\&\quad + C_4\epsilon . \end{aligned} \end{aligned}$$

We already have that $\delta _f\left( x,t_0\right) = \delta _f\left( x,\tau _0\right) \le \epsilon$ and $\delta _g\left( x,t_0\right) = \delta _g\left( x,\tau _0\right) \le \epsilon _0$ for $\epsilon _0$ small enough. Then, by the compactness of $\bar{\Omega }$, for $\tau \in \left[ 0,\tau _1 - \tau _0\right]$, $\left\| \delta _f\left( \tau \right) \right\| ^2 \le O\left( \epsilon \right)$, for all $x \in \Omega$. Make a process similarly and successively for $k = 1,2,\dots$, we have that $\left\| \delta _f\right\| ^2 \le O\left( \epsilon \right)$ for all $x\in \Omega$. Analogously, we can also prove that $\left\| \delta _g\right\| ^2 \le O\left( \epsilon \right)$.

Therefore, $\int _{\Omega }\left| f(x,t) - f_0(x,t)\right| ^2dx \le C\epsilon$ and $\int _{\Omega }|g(x,t) - g_0(x,t)|^2dx \le C\epsilon$, and we have the conclusion of the theorem. $\square$

Appendix B Proof for theorem 8

Proof

Note that $\left\| F\left( u_1,x\right) - F\left( u_2,x\right) \right\| \le C\left\| u_1 - u_2\right\| ,\forall u_1,u_2 \in D\left( F\right)$ and $|G\left( u,x\right) v|$ is bounded, $\forall u,v$ bounded due to the continuous differentiability of G in a bounded domain. Consider

$$\begin{aligned} \begin{aligned} \dfrac{1}{2}\dfrac{\partial }{\partial t} |u -v|^2&= \left( u-v\right) \dfrac{\partial }{\partial t}\left( u-v\right) \\&= \left( u -v\right) \left[ F\left( u,x\right) - F\left( v,x\right) \right] + \epsilon \left( u -v\right) G(u,x) + \epsilon \left( u - v\right) \Delta \left( u - v\right) \\&\le C|u - v|^2 + O\left( \epsilon \right) + \epsilon \left( u - v\right) \Delta \left( u - v\right) . \end{aligned} \end{aligned}$$

(B1)

Taking the integral of (B1) over $\Omega$ and using the Neumann boundary condition implies that

$$\begin{aligned} \begin{aligned} \dfrac{1}{2}\dfrac{\partial }{\partial t}\int _{\Omega } |u -v|^2dx \le C\int _{\Omega }|u-v|^2dx + O\left( \epsilon \right) - \epsilon \int _{\Omega }\left\| \nabla \left( u - v\right) \right\| ^2dx, \end{aligned} \end{aligned}$$

which leads to

$$\begin{aligned} \dfrac{\partial }{\partial t}\int _{\Omega } |u -v|^2dx \le C\int _{\Omega }|u-v|^2dx + O\left( \epsilon \right) . \end{aligned}$$

Applying Gronwall’s inequality, and we have that

$$\begin{aligned} \int _{\Omega } |u -v|^2dx \le O\left( \epsilon \right) e^{Ct^2}, \end{aligned}$$

which implies $\int _{\Omega } |u -v|^2dx = O\left( \epsilon \right)$ for all $t < \mathcal {T}$ with given $\mathcal {T} > 0$, by the compactness of $\bar{\Omega }$. $\square$