Trend to Equilibrium for a Reaction–Diffusion System Modelling Reversible Enzyme Reaction

Abstract

A spatio-temporal evolution of chemicals appearing in a reversible enzyme reaction and modelled by a four-component reaction–diffusion system with the reaction terms obtained by the law of mass action is considered. The large time behaviour of the system is studied by means of entropy methods.

Appendices

Appendix A: Duality Principle

We recall a duality principle (Pierre 2010; Pierre and Schmitt 1997) which is used to show $L^2(\log L)^2$ and $L^2$ bounds, respectively, for the solution of the system (4)–(6). Note that a more general result is proved in Pierre (2010), Chap. 6, than presented here.

Lemma 4.1

(Duality principle) Let $0<T<\infty$ and $\Omega$ be a bounded, open and regular (e.g. $C^2$) subset of $\mathbb {R}^d$. Consider a nonnegative weak solution u of the problem

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t u - \Delta (A u) \le 0, \\&\nabla (A u) \cdot \nu = 0, \quad \forall t \in I, \; x \in \partial \Omega , \\&u(0,x) = u_0(x), \\ \end{aligned} \right. \end{aligned}$$

(53)

where we assume that $0< A_1 \le A = A(t,x) \le A_2 < \infty$ is smooth, $A_1$ and $A_2$ are strictly positive constants, $u_0 \in L^2(\Omega )$ and $\int u_0 \ge 0$. Then,

$$\begin{aligned} \Vert u \Vert _{L^2(Q_T)} \le C \Vert u_0 \Vert _{L^2(\Omega )} \end{aligned}$$

(54)

where $C=C(\Omega , A_1, A_2, T)$.

Proof

Let us consider an adjoint problem: find a nonnegative function $v \in C(I;L^2(\Omega ))$ which is regular in the sense that $\partial _t v, \Delta v \in L^2(Q_T)$ and satisfies

$$\begin{aligned} \left\{ \begin{aligned}&- \partial _t v - A \Delta v = F, \\&\nabla v \cdot \nu = 0, \quad \forall t \in I, \; x \in \partial \Omega , \\&v(T,x) = 0, \\ \end{aligned} \right. \end{aligned}$$

(55)

for $F = F(t,x) \in L^2(Q_T)$ nonnegative. The existence of such v follows from the classical results on parabolic equations (Ladyzhenskaya et al. 1968).

By combining equations for u and v, we can readily check that

$$\begin{aligned} -\dfrac{\text {d}}{\, \text {d}t} \int _{\Omega } u v \ge \int _{\Omega } u F \end{aligned}$$

which, by using $v(T)=0$, yields

$$\begin{aligned} \int _{Q_T} u F \le \int _{\Omega } u_0 v_0. \end{aligned}$$

(56)

By multiplying equation for v in (55) by $-\Delta v$, integrating per partes and using the Young inequality, we obtain

$$\begin{aligned} -\dfrac{1}{2}\dfrac{\text {d}}{\, \text {d}t} \int _{\Omega } \vert \nabla v \vert ^2 + \int _{\Omega } A (\Delta v)^2 = - \int _{\Omega } F \Delta v \le \int _{\Omega } \dfrac{F^2}{2A} + \dfrac{A}{2}(\Delta v)^2, \end{aligned}$$

i.e.

$$\begin{aligned} -\dfrac{\text {d}}{\, \text {d}t} \int _{\Omega } \vert \nabla v \vert ^2 + \int _{\Omega } A (\Delta v)^2 \le \int _{\Omega } \dfrac{F^2}{A}. \end{aligned}$$

Integrating this over [0, T] and using $v(T)=0$ gives

$$\begin{aligned} \int _{\Omega } \vert \nabla v_0 \vert ^2 + \int _{Q_T} A (\Delta v)^2 \le \int _{Q_T} \dfrac{F^2}{A}. \end{aligned}$$

Thus, we obtain the a-priori bounds

$$\begin{aligned} \Vert \nabla v_0 \Vert _{L^2(\Omega ,\mathbb {R}^d)} \le \left\| \dfrac{F}{\sqrt{A}} \right\| _{L^2(Q_T)} \quad \text {and} \quad \Vert \sqrt{A} \Delta v \Vert _{L^2(\Omega )} \le \left\| \dfrac{F}{\sqrt{A}} \right\| _{L^2(Q_T)}. \end{aligned}$$

(57)

From the equation for v, we can write (again, by integrating this equation over $\Omega$ and [0, T] and using $v(T)=0$)

$$\begin{aligned} \int _{\Omega } v_0 = \int _{Q_T} A \Delta v + F. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} \int _{\Omega } v_0 = \int _{Q_T} \sqrt{A} \left( \sqrt{A} \Delta v + \dfrac{F}{\sqrt{A}}\right)&\le \Vert \sqrt{A} \Vert _{L^2(Q_T)} \left\| \sqrt{A}\Delta v + \dfrac{F}{\sqrt{A}} \right\| _{L^2(Q_T)} \\&\le 2 \Vert \sqrt{A} \Vert _{L^2(Q_T)} \left\| \dfrac{F}{\sqrt{A}} \right\| _{L^2(Q_T)}, \end{aligned} \end{aligned}$$

(58)

which follows from the Hölder inequality and (57).

To conclude the proof, let us return to (56) and write

$$\begin{aligned} \begin{aligned} 0 \le \int _{Q_T} u F \le \int _{\Omega } u_0 v_0&= \int _{\Omega } u_0( v_0 - \overline{v_0}) + u_0\overline{v_0}\\&\le \Vert u_0 \Vert _{L^2(\Omega )}\Vert v_0 - \overline{v_0} \Vert _{L^2(\Omega )} + \int _{\Omega } \overline{u_0} v_0 \\&\le C(\Omega ) \Vert u_0 \Vert _{L^2(\Omega )} \Vert \nabla v_0 \Vert _{L^2(\Omega ,\mathbb {R}^d)} + \overline{u_0} \int _{\Omega } v_0, \end{aligned} \end{aligned}$$

where we have used the Hölder and Poincaré-Wirtinger inequalities, respectively. Recall that $\overline{v} = \dfrac{1}{\vert \Omega \vert } {\displaystyle \int _{\Omega } v \, \text {d}x}.$ The norm of the gradient $v_0$ can be estimated by (57) and the last remaining integral by (58) so that we obtain

$$\begin{aligned} \int _{Q_T} u F \le \left( C(\Omega ) \Vert u_0 \Vert _{L^2(\Omega )} + 2\overline{u_0} \Vert \sqrt{A} \Vert _{L^2(Q_T)} \right) \left\| \dfrac{F}{\sqrt{A}} \right\| _{L^2(Q_T)}, \end{aligned}$$

(59)

which holds true for any $F \in L^2(Q_T)$. Thus, for $F = Au$ we can finally write

$$\begin{aligned} \Vert \sqrt{A}u\Vert _{L^2(Q_T)} \le C(\Omega ) \Vert u_0 \Vert _{L^2(\Omega )} + 2 \overline{u_0} \Vert \sqrt{A} \Vert _{L^2(Q_T)} \end{aligned}$$

(60)

and deduce (54) by using the boundedness of A, i.e. $A_1 \le A(t,x) \le A_2$. $\square$

We remark that, by construction, A(t, x) in (23) is not smooth but $L^{\infty }$ only. We refer to Breden et al. (2017) for the corresponding existence and regularity result for the adjoint problem (55).

Appendix B: Inequality (48)

In this section, we give a full proof of the inequality (48). We recall that all the summations below go for $i \in \{S,E,C,P\}$, i.e. $\sum = \sum _{i \in \{S,E,C,P\}}$.

We start by noting that that the conservation law (39), reflecting the ansatz (40) and the substitution (45), namely

$$\begin{aligned} N_{E,\infty }^2 + N_{C,\infty }^2= & {} N_{E,\infty }^2(1+\mu _E)^2 + \overline{\delta _E^2}+ N_{C,\infty }^2(1+\mu _C)^2 + \overline{\delta _C^2}, \end{aligned}$$

(61)

$$\begin{aligned} N_{S,\infty }^2 + N_{C,\infty }^2 + N_{P,\infty }^2= & {} N_{S,\infty }^2(1+\mu _S)^2 + \overline{\delta _S^2}+ N_{C,\infty }^2(1+\mu _C)^2 + \overline{\delta _C^2}\nonumber \\&+\, N_{P,\infty }^2(1+\mu _P)^2 + \overline{\delta _P^2}, \end{aligned}$$

(62)

possesses restrictions on the signs of $\mu _i$’s. In particular, we remark that

• i) $\forall i \in \{S,E,C,P\}$, $-1 \le \mu _i \le \mu _{i,max}$ where $\mu _{i,max}$ depends on $\mathbf {n_{\infty }}$;

• ii) the conservation law (61) excludes the case when $\mu _E >0$ and $\mu _C >0$, since in this case $\overline{N_E}> N_{E,\infty }$ and $\overline{N_C}> N_{C,\infty }$ and we deduce from (39), (61) and the Jensen inequality $\overline{N_i^2} \ge \overline{N_i}^2$, that

  $$\begin{aligned} M_1 = \overline{N_E^2} + \overline{N_C^2} \ge \overline{N_E}^2 + \overline{N_C}^2 > N_{E,\infty }^2 + N_{C,\infty }^2 = M_1, \end{aligned}$$

  which is a contradiction;

• iii) analogously, the conservation law (62) excludes the case when $\mu _S >0$, $\mu _C >0$ and $\mu _P >0$;

• iv) for $-1 \le \mu _E, \mu _C \le 0$, the conservation law (61) implies $N_{E,\infty }^2 \mu _E^2 + N_{C,\infty }^2 \mu _C^2 \le \sum \overline{\delta _i^2}$, since for any $s \in [-1,0]$ we have $-1 \le s \le -s^2 \le 0$ and we can deduce from (61) that

  $$\begin{aligned} \begin{aligned} 0&= N_{E,\infty }^2\left( 2\mu _E + \mu _E^2\right) + N_{C,\infty }^2\left( 2\mu _C + \mu _C^2\right) + \overline{\delta _C^2}+ \overline{\delta _E^2}\\&\le - N_{E,\infty }^2 \mu _E^2 - N_{C,\infty }^2 \mu _C^2 + \sum \overline{\delta _i^2}; \end{aligned} \end{aligned}$$

• v) analogously, for $-1 \le \mu _S, \mu _C, \mu _P \le 0$, the conservation law (62) implies that $N_{S,\infty }^2 \mu _S^2 + N_{C,\infty }^2 \mu _C^2 + N_{P,\infty }^2 \mu _P^2 \le \sum \overline{\delta _i^2}$.

To find explicitly $C_3$ and $C_4$ in (48), we have to consider all possible configurations of $\mu _i$’s in (48), that is all possible quadruples $(\mu _E, \mu _C, \mu _S, \mu _P)$ depending on their signs. The remarks (ii) and (iii) reduce the total number of quadruples by five, and the remaining 11 quadruples are shown in Table 1.

Ad (I). The remarks (iv) and (v) imply that $\sum N_{i,\infty }^2 \mu _i^2 \le 2 \sum \overline{\delta _i^2}$ and, therefore, (48) is satisfied for $C_3=3$ and $C_4=0$.

Ad (II) and (III). We prove (48) for $-1 \le \mu _E, \mu _C \le 0$ and $\mu _S$ and $\mu _P$ having opposite signs. Firstly, let us remark that (61) implies that

$$\begin{aligned} N_{E,\infty }^2 = N_{E,\infty }^2(1+\mu _E)^2 + N_{C,\infty }^2(2\mu _C + \mu _C^2) + \overline{\delta _E^2}+ \overline{\delta _C^2}, \end{aligned}$$

i.e.

$$\begin{aligned} \begin{aligned} (1+\mu _E)^2&= 1 - \frac{N_{C,\infty }^2}{N_{E,\infty }^2}\left( 2\mu _C + \mu _C^2\right) - \frac{1}{N_{E,\infty }^2} \left( \overline{\delta _E^2}+ \overline{\delta _C^2}\right) \\&\ge 1 - \frac{1}{N_{E,\infty }^2} \sum \overline{\delta _i^2}\end{aligned} \end{aligned}$$

(63)

since for $-1 \le \mu _C \le 0$ there is $-1 \le 2\mu _C + \mu _C^2 \le 0$. Then, by using an elementary inequality $a^2+b^2 \ge (a-b)^2/2$ we obtain that

$$\begin{aligned} I_1 + I_2 = ((1+\mu _S)(1+ \mu _E) - (1+\mu _C))^2 + ((1+\mu _P)(1+ \mu _E) - (1+ \mu _C))^2 \end{aligned}$$

satisfies

$$\begin{aligned} \begin{aligned} I_1 + I_2&\ge \frac{1}{2}(\mu _S-\mu _P)^2(1+\mu _E)^2 \\&\ge \frac{1}{2} K_4 \left( N_{S,\infty }^2 \mu _S^2 + N_{P,\infty }^2 \mu _P^2\right) - K_5 \sum \overline{\delta _i^2}\end{aligned} \end{aligned}$$

(64)

where we have used (63) and the fact that $\mu _S$ and $\mu _P$ have opposite signs and are bounded above by $\mu _{S,max}$ and $\mu _{P,max}$ (by the remark (i)). In (64), $K_4 = \min \left\{ 1/N_{S,\infty }^2, 1/N_{P,\infty }^2 \right\}$ is sufficient, nevertheless, we will take

$$\begin{aligned} K_4 = \min _{i \in \{S,E,C,P\}} \left\{ \dfrac{1}{N_{i,\infty }^2} \right\} \quad \text {and} \quad K_5 = \dfrac{1}{N_{E,\infty }^2} \left( \mu _{S,max}^2 + \mu _{P,max}^2\right) , \end{aligned}$$

(65)

since $K_4$ in (65) will appear several times elsewhere. We deduce from (64) and the remark (iv) that

$$\begin{aligned} \sum N_{i,\infty }^2 \mu _i^2 + \sum \overline{\delta _i^2}\le 2 \left( 1 + \frac{K_5}{K_4}\right) \sum \overline{\delta _i^2}+ \frac{2}{K_4} (I_1 + I_2), \end{aligned}$$

(66)

and we see that (48) is satisfied for

$$\begin{aligned} C_4 = \frac{2}{K_3 K_4} \quad \text {and} \quad C_3 = 2\left( 1 + \frac{K_5}{K_4}\right) + C_4 \left( \sqrt{k_{+}} K_1 + \sqrt{k_{p-}} K_2 \right) , \end{aligned}$$

when (48) is compared with the r.h.s. of (66).

Ad (IV). Assume $-1 \le \mu _E, \mu _C \le 0$ and $\mu _S, \mu _P > 0$. A combination of (61) and (62) gives

$$\begin{aligned} \begin{aligned} N_{E,\infty }^2 - N_{S,\infty }^2 - N_{P,\infty }^2&= \overline{N_E}^2 + \overline{\delta _E^2}- \overline{N_S}^2 - \overline{\delta _S^2}- \overline{N_P}^2 - \overline{\delta _P^2}\\&\le N_{E,\infty }^2 - \overline{N_S}^2 - \overline{N_P}^2 + \overline{\delta _E^2}- \overline{\delta _S^2}- \overline{\delta _P^2}, \end{aligned} \end{aligned}$$

(67)

since $\overline{N_E}^2 \le N_{E,\infty }^2$ for $-1 \le \mu _E \le 0$, where $\overline{N_E}^2 = N_{E,\infty }^2(1+\mu _E)^2$. We deduce from (67) that

$$\begin{aligned} - N_{S,\infty }^2 - N_{P,\infty }^2 \le - N_{S,\infty }^2(1+\mu _S)^2 - N_{P,\infty }^2(1+\mu _P)^2 + \overline{\delta _E^2}- \overline{\delta _S^2}- \overline{\delta _P^2}\end{aligned}$$

and

$$\begin{aligned} N_{S,\infty }^2\left( 2\mu _S + \mu _S^2\right) + N_{P,\infty }^2\left( 2\mu _P + \mu _P^2\right) \le \overline{\delta _E^2}- \overline{\delta _S^2}- \overline{\delta _P^2}\le \sum \overline{\delta _i^2}. \end{aligned}$$

Since $\mu _S, \mu _P > 0$, then $N_{S,\infty }^2 \mu _S^2 + N_{P,\infty }^2 \mu _P^2 \le \sum \overline{\delta _i^2}$. This estimate together with the remark (iv) yields $\sum N_{i,\infty }^2 \mu _i^2 \le 2\sum \overline{\delta _i^2}$. Similarly as in the case (I), (48) is satisfied for $C_3=3$ and $C_4=0$.

Ad (V). Let us now consider the case when $-1 \le \mu _S, \mu _C, \mu _P \le 0$ and $\mu _E>0$. As in the case (IV), a combination of (61) and (62) gives

$$\begin{aligned} N_{S,\infty }^2 + N_{P,\infty }^2 - N_{E,\infty }^2&= \overline{N_S}^2 + \overline{\delta _S^2}+ \overline{N_P}^2 + \overline{\delta _P^2}- \overline{N_E}^2 - \overline{\delta _E^2}\nonumber \\&\le N_{S,\infty }^2 + N_{P,\infty }^2 - \overline{N_E}^2 + \overline{\delta _S^2}+ \overline{\delta _P^2}- \overline{\delta _E^2}, \end{aligned}$$

(68)

since, again, $\overline{N_i}^2 \le N_{i,\infty }^2$ for $-1 \le \mu _i \le 0$ and $i = S,P$. Hence, for $\mu _E>0$ we deduce from (68) that $N_{E,\infty }^2 \mu _E^2 < \sum \overline{\delta _i^2}$, which with the remark (v) gives $\sum N_{i,\infty }^2 \mu _i^2 < 2\sum \overline{\delta _i^2}$. Thus, (48) is satisfied for $C_3=3$ and $C_4=0$.

Ad (VI) and (VII). Assume that $\mu _E > 0$, $-1 \le \mu _C \le 0$ and $\mu _S$ and $\mu _P$ have opposite signs. Then, using an elementary inequality $a^2+b^2 \ge (a+b)^2/2$, we obtain

$$\begin{aligned} I_1+I_2 \ge \dfrac{1}{2}(\mu _S-\mu _P)^2(1+\mu _E)^2 > \dfrac{1}{2}(\mu _S-\mu _P)^2 \ge \dfrac{1}{2}\left( \mu _S^2 + \mu _P^2\right) , \end{aligned}$$

since $(1+\mu _E)^2>1$ and $\mu _S$ and $\mu _P$ have opposite signs. Further, it holds that $(1+\mu _k)(1+\mu _E) > (1+\mu _E)$ for $\mu _k$ being either $\mu _S>0$ or $\mu _P>0$ (one of them is positive). This implies $(1+\mu _k)(1+\mu _E) - (1+\mu _C)> \mu _E - \mu _C > 0$ and thus ($\mu _E$ and $\mu _C$ have opposite signs)

$$\begin{aligned} I_1 + I_2 > (\mu _E-\mu _C)^2 \ge \mu _E^2 + \mu _C^2. \end{aligned}$$

Altogether, we obtain for both cases that $I_1 + I_2 > \sum \mu _i^2/4 \ge K_4/4 \sum N_{i,\infty }^2 \mu _i^2$ where $K_4$ is defined in (65). We deduce that (48) is satisfied for

$$\begin{aligned} C_4 = \frac{4}{K_3 K_4} \quad \text {and} \quad C_3 = 1 + C_4 \left( \sqrt{k_{+}} K_1 + \sqrt{k_{p-}} K_2\right) . \end{aligned}$$

(69)

Ad (VIII). Assume that $\mu _E, \mu _S, \mu _P > 0$ and $-1 \le \mu _C \le 0$. Using the similar arguments as in the previous case, in particular, $(1+\mu _S)(1+\mu _E) > (1+\mu _S)$, $(1+\mu _S)(1+\mu _E) > (1+\mu _E)$, $(1+\mu _P)(1+\mu _E) > (1+\mu _P)$ and $(1+\mu _P)(1+\mu _E) > (1+\mu _E)$ and since $\mu _i-\mu _C > 0$ for each $i \in \{S,E,P\}$, we can write

$$\begin{aligned} \begin{aligned} I_1 + I_2&= ((1+\mu _S)(1+ \mu _E)-(1+\mu _C))^2 + ((1+\mu _P)(1+ \mu _E)-(1+\mu _C))^2 \\&\ge \dfrac{1}{2}(\mu _S-\mu _C)^2 + (\mu _E-\mu _C)^2 + \dfrac{1}{2}(\mu _P-\mu _C)^2 \ge \dfrac{1}{2} \sum \mu _i^2 \\&\ge \dfrac{K_4}{2} \sum N_{i,\infty }^2 \mu _i^2. \end{aligned} \end{aligned}$$

Hence, (48) is satisfied for $C_4 = 2/K_3K_4$ and $C_3$ defined in (69).

Ad (IX). The case when $-1 \le \mu _E, \mu _S, \mu _P \le 0$ and $\mu _C > 0$ is similar to the case (VIII). Now we observe that $\mu _C-\mu _i > 0$ for each $i \in \{S,E,P\}$ and that $(1+\mu _S)(1+\mu _E) \le (1+\mu _S)$, $(1+\mu _S)(1+\mu _E) \le (1+\mu _E)$, $(1+\mu _P)(1+\mu _E) \le (1+\mu _P)$ and $(1+\mu _P)(1+\mu _E) \le (1+\mu _E)$ which can be used to conclude $I_1 + I_2 \ge \sum \mu _i^2/2\ge K_4/2 \sum N_{i,\infty }^2 \mu _i^2$. The constants $C_3$ and $C_4$ are the same as in the case (VIII).

Ad (X). Assume that $-1 \le \mu _E \le 0$, $\mu _C > 0$, $-1 \le \mu _S \le 0$ and $\mu _P > 0$. By the same argument as in (IX), we can write

$$\begin{aligned} I_1 + I_2 \ge I_1 \ge (\mu _C-\mu _E)^2 \ge \mu _C^2 + \mu _E^2. \end{aligned}$$

(70)

Using the same elementary inequality as in (II) and (VI), we obtain

$$\begin{aligned} I_1 + I_2 \ge \dfrac{1}{2}(\mu _S-\mu _P)^2(1+\mu _E)^2, \end{aligned}$$

(71)

where $-1 \le \mu _E \le 0$, and thus we cannot proceed in the way as in the cases (VI) and (VII) nor in the cases (II) and (III), since $\mu _C$ is positive now. Nevertheless, we distinguish two subcases when $-1 < \eta \le \mu _E \le 0$ and $-1 \le \mu _E < \eta$. For example, $\eta = -1/2$ works well, however, a more suitable constant $\eta$ could be possibly found. For $\eta = -1/2$ and $\eta \le \mu _E \le 0$, we obtain from (71) that

$$\begin{aligned} I_1 + I_2 \ge \dfrac{1}{8}(\mu _S-\mu _P)^2 \ge \dfrac{1}{8}\left( \mu _S^2+\mu _P^2\right) . \end{aligned}$$

(72)

This with (70) implies that $I_1 + I_2 \ge \sum \mu _i^2/16 \ge K_4/16 \sum N_{i,\infty }^2 \mu _i^2$ and we conclude that (48) is satisfied for $C_4 = 16/K_3K_4$ and $C_3$ defined in (69).

For $\eta = -1/2$ and $-1 \le \mu _E < \eta$ we obtain, by using an elementary inequality $(a-b)^2 \ge a^2/2-b^2$, that

$$\begin{aligned} \begin{aligned} I_1 + I_2 \ge I_1&= ((1+\mu _C) - (1+\mu _S)(1+\mu _E))^2 \\&\ge \dfrac{1}{2} (1+\mu _C)^2 - (1+\mu _S)^2(1+\mu _E)^2 > \dfrac{1}{4}, \end{aligned} \end{aligned}$$

(73)

since $(1+\mu _C)^2 > 1$ for $\mu _C > 0$ and $(1+\mu _S)^2(1+\mu _E)^2 < 1/4$ for $-1 \le \mu _S \le 0$ and $-1 \le \mu _E < -1/2$. On the other hand, $\sum N_{i,\infty }^2 \mu _i^2 \le \sum N_{i,\infty }^2 \mu _{i,\max }^2 =: K_6$ by the remark (i). We see that (48) is satisfied for $C_4 = K_6/4K_3$ and $C_3$ as in (69).

Ad (XI). Finally, assume that $-1 \le \mu _E \le 0$, $\mu _C > 0$, $\mu _S > 0$ and $-1 \le \mu _P \le 0$. This case is symmetric to the previous case (X), and thus the same procedure can be applied again (it is sufficient to exchange superscripts S and P everywhere they appear in (X)) to deduce the constants $C_3$ and $C_4$ in (48). In particular, we take $C_4 = 16/K_3K_4$ for $-1/2 \le \mu _E \le 0$ and $C_4 = K_6/4K_3$ for $-1 \le \mu _E < -1/2$. In both subcases $C_3$ is as in (69).

From the eleven cases (I)–(XI), we need to take

$$\begin{aligned} C_4 = \frac{1}{K_3} \max \left\{ \frac{16}{K_4}, \frac{K_6}{4} \right\} \end{aligned}$$

(74)

and

$$\begin{aligned} C_3 = \max \left\{ 3, 2\left( 1 + \frac{K_5}{K_4}\right) \right\} + C_4 \left( \sqrt{k_{+}} K_1 + \sqrt{k_{p-}} K_2 \right) \end{aligned}$$

(75)

to find (48) true.