Travelling waves of a reaction-diffusion model
for the acidic nitrate-ferroin reaction
Department of Mathematical Sciences, National Chengchi University, Taipei 116, Taiwan
Sheng-Chen Fu
1Abstract
In this paper we consider a reaction-diffusion system which describes the acidic nitrate-ferroin reaction. The existence of travelling wave solutions for this system is investigated. Our proofs are rigorous.
Key words: reaction-diffusion system; travelling wave; acidic nitrate-ferroin reaction
1 Introduction
Travelling waves in the acidic nitrate-ferroin reaction have drawn a lot of attention for many researchers; see, for example, [1]–[4]. In this paper we study the existence of travelling waves for the following reaction-diffusion system
∂u ∂t = Du ∂2u ∂x2 − 2kuv α + u, ∂v ∂t = Dv ∂2v ∂x2 + kuv α + u, (1)
which was derived in [1] to model the acidic nitrate-ferroin reaction. Here α and k are positive constants, u and v represent the concentrations of the ferroin and acidic nitrate respectively, and Du and Dv denote the constant
diffusion rates of the ferroin and acidic nitrate respectively.
Email address: [email protected] (Sheng-Chen Fu).
J.H. Merkin and M.A. Sadiq in [4] studied (1) together with the initial and boundary conditions u(x, 0) = u0, −∞ < x < ∞, v(x, 0) = v0g(x), |x| < l, 0, |x| > l, u → u0, v → 0 as |x| → ∞, t ≥ 0, (2)
where u0 and v0 are constants and g(x) is a continuous and differentiable
function on −l < x < l with a maximum value of unity.
For convenience, we make a change of variables
¯ u = u u0 , ¯v = v u0 , ¯t = kt and ¯x = s k Dv x.
After dropping the bars, the initial and boundary value problem (1)-(2) be-comes the dimensionless form
∂u ∂t = δ ∂2u ∂x2 − 2uv β + u, ∂v ∂t = ∂2v ∂x2 + uv β + u, (3)
together with the initial and boundary conditions
u(x, 0) = 1, −∞ < x < ∞, v(x, 0) = v0∗g(x), |x| < l∗, 0, |x| > l∗, u → 1, v → 0 as |x| → ∞, t ≥ 0, (4)
where δ = Du/Dv, β = α/u0, v∗0 = v0/u0 and l∗ =
q
kl2/D v.
By a travelling wave solution of (3), we mean a solution of the form
(u(x, t), v(x, t)) = (U (y), V (y)), (5)
where y = x − ct, for some nonnegative functions U and V satisfying
U (−∞) = 0, V (−∞) = 1/2, U (+∞) = 1, and V (+∞) = 0. (6)
that (U, V ) solves the system δUyy+ cUy − 2U V β + U = 0, Vyy+ cVy + U V β + U = 0. (7)
J.H. Merkin and M.A. Sadiq in [4] established a necessary condition for the existence of nonnegative solutions of (6)–(7). Indeed they showed the following theorem.
Theorem 1 There exists no nonnegative solutions of (6)–(7) if c < 2/√β + 1. They also showed that the large time structure of the initial value problem (3)-(4) is a travelling wave with a constant speed. In addition, the asymptotic wave speed is exactly the minimum possible speed 2/√β + 1.
An important guestion in the study of (3) is the existence of a minimum speed travelling wave solution, the estimate of the minimum speed, as well as the range of c such that travelling wave solutions exist. For the case δ = 1, it was proved in [4] that any nonnegative solution (U, V ) of the problem (6)–(7) satisfies U = 1 − 2V . Thus the system (7) can be reduced to a single equation
Vyy+ cVy+ (1 − 2V )V /(β + 1 − 2V ).
Hence A = 2V solves
Ayy+ cAy+ (1 − A)A/(β + 1 − A),
A(−∞) = 1, A(+∞) = 0.
Using the result in [5], we obtain the sufficient and necessary condition for the existence of nonnegative solutions of (6)–(7) in the following theorem.
Theorem 2 Let δ = 1. The problem (6)–(7) possesses a nonnegative solution iff c ≥ 2/√β + 1. For a fixed c, this solution is unique up to translation. Modifying the method in [6] for a different system, we shall answer the question for the case δ 6= 1. Our results are stated as follows.
Theorem 3 Let 0 < δ < 1. There exists a number cmin > 0 such that (6)–
(7) possesses a nonnegative solution iff c ≥ cmin. For a fixed c, this solution
is unique up to translation. Moreover, 2/√β + 1 ≤ cmin ≤ (2/
√
β)[(√2 + 1)/
q
δ + 2(√2 + 1)].
From Theorem 2 and Theorem 3, we know that there exists a minimun speed travelling wave solution of (3) and the set of admissible wave speed is an
interval [cmin, ∞) if δ ≤ 1. Unfortunately, our result is incomplete for δ > 1.
Indeed, we can only give a sufficient condition for the existence of nonnegative solutions of (6)–(7) in this case.
Theorem 4 Let δ > 1. There exists a nonnegative solution of (6)–(7) if c ≥ 2/√β. For a fixed c, this solution is unique up to translation.
Combining Theorem 1 and Theorem 4, we get a upper bound and lower bound of the minimun speed. Whether the set of admissible wave speed is [cmin, ∞)
is still unknown.
The rest of this paper is organized as follows. Priminary results required for the study of existence of travelling waves are contained in Section 2. We prove the existence of travelling wave solutions in Section 3.
2 Preliminary
From now on, we always assume that c > 2/√β + 1. In [4], the following properties for travelling wave solutions were proved.
Proposition 5 A nonnegative solution (U, V ) of (6)–(7) has the following properties: (a) U > 0 and V > 0 on (−∞, ∞). (b) δUy+ 2Vy+ c(U + 2V − 1) = 0 on (−∞, ∞). (c) Uy > 0 and Vy < 0 on (−∞, ∞). (d) c = 2R∞ −∞U V /(β + U )dy > 0.
Putting W = Vy and using Proposition 5(b), we may write (7) as the equivalent
third-order system Uy = − 1 δ[2W + c(U + 2V − 1)], Vy = W, Wy = −cW − U V β + U. (8)
Thus a nonnegative solution of (6)–(7) is a solution of (8) satisfying the con-ditions
(U, V, W ) → (0, 1/2, 0), as y → −∞,
(U, V, W ) → (1, 0, 0), as y → +∞.
(9)
Note that the system (8) has just two equilibrium points (1, 0, 0) and (0, 1/2, 0). To see the behavior of the integral curves about the point (1, 0, 0), we linearize
the system (8). The eigenvalues and associated eigenvectors are λ1 = −c/δ, eλ1 = (1, 0, 0) T, λ2 = −(c + q c2− 4/(β + 1)/2, e λ2 = (−2(c + λ2), c + δλ2, λ2(c + δλ2)) T, λ3 = −(c − q c2− 4/(β + 1)/2, e λ3 = (−2(c + λ3), c + δλ3, λ3(c + δλ3)) T.
Since all eignvalues are negative, it follows that the point (1, 0, 0) is a stable node. By a similar way, since the eigenvalues and associated eigenvectors for the linearized system about the point (0, 1/2, 0) are
λ1 = −c, eλ1 = (0, 1, −c) T, λ2 = −(c + q c2 + 4δ/β)/(2δ), e λ2 = (2β(c + λ2), −1/λ2, −1)) T, λ3 = −(c − q c2+ 4δ/β)/(2δ), e λ3 = (2β(c + λ3), −1/λ3, −1)) T, (10)
it follows that the point (0, 1/2, 0) is a saddle point with a two-dimensional stable manifold and a one-dimensional unstable manifold. Therefore, a non-negative solution of (6)–(7), if it exists, must go out of the unstable manifold of the point (0, 1/2, 0) and finally reaches the point (1, 0, 0). In addition, this solution is unique up to translation for a fixed c.
Introducing new variables ξ = 1 − 2V , y = z/c, R = U/c2 and using Proposi-tuon 5(b), the system (7) becomes
ξzz+ ξz = R(1 − ξ) β + c2R, δRz = 1 c2(ξz+ ξ) − R.
Setting ξz = P (ξ), we then consider the following initial value problem
P0P + P = R(1 − ξ) β + c2R, ξ > 0, δR0P = 1 c2(P + ξ) − R, ξ > 0, R(0) = 0, P (0) = 0, P > 0, R > 0 for ξ > 0. (11)
Following the proof of Lemma 2.2 in [6] and using (10), we can easily prove the following theorem. We omit the proof here.
unique solution on [0, h] for some h > 0. In addition,
P (ξ) = λξ + µξ2+ O(ξ3),
R(ξ) = βλ(λ + 1)ξ + βλ
2λδ + 1(1 − δ)µξ
2+ O(ξ3),
as ξ → 0+, where λ = [q1 + 4δ/(βc2) − 1]/(2δ) is the unique positive root of
the equation δλ2 + λ = 1/(βc2) and µ = −[c2λ2(1 + λ)2 + λ(1 + λ)](2δλ + 1)/[6δλ2+ (3δ + 2)λ + 1].
Lemma 7 For any c > 0 and δ > 0, the solution (P, R) of (11) can be continued to [0, 1) and P (1−) exists. In addition, there exists a nonnegative solution (unique up to translation) to (6)–(7) iff P (1−) = 0.
Proof. Note that R cannot hit zero before P does as long as (P, R) exists, since otherwise, there exists ξ1 > 0 such that P (ξ) > 0, R(ξ) > 0, for all 0 < ξ < ξ1
and R(ξ1) = 0. Then R0(ξ1) = (1 + ξ1/P (ξ1))/(δc2) > 0, a contradition. We
claim that P > 0 and R > 0 as long as (P, R) exists and 0 < ξ < 1. For contradiction, we assume that P > 0 in (0, ξ2) and P (ξ2−) = 0 for some
ξ2 ∈ (0, 1). Then (P0P )(ξ2−) ≤ 0. Hence, by (11), we get R(ξ2−) = 0 and so
R0(ξ2−) ≤ 0. On the other hand, δ(R0P )(ξ2−) = ξ2/c2 > 0, a contradiction.
Since P and R remain positive and bounded as long as (P, R) exists and 0 < ξ < 1, the solution (P, R) can be continued to [0, 1).
By Lemma 6, R0(0) = βλ(λ + 1) > 0. Thus, by continuity, R0 > 0 near ξ = 0. Set φ(ξ) = (P + ξ)/c2 − R. Then φ = δR0P > 0 near ξ = 0 and
φ0+ 1/(δP )φ = R(1 − ξ)/[c2P (β + c2R)] > 0 on (0, 1). So we can easily deduce that φ > 0 on (0, 1). Hence R0 > 0 on [0, 1). Since (P + ξ)0 = P0 + 1 = R(1 − ξ)/[P (β + c2R)] > 0, for all 0 < ξ < 1, the function P + ξ is increasing
on (0, 1). Hence lim
ξ→1−(P + ξ) exists snd so P (1−) exists. Now, there are two
cases: P (1−) = 0 or P (1−) > 0.
Suppose P (1−) = 0. To show the existence of nonnegative solutions to (6)– (7), it suffices to show that R(1−) = 1/c2 (i.e. U → 1 and V → 0 as ξ → 1−). Since P + ξ is increasing on (0, 1), we have P (ξ) + ξ < P (1−) + 1 = 1 on [0, 1) and so P (ξ) < 1 − ξ on [0, 1). Therefore, R(1−) = Z 1 0 1 δP[ (P + ξ) c2 − R]dξ ≥ 1 δc2 Z 1 0 1 − c2R 1 − ξ − 1dξ. (12) Since φ(1−) ≥ 0, it follows that R(1−) ≤ 1/c2. Together with the fact that R is increasing on [0, 1), we get
Combinining (12) and (13), we can deduce that R(1−) = 1/c2. Transfering
back to the original variables, we get a nonnegative solution to (6)–(7).
Suppose P (1−) > 0. Then the solution (P, R) can be continued beyond ξ = 1. Since V < 0 when ξ > 1, we conclude that there exists no nonnegative solutions to (6)–(7). 2
Lemma 8 (i) For 0 < δ < 1, R > ξ/c2, for all ξ ∈ (0, 1). (ii) For δ > 1, R < ξ/c2, for all ξ ∈ (0, 1).
Proof. Since the proofs of (i) and (ii) are similar, we shall only prove (i). Suppose 0 < δ < 1. By Lemma 6, we get
R − ξ
c2 = [βλ(λ + 1) −
1
c2]ξ + O(ξ
2) = β(1 − δ)λ2ξ + O(ξ2) > 0,
if ξ > 0 is sufficiently small. In addition, for all 0 < ξ < 1, we have
δ(R − ξ c2) 0P = δR0P − δ c2P = 1 c2(P + ξ) − R − δ c2P = 1 − δ c2 P + ξ c2 − R > −(R − ξ c2).
Therefore we can easily deduce that R − ξ/c2 > 0 for all ξ ∈ (0, 1). 2
3 The existence of travelling wave solutions
For the case 0 < δ < 1, we shall give a sufficient condition for the existence of travelling solutions in the following lemma.
Lemma 9 Let 0 < δ < 1. There exists a nonnegative solution to (6)–(7) if c > (2/√β)[(√2 + 1)/
q
δ + 2(√2 + 1)].
Proof. By lemma 6, we know that P (ξ) < λξ and R(ξ) < βλ(λ + 1)ξ if ξ is sufficiently small. Thus, Setting B = sup{η ∈ (0, 1) | P (ξ) < λξ and R(ξ) < βλ(λ + 1)ξ, ∀ξ ∈ (0, η)}, we obtain B > 0. We will show that B = 1. For contradiction, we assume that 0 < B < 1. Then either P (B) = λB or R(B) = βλ(λ + 1)B. Since
δP [R − βλ(λ + 1)ξ]0= δR0P − δβλ(λ + 1)P = 1 c2(P + ξ) − R − δβλ(λ + 1)P = β(δλ2+ λ)(P + ξ) − R − δβλ(λ + 1)P = [β(1 − δ)λ](P − λξ) − [R − βλ(λ + 1)ξ] ≤ −[R − βλ(λ + 1)ξ], ∀ξ ∈ (0, B], and (P − λξ)0P = P0P − λP =R(1 − ξ) β + c2R − (λ + 1)P = −(λ + 1)(P − λξ) +R(1 − ξ) β + c2R − λ(λ + 1)ξ < −(λ + 1)(P − λξ), ∀ξ ∈ (0, B],
we can easily obtain that P (ξ) < λξ and R(ξ) < βλ(λ + 1)ξ on (0, B]. In particular, P (B) < λB and R(B) < βλ(λ + 1)B, a contradiction. Hence P (ξ) < λξ and R(ξ) < βλ(λ + 1)ξ on (0, 1).
Now suppose c > (2/√β)[(√2 + 1)/
q
δ + 2(√2 + 1)]. Then we have λ(λ + 1) < 1/4. Thus if we choose ˆk such that λ(λ + 1) < ˆk < 1/4 then
P0P + P < βλ(λ + 1)ξ(1 − ξ)
β + c2R < λ(λ + 1)ξ(1 − ξ) < ˆkξ(1 − ξ), 0 < ξ < 1.
Let Q(ξ) be the unique solution of
QQ0+ Q = ˆkξ(1 − ξ), Q(0) = 0, Q > 0 on (0, 1).
Then Q(ξ) = γξ + O(ξ2), as ξ → 0+, where γ is the unique positive root of
γ2 + γ = ˆk. It is easy to see that γ > λ. Thus P < Q near ξ = 0. Hence it
follows from comparison principle that P ≤ Q on [0, 1). Using Lemma 2.1 of [6], we have Q(1−) = 0. Hence we conclude that P (1−) = 0. By Lemma 7, there exists a nonnegative solution to (6)–(7). 2
Proof of Theorem 3. By using continuous dependence, we can see that the set of admissible speed is closed. For each i = 1, 2, let (Pi, Ri) be the solution of
(11) on [0, 1) with c = ci, where c1 > c2 > 0. To show the existence of cmin, it
suffices to show that P1(1−) = 0 if P2(1−) = 0.
Now, we suppose P2(1−) = 0. Let λi be the positive root of δλ2+ λ = 1/(βc2i)
R1(ξ) < R2(ξ) for sufficiently small ξ. Set Ui = c2iRi, for each i = 1, 2. Then, by Lemma 6, we have U1− U2 = c21R1− c22R2 = [c2 1βλ1(λ1+ 1) − c22βλ2(λ2+ 1)]ξ + O(ξ2) = (1 − δ)(λ1− λ2) (δλ1+ 1)(δλ2+ 1) ξ + O(ξ2) < 0,
if ξ is sufficiently small. We claim that P1 < P2 on (0, 1). For contradiction,
we suppose η∗ = sup{η > 0 | P1 < P2 in (0, η)} < 1. So P1(η∗) = P2(η∗) > 0.
Now, we claim that R1 < R2 in (0, η∗]. For contradiction, we suppose there
exists ξ∗ ∈ (0, η∗] such that R
1(ξ∗) = R2(ξ∗). Then U1 > U2 at ξ = ξ∗. Recall
that U1 < U2 if ξ is small. Hence, there exists ξ∗ ∈ (0, ξ∗) such that U1 = U2
and (U1− U2)0 ≥ 0 at ξ = ξ∗. On the other hand, at ξ = ξ∗,
δ(U1− U2)0 = ( ξ∗ P1 − U1 P1 ) − (ξ∗ P2 − U2 P2 ) = ξ∗( 1 P1 − 1 P2 ) − U1( 1 P1 − 1 P2 ) = (ξ∗− U1)( 1 P1 − 1 P2 ) = (ξ∗− c21R1(ξ∗))( 1 P1 − 1 P2 ).
Hence it follows from Lemma 8(i) that δ(U1 − U2)0 < 0, a contradiction. For
0 < ξ ≤ η∗, 1 2(P 2 1 − P22) 0 = P1P10 − P2P20 = −P1+ R1(1 − ξ) β + c2 1R1 + P2− R2(1 − ξ) β + c2 2R2 = −(P1− P2) + (1 − ξ)[β(R1− R2) + (c21− c22)R1R2] (β + c2 1R1)(β + c22R2) < −(P1− P2) = −P 2 1 − P22 P1+ P2 .
Since P1 < P2 for small ξ, we conclude that P1 < P2 on (0, η∗]. In
partic-ular, P1(η∗) < P2(η∗), a contradiction. Thus P1(1−) ≤ P2(1−) = 0. Hence
P1(1−) = 0.
Finally, by applying Theorem 1 and Lemma 9, we can get the estimate of cmin. 2
Proof of Theorem 4. By Lemma 8,
if we choose ˆk such that 1/(βc2) ≤ ˆk ≤ 1/4. Arguing as the proof of Lemma 9, we get the proof. 2
Acknowledgements
This work was partially supported by National Science Council of the Repub-lic of China under the contract NSC 96-2115-M-004-002. The author thanks professor Je-Chiang Tsai for many helpful comments.
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