(3b−a2
9
)3
. Then the following assertions hold:
Case 1: When △ > 0, (19) has one real root and one pair of conjugate virtual roots;
Case 2: When △ = 0, (19) has three real roots, at least two of which are repeated roots;
Case 3: When △ < 0, (19) has three different real roots.
Moreover, the roots of (19) are
z1 =√3
ρ1+√3 ρ2+ a
3, z2 = ψ√3
ρ1+ ¯ψ√3 ρ2+ a
3, z3 = ¯ψ√3
ρ1+ ψ√3 ρ2+ a
3, where ρ1 =−2a3− 9ab + 27c
54 +√
△, ρ2 =−2a3− 9ab + 27c
54 −√
△, ψ = −1 2+ i
√3 2 and i2 =−1.
By Lemma 2, if △ ≤ 0, the basic reproduction number R0 = max{|z1|, |z2|, |z3|}; if
△ > 0, we obtain the reproduction number R0 = z1. We conclude our local stability in the following theorem.
Theorem 2. For the disease transmission model (5) (equivalently, (6)), if R0 < 1, then the disease-free equilibrium ξ0 (x0) is locally asymptotically stable. On the other hand, if R0 > 1, then ξ0 (x0) is unstable.
4 Global stability analysis
In this section, the qualitatively global analysis of model (6) (equivalently, (5)) is pre-sented. We first show that set△n+m := [0, 1]n+m is positive invariant.
Lemma 3. Set △n+m is positively invariant under the flow determined by equation (6).
That is, x(t)∈ △n+m for all t > 0 and x(0)∈ △n+m.
Proof. Denote the boundaries of set △n+m by ∂△n+m. Then it consists of the two parts
∂△1n+m and ∂△2n+m, where
∂△1n+m :={x ∈ △n+m: xk = 0 for some k}, and
∂△2n+m :={x ∈ △n+m: xk = 1 for some k}.
Then to prove that △n+m is positively invariant, it suffices to prove that the assignment vector at any boundary point of the vector field yielded by equation (6) is tangent or pointing into the set △n+m. Since △n+m is a rectangle in Rn+m, the “outer normals”
at boundaries ∂△1n+m and ∂△2n+m are ⃗n1k := −ek and ⃗n2k := ek, respectively, where ek denotes the standard unit vector in Rn+m. Moreover, we compute that
(I). For x∈ ∂△1n+m with xk = 0 for some k = 1, . . . , n + m:
˙
x· ⃗n1k =
− [
αkΘ(x1) + qk
n+m∑
j=n+1
βjxj ]
, for k = 1,· · · , n,
−rk
∑n j=1
Φ(x1), for k = n + 1, 2,· · · , n + m.
Thus, ˙x· ⃗n1k ≤ 0 for all k = 1, . . . , n + m.
(II). For x∈ ∂△2n+m with xk = 1 for some k = 1, . . . , n + m:
˙
x· ⃗n2k =
−γ, for k = 1, · · · , n,
−µk, for k = n + 1, 2,· · · , n + m.
Thus, ˙x· ⃗n2k < 0 for all k = 1, . . . , n + m.
By (I) and (II), we have that the assignment vector at any boundary point of the vector field yielded by equation (6) is tangent or pointing into the set △n+m. Hence, the proof of Lemma3 is complete.
Next, we show that DFE ξ0 defined in (7) is globally asymptotically stable in model (5) if R0 < 1. Equivalently, equilibrium x0 = 0∈ Rn+m is globally asymptotically stable
in model (6) if R0 < 1. (For simplification, we also call x0 to be a DEF for model (6).) Before it, we first show that no equilibrium lies in △n+m expect x0 and recall a general qualitative analytic result in the differential equations proposed by Lajmanovich and Yorke [12].
Lemma 4. The only equilibrium for (6) in ∂△n+m, the boundary of △n+m, is DFE x0.
Proof. Suppose the statement of Lemma 4 is false. Then there exists an equilibrium
¯
x := (¯x1,· · · , ¯xn+m) for (6) in ∂ △n+m −{x0}. Then by definition, ¯xk = 0 for some k∈ {1, · · · , n + m}.
(I). If k∈ {1, · · · , n}, then by (6), since dxdtk = 0, we have that αkΘ( ¯x1)+qk∑n+m
j=n+1βjx¯j = 0. Hence Θ( ¯x1), ∑n+m
j=n+1βjx¯j = 0. Consequently, ¯xj = 0 for all j = 1, . . . , n + m, i.e.,
¯
x = x0(= 0), a contradiction.
(II). If k ∈ {n + 1, · · · , n + m}, then by (6), since dxdtk = 0, we have that Φ( ¯x1) = 0 and hence ¯xj = 0 for all j = 1, . . . , n. Moreover, for j = n + 1, . . . , n + m, since dxdtj = 0, we have that ¯xj = 0. Thus, ¯x = x0(= 0), a contradiction.
By (I) and (II), we conclude that the only equilibrium for (6) in ∂△n+m, is the DFE x0. It completes the proof of Lemma 4.
Lemma 5. ( [12]) Consider the system dx
dt = Ax + H(x), (20)
where A is an ¯m× ¯m matrix and H(x) is continuously differentiable in a region D ∈ Rm¯. Suppose that the following assumptions hold:
(A1) The compact convex set C ⊂ D is positively invariant under the flow determined by equation (20).
(A2) lim
→0
∥H(x)∥
∥x∥ = 0.
(A3) There exist some r > 0 and eigenvector ν ∈ Rm¯ of AT such that ν · x ≥ r∥x∥ for all x∈ C.
(A4) ν· H(x) ≤ 0 for all x ∈ C.
(A5) {0} is the largest positively invariant set for (20) contained in set M := {x ∈ C : ν· H(x) = 0}.
Then either (i) x = 0 is globally asymptotically stable in C, or (ii) for any initial value x˜0 ∈ C − {0}, the solution ϕ(t, ˜x0) of (20) satisfies lim inft→∞∥ϕ(t, ˜x0)∥ ≥ m, where m > 0 is independent of ˜x0. Moreover, there exists a nontrivial equilibrium x∗ of (20) in C.
Theorem 3. If R0 < 1, then x0 is globally asymptotically stable in △n+m. On the other hand, if R0 > 1, then there exists an epidemic equilibrium x∗(> 0) in △n+m. Moreover, for any initial value ˜x0 ∈ △n+m− {x0}, the solution ϕ(t, ˜x0) of (6) satisfies lim inft→∞∥ϕ(t, ˜x0)∥ ≥ m, where m > 0 is independent of ˜x0.
Proof. Notice first that equation (6) can be rewritten in the form of (20):
dx
and
Then (i) by Lemma 3, △n+m is a positive invariant (compact) set for equation (6) in Rn+m; (ii) Since each term of H(x) has degree equal to 2, we have that lim
x→0
∥H(x)∥
∥x∥ = 0.
(iii) Let A1 := AT + aI where a = max{γ, µn+1,· · · , µn+m}. Then A1 is a nonnegative, irreducible matrix. Hence, by Perron-Frobenius theorem, there exists an eigenvalue λ∈ R of A1 such that λ = ρ(A1)(> 0). Moreover, it has a corresponding eigenvector ν > 0.
Consequently, λ− a is an eigenvalue of AT and ν is its corresponding eigenvector. Then for all x∈ △n+m = [0, 1]n+m, we have that ν· x ≥ ν0∥x∥1 ≥ ν0∥x∥2 where ν0 > 0 takes (22). This means that initial value starting at point x will leave M immediately under the flow determined by (21). Hence, we conclude that the largest invariant set for (21) contained in M is {0}. Thus, all assumptions in Lemma 5 hold and hence either one of the following cases hold: Case 1: Equilibrium x (= 0) is globally asymptotically stable
in △n+m. Case 2: For any initial value ˜x0 ∈ △n+m− {x0}, the solution ϕ(t, ˜x0) of (6) satisfies lim inft→∞∥ϕ(t, ˜x0)∥ ≥ m, where m > 0 is independent of ˜x0. Moreover, there exists a nontrivial equilibrium x∗ of (6) in △n+m. In fact, by Lemma 4, x∗(> 0) is an epidemic equilibrium. By Theorem 2, Case 1 occurs iff R0 < 1 and Case 2 occurs iff R0 > 1. This completes the proof of Theorem3.
Theorem 4. If R0 > 1, then there exists a unique endemic equilibrium x∗(> 0) of (6) such that x∗ is globally asymptotically stable in △n+m− {x0}.
Proof. Note that the existence of the endemic equilibrium x∗(> 0) is guaranteed by Theorem 3. Then we aim to show that such endemic equilibrium is unique and globally asymptotically stable.
(I). We show that the existence of the endemic equilibrium x∗ is unique. Suppose that x∗ = (x∗1,· · · , x∗n+m)T and z∗ = (z1∗,· · · , zn+m∗ )T are two distinct endemic equilibria of and z∗ are two equilibria of (6), we have that:
(i) If k0 ∈ {1, · · · , n}, then
x∗k0 on the left-hand side of the first equality. Hence, (1− x∗k0)
(ii) If k0 ∈ {n + 1, · · · , n + m}, then
−µk0x∗k0 + rk0(1− x∗k0)Φ(x∗1) =−µk0z∗k0 + rk0(1− zk∗0)Φ(z1∗) = 0.
By timing xzk0∗∗
k0
on the left-hand side of the above first equality, we have that, after some simple reduction,
(1− x∗k0)Φ(zk∗0 x∗k
0
x∗1) = (1− zk∗0)Φ(z∗1).
But this make a contradiction with x∗k0 > zk∗0 and zk∗ ≥ zk∗0 x∗k
0
x∗k for all k = 1, . . . , n + m.
By (i) and (ii), we have the result that model (6) has a unique endemic equilibrium.
(II). We show that the endemic equilibrium x∗ is globally asymptotically stable in
△n+m− {x0}. Define G and g be two real-valued functions in △n+m by
G(x) = max
1≤k≤n+m
{xk x∗k
}
and g(x) = min
1≤k≤n+m
{xk x∗k
}
. (24)
Then G(x) and g(x) are continuous and their right-hand derivatives exist along solutions of (6).
Let x(t) be a solution of (6). Then for any given t0 ≥ 0, there is some sufficiently small ϵ > 0 such that G(x(t)) = xk0x∗(t)
k0 , for some k0 ∈ {1, . . . , n + m} in t ∈ [t0, t0+ ϵ], and hence
G′|(6)(x(t0)) = x′k0(t0) x∗k0 , where G′|(6) is define as
G′|(6) = lim sup
h→0+
G(x(t + h))− G(x(t))
h .
Note that by the definition of G, we have that for t∈ [t0, t0+ ϵ], xk0(t0)
x∗k
0
≥ xk(t0)
x∗k (or x∗k≥ x∗k
0
xk0(t0)xk(t0)), k = 1, . . . , n + m. (25) In the following, we will show that if G(x(t0)) > 1 (i.e., xk0(t0) > x∗k0), then G′|(6)(x(t0)) <
0. Indeed,
(i) If k0 ∈ {1, · · · , n}, then
x∗k0x′k0(t0) xk0(t0) =
{
−γxk0(t0) + [1− xk0(t0)]
[
αk0Θ(x1(t0)) + qk0
n+m∑
j=n+1
βjxj(t0)
]} x∗k0 xk0(t0)
=−γx∗k0 + [1− xk0(t0)]
[
αk0Θ( x∗k
0
xk0(t0)x1(t0)) + qk0
n+m∑
j=n+1
βj
( x∗k
0
xk0(t0)xj(t0) )]
<−γx∗k0 + [1− x∗k0] [
αk0Θ(x∗1) + qk0
n+m∑
j=n+1
βjx∗j ]
(by (25), x∗ > 0 and xk0(t0) > x∗k0)
= 0,
since x∗ is an equilibrium. Hence, we have that G′|(6)(x(t0)) < 0.
(ii) If k0 ∈ {n + 1, · · · , n + m}, then
x∗k0x′k0(t0)
xk0(t0) ={−µk0xk0(t0) + rk0[1− xk0(t0)]Φ(x1(t0))} x∗k0 xk0(t0)
=−µk0x∗k0 + rk0[1− xk0(t0)]Φ( x∗k0
xk0(t0)x1(t0))
<−µk0x∗k0 + rk0[1− x∗k0]Φ(x∗1) (by (25), x∗ > 0 and xk0(t0) > x∗k0)
= 0,
since x∗ is an equilibrium. Hence, we have that G′|(6)(x(t0)) < 0.
By (i) and (ii), we showed that if G(x(t0)) > 1, then G′|(6)(x(t0)) < 0. By the similar argument, it can be showed that if g(x(t0)) < 1, then g′|(6)(x(t0)) > 0. Moreover, if G(x(t0)) = 1, then G′|(6)(x(t0)) ≤ 0, and if g(x(t0)) = 1, then g′|(6)(x(t0)) ≥ 0. The proof of these assertions are omitted here since the similarity.
Define the Lyapunov candidate functions U and V in △n+m by U (x) = max{G(x) − 1, 0},
V (x) = max{1 − g(x), 0}.
Then U and V are continuous and nonnegative functions in△n+m. Moreover,
By (I) and (II), the proof of Theorem 4 is complete.