Figures & data
Table 1. Parameters value.
Figure 1. For the verification of the hepatitis B extinction i.e. Theorem 4.1, we assume the parameter values as: b = 0.5, ,
,
, v = 0.01,
,
,
,
. It is easy to ensure the condition (a) i.e.
,
. This indicates the extinction of the hepatitis B. On the other hand, for the corresponding deterministic model (Equation2
(2)
(2) ),
then the endemic equilibrium (
) is globally asymptotically stable. (a) Test 1: Realization 1 with time in days. (b) Test 1: Realization 2 with time in days. (c) Test 1: Mean Solution with time in days.
![Figure 1. For the verification of the hepatitis B extinction i.e. Theorem 4.1, we assume the parameter values as: b = 0.5, β=0.1, β2=0.5, μ0=0.5, v = 0.01, γ1=0.2, γ2=0.4, μ1=0.2, σ=0.5. It is easy to ensure the condition (a) i.e. R0=0.0.734729<1, α(μ0+v)=0.306>0.125=σ2b. This indicates the extinction of the hepatitis B. On the other hand, for the corresponding deterministic model (Equation2(2) dS(t)=(b−αS(t)C(t)−(v+μ0)S(t))dt,dA(t)=(αS(t)C(t)−(γ1+μ0+β)A(t))dt,dC(t)=(βA(t)−(γ2+μ0+μ1)C(t))dt,dR(t)=vS(t)+(γ2C(t)−μ0R(t)+γ1A(t))dt.(2) ), R0d=1.2 then the endemic equilibrium (S∗,A∗,C∗,R∗) is globally asymptotically stable. (a) Test 1: Realization 1 with time in days. (b) Test 1: Realization 2 with time in days. (c) Test 1: Mean Solution with time in days.](/cms/asset/5711f63e-5ad1-4d88-a5a9-32160a1ff4bf/tjbd_a_1833993_f0001_oc.jpg)
Figure 2. The plot shows the extinction of the proposed model for
. Comparing with Figure (1), with the noise getting smaller, the fluctuation of the solution of system (Equation1
(1)
(1) ) is getting weaker. (a) Test 2: Realization 1 with time in days. (b) Test 2: Realization 2 with time in days. (c) Test 2: Mean Solution with time in days.
![Figure 2. The plot shows the extinction of the proposed model (1) for σ=0.35. Comparing with Figure (1), with the noise getting smaller, the fluctuation of the solution of system (Equation1(1) dS(t)=(b−αS(t)C(t)−(v+μ0)S(t))dt−σS(t)C(t)dB(t),dA(t)=(αS(t)C(t)−(γ1+μ0+β)A(t))dt+σS(t)C(t)dB(t),dC(t)=(βA(t)−(γ2+μ0+μ1)C(t))dt,dR(t)=vS(t)+(γ2C(t)−μ0R(t)+γ1A(t))dt.(1) ) is getting weaker. (a) Test 2: Realization 1 with time in days. (b) Test 2: Realization 2 with time in days. (c) Test 2: Mean Solution with time in days.](/cms/asset/fc171bf7-30f5-4b2d-82ce-6eb5f34a76c4/tjbd_a_1833993_f0002_oc.jpg)