Figures & data
Figure 1. The division of the -plane depending on the signs of
,
and
. It shows that the curves
,
and
divide the
-plane into six subregions:
,
,
,
,
and
. There exist two positive equilibria in subregion
(yellow), a unique positive equilibrium in subregions
and two curves
for
, and
for
(red), and no positive equilibria in subregions
together with the curve
for
(blue).
![Figure 1. The division of the μα-plane depending on the signs of D(μ,α), f(0,α) and Γx(α). It shows that the curves α1∗(μ), α2∗(μ) and α3∗ divide the μα-plane into six subregions: Ω1={(μ,α):D(μ,α)>0,f(0,α)>0,Γx(α)>0}, Ω2={(μ,α):D(μ,α)>0,f(0,α)<0,Γx(α)>0}, Ω3={(μ,α):D(μ,α)>0,f(0,α)<0,Γx(α)<0}, Ω4={(μ,α):D(μ,α)>0,f(0,α)>0,Γx(α)<0}, Ω5={(μ,α):D(μ,α)<0,f(0,α)>0,Γx(α)<0} and Ω6={(μ,α):D(μ,α)<0,f(0,α)>0,Γx(α)>0}. There exist two positive equilibria in subregion Ω1 (yellow), a unique positive equilibrium in subregions Ω2∪Ω3 and two curves α=α2∗(μ) for μ∈(0,μ2∗/2), and α=α1∗(μ) for μ∈(μ1∗,μ2∗/2) (red), and no positive equilibria in subregions Ω4∪Ω5∪Ω6 together with the curve α=α2∗(μ) for μ∈[μ2∗/2,μ2∗) (blue).](/cms/asset/f260e052-e8eb-494b-9a4b-b42815583024/tjbd_a_1977400_f0001_oc.jpg)
Figure 2. Given and
, we have
. For the case
, we get
,
. Numerical trials imply that
. Taking
, we have
,
and
. At
,
and
. Furthermore, when increasing β to 0.004, both
and
vanish, and
.
![Figure 2. Given sf=0.1 and sh=0.9, we have μ1∗≈0.1975. For the case μ=0.15<μ1∗, we get x1∗(μ)≈0.3375, x2∗(μ)≈0.7736. Numerical trials imply that β1∗≈0.0255. Taking β=0.01<β1∗, we have x1∗(μ,β)≈0.0367, x2∗(μ,β)≈0.2987 and x3∗(μ,β)≈0.7758. At β1∗, x1∗(μ,β1∗)=x2∗(μ,β1∗)≈0.1661 and x3∗(μ,β1∗)≈0.7788. Furthermore, when increasing β to 0.004, both x1∗(μ,β) and x2∗(μ,β) vanish, and x3∗(μ,β)≈0.7815.](/cms/asset/4e09a53c-bfec-40a6-9603-017d32886bc8/tjbd_a_1977400_f0002_oc.jpg)
Figure 3. Given and
, we take
. When
,
and
coincide to
. Numerical trials offer
. When
,
,
and
. When
, both
and
coincide to
and
. For
,
.
![Figure 3. Given sf=0.1 and sh=0.9, we take μ=μ1∗≈0.1975. When β=0, x1∗(μ1∗) and x2∗(μ1∗) coincide to x1∗(μ1∗)=x2∗(μ1∗)≈0.5556. Numerical trials offer β2∗≈0.0426. When β=0.02<β2∗, x1∗(μ1∗,β)≈0.0606, x2∗(μ1∗,β)≈0.4209 and x3∗(μ1∗,β)≈0.6296. When β=β2∗, both x1∗(μ1∗,β2∗) and x2∗(μ1∗,β2∗) coincide to x1∗(μ1∗,β2∗)=x2∗(μ1∗,β2∗)≈0.2280 and x3∗(μ1∗,β2∗)≈0.6539. For β=0.08>β2∗, x3∗(μ1∗,β)≈0.6772.](/cms/asset/de78c92d-2c15-47e8-810a-a0b69dc92ea3/tjbd_a_1977400_f0003_oc.jpg)
Figure 4. Given and
, we take
. Numerical simulations show that
,
. The number of zeros of
lying in
goes from 1, passing 2, 3, 2, and finally to 1 as β increases from 0 to the β with
.
![Figure 4. Given sf=0.1 and sh=0.9, we take μ=0.2>μ1∗≈0.1975. Numerical simulations show that β3∗≈0.0057, β4∗≈0.0437. The number of zeros of g(x,β) lying in (0,1) goes from 1, passing 2, 3, 2, and finally to 1 as β increases from 0 to the β with β>β4∗.](/cms/asset/5ff986d7-878f-4629-9f83-cbb317e4165b/tjbd_a_1977400_f0004_oc.jpg)
Figure 5. Distable dynamics driven by model (Equation4(4)
(4) ) and model (Equation16
(16)
(16) ). Panel (A) is for model (Equation4
(4)
(4) ) and Panel (B) is for model (Equation16
(16)
(16) ).
![Figure 5. Distable dynamics driven by model (Equation4(4) xn+1=(1−μ)(1−sf)(1+α)xnshxn2−[sf+sh+α(sf−sh)]xn+1+α(1−sh),n=0,1,2,….(4) ) and model (Equation16(16) xn+1=(1−μ)(1−sf)(β+xn)shxn2−(sf+sh)xn+1+β(1−sf),n=0,1,2,….(16) ). Panel (A) is for model (Equation4(4) xn+1=(1−μ)(1−sf)(1+α)xnshxn2−[sf+sh+α(sf−sh)]xn+1+α(1−sh),n=0,1,2,….(4) ) and Panel (B) is for model (Equation16(16) xn+1=(1−μ)(1−sf)(β+xn)shxn2−(sf+sh)xn+1+β(1−sf),n=0,1,2,….(16) ).](/cms/asset/4095c98a-5a3d-450d-826d-5d0e5a59fb94/tjbd_a_1977400_f0005_oc.jpg)
Figure 6. Comparisons on the infection frequency thresholds (A) and the polymorphic states (B) driven by (Equation3(3)
(3) ), (Equation16
(16)
(16) ) and (Equation4
(4)
(4) ) on different maternal leakage rates μ lying in
.
![Figure 6. Comparisons on the infection frequency thresholds (A) and the polymorphic states (B) driven by (Equation3(3) xn+1=(1−μ)(1−sf)(1+r)(xn+r)shxn2−sf+sh+r(sf−sh)xn+1+(2−sf−sh)r+(1−sf)r2,n=0,1,2,…(3) ), (Equation16(16) xn+1=(1−μ)(1−sf)(β+xn)shxn2−(sf+sh)xn+1+β(1−sf),n=0,1,2,….(16) ) and (Equation4(4) xn+1=(1−μ)(1−sf)(1+α)xnshxn2−[sf+sh+α(sf−sh)]xn+1+α(1−sh),n=0,1,2,….(4) ) on different maternal leakage rates μ lying in (0,μ1∗).](/cms/asset/1a54df55-63d4-4227-89d5-4fcc3e61ca7f/tjbd_a_1977400_f0006_oc.jpg)