We have found that the infected steady state branch intersects both the uninfected and the pure infection branches of steady states at transcritical bifurcation points. We next consider what happens to the infected branch in between these two bifurcations.

**Proof.** Substituting for

D using (

40) into (

46), we obtain an equation involving

z and

$\u03f5$ to be solved for the infected branch of solutions given by:

To find limit points on the infected steady state branch [

21], we must solve the two equations:

and check the non-degeneracy conditions:

Since

$g(z,\u03f5)$ is quadratic in

z, then clearly,

${g}_{z}(z,\u03f5)$ is linear in

z, and the

z coefficient is strictly positive. Thus, the second equation can be solved for

z and substituted back into the first equation, which gives the quadratic equation in

$\u03f5$:

where:

The discriminant of the quadratic is given by:

where:

We have assumed that (

39) holds, and this implies (

41) also; this gives

${\Delta}_{1}>0$. We also note that

${\Delta}_{2}$ can be expressed as:

Assumption (

39) implies that (

42) holds, which implies that the second term is positive. Moreover,

using (

37), and this implies that

${\Delta}_{2}>0$ also. Therefore,

$\Delta >0$, and so, the quadratic Equation (

64) will have two distinct solutions. We therefore expect to have two limit points on the infected solution branch(es). Now,

${g}_{zz}(z,\u03f5)=2BCR(B+C){(1-\alpha \u03f5)}^{2}>0$ using (

37), (

38), and so, the first non-degeneracy condition in (

63) is satisfied. However, it is not easy to verify the second non-degeneracy condition. We now use a different approach to get further information regarding the infected solution branches, which will also confirm the existence of two limit points.

To get a more complete picture of the solutions, we substitute:

into (

62). The parameters

${z}_{0}$,

${z}_{1}$,

${\tilde{\u03f5}}_{0}$ and

${\tilde{\u03f5}}_{1}$ can be solved for in terms of the parameters so that our equation reduces to:

where:

Clearly,

${h}_{1},{h}_{2}>0$, and we showed above that

${\Delta}_{1}$,

${\Delta}_{2}>0$; so,

${h}_{0}>0$ also. Thus, (

69) is the equation of a hyperbola, and the two solution branches are given in parametric form by:

These solutions exist for all

$\delta z$ and for

$\left|\delta \u03f5\right|\ge \sqrt{{h}_{0}/{h}_{1}}$. This gives rise to parametric solutions of (

62) given by:

Limit points on these branches occur when:

Now,

$\beta =0$ is the unique solution of the equation

$d\u03f5/d\beta =0$ and:

For this derivative to be finite, we clearly require

${\tilde{\u03f5}}_{1}\pm \sqrt{{h}_{0}/{h}_{1}}\ne 0$. Now,

${\tilde{\u03f5}}_{1}<0$, and so,

${\tilde{\u03f5}}_{1}-\sqrt{{h}_{0}/{h}_{1}}<0$; however, it is possible that

${\tilde{\u03f5}}_{1}+\sqrt{{h}_{0}/{h}_{1}}=0$. However, we will show later that the right-hand (+) branch is outside our range of interest, and so, it does not matter whether this quantity is zero or non-zero. Thus, on the left-hand (−) branch, we have:

and so,

$d\u03f5/dz=0$ when

$\beta =0$. This point will be a quadratic limit point provided that the non-degeneracy condition

${d}^{2}\u03f5/d{z}^{2}{|}_{\beta =0}\ne 0$ is satisfied. It is a matter of calculation to show that for the left-hand branch:

again using the fact that

${\tilde{\u03f5}}_{1}-\sqrt{{h}_{0}/{h}_{1}}<0$, and so, we do indeed have a quadratic limit point when

$\beta =0$. Thus, the limit point on the left-hand branch of solutions occurs at:

and two solutions exist for each

$\u03f5<\u03f5\left(0\right)$, which is also confirmed by the negative second derivative.

We note that there are no solutions of (

69) when

$\delta \u03f5=0$, and this corresponds to

$\u03f5={\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1}$. We now show that

${\u03f5}_{2}<{\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1}<1/\alpha $. The right-hand inequality is clearly satisfied since

${\tilde{\u03f5}}_{0}=1/\alpha $ and

${\tilde{\u03f5}}_{1}<0$. To verify the left-hand inequality, we consider the quadratic equation (

55). We have already shown in the proof of Lemma 3 that the two solutions

${\u03f5}_{2}$ and

${\u03f5}_{3}$ of this equation satisfy

${\u03f5}_{2}<1<{\u03f5}_{3}$. Since the coefficient of

${\u03f5}^{2}$ is negative, this implies that the quadratic function

$p\left(\u03f5\right)$ is positive if and only if

${\u03f5}_{2}<\u03f5<{\u03f5}_{3}$. If

$p({\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1})>0$, then this implies that

${\u03f5}_{2}<{\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1}$ as required. It can be shown that:

where

${a}_{0},{a}_{1},{a}_{3}>0$, using (

37) and (

41). The remaining coefficient,

${a}_{2}$, is given by:

where

${b}_{0},{b}_{1}>0$ and:

Now,

${b}_{2}$ is negative for

$\alpha >0$ and sufficiently small

R. Substituting

$R=(\sqrt{2}-1)/2+\tilde{R}$ into

${b}_{2}$ gives:

The first and second order coefficients of $\tilde{R}$ are positive, and the constant coefficient is non-negative for all $\alpha \in [0,1]$. Thus, ${b}_{2}>0$ for $\tilde{R}>0$ and for all $\alpha \in [0,1]$, or equivalently, for all $R>(\sqrt{2}-1)/2$. In this case, ${a}_{2}>0$ also, which then implies that $p({\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1})>0$, as required.

Thus, we have proved that

$p({\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1})>0$ using Condition (

61). However, even if

${b}_{2}<0$, there are still many other positive terms in

$p({\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1})$, and so, it may be the case that Condition (

61) is not necessary. We note that the minimum of

${b}_{2}$ in the region

$R\ge 0$,

$\alpha \in [0,1]$ occurs at

$R=0$,

$\alpha =1$. Substituting

$R=0$,

$\alpha =1$ into the cubic polynomial in

P in (

71) gives:

Using assumption (

42), this is positive. Moreover, expanding the cubic polynomial in a Taylor series about the point

$(R,\alpha )=(0,1)$, we find that the first order terms are positive, and so, for sufficiently small

$R>0$ and

$\alpha <1$, the cubic polynomial in

P will be positive. However, this does not guarantee that it is positive for all

$R>0$ and

$\alpha \in [0,1]$, although we conjecture that this is in fact the case.

Now, the two branches of solutions (

70) occur on either side of the gap in

$\u03f5$, and so, one branch exists for

$\u03f5<{\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1}$ and the other for

$\u03f5>{\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1}$. The left-hand branch is therefore the only branch of infected solutions for

$\u03f5<{\tilde{\u03f5}}_{0}+{\tilde{\u03f5}}_{1}$, and so, the two bifurcating branches of infected solutions arising from the two bifurcations described above must be part of this single branch.

Thus, there is a single branch of infected solutions that connects the two bifurcation points, as claimed. There must also be a limit point on this branch, which occurs for $\u03f5>{\u03f5}_{2}$.

The only steady state solutions with

$x=0$ are the uninfected or pure infection solutions. Thus, all other solutions, in particular the infected solutions, have

$x\ne 0$. In the case of

${c}_{4}>0$ (

Figure 4 (top)), the only way that the valid infected branches arising from the two bifurcations can connect on a curve with a single limit point is for the limit point to occur on the invalid solutions after the bifurcation on the pure infection branch, as shown in

Figure 5a. When

${c}_{4}<0$ (

Figure 4 (middle, bottom)), the valid infected solution emanates from the pure infection branch to the right of the bifurcation point, and the only way that this can connect to the bifurcation on the uninfected branch at

$\u03f5=0$ is for there to be a limit point in between the two bifurcations.

We note that the left branch of (

69) exists for all

$\delta z$, but does not exist for all

z. As

$\delta \u03f5\to -\infty $ (

$\beta \to \pm \infty $), the left branch of the hyperbola asymptotes to the straight lines

$\delta z=\pm \sqrt{{h}_{1}/{h}_{2}}\phantom{\rule{0.166667em}{0ex}}\delta \u03f5$. Substituting these into

$z\left(\beta \right)$ given by (

70), we see that

z converges to the constant values:

as

$\u03f5\to -\infty $. Clearly one of these asymptotes is negative and the other positive, as we would expect. ☐

Finally, we recall the assumption we made in Theorem 4 that

D is defined by (

40). If this is not the case, then the same method described above can be used to derive Equation (

69), and

${z}_{0}$,

${\tilde{\u03f5}}_{0}$,

${h}_{1}$ and

${h}_{2}$ are unchanged. However, the other coefficients now involve the parameter

D. In this case, it is not possible to determine the sign of the coefficient

${h}_{0}$ in (

69). As long as

${h}_{0}$ remains positive, the same picture as described above will hold qualitatively. However, if

${h}_{0}$ changes sign and becomes negative, then the structure of the solutions of (

69) changes, so that there are no limit points occurring. Therefore, if (

40) does not hold, then there is an extra condition

${h}_{0}>0$ that is required to give the same solution structure.