We can think each variable C, D and N as depending on the interferon level I, the upregulation level U and the free virus level A. Thus, we have
$$\begin{array}{@{}rcl@{}} F_{1}(\boldsymbol{C}(\boldsymbol{I,U,A}), \boldsymbol{D}(\boldsymbol{I,U,A}), \boldsymbol{N}(\boldsymbol{I,U,A})) &=& H_{1}(\boldsymbol{I,U,A})\\ F_{2}(\boldsymbol{C}(\boldsymbol{I,U,A}), \boldsymbol{D}(\boldsymbol{I,U,A}), \boldsymbol{N}(\boldsymbol{I,U,A})) &=& H_{2}(\boldsymbol{I,U,A})\\ F_{3}(\boldsymbol{C}(\boldsymbol{I,U,A}), \boldsymbol{D}(\boldsymbol{I,U,A}), \boldsymbol{N}(\boldsymbol{I,U,A})) &=& H_{3}(\boldsymbol{I,U,A}) \end{array} $$
We assume the dynamics here are then
$$\begin{array}{@{}rcl@{}} \boldsymbol{C}^{\prime} &=& H_{1}(\boldsymbol{I,U,A}) \\ \boldsymbol{D}^{\prime}&=& H_{2}(\boldsymbol{I,U,A}) \\ \boldsymbol{N}^{\prime}&=& H_{3}(\boldsymbol{I,U,A}) \end{array} $$
As before assume C, D and C have achieved the same optimal values C
0
=0, D
0
=0 and N
0
=0 prior to the moment of infection with virus dose S
0
. These correspond to the starting values of prior to infection for base values I
0
, U
0
and A
0
. Initially, we don’t expect IFN- γ signals so I
0
=0. Eventually, we do expect some level of upregulation due to this initial dose and from experimental data, we expect the upregulation to be proportional to the level of the dose S
0
; we will assume this is a simple scaling factor, i.e. U
0
=q
1
S
0
for some suitable parameter q
1. Also, once the viral dose is administered, we would expect some fraction of it to remain as free virus which as discussed in [10] is modeled as A
0
=(1−p
0)S
0
. We will deal with these initial values after infection later in the Section “The IFN- γ, upregulation and free virus model”. But now, we think of all the initial values as zero; i.e. I
0
=0, U
0
=0 and A
0
=0. We still don’t expect to have any contribution to C
′(0), D
′(0) and N
′(0); i.e. H
1(I
0
,U
0
,A
0
)=0, H
2(I
0
,U
0
,A
0
)=0 and H
3(I
0
,U
0
,A
0
)=0. We are interested in the deviation of C, D and N from their optimal values C
0
, D
0
and N
0
due to the changes i, u and a from the base IFN- γ, upregulation and virus values. So let c=C−C
0
, d=D−D
0
and n=N−N
0
. We can then write C=C
0
+c, D=D
0
+d and N=N
0
+n The model can then be rewritten as
$$\begin{array}{@{}rcl@{}} (\boldsymbol{C_{0}} + \boldsymbol{c})^{\prime}(t) &=& H_{1}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) \\ (\boldsymbol{D_{0}} + \boldsymbol{d})^{\prime}(t) &=& H_{2}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a})\\ (\boldsymbol{D_{0}} + \boldsymbol{d})^{\prime}(t) &=& H_{3}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) \end{array} $$
which, as usual, implies
$$\begin{array}{@{}rcl@{}} \boldsymbol{c}^{\prime}(t) &=& H_{1}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) \\ \boldsymbol{d}^{\prime}(t) &=& H_{2}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a})\\ \boldsymbol{d}^{\prime}(t) &=& H_{3}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) \end{array} $$
Next, we again perform a tangent plane approximation on the nonlinear dynamics functions H
1, H
2 and H
3 just as we did in Section “Linearization details” for the F
1, F
2 and F
3 we use for the CDN model.
Linearization details
Once again, it seems reasonable to assume that since we are so close to ordinary operating conditions, the tangent plane errors are negligible. Letting ()o denote that we are evaluating the partial derivatives at the point (I
0
,U
0
,A
0
), we find
$$\begin{array}{@{}rcl@{}} H_{1}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) &\approx& \left(\frac{\partial H_{1}}{\partial \boldsymbol{i}}\right)^{o} \boldsymbol{i}+ \left(\frac{\partial H_{1}}{\partial \boldsymbol{u}}\right)^{o} \boldsymbol{u}\\&&+ \left(\frac{\partial H_{1}}{\partial \boldsymbol{a}}\right)^{o} \boldsymbol{a}\\ H_{2}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) &\approx& \left(\frac{\partial H_{2}}{\partial \boldsymbol{i}}\right)^{o} \boldsymbol{i}+ \left(\frac{\partial H_{2}}{\partial \boldsymbol{u}}\right)^{o} \boldsymbol{u}\\&&+ \left(\frac{\partial H_{2}}{\partial \boldsymbol{a}}\right)^{o} \boldsymbol{a}\\ H_{3}(\boldsymbol{I_{0}} + \boldsymbol{i},\boldsymbol{U_{0}} + \boldsymbol{u}, \boldsymbol{A_{0}} + \boldsymbol{a}) &\approx& \left(\frac{\partial H_{3}}{\partial \boldsymbol{i}}\right)^{o} \boldsymbol{i}+ \left(\frac{\partial H_{3}}{\partial \boldsymbol{u}}\right)^{o} \boldsymbol{u}\\&&+ \left(\frac{\partial H_{3}}{\partial \boldsymbol{a}}\right)^{o} \boldsymbol{a}\\ \end{array} $$
The corresponding nonlinear dynamics approximation in matrix - vector form is then
$$\begin{array}{@{}rcl@{}} \left[\begin{array}{l} \boldsymbol{c}^{\prime}\\ \boldsymbol{d}^{\prime}\\ \boldsymbol{n}^{\prime} \end{array}\right] \approx \left[\begin{array}{lll} H_{1\boldsymbol{i}}^{o} & H_{1\boldsymbol{u}}^{o} & H_{1\boldsymbol{a}}^{o}\\ H_{2\boldsymbol{i}}^{o} & H_{2\boldsymbol{u}}^{o} & H_{2\boldsymbol{a}}^{o}\\ H_{3\boldsymbol{i}}^{o} & H_{3\boldsymbol{u}}^{o} & H_{3\boldsymbol{a}}^{o} \end{array}\right] \left[\begin{array}{l} \boldsymbol{i}\\ \boldsymbol{u}\\ \boldsymbol{a} \end{array}\right] \end{array} $$
where we now use a standard subscript scheme to indicate the partials. We therefore find the nonlinear dynamics approximation is
$$\begin{array}{@{}rcl@{}} \left[\begin{array}{l} \boldsymbol{c}^{\prime}\\ \boldsymbol{d}^{\prime}\\ \boldsymbol{n}^{\prime} \end{array}\right] \approx \left[\begin{array}{lll} H_{1\boldsymbol{i}}^{o} & H_{1\boldsymbol{u}}^{o} & H_{1\boldsymbol{a}}^{o}\\ H_{2\boldsymbol{i}}^{o} & H_{2\boldsymbol{u}}^{o} & H_{2\boldsymbol{a}}^{o}\\ H_{3\boldsymbol{i}}^{o} & H_{3\boldsymbol{u}}^{o} & H_{3\boldsymbol{a}}^{o} \end{array}\right] \left[\begin{array}{l} \boldsymbol{i}\\ \boldsymbol{u}\\ \boldsymbol{a} \end{array}\right] \end{array} $$
where we now use a standard subscript scheme to indicate the partials.
The algebraic signs of the linearization matrix
If we hold everything constant except i which we increase to i+δ
i, what happens? Increasing the IFN- γ level should increase collateral damage. Hence, \(H_{1\boldsymbol {i}}^{o} = +\). What about the number of dividing cells that are infected? We do not think this should have an effect; hence, \(H_{2\boldsymbol {i}}^{o} = 0\) too. A similar argument suggests \(H_{3\boldsymbol {i}}^{o} = 0\) as well. Thus, the coefficient matrix above which we call Ψ so far looks like
$$\begin{array}{@{}rcl@{}} \boldsymbol{\Psi} &=&\left[\begin{array}{lll} + & H_{1\boldsymbol{u}}^{o} & H_{1\boldsymbol{a}}^{o} \\ 0 & H_{2\boldsymbol{u}}^{o} & H_{2\boldsymbol{a}}^{o} \\ 0 & H_{3\boldsymbol{u}}^{o} & H_{3\boldsymbol{a}}^{o} \end{array}\right] \end{array} $$
Now hold everything constant except the upregulation u and increase u to u+δ
u. What happens? Increasing the upregulation level should increase collateral damage. Hence, \(H_{1\boldsymbol {u}}^{o} = +\). Now infected cells are lysed once their IFN- γ upregulation hits a certain level and the increase in upregulation that makes the infected cells more visible certainly effects this as it will take fewer additional IFN- γ signals to trigger lysis. In general, there is a lag of say \(T_{L}^{D}\) and \(T_{L}^{N}\) time steps before this happens. With the increase δ
u, we expect \(T_{L}^{D}\) and \(T_{L}^{N}\) to both go down. Hence D and N decrease and we have \(H_{2\boldsymbol {u}}^{o} = -\) and \(H_{3\boldsymbol {u}}^{o} = -\) for each type of infected cell. Thus, the coefficient matrix looks like
$$\begin{array}{@{}rcl@{}} \boldsymbol{\Psi} &=&\left[\begin{array}{lll} + & + & H_{1\boldsymbol{a}}^{o} \\ 0 & - & H_{2\boldsymbol{a}}^{o} \\ 0 & - & H_{3\boldsymbol{a}}^{o} \end{array}\right] \end{array} $$
Next hold everything constant except the virus level a and increase a to a+δ
a. What happens? Let’s think of the virus increase δ
a as giving rise to an increase in the amount of virus stored inside a dividing cell or a non dividing cell. Now if the amount of virus in the cell goes up, that means when the cell is lysed, there is more virus available to infect cells which means more cells will be infected in later times. An increase in virus means an increase in collateral damage in general, so \(H_{1\boldsymbol {a}}^{o} = +\). At a given time then, A is the virus level. We can write A=A
F
+A
D
+A
N
where A
F
is the free virus, A
D
is the virus inside the dividing cells and A
N
is the virus inside the nondividing cells. So if A
D
goes up, we expect the amount of A and A
F
to remain constant. Hence, if A
D
goes up, A
N
goes down.
The increase in A
D
would mean more virus is released on lysis and hence an increase in D. However, that increase in A
D
is a concomitant decrease in N. So we should have \(H_{2\boldsymbol {a}}^{o} = +\) and \(H_{3\boldsymbol {a}}^{o} = -\). Thus, the coefficient matrix now looks like
$$\begin{array}{@{}rcl@{}} \boldsymbol{\Psi} &=&\left[\begin{array}{lll} + & + & + \\ 0 & - & +\\ 0 & - & - \end{array}\right] \end{array} $$
Or letting \(H_{2\boldsymbol {u}}^{o} = -a\), \(H_{3\boldsymbol {u}}^{o} = -b\), \(H_{2\boldsymbol {a}}^{o} = c\), \(H_{3\boldsymbol {a}}^{o} = d\), the coefficient matrix now looks like
$$\begin{array}{@{}rcl@{}} \boldsymbol{\Psi} &=&\left[\begin{array}{lll} + & + & + \\ 0 & - a & c\\ 0 & - b & -d \end{array}\right] \end{array} $$
Thus, we have the changes in collateral damage and infection levels
$$\begin{array}{@{}rcl@{}} \boldsymbol{c}^{\prime} &=& H_{1\boldsymbol{i}}^{o} \: \boldsymbol{i} + H_{1\boldsymbol{u}}^{o} \: \boldsymbol{u} + H_{1\boldsymbol{a}}^{o} \: \boldsymbol{a}\\ \boldsymbol{d}^{\prime} &=& H_{2\boldsymbol{u}}^{o} \: \boldsymbol{u} + H_{2\boldsymbol{a}}^{o} \: \boldsymbol{a}\\ \boldsymbol{n}^{\prime} &=& H_{3\boldsymbol{u}}^{o} \: \boldsymbol{u} + H_{3\boldsymbol{a}}^{o} \: \boldsymbol{a} \end{array} $$
Now we need to estimate i, u and a.