Virus-immune dynamics determined by prey-predator interaction network and epistasis in viral fitness landscape

Abstract

Population dynamics and evolutionary genetics underly the structure of ecosystems, changing on the same timescale for interacting species with rapid turnover, such as virus (e.g. HIV) and immune response. Thus, an important problem in mathematical modeling is to connect ecology, evolution and genetics, which often have been treated separately. Here, extending analysis of multiple virus and immune response populations in a resource—prey (consumer)—predator model from Browne and Smith (2018), we show that long term dynamics of viral mutants evolving resistance at distinct epitopes (viral proteins targeted by immune responses) are governed by epistasis in the virus fitness landscape. In particular, the stability of persistent equilibrium virus-immune (prey-predator) network structures, such as nested and one-to-one, and bifurcations are determined by a collection of circuits defined by combinations of viral fitnesses that are minimally additive within a hypercube of binary sequences representing all possible viral epitope sequences ordered according to immunodominance hierarchy. Numerical solutions of our ordinary differential equation system, along with an extended stochastic version including random mutation, demonstrate how pairwise or multiplicative epistatic interactions shape viral evolution against concurrent immune responses and convergence to the multi-variant steady state predicted by theoretical results. Furthermore, simulations illustrate how periodic infusions of subdominant immune responses can induce a bifurcation in the persistent viral strains, offering superior host outcome over an alternative strategy of immunotherapy with strongest immune response.

Appendix

1.1 Proofs of Theorems

Proof of Proposition 2

Let ${\mathcal {C}}\subset \left\{ 0,1\right\} ^n$ be a circuit and suppose by way of contradiction that $\mathcal E^*=(x^*,\textbf{y}^*,\textbf{z}^*)$ is an equilibrium of (4) with $y_{{\textbf{k}}}^*>0$ for corresponding sequences ${\textbf{k}}\in {\mathcal {C}}$. Adding up the relative growth rates of these nonzero components at equilibrium in differential Eq. (4), we find the following:

$$\begin{aligned} 0&=\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} \frac{\dot{y}_{{\textbf{k}}}}{\gamma _{\textbf{k}} y_{\textbf{k}}^*} \\&=x^*\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} {\mathcal {R}}_{{\textbf{k}}} -\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} - \sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} \sum _{j=1}^n (1-{\textbf{k}}_j) z_{j}^* \\&=x^*\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} {\mathcal {R}}_{{\textbf{k}}} -\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}}- \sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} \left( {\textbf{1}}- {\textbf{k}}\right) \cdot \textbf{z}^* \\&=x^*\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} {\mathcal {R}}_{{\textbf{k}}} -\left( 1+{\textbf{1}} \cdot \textbf{z}^* \right) \sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} + \left( \sum _{{\textbf{k}}\in {\mathcal {C}}} a_{\textbf{k}} \textbf{k} \right) \cdot \textbf{z}^* \\&=x^*\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} \mathcal R_{{\textbf{k}}}, \end{aligned}$$

because $\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} \textbf{k}=0$ and $\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} =0$ for a circuit. This contradicts assumption $\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} {\mathcal {R}}_{{\textbf{k}}} \ne 0$ and thus proves the first statement. The next statement follows from Proposition 1 upon assuming $\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} R_{{\textbf{k}}}=0$. Indeed, uniqueness of equilibrium in a certain positivity class is equivalent to $\textrm{Ker}(A')^T\cap \mathbf {{\mathcal {R}}}'^{\perp }=\left\{ {\textbf{0}}\right\}$, which is equivalent to the condition that the augmented matrix $C=\begin{pmatrix} A'&\mathbf {{\mathcal {R}}'}\end{pmatrix}^T$ has trivial kernel (Browne and Smith 2018). Here $A'$ is the $m'\times n'$ interaction matrix consisting of the $m'$ strains comprising the circuit and $n'$ (positive component) immune responses. Consider the vector $\varvec{\alpha }$ consisting of the circuit weights. Then from the previous points, we find that $C \varvec{\alpha }={\textbf{0}}$. Thus, there cannot be a unique equilibrium with $y^*_{{\textbf{k}}}>0$ for all ${\textbf{k}}\in \mathcal C$. If such an equilibrium exists, then the virus component vector denoted by $\bar{{\textbf{y}}}$ which satisfies $\bar{\textbf{y}}-{\textbf{y}}^*\in \textrm{Ker}(A')^T\cap \mathbf {{\mathcal {R}}}'^{\perp }$, also satisfies equilibrium equations, and thus there are infinitely many equilibria with component $\bar{{\textbf{y}}}= {\textbf{y}}^* + \beta \varvec{\alpha }$ for $\beta \in {\mathbb {R}}$. $\square$

Proof I of Theorem 2

In order to prove the theorem, we show that the general saturated equilibria inequalities (11) reduce to vanishing linear forms of invasion circuits (Definition 3) when evaluated at the nested network equilibria (14). Then we will apply Theorem 1 to show stability of nested equilibrium and uniform persistence of associated positive component strains.

Let ${\mathcal {R}}_0> {\mathcal {Q}}_0:=1$ and k be the largest integer in [1, n] such that ${\mathcal {R}}_{k-1} > {\mathcal {Q}}_{k}$. First, we look at the case $k=n$ and ${\mathcal {R}}_n>{\mathcal {Q}}_n$ when $\widetilde{{\mathcal {E}}}_{n}$ is non-negative equilibrium with positive components at the $n+1$ nested strains $y_0,y_1,\dots ,y_n$. Consider a given missing viral strain $y_{i}$ ($i\in [n+1,2^n-1]$) with sequence ${\textbf{i}}$. We define a linear form, ${\mathcal {A}}_i$, based on it’s invasion rate as follows:

$$\begin{aligned} \frac{\dot{y}_i}{\gamma _i y_i} = -\frac{{\mathcal {A}}_{i}}{\mathcal R_n}, \ \ \text {where} \ \ -{\mathcal {A}}_{i}:= {\mathcal {R}}_{i}-\mathcal R_{n} -\sum _{j=1}^n (1-i_j) \left( {\mathcal {R}}_{j-1}- {\mathcal {R}}_{j} \right) . \end{aligned}$$

(22)

The telesco** sum above is determined by the following sequence: $\left( \alpha _{j}\right) , \ j=0,1,\dots ,n$, where $\alpha _{0}=1-i_{1}$, $\alpha _{j}=i_{j}-i_{j+1}$ for $j=2,\dots ,n-1$, $\alpha _{n}=i_{n}$. In this way, $-{\mathcal {A}}_{i}:= {\mathcal {R}}_{i}-\sum _{j=0}^{n}\ \alpha _j {\mathcal {R}}_j$. In order to prove that this is a vanishing linear form of a circuit, we show that it is the linear form of a minimally linearly dependent collection of extended binary sequences. Denote the binary sequences of nested network as ${\textbf{k}}_0,\dots ,{\textbf{k}}_n$ corresponding to ordered strains $y_0,\dots , y_n$. Let $\mathcal N\subset \left\{ 0,1\right\} ^{n+1}$ denote the subset of nested extended binary sequences, where $\hat{{\textbf{i}}}={\textbf{i}}1 \in \left\{ 0,1\right\} ^{n+1}{\setminus } {\mathcal {N}}$ and $\hat{\textbf{k}}={\textbf{k}}1\in {\mathcal {N}}$ represent binary sequences extended by digit 1. Notice that ${\mathcal {N}}$ forms a basis of $\mathbb R^{n+1}$ (since the $n+1\times n+1$ matrix $\left( \textbf{k}_n,{\textbf{k}}_{n-1},\dots ,{\textbf{k}}_0 \right)$ has a triangular row reduced eschelon form with values $\pm 1$ on diagonal). Thus for $\hat{{\textbf{i}}} \in \left\{ 0,1\right\} ^{n+1}{\setminus } {\mathcal {N}}$, there is a unique set of coefficients $\alpha _{j}$, $j=0,1,2,\dots ,n$, yielding $\hat{{\textbf{i}}}$ as a linear combination of the nested network vectors:

$$\begin{aligned} \hat{{\textbf{i}}}=\alpha _{0}\hat{{\textbf{k}}}_{0}+\alpha _{1}\hat{{\textbf{k}}}_{1}+\dots +\alpha _{n}\hat{{\textbf{k}}}_{n}. \end{aligned}$$

The above linear system resolves as follows:

$$\begin{aligned}&\alpha _{1}+\alpha _{2}+\dots +\alpha _{n}=i_{1}\\&\alpha _{2}+\alpha _{3}+\dots +\alpha _{n}=i_{2}\\&\qquad \qquad \vdots \\&\alpha _{k}+\alpha _{k+1}+\dots +\alpha _{n}=i_{k}\\&\qquad \qquad \vdots \\&\alpha _{n}=i_{n}\\&\alpha _{0}+\alpha _{1}+\dots +\alpha _{n}=1 \end{aligned}$$

which leads us to the set of coefficients $\alpha _{k}$ where $k=0,1,\dots ,n$ defined by the following:

$$\begin{aligned}{} & {} \alpha _{0}=1-i_{1}\\{} & {} \alpha _{k}=i_{k}-i_{k+1} \quad \text {for} \quad k=1,\dots ,n-1\\{} & {} \alpha _{n}=i_{n}\end{aligned}$$

Therefore the set $\hat{{\textbf{i}}}\cup \left\{ \hat{\textbf{k}}\right\} _{\hat{{\textbf{k}}}\in {\mathcal {N}}}$ is linearly dependent. Let $\alpha _i$ be the nonzero terms in sequence $(\alpha _j)$, i.e. $\varTheta _i:=\left\{ j\in [0,n]: \alpha _j\ne 0\right\}$, where $\alpha _j=\pm 1$ for $\alpha _j\in \varTheta _i$. Since $(\alpha _j)$ is unique linear combination with respect to basis ${\mathcal {N}}$, the set $\hat{{\textbf{i}}}\cup \left\{ \hat{\textbf{k}}\right\} _{\hat{{\textbf{k}}}\in \varTheta _i}$ is a minimal linearly dependent set. Thus we obtain the following circuit and corresponding vanishing linear form:

$$\begin{aligned} {\mathcal {C}}_i&=y_i \cup \left\{ y_j \right\} _{j\in \varTheta _i}, \qquad {\mathcal {A}}_i= -{\mathcal {R}}_i -\sum _{j\in \varTheta _i}\alpha _j \mathcal R_{j}. \end{aligned}$$

Furthermore, if ${\mathcal {A}}_i>0$, then $\frac{\dot{y}_i}{\gamma _i y_i} <0$ for $i\in [n+1,2^n-1]$. For this case $k=n$, the nested equilibrium (14) include all immune response $z_1,\dots ,z_n$ as positive components. If ${\mathcal {R}}_n>\mathcal Q_n$, then all missing species of non-negative equilibrium, $\widetilde{{\mathcal {E}}}_{n}$ are the $y_{i}$ with index $i\in [n+1,2^n-1]$. Thus, inequalities (11) hold strictly and the conclusions of Theorem 1 follow, in particular, $\widetilde{{\mathcal {E}}}_{n}$ is stable, $y_0,\dots ,y_n$ are uniformly persistent, and missing strains $y_{i}$, $i\in [n+1,2^n-1]$, go extinct. Notice in this case of $\mathcal R_n>{\mathcal {Q}}_n$, that $\widetilde{{\mathcal {E}}}_{n}$ becomes unstable if any of the invasion circuits’ linear forms become non-positive. In particular, if ${\mathcal {A}}_i\le 0$, then $\frac{\dot{y}_i}{\gamma _i y_i} \ge 0$, yielding the necessary condition of stability of $\widetilde{{\mathcal {E}}}_{n}$ based on sign of ${\mathcal {A}}_i$ which is stated in Corollary 1. Furthermore, by Proposition 2, in the critical case of ${\mathcal {A}}_i=0$, there exists a continuum of equilibria with positive components including the circuit elements consisting of invading strain, $y_i$ and subset of nested strains ${\mathcal {S}}_i$, ${\mathcal {C}}_i=\left\{ y_i\right\} \cup {\mathcal {S}}_i$, which connects to nested equilibrium $\widetilde{{\mathcal {E}}}_{n}$.

Next, consider the case $k=n$ but ${\mathcal {R}}_n\le {\mathcal {Q}}_n$, then we consider non-negative equilibrium $\overline{\mathcal E}_{n}$. Notice that

$$\begin{aligned} \frac{\dot{y}_i}{\gamma _i y_i}&= -\frac{1}{{\mathcal {Q}}_n}\left[ {\mathcal {R}}_{i}-{\mathcal {Q}}_{n} -\sum _{j=1}^{n-1} (1-i_j) \left( {\mathcal {R}}_{j-1}- {\mathcal {R}}_{j}\right) -(1-i_n)\left( {\mathcal {R}}_{n-1}- {\mathcal {Q}}_{n}\right) \right] \\&= -\frac{1}{{\mathcal {Q}}_n}\left[ {\mathcal {R}}_{i}-{\mathcal {R}}_{n-1} -\sum _{j=1}^{n-1} (1-i_j) \left( {\mathcal {R}}_{j-1}- {\mathcal {R}}_{j}\right) +i_n\left( {\mathcal {R}}_{n-1}- {\mathcal {Q}}_{n}\right) \right] \\&\le -\frac{1}{{\mathcal {Q}}_n}\left[ {\mathcal {R}}_{i}-{\mathcal {R}}_{n-1} -\sum _{j=1}^{n-1} (1-i_j) \left( {\mathcal {R}}_{j-1}- {\mathcal {R}}_{j}\right) +i_n\left( {\mathcal {R}}_{n-1}- {\mathcal {R}}_{n}\right) \right] \\&= -\frac{{\mathcal {A}}_{i}}{{\mathcal {Q}}_n}, \end{aligned}$$

where ${\mathcal {A}}_{i}$ is the same vanishing linear form (22) with corresponding circuit ${\mathcal {C}}_i$ as prior case. Therefore, if ${\mathcal {A}}_{i}>0$ (invasion circuit has positive epistasis), then $\frac{\dot{y}_i}{\gamma _i y_i}\le -\frac{\mathcal A_{i}}{{\mathcal {Q}}_n}<0$. The only other missing species additional to $y_{i}$, $i\in [n+1,2^n-1]$, is $y_n$. It is not hard to see that $\frac{\dot{y}_n}{\gamma _n y_n}=\frac{{\mathcal {R}}_n -\mathcal Q_n}{{\mathcal {Q}}_n} \le 0$. Thus, all saturated inequalities (11) are satisfied strictly (except when $\mathcal R_n={\mathcal {Q}}_n$), and stability of $\overline{{\mathcal {E}}}_{n}$ (and other conclusions) from Theorem 1 follow. In the case ${\mathcal {R}}_n={\mathcal {Q}}_n$, arguments from Browne (2017) give same result. Finally, suppose $k<n$, so there are k positive component immune responses, $z_1,\dots ,z_k$, in feasible nested equilibrium $\widetilde{{\mathcal {E}}}_{k}$ or $\bar{{\mathcal {E}}}_k$. Consider the missing strains $y_{i}$, $i\in [n+1,2^n-1]$, which are non-nested strains. Let $y_i$ be contained on the k dimensional hypercube. It necessarily has binary sequence where $i_{k+1}=\dots =i_n=0$. Observe from above that the invasion circuit involving $y_i$ will contain nested strains on the k dimensional (sub-)hypercube; $y_0,\dots ,y_k$. It follows that if ${\mathcal {A}}_i>0$, $\frac{\dot{y}_i}{\gamma _i y_i}\le -\frac{{\mathcal {A}}_{i}}{ C_k}<0$, where $C_k={\mathcal {R}}_k$ when ${\mathcal {R}}_k>{\mathcal {Q}}_k$ and $C_k={\mathcal {Q}}_k$ ${\mathcal {R}}_k\le {\mathcal {Q}}_k$. For a strain $y_i$ not on k dimensional hypercube, then we can find another strain $y_{\ell }$ with same sequence in first k bits, and less mutations overall, so that ${\mathcal {R}}_{\ell }>{\mathcal {R}}_k$. Then it is not hard to see that

$$\begin{aligned} \frac{\dot{y}_i}{\gamma _i y_i} = {\mathcal {R}}_ix^*-1-\sum _{j=1}^{k}z_k^*<{\mathcal {R}}_jx^*-1-\sum _{j=1}^{k}z_k^*=\frac{\dot{y}_i}{\gamma _i y_i}\le 0. \end{aligned}$$

Arguments from Browne (2017) show that $\frac{s_j\dot{z}_j}{\sigma _jz_j}_{|_{{\mathcal {E}}^*}}<0$ for $j>k$. Applying Theorem 1 gives the desired result. $\square$

Proof II of Theorem 2

We prove the case where ${\mathcal {R}}_n>{\mathcal {Q}}_n$, so that $\widetilde{{\mathcal {E}}}_{n}$ is feasible equilibrium and, by same arguments in Proof I above, the other cases follow. Consider a given missing viral strain $y_{i}$ ($i\in [n+1,2^n-1]$) with sequence ${\textbf{i}}$. Define the following linear form based on it’s invasion rate:

$$\begin{aligned} \frac{\dot{y}_i}{\gamma _i y_i} = -\frac{{\mathcal {A}}_{i}}{\mathcal R_n}, \ \ \text {where} \ \ -{\mathcal {A}}_{i}:= {\mathcal {R}}_{i}+\mathcal R_{n} +\sum _{j=1}^n (1-i_j) \left( {\mathcal {R}}_{j-1}- {\mathcal {R}}_{j} \right) , \end{aligned}$$

and $C_n={\mathcal {R}}_n$ when ${\mathcal {R}}_n>{\mathcal {Q}}_n$ and $C_n={\mathcal {Q}}_n$ ${\mathcal {R}}_n\le {\mathcal {Q}}_n$. We claim that ${\mathcal {A}}_{i}=0$ in additive case, and furthermore $\mathcal A_{i}\ne 0$ if any (non-zero) viral fitness is removed from ${\mathcal {A}}_{i}$ in the resulting sum. In other words we claim that ${\mathcal {A}}_{i}$ defines a circuit ${\mathcal {C}}$ containing strain i and other strains on nested network. To test additivity, it suffices to consider the linear form on the binary sequences:

$$\begin{aligned} -f_{i}&:= {\textbf{i}}-1^n - \sum _{j=1}^n (1-i_j) \left( 1^{j-1}0^{n-j+1}- 1^{j}0^{n-j} \right) \end{aligned}$$

Since ${\textbf{i}}$ is not in nested network ($i\in [n+1,2^n-1]$), there exists $p\in [1,n-1]$ such that $i_p=0,i_{p+1}=1$. In other words, there exists a 01 string in the binary sequence $\textbf{i}$. We prove that ${\mathcal {A}}_i$ defines a circuit by induction on the number of 01 strings, s. First suppose that $s=1$. Let $0\le m_1 <p$ be maximal such that $i_{p}=1$ and $p+1\le m_2\le n$ be maximal such that $i_{m_2}=0$. With these conditions, ${\textbf{i}}=1^{m_1}0^{p-m_1}1^{m_2-p}0^{n-m_2}$. Then

$$\begin{aligned} -f_{i}&:= {\textbf{i}}- 1^n - \sum _{j=1}^n (1-i_j) \left( 1^{j-1}0^{n-j+1}- 1^{j}0^{n-j} \right) \nonumber \\&={\textbf{i}}-1^n - \left( 1^{m_1}0^{n-m_1}-1^{p}0^{n-p}\right) - \left( 1^{m_2}0^{n-m_2}-1^n \right) \nonumber \\ \Rightarrow&f_i= {\textbf{i}} - 1^{m_1}0^{n-m_1}+1^{p}0^{n-p} - 1^{m_2}0^{n-m_2} \nonumber \\&= 0^p1^{m-p-1}0^{n-m_2}-0^p1^{m-p-1}0^{n-m_2} \nonumber \\&=0 . \end{aligned}$$

(23)

Furthermore $f_i= {\textbf{i}} - 1^{m_1}0^{n-m_1}+1^{p}0^{n-p} - 1^{m_2}0^{n-m_2}$ contains the viral sequences corresponding the non-zero fitness quantities in ${\mathcal {A}}_i$. Thus ${\mathcal {A}}_i$ defines a circuit since the minimal circuit size is 4. Now for the induction step, consider $s>1$. Assume that ${\mathcal {A}}_{\ell }$ defines a circuit for any sequence $\mathbf {\ell }$ with $s-1$ or less (01) strings, and suppose the sequence ${\textbf{i}}$ has s (01) strings. Let $p_1<p_2<\dots <p_s$ be locations of the 01 strings (with $i_{p_j}=0,i_{p_j+1}=1$). Let $0\le m_1 <p_1$ be maximal such that $i_{m_1}=1$ and $p_1+1\le m_2\le p_2$ be maximal such that $i_{m_2}=1$. So $\textbf{i}=1^{m_1}0^{p_1-m_1}1^{m_2-p_1}i_{p_2}\dots i_n$. Then

$$\begin{aligned} f_{i}&:= {\textbf{i}}- 1^n - \left( 1^{m_1}0^{n-m_1}-1^{p_1}0^{n-p_1}\right) - \sum _{j=p_1+2}^n (1-i_j) \left( 1^{j-1}0^{n-j+1}- 1^{j}0^{n-j} \right) \nonumber \\&=1^{m_1-1}0^{p_2-m_1}i_{p_2}\dots i_n-1^n - \sum _{j=p_1+2}^n (1-i_j) \left( 1^{j-1}0^{n-j+1}- 1^{j}0^{n-j} \right) \nonumber \\&={\tilde{\textbf{i}}} - 1^n - \sum _{j=1}^n (1-{\tilde{i}}_j) \left( 1^{j-1}0^{n-j+1}- 1^{j}0^{n-j} \right) \nonumber \\&= f_{{\tilde{i}}}, \end{aligned}$$

(24)

where ${\tilde{\textbf{i}}}=1^{m_1}0^{p_2-m_1}1i_{p_2+2}\dots i_n$ has $s-1$ (01) strings. Thus by induction hypothesis, we obtain $f_i=f_{{\tilde{i}}}=0$. Let ${\mathcal {C}}_i$ denote the collection of viral sequences corresponding the non-zero fitness quantities in ${\mathcal {A}}_i$. Notice that it is not hard to ascertain from the above calculations that

$$\begin{aligned} {\mathcal {C}}_i&={\textbf{i}} \cup \left\{ 1^{m_j}0^{n-m_j},1^{p_j}0^{n-p_j}\right\} _{j=1}^s \cup 1^{m_{s+1}}0^{n-m_{s+1}}, \\ {\mathcal {A}}_i&={\mathcal {R}}_i -\sum _{j=1}^{s+1}{\mathcal {R}}_{m_j} + \sum _{j=1}^{s}{\mathcal {R}}_{p_j}, \quad (0\le m_1< p_1< m_2< \dots< p_s < m_{s+1} \le n+1 ). \end{aligned}$$

Consider an arbitrary proper subset ${\mathcal {B}}$ of ${\mathcal {C}}_i$. First, we claim that there can not be a circuit consisting solely of sequences in the nested network. Suppose by way of contradiction that there exists a linear form with $g:=\sum _{j=1}^{n+1} b_j {\textbf{j}}=0$. Let $k=\max \left\{ 1\le j\le n+1 | b_j\ne 0 \right\}$. Then for the $k^{th}$ digit in the binary sequence of $g_N$, we find $(g)_k\ne 0$. So there are no vanishing linear forms on the nested network. Thus it suffices to consider the case where ${\textbf{i}}\in {\mathcal {B}}$. Motivated from calculations above, define

$$\begin{aligned} {\tilde{\textbf{i}}}={\textbf{i}} + \sum _{{\mathcal {C}}_i\setminus \mathcal B} \left( -1^{m_j}0^{n-m_j}+1^{p_j}0^{n-p_j}\right) , \end{aligned}$$

where ${\tilde{\textbf{i}}}$ is not in nested network since $\mathcal B\ne \emptyset$. Furthermore because ${\mathcal {C}}_i{\setminus } {\mathcal {B}}\ne \emptyset$, we obtain that ${\tilde{\textbf{i}}}$ has less than s (01) strings. By induction hypothesis, $\mathcal A_{{\tilde{i}}}$ defines a circuit ${\mathcal {C}}_{{\tilde{i}}}$ for the sequence ${\tilde{\textbf{i}}}$, where ${\mathcal {C}}_{\tilde{i}}=\left\{ {\tilde{\textbf{i}}}\right\} \cup \mathcal B{\setminus }\left\{ \textbf{i}\right\}$. Denote the vanishing linear form as $f_{{\tilde{i}}}=\sum _{\ell \in {\mathcal {C}}_{{\tilde{i}}}} a_{\ell } \mathbf {\ell }$. Now for arbitrary coefficients $b_j$,

$$\begin{aligned} \sum _{j\in {\mathcal {B}}} b_j {\textbf{j}}&= b_i {\textbf{i}} + \sum _{ {\mathcal {B}}\setminus \left\{ \textbf{i}\right\} } b_j {\textbf{j}} \\&= b_i \left( {\tilde{\textbf{i}}}-\sum _{{\mathcal {C}}_i\setminus {\mathcal {B}}} \left( -1^{m_j}0^{n-m_j}+1^{p_j}0^{n-p_j}\right) \right) + \sum _{ {\mathcal {B}}\setminus \left\{ \textbf{i}\right\} } b_j {\textbf{j}} \\&= \sum _{ {\mathcal {B}}\setminus \left\{ \textbf{i}\right\} } (b_j-b_ia_j) {\textbf{j}} -b_i \sum _{j\in {\mathcal {C}}_{{\tilde{i}}}} a_{j} {\textbf{j}} , \end{aligned}$$

The above sum consists solely of sequences in the nested network and thus there are no vanishing linear forms. This implies that the above sum is zero only if $b_i=0$, which further leads to conclusion that $b_j=0$ for $j\in {\mathcal {B}}{\setminus }\left\{ \textbf{i}\right\}$. Thus the proper subset ${\mathcal {B}}$ can not be a circuit for any linear form. $\square$

Proof of Theorem 3

By Proposition 6 in Browne and Smith (2018), equivalent conditions for stability of (i) ${\mathcal {E}}^{\dagger }_n$ or (ii) $\mathcal E^{\ddagger }_{n+1}$ are the following:

• i. ${\mathcal {R}}_{n+1}\le {\mathcal {P}}_n$ and $\left( |\varLambda _{i}| - 1 \right) {\mathcal {P}}_n + {\mathcal {R}}_{i} \le \sum \limits _{j\in \varLambda _{i}} {\mathcal {R}}_j \quad \forall i \in [n+2, 2^n]$, in which case $\varOmega _y=\varOmega _z= [1,n]$.

• ii. ${\mathcal {R}}_{n+1}> {\mathcal {P}}_n$ and $\left( |\varLambda _{i}| - 1 \right) {\mathcal {R}}_{n+1} + {\mathcal {R}}_{i} \le \sum \limits _{j\in \varLambda _{i}} {\mathcal {R}}_j \quad \forall i \in [n+2, 2^n]$, in which case $\varOmega _y=[1,n+1]$ and $\varOmega _z= [1,n]$.

Furthermore, (i) and (ii) are the only possible stable equilibria ${\mathcal {E}}^*$ with a strain-specific subgraph, i.e. $\varOmega _y\subseteq [1,n+1]$, and strains $y_1,\dots ,y_n$ ($y_{n+1}$ also for (ii)) are uniformly persistent. We remark that these prior results follow from Theorem 1. Now to prove Theorem 3, we will show that positive epistasis of corresponding invasion circuits are sufficient (also necessary in case (ii)) for inequalities in (i) and (ii) to hold. Fix an invading strain $y_i$, $i\in [n+2, 2^n]$, with binary sequence. First, note that a sufficient condition for inequalities in cases (i) and (ii) to be satisfied is $\frac{{\mathcal {A}}_{i}}{K_n}\ge 0$ where ${\mathcal {A}}_i= -{\mathcal {R}}_i -\left( |\varLambda _{i}| - 1 \right) {\mathcal {R}}_{n+1} + \sum _{j\in \varLambda _i} {\mathcal {R}}_{j}$, and $K_n={\mathcal {P}}_n$ if ${\mathcal {R}}_{n+1}\le {\mathcal {P}}_n$ and $K_n={\mathcal {R}}_{n+1}$ if ${\mathcal {R}}_{n+1}> {\mathcal {P}}_n$. In particular, inequality in case (ii) is equivalent to $\frac{{\mathcal {A}}_{i}}{K_n}\ge 0$. To show that ${\mathcal {C}}_i=y_i \cup \left\{ y_j \right\} _{j\in \varLambda _i}$ is a circuit with linear form ${\mathcal {A}}_i$, we proceed with a similar approach to our first proof of Theorem 2. Denote the binary sequences of one-to-one network as ${\textbf{k}}_1,\dots ,\textbf{k}_{n+1}$ corresponding to ordered strains $y_1,\dots , y_{n+1}$. Let ${\mathcal {S}}\subset \left\{ 0,1\right\} ^{n+1}$ denote the subset of strain-specific extended binary sequences, where $\hat{\textbf{i}}={\textbf{i}}1 \in \left\{ 0,1\right\} ^{n+1}{\setminus } {\mathcal {S}}$ and $\hat{{\textbf{k}}_j}={\textbf{k}}_j1\in {\mathcal {S}}$ represent binary sequences extended by digit 1. Notice that ${\mathcal {S}}$ forms a basis of ${\mathbb {R}}^{n+1}$. Indeed, it is not hard to show the row reduced echelon form of $n+1 \times n+1$ matrix is triangular. Thus for $\hat{{\textbf{i}}} \in \left\{ 0,1\right\} ^{n+1}{\setminus } {\mathcal {N}}$, there is a unique set of coefficients $\alpha _{j}$, $j=1,2,\dots ,n+1$, yielding $\hat{{\textbf{i}}}$ as a linear combination of the nested network vectors:

$$\begin{aligned} \hat{{\textbf{i}}}=\alpha _{1}\hat{{\textbf{k}}}_{1}+\dots +\alpha _{n+1}\hat{{\textbf{k}}}_{n+1}. \end{aligned}$$

The above linear system resolves as follows:

$$\begin{aligned} \alpha _{2}+\dots +\alpha _{n+1}&=i_{1}\\&\vdots \\ \sum _{j\ne k}\alpha _{j}&=i_{k}\\&\vdots \\ \alpha _{1}+\dots +\alpha _{n+1}&=1 \end{aligned}$$

which leads us to the set of coefficients $\alpha _{k}$ where $k=1,\dots ,n+1$ defined by the following:

$$\begin{aligned}\alpha _{k}=1-i_{k}\quad \text {for} \quad k=1,\dots ,n\\\alpha _{n+1}=1-(n-\sum _{k=1}^n i_k)= -\left( |\varLambda _{i}| - 1 \right) \end{aligned}$$

Thus, with analogous argument as before, we obtain the indicated circuit ${\mathcal {C}}_i$ and corresponding linear form ${\mathcal {A}}_i$. $\square$

Proof of Proposition 3

Let ${\textbf{i}} \in \left\{ 0,1\right\} ^n \setminus {\mathcal {S}}$ with integer coordinates $\left( \alpha _{{\textbf{k}}}\right) _{{\textbf{k}}\in {\mathcal {S}}}$ with respect to ${\mathcal {S}}\times \left\{ 1\right\}$ as a basis of ${\mathbb {R}}^{n+1}$ and addended binary sequence ${\textbf{i}}1\in \left\{ 0,1\right\} ^n\times \left\{ 1\right\}$. Clearly ${\mathcal {C}} \times \left\{ 1\right\} = {\mathcal {S}}\times \left\{ 1\right\} \cup \left\{ {\textbf{i}} 1\right\}$ is a linearly dependent set ${\mathbb {R}}^{n+1}$ with linear form on fitnesses given by ${\mathcal {A}}_{{\textbf{i}}}=-{\mathcal {R}}_{{\textbf{i}}}-\sum _{{\textbf{k}}\in {\mathcal {S}}} \alpha _{{\textbf{k}}} {\mathcal {R}}_{{\textbf{k}}}$. Furthermore any proper subset is linearly independent since ${\mathcal {S}}\times \left\{ 1\right\}$ is a basis of ${\mathbb {R}}^{n+1}$. Thus ${\mathcal {C}}$ is a circuit with linear form ${\mathcal {A}}_{{\textbf{i}}}$. By proof of Prop 2,

$$\begin{aligned} \frac{\dot{y}_i}{\gamma _i y_i}&= \frac{\dot{y}_i}{\gamma _i y_i} + \sum _{\ell =1}^{n+1} \alpha _{\ell } \frac{\dot{y}_{\ell }}{\gamma _{\ell } y_{\ell }} \\ {}&= \sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} \frac{\dot{y}_{{\textbf{k}}}}{\gamma _{{\textbf{k}}} y_{{\textbf{k}}}} \\ {}&=x^*\sum _{{\textbf{k}}\in {\mathcal {C}}} \alpha _{{\textbf{k}}} {\mathcal {R}}_{{\textbf{k}}} \\&= x^*\left[ {\mathcal {R}}_i + \sum _{\ell =1}^{n+1} \alpha _{\ell }{\mathcal {R}}_{\ell }\right] \\&= -x^*{\mathcal {A}}_{{\textbf{i}}} . \end{aligned}$$

Thus, by Theorem 1, the stability of ${\mathcal {E}}^*$ is determined by the sign of ${\mathcal {A}}_{{\textbf{i}}}$. $\square$

Proof of Proposition 4

We apply Proposition 3. Consider the network with at most one mutation, $\tilde{{\mathcal {S}}}_1$, consisting of wild-type and 1-mutation viral strains $y_0,y_1,\dots ,y_n$ where the sequence of $y_j$ is ${\textbf{j}}=\left( \delta _{\ell j}\right) _{\ell =1}^{n}$ for $j=1,\dots ,n$. First, notice that binary sequences of $y_0,y_1,\dots ,y_n$ form basis of ${\mathbb {R}}^{n+1}$. For missing strain, $y_i$, $i=n+1,\dots 2^n-1$, ${\textbf{i}} \in \left\{ 0,1\right\} ^n {\setminus } \tilde{{\mathcal {S}}}_1$, the coordinates of addended binary sequence ${\textbf{i}}1$ with respect to this basis can be seen to be $\alpha _j=i_j, \ j=1,\dots ,n, \ \alpha _0=1-\sum _{j=1}^n i_j$. Thus, by Proposition 3, $\frac{\dot{y}_i}{\gamma _i y_i} = -x^*{\mathcal {A}}_{{\textbf{i}}}$, where ${\mathcal {A}}_{{\textbf{i}}}=-{\mathcal {R}}_{{\textbf{i}}}-\sum _j=1^n i_j {\mathcal {R}}_j - \left( 1-\sum _{j=1}^n i_j\right) {\mathcal {R}}_0= -{\mathcal {R}}_{{\textbf{i}}} -\left( n- |\varLambda _{i}| - 1 \right) \mathcal R_{0} + \sum _{j\notin \varLambda _i} {\mathcal {R}}_{j}$. $\square$

Proof of Theorem 4

First assume that pairwise interaction matrix B is positive and consider the stability of the nested equilibrium, $\widetilde{{\mathcal {E}}}_{n}$ (or $\overline{{\mathcal {E}}}_{n}$), as characterized by circuits in Corollary 1 (ii). We proceed by induction on the number of (01) strings denoted by s for the invading strain. Suppose $s=1$ and the invading strain is written as in prior proof as $\textbf{i}=1^{m_1}0^{p-m_1}1^{m_2-p}0^{n-m_2}$ and the collection of strains in the circuit is given by ${\mathcal {C}}_i={\textbf{i}} \cup \left\{ 1^{m_1}0^{n-m_1},1^{p_1}0^{n-p_1}\right\} \cup 1^{m_{2}}0^{n-m_{2}}$. Then since the additive elements will sum to zero in the linear form ${\mathcal {A}}_i$, the only remain terms come from pairwise interactions in B and can be calculated as:

$$\begin{aligned} -{\mathcal {A}}_i&=\sum _{j=1}^{m_1}\sum _{k>j}^{m_1}B_{jk}+\sum _{j=1}^{m_1}\sum _{k=p_1+1}^{m_2}B_{jk}+\sum _{j=p_1+1}^{m_2}\sum _{k>j}^{m_2}B_{jk} \\&\qquad -\sum _{j=1}^{m_1}\sum _{k>j}^{m_1}B_{jk}+\sum _{j=1}^{p_1}\sum _{k>j}^{p_1}B_{jk}-\sum _{j=1}^{m_2}\sum _{k>j}^{m_2}B_{jk} \\&= \sum _{j=1}^{p_1}\sum _{k>j}^{p_1}B_{jk}- \sum _{j=1}^{m_1}\sum _{k>j}^{p_1}B_{jk}-\sum _{j=m_1+1}^{p_1}\sum _{k>j}^{m_2}B_{jk} \\&= -\sum _{j=m_1+1}^{p_1}\sum _{k=p_1+1}^{m_2}B_{jk} <0 \end{aligned}$$

Now for the induction step, suppose that ${\textbf{i}}$ has s (01) strings. It is not hard to see that ${\mathcal {A}}_i=\mathcal A_{{\tilde{i}}}$, for invading strain ${\tilde{\textbf{i}}}$, where ${\tilde{\textbf{i}}}=1^{m_1}0^{p_2-m_1}1i_{p_2+2}\dots i_n$ has $s-1$ (01) strings. Thus by induction hypothesis $-{\mathcal {A}}_i<0$, or ${\mathcal {A}}_i>0$ giving positive epistasis and stability of nested network.

Next suppose that matrix B is negative and consider the stability of ${\mathcal {E}}^{\dagger }_n$ and ${\mathcal {E}}^{\ddagger }_{n+1}$ consisting of strains $y_1,\dots ,y_n,y_{n+1}$ with binary sequences ${\textbf{k}}_1,\dots ,{\textbf{k}}_{n+1}$, where $\varLambda _j=\left\{ j\right\}$ for $j=1,\dots ,n$ and $\varLambda _{n+1}=\emptyset$. We inspect the invasion circuit of a strain with sequence ${\textbf{i}}$ outside the one-to-one network. Let s be the number of 1s in sequence ${\textbf{i}}$, located at loci $\ell _1,\dots ,\ell _s$, where $0\le s\le n-2$. Again the additive terms in ${\mathcal {A}}_{{\textbf{i}}}$ are zero and thus we have:

$$\begin{aligned} {\mathcal {A}}_i&= -{\mathcal {R}}_i -\left( |\varLambda _{i}| - 1 \right) {\mathcal {R}}_{n+1} + \sum _{j\in \varLambda _i} {\mathcal {R}}_{j} \\&=-\sum _{j=1}^{s}\sum _{k>\ell _j} B_{\ell _j k}-(n-1-s)\sum _{j=1}^{n}\sum _{k>j}B_{jk}+(n-s)\sum _{j=1}^{n}\sum _{k>j}B_{jk}\\&\qquad -\sum _{m=1}^{n}\left[ \sum _{k>m}B_{mk}+\sum _{k=1}^{m-1}B_{km}\right] + \sum _{j=1}^{s}\left[ \sum _{k>\ell _j}B_{\ell _j k}+\sum _{k=1}^{\ell _j-1}B_{k \ell _j}\right] \\&= -\sum _{j=1}^{n}\sum _{k=1}^{j-1}B_{kj} + \sum _{j=1}^{s}\sum _{k=1}^{\ell _j-1}B_{k \ell _j} \\&>0 \quad \text {since} \quad B<0, \ s\le n-2. \end{aligned}$$

Finally, for the network with at most one mutation, only the invading strain ${\textbf{i}}$ will have $\ge 2$ mutations, so

$$\begin{aligned} {\mathcal {A}}_{{\textbf{i}}}&= -{\mathcal {R}}_{{\textbf{i}}} -\left( n-|\varLambda _{i}| - 1 \right) {\mathcal {R}}_{0} + \sum _{j\notin \varLambda _i} {\mathcal {R}}_{j} \\&=-\sum _{j=1}^{n}\sum _{k>j} B_{j k} >0 \quad \text {since} \quad B<0. \end{aligned}$$

$\square$

Proof of Theorem 5

Let $0<f_j<1, j=1,\dots ,n$ represent the multiplicative fitness costs for each epitope. Similar to proof of Proposition 4, we use stability formulation by circuits in Corollary 1 (ii) and prove by induction on the number of (01) strings denoted by s for the invading strain. Suppose $s=1$ and the invading strain is written as in prior proof as ${\textbf{i}}=1^{m_1}0^{p-m_1}1^{m_2-p}0^{n-m_2}$ and the collection of strains in the circuit is given by ${\mathcal {C}}_i={\textbf{i}} \cup \left\{ 1^{m_1}0^{n-m_1},1^{p_1}0^{n-p_1}\right\} \cup 1^{m_{2}}0^{n-m_{2}}$. Then the linear form ${\mathcal {A}}_i$ can be calculated as:

$$\begin{aligned} -{\mathcal {A}}_i&={\mathcal {R}}_0\left( f_1\cdots f_{m_1}f_{p_1+1}\cdots f_{m_2}-f_1\cdots f_{m_1}+f_1\cdots f_{p_1}-f_1\cdots f_{m_2}\right) \\&={\mathcal {R}}_0f_1\cdots f_{m_1}(1-f_{p_1+1}\cdots f_{m_2})(f_{m_1+1}\cdots f_{p_1}-1) \\&<0 \end{aligned}$$

Now for the induction step, suppose that ${\textbf{i}}$ has s (01) strings. It is not hard to see that ${\mathcal {A}}_i=\mathcal A_{{\tilde{i}}}$, for invading strain ${\tilde{\textbf{i}}}$, where ${\tilde{\textbf{i}}}=1^{m_1}0^{p_2-m_1}1i_{p_2+2}\dots i_n$ has $s-1$ (01) strings. Thus by induction hypothesis $-{\mathcal {A}}_i<0$, proving nested network is stable. $\square$

1.2 The “at most one mutation” network equilibria

Consider the network with at most one mutation, $\tilde{\mathcal S}_1$, consisting of wild-type and 1-mutation viral strains $y_0,y_1,\dots ,y_n$ where the sequence of $y_j$ is $\textbf{j}=\left( \delta _{\ell j}\right) _{\ell =1}^{n}$ for $j=1,\dots ,n$. First it is simpler to look at the n strain equilibrium $\mathcal E^{1*}$ containing positive components for $y_1^*,\dots ,y_n^*$, where $y_0^*=0$, i.e. leaving out the wild-type strain. By Proposition 1 and (10), such a positive equilibrium ${\mathcal {E}}^{1*}=(x^*,y^{1*},z^*)$ of system (4) satisfies

$$\begin{aligned} x^*&=\frac{1}{\sum _{j=1}^n{\mathcal {R}}_j - (n-1){\mathcal {R}}_0}, \quad A y^{1*} = \textbf{s}, \quad Az^*=\mathbf {{\mathcal {R}}}^1x^*-\textbf{1}, \quad \text {where} \quad A=\textbf{1} \left( \textbf{1}\right) ^T - I_n, \\ A^{-1}&=\frac{1}{n-1}\textbf{1} \left( \textbf{1}\right) ^T - I_n, \quad y^{1}=\left( y_1,y_2,\dots ,y_n\right) ^T, \quad \mathbf {\mathcal R}^1=\left( {\mathcal {R}}_1,{\mathcal {R}}_2,\dots ,{\mathcal {R}}_n\right) ^T \end{aligned}$$

with $I_n$ is the $n\times n$ identity matrix. Here we find that:

$$\begin{aligned} y_i^{*}&= \frac{1}{n-1}\left( -(n-2)s_i+\sum _{j\ne i}s_j\right) , \quad x^*=\frac{n-1}{n-1+ \sum _i {\mathcal {R}}_is_i}, \quad z^*_i=\frac{1}{n-1}({\mathcal {R}}_ix^*-1) \end{aligned}$$

With the immunodominance hierarchy $s_i\le s_{i+1}$, then $y_i^{*}>0$ if $s_1>\sum _{i>1}(s_n-s_i)$ and $z_i^*>0$ if $\mathcal R_i\left( n-1-\sum _i s_i\right) >n-1$. If these conditions are satisfied, then the equilibrium ${\mathcal {E}}^{1*}$ is saturated in the subsystem restricted to ${\mathcal {S}}_1$. In Browne and Smith (2018) we showed that in the larger network of viral strains, the equilibrium ${\mathcal {E}}^{1*}$ is always unstable in the case with equal reproduction numbers ${\mathcal {R}}_1=\mathcal R_2=\dots ={\mathcal {R}}_n$.

Now consider invasion by the wild strain $y_0$, which can result in an $n+1$ strain equilibrium $\widetilde{{\mathcal {E}}}^{1*}$ consisting of the viral strain network $\tilde{{\mathcal {S}}}_1$. By Proposition 1, the positive components $x^*, {\widetilde{y}}^{1*},z^*$ of $\widetilde{{\mathcal {E}}}^{1*}$ satisfies:

$$\begin{aligned} x^*&=\textbf{1}^{\,T} C^{-1}_{(n+1)}, \quad \text {here} \quad C=\begin{pmatrix} A &{} \mathbf {{\mathcal {R}}} \\ \textbf{1}^{\,T} &{} {\mathcal {R}}_0 \end{pmatrix}, \\ \Rightarrow x^*&=\frac{1}{\sum _{j=1}^n {\mathcal {R}}_j - (n-1) {\mathcal {R}}_0}, \\ {\widetilde{y}}^{1*}&=\begin{pmatrix} \textbf{s}^{\,T}&\frac{1}{x^*}-1 \end{pmatrix} C^{-1} \\ A z^*&= \mathbf {{\mathcal {R}}} x^* -1, \quad \sum _{i=1}^n z^*_i = \mathcal R_0 x^*-1 . \end{aligned}$$

The above equations are difficult to analyze in general, but when $x^*>0, {\widetilde{y}}^{1*}> {\textbf{0}}, z^*> {\textbf{0}}$, the $n+1$ strain “at most one mutation” equilibrium will be positive. Furthermore, if the linear forms of invasion circuits (21) are positive, then by Proposition 4, $\widetilde{{\mathcal {E}}}^{1*}$ will be stable.