key: cord-0158921-wqdh171i
authors: Hynd, Ryan; Ikpe, Dennis; Pendleton, Terrance
title: Two critical times for the SIR model
date: 2020-09-21
journal: nan
DOI: nan
sha: 2ae96cc38282cbde5cc20163b5b4a30ba2e77f89
doc_id: 158921
cord_uid: wqdh171i

We consider the SIR model and study the first time the number of infected individuals begins to decrease and the first time this population is below a given threshold. We interpret these times as functions of the initial susceptible and infected populations and characterize them as solutions of a certain partial differential equation. This allows us to obtain integral representations of these times and in turn to estimate them precisely for large populations.

The susceptible, infected, and recovered (SIR) model in epidemiology involves the system of ODE Ṡ = −βSİ I = βSI − γI. is the recovered compartment of the population at time t. The parameters β > 0 and γ > 0 are the infected and recovery rates per unit time, respectively. We note that for given initial conditions S(0), I(0) ≥ 0, there is a unique solution of the SIR ODE. Indeed, a local solution pair S, I : [0, T ) → R of (1.1) exists by standard ODE theory (see for example section I.2 [3] ). As for each time t ≥ 0. That is, the path t → (S(t), I(t)) belongs to a level set of the function ψ(x, y) := x + y − (γ/β) ln x.

In particular, we may consider I as a function of S.

Decay of infected individuals. The number of infected individuals tends to 0 as t → ∞ lim t→∞ I(t) = 0.

In particular, for any given threshold µ > 0, there is a finite time t such that the number of infected individuals falls below µ I(t) ≤ µ.

In what follows, we will write u ≥ 0 for the first time in which I falls below µ.

Infected individuals eventually decrease. There is a time v ≥ 0 for which I is decreasing on [v, ∞).

Since I is a positive function andİ(t) = (βS(t) − γ)I(t), v can be taken to be the first time t that the number of susceptible individuals falls below the ratio of the recovery and infected rates S(t) ≤ γ β .

In this note, we will study the times u and v mentioned above as functions of the initial conditions S(0) and I(0). First, we will fix a threshold µ > 0 Figure 1 : Plot of the solution pair S, I of (1.1) with S(0) = x and I(0) = y. S(t) is shown in blue and I(t) is shown in red. Note that u(x, y) is the first time t such that I(t) ≤ µ.

u(x, y) := min{t ≥ 0 :

for x, y ≥ 0. Here S, I are solutions of (1.1) with S(0) = x and I(0) = y. As u(x, y) = 0 when 0 ≤ y < µ, we will focus on the values of u(x, y) for x ≥ 0 and y ≥ µ.

We'll see that u is a smooth function on (0, ∞) × (µ, ∞) which satisfies the PDE βxy∂ x u + (γ − βx)y∂ y u = 1 (1.4) and boundary condition u(x, µ) = 0, x ∈ [0, γ/β].

(1.5)

Moreover, we will also use ψ to write a representation formula for u. To this end, we note that for x > 0, y ≥ µ and there is a unique a(x, y) ∈ (0, γ/β] such that ψ(x, y) = ψ(a(x, y), µ).

Further, a(x, y) < x when x > γ/β or y > µ. These fact follows easily from the definition of ψ; Figure 3 below also provides a schematic. A basic result involving u is as follows.

Theorem 1.1. The function u defined in (1.3) has the following properties.

(i) u is continuous on [0, ∞) × [µ, ∞) and is smooth in (0, ∞) × (µ, ∞).

(ii) u is the unique solution of (1.4) in (0, ∞)×(µ, ∞) which satisfies the boundary condition (1.5).

(iii) For each x > 0 and y ≥ µ, y) ) .

Next, we will study v(x, y) := min{t ≥ 0 : S(t) ≤ γ/β} (1.6) for x ≥ 0 and y ≥ 0. Above, we are assuming that S, I is the solution pair of the SIR ODE (1.1) with S(0) = x and I(0) = y. Note that v(x, y) records the first time t that I(t) starts to decrease. Since

we will focus on the values of v(x, y) for x > γ/β and y > 0. The methods we use to prove Theorem 1.1 extend analogously to v. In particular, v satisfies the same PDE as u with the boundary condition v(γ/β, y) = 0, y ∈ (0, ∞).

(1.7)

Almost in parallel with Theorem 1.1, we have the subsequent assertion.

Theorem 1.2. The function v defined in (1.6) has the following properties.

(i) v is continuous on [γ/β, ∞) × (0, ∞) and is smooth in (γ/β, ∞) × (0, ∞).

(ii) v is the unique solution of (1.4) in (γ/β, ∞) × (0, ∞) which satisfies the boundary condition (1.7).

(iii) For each x ≥ γ/β and y > 0,

v(x, y) =

x γ/β dz βz ((γ/β) ln z − z + ψ(x, y)) .

We will then use the above representation formulae to show how to precisely estimate the u(x, y) and v(x, y) when x + y is large. 

· v(x, y) = 1.

(1.9)

The limit (1.8) implies that u(x, y) tends to ∞ as x + y → ∞. This limit also reduces to an exact formula when x = 0. In this case, S(t) = 0 and I(t) = ye −γt , so u(0, y) = 1 γ ln y µ .

Alternatively, the limit (1.9) implies that v(x, y) tends to 0 as x + y → ∞ with y ≥ δ for each δ > 0. This will be a crucial element of our proof of (1.8). This paper is organized as follows. In section 2, we will study u and prove Theorem 1.1. Then in section 3, we will indicate what changes are necessary so that our proof of Theorem 1.1 adapts to Theorem 1.2. Finally, we will prove Theorem 1.3 in section 4. We also would like to acknowledge that this material is based upon work supported by the NSF under 

This section is dedicated to proving Theorem 1.1. We will begin with an elementary upper bound on u.

Proof. Let S, I be the solution pair of the SIR ODE (1.1) with S(0) = x and I(0) = y. Since

Choosing t = u(x, y) and noting that

Next we observe that I is always decreasing at the time it reaches the threshold µ.

Lemma 2.2. Let x > 0, y > µ, and suppose S, I is the solution of (1.1) with S(0) = x and

If βx ≤ γ, then βS(t) < γ for all t > 0 since S is decreasing. Consequently, (2.1) holds for t = u(x, y). Alternatively, if βx > γ, then I is nonincreasing on the interval [0, v(x, y)] and βS(v(x, y)) − γ = 0.

In particular, I(v(x, y)) ≥ y > µ. Thus, v(x, y) < u(x, y). As S is decreasing,

for each x ≥ 0 and y ≥ µ.

We will also need to verify that solutions of the SIR ODE (1.1) depend continuously on their initial conditions. Lemma 2.4. Suppose x k ≥ 0 and y k ≥ 0, and let S k , I k be the solution of (1.1) with S k (0) = x k and I k (0) = y k for each k ∈ N. If x k → x and y k → y as k → ∞, then

for each sequence t k ≥ 0 such that t k → t. Here S, I is the solution of (1.1) with S(0) = x and I(0) = y.

Proof. We note that S k and I k are nonnegative functions with

for all t ≥ 0. Thus, S k and I k are uniformly bounded independently of k ∈ N. In view of (1.1), we additionally have

The Arzelà-Ascoli theorem then implies there are subsequences S k j and I k j converging uniformly on any bounded subinterval of [0, ∞) to continuous functions S and I, respectively. Observe

for each t ≥ 0. Letting k = k j → ∞ and employing the uniform convergence of S k j and I k j on [0, t] for each t ≥ 0, we see that S, I is the solution of (1.1) with S(0) = x and I(0) = y. As this limit is independent of the subsequence, it must be that S k and I k converge to S and I, respectively, locally uniformly on [0, ∞). As a result, we conclude (2.2).

In our proof of Theorem 1.1 below, we will employ the flow of the SIR ODE (1.1). This is the mapping

where S, I is the solution pair of (1.1) with S(0) = x and I(0) = y. We will also write Φ = (Φ 1 , Φ 2 ) so that Φ 1 (x, y, t) = S(t) and Φ 2 (x, y, t) = I(t).

A direct corollary of Lemma 2.4 is that Φ is a continuous mapping. With a bit more work, it can also be shown that Φ : (0, ∞) 3 → [0, ∞) 2 is smooth (exercise 3.2 in Chapter 1 of [3] , Chapter 1 section 7 of [2] ).

Proof of Theorem 1.1. (i) Suppose x k ≥ 0 and y k ≥ µ with x k → x and y k → y as k → ∞. By Lemma (2.1), u(x k , y k ) is uniformly bounded. We can then select a subsequence u(x k j , y k j ) which converges to some t ≥ 0. From the definition of u, we also have

As y ≥ µ, the limit t is equal to u(x, y). Because this limit is independent of the subsequence u(x k j , y k j ), it must be that u(x, y) = lim k→∞ u(x k , y k ).

It follows that u is continuous on [0, ∞) × [µ, ∞). Let x > 0 and y > µ and recall that Φ 2 (x, y, u(x, y)) = µ. By Lemma 2.2,

Since Φ 2 is smooth in a neighborhood of (x, y, u(x, y)), the implicit function theorem implies that u is smooth in a neighborhood of (x, y). We conclude that u is smooth in (0, ∞)×(µ, ∞).

(ii) Fix x > 0 and y > µ, and let S, I be the solution of (1.1) with S(0) = x and I(0) = y. Observe that for each 0 ≤ t < u(x, y), u(S(t), I(t)) = min{τ ≥ 0 :

Therefore,

We conclude that u satisfies (1.4). Now suppose w is a solution of (1.4) which satisfies the boundary condition (1.5). Note d dt w(S(t), I(t)) = − βS(t)I(t)∂ x w(S(t), I(t)) + (γ − βS(t))I(t)∂ y w(S(t), I(t)) = −1 for 0 ≤ t < u(x, y). Integrating this equation from t = 0 to t = u(x, y) gives w(S(u(x, y)), I(u(x, y))) − w(x, x) = −u(x, y).

Since I(u(x, y)) = µ, βS(u(x, y)) ≤ γ, and w(S(u(x, y)), µ) = 0, it follows that w(x, y) = u(x, y). Therefore, u is the unique solution of the PDE (1.4) which satisfies the boundary condition (1.5).

(iii) Suppose either x > 0 and y > µ or x > γ/β and y = µ. We recall that a = a(x, y) ∈ (0, γ/β] is the unique solution of ψ(x, y) = ψ(a, µ).

It is also easy to check that γ β ln z − z + ψ(x, y) > µ for a < z < x. See Figure 3 for an example. Figure 3 : Graph of w = (γ/β) ln z − z + ψ(x, y) for a ≤ z ≤ x. Here x > 0, y > µ, and a = a(x, y) ∈ (0, γ/β] is the unique solution of ψ(x, y) = ψ(a, µ). Also note that this graph is a subset of the level set {(z, w) ∈ (0, ∞) 2 : ψ(z, w) = ψ(x, y)}.

Since u solves the PDE (1.4), d dz u z, γ β ln z − z + ψ(x, y) = βzy∂ x u(z, y) + (γ − βz)y∂ y u(z, y) βzy y= γ β ln z−z+ψ(x,y) = 1 βz γ β ln z − z + ψ(x, y)

for a < z < x. Integrating from z = a to z = x and using the boundary condition (1.5) gives u(x, y) = x a dz βz ((γ/β) ln z − z + ψ(x, y)) . 

We will briefly point out what needs to be adapted from the previous section so that we can conclude Theorem 1.2 involving v. We first note that v is locally bounded in [γ/β, ∞) × (0, ∞).

Proof. Let S, I denote the solution of (1.1) with S(0) = x and I(0) = y. Since I(t) is increasing on t ∈ [0, v(x, y)], I(t) ≥ y for t ∈ [0, v(x, y)]. In view of (1.2)

Taking the natural logarithm and rearranging leads to v(x, y) ≤ (ln x − ln(γ/β))/βy, which is what we wanted to show.

The next assertion follows since S is decreasing whenever I is initially positive. The main point of stating this lemma is to make an analogy with Lemma 2.2.

Lemma 3.2. Let x ≥ γ/β and y > 0, and suppose S, I is the solution of (1.1) with S(0) = x and I(0) = y. ThenṠ

(v(x, y)) < 0.

Having established Lemmas 3.1 and 3.2, we can now argue virtually the same way we did in the previous section to conclude Theorem 1.2. Consequently, we will omit a proof. 

In this final section, we will derive a few estimates on u(x, y) and v(x, y) that we will need to prove Theorem 1.3. First, we record an upper and lower bound on u. 

If x ∈ [0, γ/β) and y ≥ µ, then

Observe that for each x ≥ 0 and y ≥ µ,

Now fix x ≥ 0 and y ≥ µ and suppose S, I is the solution of (1.1) with S(0) = x and I(0) = y. By our computation above, d dt w(S(t), I(t)) = − βS(t)I(t)∂ x w(S(t), I(t)) + (γ − βS(t))I(t)∂ y w(S(t), I(t)) ≥ −1 for 0 ≤ t ≤ u(x, y). And integrating this inequality from t = 0 to t = u(x, y) gives w(S(u(x, y)), µ) − w(x, y) ≥ −u(x, y). Since S(u(x, y)) ≤ γ/β,

Combined with (4.3) this implies u(x, y) ≥ w(x, y). We conclude (4.1). Now suppose βx < γ and y > µ. By (1.2),

Taking the natural logarithm and rearranging gives (4.2).

Likewise, we can identify convenient upper and lower bounds for v(x, y). To this end, we will exploit the fact that for each x > γ/β and y > 0

is concave on the interval γ/β ≤ z ≤ x. This implies

and

for γ/β ≤ z ≤ x. Then

and

v(x, y) ≥

.

(4.8)

Proof. We will appeal to part (iii) of Theorem 1.2 which asserts v(x, y) =

x γ/β dz βzg(z) .

Here g(z) is defined in (4.4). We will also employ the identity d dz

.

(4.9) By (4.5) and (4.9),

Similarly, (4.6) and (4.9) give Proof. Choose sequences x k ≥ 0 and y k ≥ δ with x k + y k → ∞ such that lim sup

If x k ≤ γ/β for infinitely many k ∈ N, then v(x k , y k ) = 0 for infinitely many k and lim k→∞ v(x k , y k ) = 0.

(4.11)

Otherwise, we may as well suppose that x k > γ/β for all k ∈ N. In this case,

x k for all large enough k. Therefore, (4.11) holds.

Alternatively, we can pass to a subsequence if necessary and suppose x k ≤ c for all k ∈ N and y k → ∞. Lemma 3.1 then gives v(x k , y k ) ≤ ln c − ln(γ/β) βy k → 0. 

Proof of (4.12). In view of (4.1),

It follows that u(x, y) → ∞ as x + y → ∞ with x ≥ 0 and y ≥ µ. In view of Corollary 4.10, we may select N ∈ N so large that v(x, y) < u(x, y)

for all x + y ≥ N with x ≥ 0 and y ≥ µ. Suppose x + y ≥ N with x ≥ 0 and y ≥ µ and choose a time t such that v(x, y) < t < u(x, y).

Note that as t > v(x, y), S(t) < γ β .

Here S, I is the solution of the SIR ODE (1.1) with S(0) = x and I(0) = y. By (4.2), we also have u(x, y) = t + u(S(t), I(t))

(4.14)

In addition, we can use (1.2) to find (x,y) ) .

Here we used that I(τ ) ≥ µ when τ ≤ u(x, y). Therefore,

Combining this inequality with (4.14) gives

As a result,

We conclude (4.12) upon sending t → ∞.

In our closing argument below, we will employ the elementary inequalities

which hold for x > γ/β. They follow as the natural logarithm is concave.

Proof of (4.13). By the upper bound (4.7),

β(x − γ/β + y) ln x − ln(γ/β) − ln y + ln (x − γ/β + y)

· v(x, y) ≤ β(x + y) ln x − ln(γ/β) − ln y + ln x − γ/β + y − γ β (ln x − ln(γ/β))

· v(x, y) We may select sequences x k > γ/β and y k > 0 with x k + y k → ∞ and lim sup x+y→∞ x>γ/β, y>0

Alternatively, x k has a bounded subsequence. Passing to a subsequence if necessary, we may assume that x k ≤ c for some constant c. In which case, y k → ∞. Employing (4.15), we find

It follows that the limit in (4.17) is 0. And in view of (4.16), lim sup x+y→∞ x>γ/β, y>0 β(x − γ/β + y) ln

x γ/β

x − γ/β y + 1

· v(x, y) ≤ 1.

By the lower bound (4.8),

β(x − γ/β + y)v(x, y) ln x γ/β x − γ/β y + 1 = β(x − γ/β + y)v(x, y) ln x − ln(γ/β) − ln y + ln (x − γ/β + y)

≥ ln x − ln(γ/β) − ln y + ln x − γ/β + y + γ β ( γ βx − 1) ln x − ln(γ/β) − ln y + ln (x − γ/β + y) = 1 + ln

x − γ/β + y + γ β ( γ βx − 1) x − γ/β + y ln x − ln(γ/β) − ln y + ln (x − γ/β + y)

(4.18)

In the last inequality, we used (4.15). We conclude the limit in (4.19) is 0. In view of inequality (4.18), lim inf x+y→∞ x>γ/β, y>0 β(x − γ/β + y) ln

x γ/β x − γ/β y + 1

· v(x, y) ≥ 1.

Mathematical models in epidemiology

Theory of ordinary differential equations

Ordinary differential equations. Robert E

An introduction to mathematical epidemiology

Let us choose sequences x k > γ/β and y k > 0 such that x k + y k → ∞ and lim inf x+y→∞ x>γ/β, y>0We recall that ln(1 + z)/z → 1 as z → 0, which impliesfor all nonpositive z sufficiently close to 0. Sincewe then have