Turning Points of Analytical Solutions to the Linear

Turning points of analytical solutions to the linear, constant-population SIR model for campylobacteriosis and cryptosporidiosis

Graham McBride and Sharleen Harper, NIWA, Hamilton, NZ, March 2014 [email protected]; [email protected]

Under assumptions stated in McBride et al. ("Projected changes in reported campylobacteriosis and cryptosporidiosis rates as a function of climate change: a New Zealand study", SERRA, submitted), the governing differential equations for a constant- size population (N) for age-dependent but time-invariant (static) SIR model compartments (Susceptibles, Ill, Recovered) are: dS/da = b – αS – cK1K2S – cK1(1 – K2)S + δR, dI/da = cK1K2S – (α + γ)I, dR/da = γI + cK1(1 – K2)S – (α + δ)R, where: a denotes age, Sa=0 = N, Ia=0 = 0, Ra=0 = 0 and N = S + I + R. Parameters are: b = specific immigration rate of susceptibles (# per annum); c = specific rate of contact with pathogen (per annum)

K1 = probability of infection given contact;

K2 = probability of illness given infection; γ = reciprocal of the shedding period (per annum); α = specific death rate in population (per annum); δ = specific immunity loss rate (per annum).

Note that summing these equations gives us b = αN (recruitment rate = death rate), which is used to remove b from the first equation.

The analytical solutions presented by McBride & French (2006) refer.2 They are cited in square brackets. A summary table of findings is given near the end of this document.

We consider three cases, based on the sign of the model's discriminant (): (i)  = 0; (ii)  > 0; (iii)  < 0, where  ν2 –y (see below).

Case (i):  = 0 The first derivative of the illness rate equation [16] is (1) where (2)

2 McBride, G.B.; French, N.P. (2006). Accounting for age-dependent susceptibility and occupation- dependent immune status: a new linear analytical model. WSEAS Transactions on Mathematics 11(5): 1241–1246. They are repeated on the last page of these notes. 1 D:\Docs\2018-04-18\093ba36ace4e19d498246df417df0b5c.docx with (3) and (4) We will also need (5) and (6) These relationships will also be useful (7) (8) (9) (10) (11) (12) Therefore, noting that  = 0  y = , we have (13) Now we need a useful expression for the steady states in terms of the variables defined above, i.e., (14) (15)

3 and to get r it seems simplest to use the fact that r = 1 – (s + i), giving (16) Therefore (17) a pretty result. So the solution is (18) and so (19) This has extrema (where di/da = 0 in the a > 0 domain) at (20) This result suggests that  will play a significant role in the form of the solutions to be obtained.4

3 It is simplest because the numerator of r in [21] is pcK1 – g, and just how this can be translated into a useful combination of the variables defined on the first page isn't immediately obvious. 4 The result for   0 in (20) follows by noting that in (19) g(1 – a)e–a = g(1 – a)/ea = g(1 – a)/[1 + (a)/1! +(a)2/2! +(a)3/3! + …] = (g/a – )/[1/a +  +a/2 +a2/6 + …]  0 as a  .

2 2 2 The extrema predicted by (20) will be a maximum iff d i/da < 0 at a = aext, and it will be a flat line if d2i/da2 = 0 (and all higher derivatives also vanish). So let's see. The second derivative is (21) and so (22) which is strictly non-positive, QED. Susceptible curve We have (from [15]) (23) So, noting that (24) we have (25) and so (26)

This equation is always negative because, from (7), we have  =  + cK1 > . Note that at the extrema (27)

Recovery curve Noting that dr/da = –(ds/da + di/da), we have (28) and so (29) As a check, we can evaluate the solution directly, i.e., from [17] (30)

Now  –  =  + cK1 –  – g = cK1(1 – K2)  0. Therefore (30) is always positive. Also at the extrema we have (31) Check: (32)

Case (ii):  > 0 The first derivative of the illness rate equation [13] is (33) where (34)

3 D:\Docs\2018-04-18\093ba36ace4e19d498246df417df0b5c.docx and (35) The steady-states are now (36) (37) and to get r we again use the fact that r = 1 – (s + i), giving (38) We can now calculate (39) another pretty result! So by defining , (40) the derivative is (41) and so (42) Note that this solution collapses to the special case [when  = 0, i.e., (19)].5 The infected proportion of the population, i(a), has maxima/minima when Equation (42) is equal to zero. This occurs as a  , and also when/if the terms in the bracket vanish.6 The latter occurs only if a value of a can be found such that ( + ) – ( – )ea = 0, that is (43) which is defined only if . Let’s examine this term.

The numerator is always positive, so we need only look at the denominator. Noting that  is strictly non-negative—but  is not—we simply need. (44)7 and so the turning points of i(a) are: (45)

Now if  < 0 and || > , aext will be negative and that is not in the solution domain of interest (which is a > 0). We therefore need to ensure that ( + )/( – ) > 1. Therefore (45) is modified to (46)

2 2 Now consider the case  >  (>0). The illness rate at aext will be a maximum iff d i/da < 0 at that age—so let's see. The second derivative is

5 As   0 we have as the approximate right-hand-side of (42): g[ – a]e–a/ (noting that e 1 +  as   0). This collapses to the right-hand-side of (19), QED [i.e., to g(1 – a)e–a]. 6 Note that (42) can be written as di/da = (g/)[( + )e–(+)a – ( – )e–(–)a]. For di/da  0 as a   we therefore require that  >  and this is always true [noting that  = (p + q)/2,  = (p – q)/2 and  = | 2  – y| where y = cK1K2 > 0].

7 Equation (44) can be shown to be equivalent to requiring that  > (1 – K2). To see that, we note from 2 2 2 2 2 the preceding equation that we want  >  . Using (6) we therefore require that  +2cK1 + (cK1) > 

– cK1K2  2 + cK1 > 2. Then, using (5),  – cK1 –  + cK1 > K2, thus  > (1 – K2).

4 (47) and so (48) thus (49) which, given that  >  (> 0), is strictly negative, QED. Let's now consider the gradients of s and r with age.

Susceptible curve We have (from [12]) (50) where (51) Now consider the expression in braces in (50); call it J. Therefore (52) The term in brackets can be multiplied out to give (53) so the solution is (54) This equation is not strictly non-positive ( a), because   . Check: as   0 the term in brackets in (54) becomes (55) Substituting (55) back into (54) we obtain (56) Which is equation (26), QED. Finally, note that (57) and so (58)

5 D:\Docs\2018-04-18\093ba36ace4e19d498246df417df0b5c.docx Recovery curve Noting that dr/da = –(ds/da + di/da) we have and so (59) This equation is not strictly non-positive ( a), because   . Check: as   0 the term in brackets becomes (60) Substitution into (59) recovers (29), QED. Finally, note that (61) and so (62)

Case (iii):  < 0 In this case, . Therefore (63) so that (64) (65) and therefore, using r = 1 – (s + r), (66) Now the first derivative of the illness rate equation [19] is (67)

Inserting (65)–(66) for the steady states, we have the cosine term [J1 =( + )i + gr] (68)

 and for the sine term [J2 =( – )i – gr] (69) and so we have (70) The extrema of i(a) occur when Equation (70) is equal to zero. This occurs at as a  , and also when cos(a) = sin(a), for example, when (71) Note that, because of the periodic nature of the tangent function, there are many possible values for aext. In fact, 8 (72)

–1 8 Only those extrema which are positive lie in the solution domain, and aext,n > 0 iff n > –[tan (/)]/.

Note too that if  = 0 and di/da = 0, (70) requires that cos(aext) = 0 [so sin(aext) = 1]. To keep the

6 The derivative of (70) is (73) Now in order to examine the sign of (73) at the extrema given by (72), we will need these identities9 for x > 0 (74) for x < 0, and (75) for all real x, (76) and also these identities (77) We must consider four possibilities for the sign of (73) at the extrema given by (72): 1.  < 0 and n odd and so (78) which is strictly negative, therefore are maxima.

second derivative negative (at aext) we must have sin(aext) > 0 in (73), giving aext = (2n + 1)/ (n = 0,

2, 4,…) . But we must also maximize [19] and so the smallest age must be take, i.e., aext = /. 9 Given at http://mathworld.wolfram.com/InverseTangent.html. 7 D:\Docs\2018-04-18\093ba36ace4e19d498246df417df0b5c.docx 2.  < 0 and n even and so (79) which is strictly positive, therefore are minima. 3.  > 0 and n odd and so (80) which is strictly positive, therefore are minima. 4.  > 0 and n even and so (81) which is strictly negative, therefore are maxima. Recovered curve We have (from [20]) (82)

The coefficient of the cosine term [J1 = i +g( – )r] can be written as (83)

 The coefficient of the sine term [J2 = i +g( + )r] can be written as (84) Substituting (83) & (84) into (82) we obtain and so (85) Also, using the relation that dr/da = –(ds/da + di/da), we have (86) Note that as   0, then (85) & (86) collapse to the zero discriminant cases (27) & (31). Furthermore, using (74–(77), we have: 1. For  < 0 and n odd and so . (87) Also, and so

8 . (88)

2. For  < 0 and n even and so . (89) Also, and so . (90)

9 D:\Docs\2018-04-18\093ba36ace4e19d498246df417df0b5c.docx 3. For  > 0 and n odd (91) and so . (92) Also, (93) and so . (94)

4. For  > 0 and n even and so . (95) Also, and so . (96)

Speculation: The maximum value of i predicted by [19] occurs using (72) with n = 0 (when  > 0) and with n = 1 (when  < 0). That this is so is now shown. The maxima/minima of i(a) are located at

For  < 0, the maxima are given by aext,n with n odd, and for  > 0 the maxima are given by aext,n with n even.

1. For  < 0 and n odd, and so (97)

Since is strictly positive, decreases with increasing n. As a result, the smallest aext,n which is positive will correspond to the absolute maximum of in the solution domain. The smallest positive aext,n is given by the lowest whole odd number n satisfying

. (98)

Now, therefore . As a result, , and so from (98). Consequently, the lowest whole odd number satisfying (98) is n = 1, and therefore aext,1 corresponds to the absolute maximum of i(a) in the solution domain.

2. For  > 0 and n even, and so

10 (99) As above, since is strictly positive, decreases with increasing n. As a result, the smallest aext,n which is positive will correspond to the absolute maximum of in the solution domain. The smallest positive aext,n is given by the lowest whole even number n satisfying (98). In this case, therefore . As a result, , and so from (98). Consequently, the lowest whole even number satisfying (98) is n = 0, and therefore aext,0 corresponds to the absolute maximum of i(a) in the solution domain.

2 2 Note that i = g( – )/( +  ), so (97) & (99) show that i(aext) > i. Special case:  = 0 (i) If  = 0, then from (19), (20), (22), (26) and (29), 2 2 aext  , di/da = d i/da = ds/da = dr/da = 0. (ii) If  > 0, then from (42), (46), (49), (54) and (59), 2 2 aext  , di/da = d i/da = ds/da = dr/da = 0.

(iii) If  < 0, then using (70) in the a > 0 domain, di/da = 0, when aext = (2n + 1)/n = 0, 1, 2,…, , n = 0, 1, 2,… (100)10 and so from (73) (101) which is strictly non-positive, QED. Finally, using (85) & (86) we have (102) (103)

The value of i at finite aext

2 (i) For aext = 1/. Noting that i = g(– )/ and i – r = –/, we have (from [16]), (104) giving (105)

2 Note that i = g( – )/ , so (105) shows that i(aext) > i.

2 2 (ii) For aext = ln[(+ )/(– )]/ ( > ). Noting that i = g(m – )/( –  ) and i

– r = –(– )/(– , by rewriting [13] at aext as (106)

and noting that exp(aext) – 1 = /( – ), we obtain (107) and so

10 Note that tan–1(x) = /2 – 1/x + 1/(3x3) – 1/(5x5) + … (x  1), so as  → 0 then / (≡ x) →  and so Eq. (71) → /(2). 11 D:\Docs\2018-04-18\093ba36ace4e19d498246df417df0b5c.docx (108)

2 2 Note that i = g( – )/( +  ), so (108) shows that i(aext) > i.

(iii) For we merely appeal to (97) and (99).

12 SUMMARY OF SOLUTIONS (in the positive a domain)

Variable Case (i):  = 0 Case (ii):  > 0 Case (iii):  < 0 aext i(aext)

2 The parameter groups are:  =  + ( + cK1 + )/2;  = ( – cK1 – )/2;  = ( + cK1 – )/2; g = cK1K2;  =  + g;  = ||, where  =  – y and y = g;  =

2. The age-at-peak-illness-rate is aext, the value of a at which di/da = 0 (which can be demonstrated from the above). Note that aext:<0,=0 = /.

13 APPENDIX: Special cases

If K2 = 0, we have (from McBride & French 2006) (A.1) (A.2) (A.3) (A.4) (A.5) So defining s = S/N and r = R/N, (A.1) becomes (A.6) where (A.7) and so (A.8) (A.9) (A.10) If  = 0 and  = 0, we have (A.11) and so (A.12) (A.13) (A.14) where the Methuselah states are now (A.15) (A.16) (A.17)

14 Analytical solutions to the linear SIR model