R₀ of a human–animal leptospirosis model: formula and derivation

By Maria Esther Leyva Borges · Last updated 20 October 2026

In the leptospirosis model of Baca-Carrasco, Olmos and Barradas (2015), with infected animals, infected humans and free bacteria in the environment, R₀ is the largest root of λ³ − R_AA λ² − (R_ABA² + R_HBH²) λ + R_AA R_HBH² − R_AHBA³ = 0. Each R is the contribution of one transmission route: animal to animal, animals through the environment, humans through the environment, and animals to humans to the environment and back. R₀ is not a single product because the routes share compartments.

R_0 = \sqrt[3]{\frac{n_{a}^{2} c_{1} c_{2} c_{5}}{6 \alpha_{2}^{2} k} + \frac{c_{1}^{3} n_{a}^{3}}{27 \alpha_{2}^{3}} - \frac{c_{1} c_{4} c_{6} n_{a} n_{h}}{3 \alpha_{1} \alpha_{2} k} + \frac{c_{2} c_{3} c_{6} n_{a} n_{h}}{2 \alpha_{1} \alpha_{2} k} - \frac{\sqrt{n_{a}^{2} \alpha_{1} k \left(9 \alpha_{1} \alpha_{2} c_{1} c_{2} c_{5} n_{a} + 2 c_{1}^{3} n_{a}^{2} \alpha_{1} k + 9 \alpha_{2}^{2} c_{6} n_{h} \left(- 2 c_{1} c_{4} + 3 c_{2} c_{3}\right)\right)^{2} - 4 \left(3 \alpha_{1} \alpha_{2} c_{2} c_{5} n_{a} + c_{1}^{2} n_{a}^{2} \alpha_{1} k + 3 \alpha_{2}^{2} c_{4} c_{6} n_{h}\right)^{3}}}{54 \alpha_{1}^{\frac{3}{2}} \alpha_{2}^{3} k^{\frac{3}{2}}}} + \sqrt[3]{\frac{n_{a}^{2} c_{1} c_{2} c_{5}}{6 \alpha_{2}^{2} k} + \frac{c_{1}^{3} n_{a}^{3}}{27 \alpha_{2}^{3}} - \frac{c_{1} c_{4} c_{6} n_{a} n_{h}}{3 \alpha_{1} \alpha_{2} k} + \frac{c_{2} c_{3} c_{6} n_{a} n_{h}}{2 \alpha_{1} \alpha_{2} k} + \frac{\sqrt{n_{a}^{2} \alpha_{1} k \left(9 \alpha_{1} \alpha_{2} c_{1} c_{2} c_{5} n_{a} + 2 c_{1}^{3} n_{a}^{2} \alpha_{1} k + 9 \alpha_{2}^{2} c_{6} n_{h} \left(- 2 c_{1} c_{4} + 3 c_{2} c_{3}\right)\right)^{2} - 4 \left(3 \alpha_{1} \alpha_{2} c_{2} c_{5} n_{a} + c_{1}^{2} n_{a}^{2} \alpha_{1} k + 3 \alpha_{2}^{2} c_{4} c_{6} n_{h}\right)^{3}}}{54 \alpha_{1}^{\frac{3}{2}} \alpha_{2}^{3} k^{\frac{3}{2}}}} + \frac{c_{1} n_{a}}{3 \alpha_{2}}

What the leptospirosis model describes

Leptospirosis is caused by Leptospira bacteria that infected animals shed in their urine and that survive for a while in water and soil. The model has five compartments: susceptible and infected animals (S_A, I_A), susceptible and infected humans (S_H, I_H), and the free bacteria in the environment (B). Animals are infected by other animals and by the bacteria; humans are infected by animals and by the bacteria, but not by other humans. Infected animals and infected humans both shed bacteria, so both are infectious. Infected animals stay infected until they die; humans recover and become susceptible again. Its parameters are:

From the leptospirosis equations to the infection diagram

The infection diagram is read off the equations of the infected compartments. With N_A = S_A + I_A, the model is:

The infected compartments are I_A, I_H and B. At the disease-free equilibrium (DFE) I_A = I_H = B = 0, and the R₀ formula uses the equilibrium values S_A* and S_H*. With β = α₂ births balance deaths, the animal population stays constant and S_A* = N_A; humans have no births or deaths, so S_H* = N_H. Each term of the I_A, I_H and B equations becomes one arrow:

Shedding (c₅, c₆) is drawn as new infection because it increases the infected set without moving anyone: an infected animal that sheds bacteria stays in I_A, and B grows. In c₁ I_A S_A the cause and the receiver are the same compartment, so the arrow is a loop. There are no transfer arrows, and all three nodes are infectious because each one originates new infections. See how to build an infection diagram for the general rules.

The leptospirosis infection diagram

The diagram has three infectious nodes: I_A, I_H and B, which is not a population of hosts but causes new infections directly. Six dashed new-infection arrows connect them: a loop on I_A, animals and humans infected by B, humans infected by animals, and shedding from I_A and I_H into B. Each node has one outflow: animal deaths α₂, human recovery α₁ and bacterial decay k. Nothing points from I_H to I_A: humans reach animals only through the environment.

How R₀ is derived

R0Calc finds four infection cycles in the diagram and computes the term of each: the product, over the nodes of the cycle, of the rate coming in divided by the rate going out, with the k-th root for a cycle through k infectious nodes.

• I_A → I_A (animal to animal): R_AA = c₁S_A*/α₂.
• I_A → B → I_A (animals through the environment): R_ABA = √(c₂c₅S_A*/(α₂k)).
• I_H → B → I_H (humans through the environment): R_HBH = √(c₄c₆S_H*/(α₁k)).
• I_A → I_H → B → I_A (animals to humans to the environment and back): R_AHBA = ∛(c₂c₃c₆S_A*S_H*/(α₁α₂k)).

The cycles share infectious nodes, so their terms cannot simply be added. R0Calc combines them into the cubic at the top of the page and returns its largest root. The constant term pairs the two cycles that do not touch, I_A → I_A and I_H → B → I_H, against the cycle through all three nodes.

The next-generation matrix method (Diekmann, Heesterbeek and Metz, 1990; van den Driessche and Watmough, 2002) gives the same result. With infected compartments (I_A, I_H, B):

The characteristic polynomial of FV⁻¹ is

which is the cubic above written in the parameters, since c₆S_A*S_H*(c₁c₄ − c₂c₃)/(α₁α₂k) = R_AAR_HBH² − R_AHBA³. R₀ is the spectral radius of FV⁻¹, its largest eigenvalue. R0Calc's closed form agrees with it: checked with SymPy on 340 random sets of positive parameters, covering cubics with one and with three real roots, with relative differences below 10⁻³².

Worked example

Take N_A = 1,000 animals that live two years on average (α₂ = 0.5 per year) and N_H = 2,000 people who recover in about a month (α₁ = 12 per year), bacteria that survive about a week (k = 52 per year), infection rates c₁ = 4 × 10⁻⁴, c₂ = c₃ = c₄ = 10⁻⁴ and shedding rates c₅ = 100 and c₆ = 50, all per year. The cycle terms are R_AA = 0.8, R_ABA² ≈ 0.385, R_HBH² ≈ 0.016 and R_AHBA³ ≈ 0.003, and the largest root of the cubic is R₀ ≈ 1.143. Neither route alone reaches the threshold (R_AA = 0.8, R_ABA ≈ 0.62), but together they do. These are illustrative values, not estimates from Baca-Carrasco et al. or for a real setting.

Special case: no human transmission routes

If humans are left out (c₃ = c₄ = c₆ = 0), the cubic has a root λ = 0 and the rest is a quadratic, so R₀ = (R_AA + √(R_AA² + 4R_ABA²))/2, with R_AA = c₁S_A*/α₂ and R_ABA² = c₂c₅S_A*/(α₂k). R0Calc returns it as:

Without the human routes, R₀ depends only on the cycles animal–animal and animal–environment–animal; the difference with the general case measures how much humans add to transmission. In the worked example it falls from 1.143 to 1.138: humans add little there because they are infectious for only about a month.

Compute it in R0Calc

1. Open the calculator, paste the YAML below into the YAML panel and choose Apply the YAML to the diagram (the ← button between the YAML and the diagram).
2. R₀ appears under the diagram as the closed form above. Open the cycle breakdown to see the four cycles and their terms.
3. Delete the arrows into and out of I_H to get the special case without humans, or set c₆ to 0 to make humans dead-end hosts.

Labels in R0Calc cannot contain an asterisk, so the YAML writes S_A* and S_H* as their DFE values N_A and N_H. R0Calc prints Latin letters in lower case, so these appear as n_a and n_h in the result.

non_infectious_infested: []
infectious_infested:
  - I_A
  - I_H
  - B
new_infections_arrows:
  - from: I_A
    to: I_A
    param: c_1 N_A
  - from: B
    to: I_A
    param: c_2 N_A
  - from: I_A
    to: I_H
    param: c_3 N_H
  - from: B
    to: I_H
    param: c_4 N_H
  - from: I_A
    to: B
    param: c_5
  - from: I_H
    to: B
    param: c_6
transfer_arrows: []
outs_arrows:
  - from: I_A
    param: \alpha_2
  - from: I_H
    param: \alpha_1
  - from: B
    param: k

Frequently asked questions

What is R₀ for leptospirosis in this model?

The largest root of λ³ − R_AA λ² − (R_ABA² + R_HBH²) λ + R_AA R_HBH² − R_AHBA³ = 0, where R_AA = c₁S_A*/α₂, R_ABA² = c₂c₅S_A*/(α₂k), R_HBH² = c₄c₆S_H*/(α₁k) and R_AHBA³ = c₂c₃c₆S_A*S_H*/(α₁α₂k). It is the spectral radius of the next-generation matrix.

Why are bacteria in the environment a node of the infection diagram?

Because free bacteria cause new infections directly: contact with contaminated water or soil infects animals and humans at rates c₂ and c₄. A variable belongs to the infection diagram because of its role in transmission, not because it is a population of hosts. Shedding (c₅, c₆) adds to B without moving anyone, so it is drawn as new-infection arrows into B.

Do humans affect R₀ of leptospirosis?

Yes, if infected humans shed bacteria (c₆ > 0). Humans enter R₀ through two cycles: I_H → B → I_H and I_A → I_H → B → I_A. Both need c₆; with c₆ = 0 humans are dead-end hosts and R₀ depends only on the animal cycles.

References

• Baca-Carrasco D., Olmos D., Barradas I. (2015). A mathematical model for human and animal leptospirosis. Journal of Biological Systems, 23(Suppl. 1), S55–S65. doi:10.1142/S0218339015400057
• Diekmann O., Heesterbeek J. A. P., Metz J. A. J. (1990). On the definition and the computation of the basic reproduction ratio R₀ in models for infectious diseases in heterogeneous populations. Journal of Mathematical Biology, 28, 365–382.
• van den Driessche P., Watmough J. (2002). Reproduction numbers and sub-threshold endemic equilibria for compartmental models of disease transmission. Mathematical Biosciences, 180, 29–48.