Additional file 1 for: Assessing the optimal virulence of malaria-targeting mosquito pathogens: a mathematical study of engineered Metarhizium anisopliae Bernhard Konrad 1, Michael Lindstrom 1, Anja Gumpinger, Jielin Zhu 1, Daniel Coombs 1, 1 Department of Mathematics and Institute of Applied Mathematics, University of British Columbia, Vancouver, BC, Canada TU München, Fakultät für Mathematik, Boltzmannstraße 3, 85748 Garching b. München, Germany E-mail: coombs@math.ubc.ca roof of theorems In this additional file we present the lengthy but straightforward proofs of some mathematical results used in the text. Main Result 1 Local Stability. 1. The trivial equilibrium h, S, I, F 0, 0, 0, 0 is locally asymptotically stable if and only if κ < µ+µ+σ µ++σ.. The malaria-free equilibrium h, S, I, F 0, S m, 0, F m exists in the positive plane and is locally asymptotically stable if and only if both of the following conditions are fulfilled: µ + γs m κ > and R 0 ρµ + H < 1. Where S m µ+σ µ+µ+σ µ++σ 1 κµ++σ and F m µ+σ S m. 3. The endemic equilibrium h, S, I, F h e, S e, I e, F e exists in the positive plane and is locally asymptotically stable if and only if both of the following conditions are fulfilled: µ + γs m κ > and R 0 ρµ + H > 1. Where h e R 0 1 R 0 +/µ+, S e 1 µ++ R 0 1 S m, I e µ++ R 0 1 S m, F e F m. roof. We first show how to get to the steady states and that the system indeed has exactly these three steady states. The trivial steady state h, S, I, F 0, 0, 0, 0 is straight forward. In the malaria-free steady state neither infected humans h m 0, nor infected mosquitoes I m 0 are present, but susceptible mosquitoes S m and mosquitoes with fungus F m are around. The equations reduce to ds dt κs + F 1 S + F µ + S df S F dt Setting the rate of change df dt to zero tells us F m µ+σ S m. lugging this into the former equation, after 1
we set ds dt to zero yields note that S m + F m µ++σ µ+σ S m 0 κs m 1 S m µ + S m κ κ µ + 0 S m κ µ + S m κ µ + 1, κ where we divided by S m since we assumed S m 0. If S m 0 we reach the trivial steady state. Therefore F m 1. S m µ + κ This brings us to the endemic steady state, where h e, S e, I e and F e are all distinct from zero. Note that S e 0 or F e 0 would lead us to the trivial steady state, since populations can not be negative. First, note that we can divide by 1 h e, since h e 1 due to the first equation in 1. Setting dh dt 0 yields I e ρhhe di γ1 h e. Setting dt 0 yields I e Sehe µ+. Combining those those two equations yields S e ρhµ+ 0 we obtain Next, setting df dt Therefore, F e S e + I e + F e S ρh e + I e γ1 h e h e + µ +. ρh µ + γ1 h e ρh γ1 h e + h e + h e + µ + µ + + h e + h e + µ + ρh γ1 h e h e + µ + lugging this into the equation ds dt 0 will lead to h e, but not before a lengthy calculation: ρh 0 κ γ1 h e h ρh e + µ + 1 γ1 h e h e + µ + multiplying by γ1 h e ρh 0 Multiplying by ρhµ + ρhµ + h e µ + γ1 h e γ1 h e yields κ h e + µ + 1 γ1 h e. ρh γ1 h e h e + µ + µ + h e + µ +. µ+σ κµ++σh e+µ+, which is positive since h e is positive, and rearranging we obtain ρhh e + µ + γ1 h e 1 µ + κ To solve this for h e we multiply by the denominator on the left hand side and expand the resulting terms: ρhh e + ρhµ + γ γh e γµ + κ + γµ + h e, κ
which we rearrange to h e ρh + µ + γ 1 κ γ 1 µ + κ We now divide by ρhµ + and recall that This results in h e + From here we can quickly find S e, I e and F e : S e I e ρhµ + γ1 h e ρhh e γ1 h e ρhµ +. γs m ρµ+h 1, or equivalently h e µ + µ+ 1 + µ+ S m 1 h e S m + R 0 1 S m + µ + F e S e + I e S m F m R 0 S m 1 S m h e S m 1 µ + 1 h e µ + + µ+ γµ+σ µ+µ+σ ρhµ+µ++σ 1 κµ++σ. 1 + /µ +. + µ + + µ+ 1 + µ+ 1 To determine the local stability of the steady states we compute the characteristic polynomial of the linearized system around the equilibrium: γ I H ρ λ 0 γ 1 H 1 h 0 det S κ1 S+I+F h µ + λ κ1 S+I+F κ1 S+I+F S h µ + λ 0 0 λ det γ I H ρ λ 0 γ 1 H 1 h 0 0 κ1 S+I+F µ + λ κ1 S+I+F µ + λ κ1 S+I+F S h µ + λ 0 0 λ det γ I H ρ λ 0 γ 1 H 1 h 0 0 κ1 S+I+F µ + λ 0 κ1 S+I+F S h µ + h λ 0 0 0 λ det γ I H ρ λ γ 1 H 1 h κ 1 S+I+F S µ + h λ det µ + λ κ 1 S+I+F }{{} λ }{{} :A :B The fourth-order polynomial factors nicely into two second-order polynomials. This is to be expected: olynomial A says if malaria can invade the population, polynomial B decides if mosquitoes can invade the environment. We now split the analysis into the three equilibria: 3
1. Trivial equilibrium h, S, I, F 0, 0, 0, 0. Then A becomes A det ρ λ γ 1 H 0 µ + λ ρ + λµ + + λ, which clearly has the two negative roots λ ρ and λ µ +. Hence, malaria can not invade a void population and the stability of the trivial steady state is solely determined by B. B becomes B κ µ + λ κ λ κ µ + λ λ κ λ + + µ + κ λ + µ + κ + The constant term is positive if and only if κ < µ + µ + 1 If this condition is not fulfilled, then there is a positive root and hence the steady state is not stable. If this condition is fulfilled, then κ < µ + < µ + + and hence the linear term is also positive. Thus only negative roots exist and the steady state is table. Hence the trivial steady state is stable if and only if the condition 1 on κ holds.. Malaria-free equilibrium h, S, I, F 0, S m, 0, F m. In the malaria-free equilibrium A becomes A det ρ λ γ 1 H S m µ + λ ρ λ µ + λ γ S m H λ + ρ + µ + λ + ρµ + γ S m H. This polynomial has only negative roots if and only if the constant term is positive, i.e. if and only if ρµ + γ S m H > 0, which is equivalent to R 0 γsm ρµ+h < 1. The polynomial B becomes B det κ 1 Sm+Fm µ + λ κ 1 Sm+Fm λ Before we go about calculating this, let us first simplify the first row a little bit: With that we obtain κ 1 S m + F m S m + F m S m + S m S m κ µ + κ µ + 1 κ κ 1 1 µ + µ + κ. κ 4
Therefore B det det κ 1 Sm+Fm µ+µ+σ µ++σ κ µ + λ µ+µ+σ µ++σ µ + λ + λ µ + λ κ 1 Sm+Fm λ κ λ κ µ + λ λ µ + κ + µ + + µ + + κ + µ + λ + λ κ + µ + 1 + 1 µ + + κ µ +. µ + κ It is now obvious that this polynomial has two roots with negative real part, if only if the constant term is positive, i.e. if κ > 3. Endemic equilibrium h, S, I, F h e, S e, I e, F e. µ +. In the endemic steady state the polynomial A becomes Ie A γ H ρ λ γ 1 H 1 h e S e µ + h e λ λ + λ γ I e H + ρ + µ + + h e + γ I e H µ + + γ I e H h e + ρµ + + ρh e γ S e H + γ S e H h e λ + λ γ I e H + ρ + µ + + h e + γ I e H µ + + ρµ + + γ I e H γ S e H 1 h e, where we used that ρh e γ Ie H 1 h e. In the constant term we now substitute S e and I e with the expressions we just found above, and also use the definition of R 0. Factoring the result yields A λ + λ γ I e + γs m H λ + λ H + ρ + µ + + h e µ + 1 + µ + + 1 + γ I e H + ρ + µ + + h e R 0 1 + µ + 1 h e 1 + µ + + ρµ R 0 1 + h e + 1 h e + µ +, 1 where we again used the definition of R 0. We now see that, if R 0 > 1, then all roots of A have negative real part. For B, note that S e + I e + F e S m + F m, and hence B is the same as in the malaria-free steady state. 5
Main Result Fungus exposure rate. The malaria prevalence in humans h e h e, σ decreases when the fungus-exposure rate increases. The total number of mosquitoes S e + I e + F e S e + I e + F e, σ also decreases when increases, unless the fungus-induced death rate σ is zero, in which case the total number of mosquitoes is unaffected by. roof. Before we prove the claim, we first show that S m S m, σ is a decreasing function of is, i.e. S m, σ < 0 for all, σ 0. A straight forward calculation yields S m µ +, σ 1 κ µ + κ κµ + 1 + κ κ µ + 1 κ + κ 1 µ + µ + + 1 < 0, κ κ since κ > µ+µ+σ µ++σ by the assumption that malaria is endemic. Now, let 0 1 < and fix σ 0. For notational convenience we omit the argument σ in the following calculation. From the calculation above we can further deduce that S m 1 > S m R 0 1 γs m 1 ρµ + 1 H > γs m ρµ + H R 0. Finally, we are now in position to show the first claim: h e < h e 1 R 0 1 R 0 + /µ + < R 0 1 1 R 0 1 + /µ + 1 0 < µ + 1 µ + R 0 1 R 0 + R 0 1 µ + 1 R 0 µ + + 1 0 < µ + 1 µ + R 0 1 R 0 + γ ρh S m 1 S m + 1. The former results show that the last line is true, hence the first claim follows. To see that the total number of mosquitoes S e + I e + F e also decreases with, we calculate that S e + I e + F e, σ S m, σ 1 S m, σ + 1 σ κ 0, S m, σ µ + κ 1 µ + κ which only vanishes when σ 0 and hence proves the second claim. + 1 µ + κ Main Result 3 Fungus-induced death rate on mosquitoes. For > 0 the total number of mosquitoes, S e + I e + F e, decreases if the fungus-induced death rate σ increases. 6
roof. This is a straight forward calculation. Recall S e + I e + F e S m + F m S m 1 + µ+µ+σ κµ++σ and calculate S e + I e + F e κµ + κµ + µ +, σ σ κ κ, which is negative for all > 0. µ+σ 1 Main Result 4 Fungus-induced death rate on malaria prevalence in humans. If 0 i.e. no fungus spraying the human malaria incidence rate h e h e 0, σ is constant in σ. If > 0 i.e. the fungus is applied the malaria prevalence in the human population h e, σ, as well as the total number of infected mosquitoes I e, σ, both have their maximum at σ κ µµ + µ κ +. roof. Let 0 be fixed. To find the local maximum of h e, σ we first calculate the derivative with respect to σ. This yields h e σ, σ 1 + /µ +, σ + /µ +, σ σ so that we see that h e, σ has a critical point whenever R 0, σ has a critical point. However, R 0, σ in turn a critical point whenever S m, σ has a critical point:, σ σ Thus we first study the critical points of S m, σ. S m µ +, σ σ 1 κ 1. µ + κ γ ρµ + H S m, σ. σ µ + µ + κ Now we see that if 0, then S m 0, σ, and hence h e 0, σ is constant for all values of σ. So from now on let > 0. Then the value σ that makes σ S m, σ 0 solves 1 µ + κ σ κ µ + µ + µ κ If κ µ +, then the equation above yields 0 µ +, which is only true for 0, since µ > 0. As we have asserted before, h is constant in this case. Therefore, if κ µ + no maximum exists. If κ µ + we obtain σ κ µµ + µ κ +, i.e. Sm σ, σ 0 and thus R 0 σ, σ 0 he σ, σ. We use the second derivative test to quickly check that the value at, σ is indeed a maximum: 0 {}}{ S m σ, σ 3 1 + µ + κ µ + µ + κ µ + 1 κ 3 µ + σ < 0. 7
Since Sm he σ, σ and and σ, σ have the same sign everywhere this implies that h e also has its maximum at, σ. Finally, I e is an increasing function of h e and hence is also maximized at, σ. 8