Continuous time population models Jaap van der Meer jaap.van.der.meer@nioz.nl Abstract Many simple theoretical population models in continuous time relate the rate of change of the size of two populations by means of a coupled set of non-linear differential equations to the population sizes. Various classical models are discussed, including a DEB model for a substrate eating V1-morph in a chemostat.
April 11, 2011 1/39 Content One population Linear model: exponential growth Non-linear model: logistic growth Two populations Uncoupled linear model Coupled linear model Non-linear model: logistic predator-prey model DEB V1-morphs Chemostat models
April 11, 2011 2/39 Exponential growth Rate of change is a linear function of population size itself, dx/dt = rx. Equilibria can be obtained by setting dx/dt = 0, which gives rise to one equilibrium x = 0. A general solution of this linear differential equation can be obtained by separation of variables and subsequent integration, 1 x dx = rdt, and is given by lnx = rt + c or x = x 0 e rt. If time t goes to infinity, population size x will either be infinitely large if r > 0 or will approach zero if r < 0, whatever the initial value x 0 is.
April 11, 2011 3/39 Exponential growth Population size x 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0 20 40 60 80 100 Time t Figure 1: Exponential growth either means unlimited growth in case the instantaneous growth rate r is positive, or extinction when the rate is negative.
April 11, 2011 4/39 Exponential growth Rate of change dx/dt 0.06 0.02 0.02 0.04 0.06 1 0 1 2 3 Population size x Figure 2: The equilibrium of the exponential growth equation, which is given by x = 0 is either unstable in case the slope r of the linear relationship between population rate of change dx/dt and population size x is positive, or stable when the slope r is negative. The slope might be called the eigenvalue of the equation.
April 11, 2011 5/39 Logistic growth The logistic equation, given by dx/dt = rx(1 x/k), has no unbounded growth. The two equilibria are x = 0 and x = k. Generally, a tangent in x = a is given by y = f(a) + f (a)(x a), where f (a) is the first derivative of the function f(x) in x = a. Here, the tangent in x = k is given by dx/dt = r(x k), and the linear approximation in the equilibrium x = k can be re-written as dy/dt = ry, where y = x k. Nothing else than the exponential growth equation.
April 11, 2011 6/39 Logistic growth Rate of change dx/dt 0.005 0.000 0.005 0.010 0.015 1 0 1 2 3 Population size x Figure 3: The logistic growth equation has two equilibria. The red and blue lines show linear approximations of the growth equation in the neighbourhood of the equilibria. One equilibrium is locally unstable, the other is locally stable.
April 11, 2011 7/39 Logistic growth Population size x 0 1 2 3 4 0 20 40 60 80 100 Time t Figure 4: Four particular solutions for the logistic growth equation are given by the black lines. The red and blue lines show various other specific solutions in the neighbourhood of the equilibria. They resemble exponential growth.
April 11, 2011 8/39 Allee effect Rate of change dx/dt 0.02 0.01 0.00 0.01 0.02 1 0 1 2 3 Population size x Figure 5: A cubic relationship between dx/dt and x may give rise to multiple equilibria, which are either stable or unstable. As a result of a slight change in the parameter values the equilibria can easily disappear, as depicted by the green lines.
April 11, 2011 9/39 Two populations If two populations grow exponentially, we may write the differential equations in matrix form [ dx0 dt dx 1 dt = [ λ0 0 0 λ 1 [ x0 x 1 (in short notation x = Ax) and plot the trajectories in a phase plane diagram, in which x 1 is plotted versus x 0. If the two growth coefficients (or eigenvalues) are negative the equilibrium is stable. The two populations are decoupled and the analysis is exactly the same as discussed for a single population.
April 11, 2011 10/39 Figure 6: Node, where λ 2 < λ 1 < 0.
April 11, 2011 11/39 Coupled populations But what do we do when the growth of the two populations is coupled, such that the differential equations in matrix form look like [ dx0 dt dx 1 dt = [ a b c d [ x0 x 1 where the elements b and c are not necessarily zero? We need some matrix algebra.
April 11, 2011 12/39 A new basis 3 2 1 0 1 2 3 (3, 2) (1,1) 2 1 0 1 2 3 4 Figure 7: A new basis.
April 11, 2011 13/39 Consider now the multiplication of a vector (expressed with respect to the standard basis) with a matrix. For example [ [ [ 1 2 3 7 Ax = = 3 2 2 13 This matrix A does with respect to the standard basis β the same thing as the matrix B does with respect to the basis η: [ [ [ 4 0 1 4 By = = 0 1 1 1 Note that [ 2 4 3 [ 1 + 1 1 = [ 7 13
April 11, 2011 14/39 Eigenvectors as the new basis 15 10 5 0 5 10 15 15 10 5 0 5 10 15 Figure 8: A new basis is shown with blue vectors. The red vectors are the standard basis. The purple arrow refers to the example. Points on a blue vector (or its elongation), such as the purple point in the upper right, will stay on that vector.
April 11, 2011 15/39 Eigenvectors and eigenvalues The fundamental equation needed is Ax = λx, where x, which is not the zero vector x = 0, is an eigenvector and λ an eigenvalue of A. Hence, if one multiplies the matrix A with its eigenvector, the result is simply a scalar multiplication of the eigenvector. A two by two matrix has at most two eigenvectors. The equation Ax = λx is re-written as (A λi)x = 0, where is called the identity matrix. I = [ 1 0 0 1
April 11, 2011 16/39 The determinant The equation Ax = [ a b c d [ x1 x 2 is true if ax 1 +bx 2 = 0 and cx 1 +dx 2 = 0, hence if x 2 = a b x 1 and if x 2 = c d x 1. A solution, other than the zero vector, only exists if a/b = c/d, which results in ad bc = 0. The term ad bc is called the determinant of the matrix A. Hence, the equation (A λi)x = 0 has only a solution, other than x = 0, if the determinant of the matrix A λi is zero, hence if (a λ)(d λ) bc = 0. = [ 0 0
April 11, 2011 17/39 The characteristic equation This leads to the so-called characteristic equation: The eigenvalues are thus λ 2 (a + d)λ + (ad bc) = λ 2 T λ + D = 0 λ 1,2 = (a + d) ± (a + d) 2 4(ad bc) 2 = T ± T 2 4D 2 where T = a + d is the trace and D = ad bc the determinant of the matrix A. For the case that the part under the square root sign, which is called the discriminant, is positive, it can easily be seen that both eigenvalues are negative if and only if T < 0 and D > 0.
April 11, 2011 18/39 Returning to the example A = [ 1 2 3 2 gives the characteristic equation λ 2 + 3λ 4 = (λ + 4)(λ 1) = 0, resulting in the eigenvalues λ 1 = 4 and λ 1 = 1. The trace T = 3, the determinant D = 4 and the discriminant equals 25. The eigenvalue λ = 4 reveals (A + 4I)e 1 = [ 3 2 3 2 e 1 = 0 Hence e 1 = [2 3 t is an accompanying eigenvector. Similarly, for the eigenvalue λ = 1.
April 11, 2011 19/39 Coupled populations Hence, when the growth of the two populations is coupled, such that [ dx0 dt dx 1 dt = [ a b c d [ x0 x 1 we can now easily re-write it in the form of [ dy0 dt dy 1 dt = [ λ0 0 0 λ 1 [ y0 y 1 for which we know the solution (or don t we?).
April 11, 2011 20/39 Figure 9: Saddle, where λ 1 < 0 and λ 2 > 0. The orange lines are the eigenvectors. The purple lines the isoclines, for which dx i /dt = 0
April 11, 2011 21/39 Complex numbers But what if the discriminant not positive? λ 1,2 = (a + d) ± (a + d) 2 4(ad bc) 2 = T ± T 2 4D 2 We enter the realm of imaginary numbers i = 1, but thanks to Euler can we re-write them in terms of sines and cosines. Noting that i 2 = 1, i 3 = i, i 4 = 1, i 5 = i, i 6 = 1, etc. gives you some intuitive understanding of this link. The practical implication is that the stability properties are determined by the real part, and that the equilibrium is either approached in spirals (if the real part is negative) or that the system moves away from the equilibrium in ever extending spirals (if the real part is positive).
April 11, 2011 22/39 A logistic predator-prey model A predator-prey model, in which the prey population follows logistic growth in the absence of predation, and where the predation term is based on the laws of mass-action, can be written (in a dimensionless form) as dx 0 dτ = x 0(1 x 0 ) qx 0 x 1 dx 1 dτ = pqx 0x 1 mx 1 where x 0 is the prey and x 1 the predator population size. ( ) The biologically interesting equilibrium is given by x 0 = m pq and x 1 = 1 q 1 m pq. Again we can use a linear approximation around the equilibrium, using the proposition that a non-linear function f(x 0, x 1 ) around the point x 0 = a and
April 11, 2011 23/39 x 1 = b can be approximated by the tangent plane, which is given by y = f(a, b) + f x 0 (x 0 a) + f x 1 (x 1 b) where the partial derivatives are taken at x 0 = a and x 1 = b. For our purpose this leads to the matrix equation: [ d(x0 x 0 ) dτ d(x 1 x 1 ) dτ = f (x,x ) [ f0 (x 0,x 1 ) x 0 f 1 (x 0,x 1 ) 0 0 1 x 1 f 1 (x 0,x 1 ) x 0 x 1 x=x [ x0 x 0 x 1 x 1 where f i (x 0, x 1 ) is the differential equation for x i.
April 11, 2011 24/39 The Jacobian equals [ x 0 qx 0 p(1 x 0) 0 = [ m pq m p p(1 m pq ) 0 which leads to a trace T = x 0 = m pq that is always negative (given that the parameters m, p and q are positive) and a determinant D = mqx 1 = m(1 m pq ) that is positive as long as x 1 > 0. Hence the equilibrium is stable. The discriminant T 2 4D = x 2 0 4m( x 0) can be either positive or negative, implying that the equilibrium is or is not approached in spirals. For example, if q = 10, p = 0.5 and m = 0.1, then T = 0.02, D = 0.098 and the discriminant equals -0.3916, implying complex eigenvalues. The stable equilibrium x 0 = 0.02 and x 1 = 0.098 is approached in spirals.
April 11, 2011 25/39 x1 0.01 0.02 0.05 0.10 0.20 1e 05 1e 04 1e 03 1e 02 1e 01 x0 Figure 10: A specific solution of the logistic predator-prey model. lines give the isoclines, which are defined by dx i /dτ = 0. The purple
April 11, 2011 26/39 V1-morphs Reserve density follows first order dynamics d[e dt = ṗa V c[e = [ṗ Amf k E [E (1) where k E has been given the name specific energy conductance. It has the physical dimension per time. The mobilization rate equals ṗ C = k E [EV [E dv dt (2)
April 11, 2011 27/39 For organisms that simply divide into two daughter cells, and which are classified as juveniles in DEB terminology, κ can be set equal to one. Hence the allocation is given by ṗ C = [E G dv dt + [ṗ MV (3) Substituting equation 2 in equation 3 gives the growth equation dv dt = k E [E [ṗ M V (4) [E + [E G
April 11, 2011 28/39 Under constant food conditions, when the reserve density is in equilibrium and proportional to the scaled functional response, the growth equation simplifies to dv dt = [ṗ Amf [ṗ M 1 [ṗ Am f k E + [E V = ṙv (5) G where ṙ is the specific growth rate. Hence, the growth rate of the structural volume is proportional to the structural volume itself. V1-morphs show exponential growth at constant food density.
April 11, 2011 29/39 A mass-mass framework One might choose to write the reserve dynamics and the growth equation in a mass-mass framework. Equation 1 becomes dm E dt = d[e dt 1 µ E [M V = j EAmf k E m E (6) and Equation 4 becomes dm V dt = dv dt [M V = k E m E j EM m E + y EV M V (7)
April 11, 2011 30/39 Table 1: Primary parameters of the DEB model for V1-morphs, specific for a mass-mass framework. The last column indicates the relationship with the parameters from the energy-length framework that has been replaced. µ E is the potential energy of the reserves expressed in energy per C-mole, and [M V is the specific density of the structural body expressed in C-moles per volume. Symbol Dimension Interpretation Relationship j EAm ## 1 t 1 Mass-specific [ṗ Am = µ E [M V j EAm maximum assimilation rate y EX ## 1 Yield of reserve µ AX = µ E y EX on food j EM ## 1 t 1 Mass-specific [ṗ M = µ E [M V j EM maintenance rate y EV ## 1 Mass-specific costs of growth [E G = µ E [M V y EV
April 11, 2011 31/39 Yield The ratio of yield on substrate use is given by the rate of increase of structural mass and reserves divided by the ingestion rate: Y = dm V /dt + d(m E M V )/dt (8) J X Assuming constant substrate availability, and hence that the reserve density is in equilibrium, it follows that Y = (1 + m E )ṙm V J X (9)
April 11, 2011 32/39 The yield on substrate can be expressed as a function of the food availability only Y = y V X y EV f + g f + g f l d f (10) where y V X is known as the true yield on substrate in microbiology. The compound parameter g is the energy investment ratio. It is defined as the ratio of the energetic costs of structure [E G relative to the maximum available energy for growth and maintenance κ[e m. For dividing V1-morphs, with κ = 1, g can be written as k E [E G /[ṗ Am within an energy-length framework. The compound parameter l d is defined as the ratio between the volume-specific maintenance rate [ṗ M and the volume-specific maximum assimilation rate [ṗ Am.
April 11, 2011 33/39 The chemostat Total food intake rate of the V1-morphs equals J X = [ J Xm fv, where [ J Xm is the maximum volume-specific intake rate, f the scaled functional response, and V the overall volume of the population of V1-morphs. The rates of change of the food density and of the biomass density of V1- morphs X 1 = M V /V C look like: dx 0 dt = ḣ (X J r X 0 ) X dx 1 V C dt = Y J X V ḣx 1 C where ḣ is the throughput rate, and V C the constant volume of the chemostat.
April 11, 2011 34/39 A dimensionless form The model can be expressed in a dimensionless form resulting in dx 0 dτ = x r x 0 αfx 1 dx 1 dτ = Y αfx 1 x 1
April 11, 2011 35/39 Alternative models Y = y V X g f l d f + g f (11) DEB V1-morphs Droop model has no maintenance: l d = 0 Monod model has no reserves: g Lotka-Volterra kind of model has no handling time: f x 0
April 11, 2011 36/39 The Lotka-Volterra kind of model Equilibria x 0 = (αy ) 1 x 1 = 1 α (x rαy 1) The Jacobian of this equilibrium equals: f (x,x ) [ f0 (x 0,x 1 ) x 0 f 1 (x 0,x 1 ) 0 0 1 x 1 f 1 (x 0,x 1 ) x 0 x 1 x=x = [ αx 1 1 αx 0 Y αx 1 Y αx 0 1 = [ x r αy Y (x r αy 1) 0 1 Y where f i (x 0, x 1 ) is the differential equation for x i. Eigenvalues equal λ 1 = (x r αy 1) and λ 2 = 1. Both eigenvalues are real and negative, pointing to a so-called node.
April 11, 2011 37/39 Figure 11: Simulated trajectories of the Lotka-Volterra kind of model. Isoclines in purple.
April 11, 2011 38/39 Table 2: A numerical example of the chemostat models. The first part gives the parameter values, the second the equilibrium, the third the elements of the Jacobian, and the fourth the eigenvalues. Lotka Monod Droop DEB x r 10 10 10 10 y V X 0.85 0.85 0.85 0.85 α 3 3 3 3 g - - 1 1 l d - - - 0.1 x 0 0.39 0.65 1.82 4.25 x 1 8.17 7.95 4.23 2.37 a -25.5-9.81-2.60-1.26 b -1.18-1.18-1.94-2.43 c 20.8 7.50 0.501 0.0734 d 0 0 0 0 λ 1-24.5-8.81-2.14-1.09 λ 2-1 -1-0.453-0.163
April 11, 2011 39/39 References [1 Gurney, M. and Nisbet, R. (1998) Ecological dynamics. Publisher, Place [2 Kooijman, S.A.L.M. (2010) Dynamic Energy Budget theory for metabolic organization. Cambridge University Press, Cambridge [3 Strang, G. (1980) Linear algebra and its applications. Second edition. Academic Press, New York