Interactions. Yuan Gao. Spring Applied Mathematics University of Washington

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1 Interactions Yuan Gao Applied Mathematics University of Washington Spring / 27

2 Nonlinear System Consider the following coupled ODEs: dx = f (x, y). dt dy = g(x, y). dt In general, when f and g are nonlinear functions of x and y, it is difficult to obtain a general solution to this system. Again we would like to study the long time behaviour of this system. The equilibrium solutions are x and y that solves f (x, y ) = g(x, y ) = 0. 2 / 27

3 Stability To find the stability of equilibrium points, we make a small perturbation from (x, y ). Let x(t) = x + u(t). y(t) = y + v(t). Since x and v do not depend on time, dx dt = d(x + u) = du dt dt. dy dt = d(y + v) = dv dt dt. 3 / 27

4 Linearization Take Taylor expansions of f (x, y) around (x, y ): f (x, y) = f (x, y ) + f x (x, y )(x x ) + f y (x, y )(y y ) + = f (x, y ) + f x (x, y )u + f y (x, y )v + f x (x, y )u + f y (x, y )v, where we use the fact that f (x, y ) = 0. Similarly for g(x, y) we have g(x, y) g x (x, y )u + g y (x, y )v +. We ended up with a coupled linear system: du dt = f (x, y) f x(x, y )u + f y (x, y )v. dv dt = g(x, y) g x(x, y )u + g y (x, y )v. 4 / 27

5 Matrix Form We can write the linear system in Matrix form: [ ] [ d u fx (x =, y ) f y (x, y ] [ ] ) u dt v g x (x, y ) g y (x, y. ) v For a system d dt z = Az, we assume solution of the form z = ke rt. Plug it in, we get or rke rt = Ake rt, rk = Ak. So r is the eigenvalue and k is the eigenvector of A. 5 / 27

6 General Solution For a 2 2 system, we have two eigenvalues/eigenvectors, so the general solution is z(t) = c 1 k 1 e r 1t + c 2 k 2 e r 2t, where (r 1, k 1 ) and (r 2, k 2 ) are the eigen-pairs of A. To find the eigenvalues, we set [ ] a11 λ a det 12 = 0. a 22 λ That gives us the equation: a 21 λ 2 (a 11 + a 22 )λ + (a 11 a 22 a 12 a 21 ) = 0, or λ 2 Tr(A)λ + det(a) = 0. 6 / 27

7 Stability and Eigenvalues Solving the quadratic equation we have λ = Tr(A) Tr(A) ± 2 4det(A). 2 2 In order for the equilibrium point (x, y ) to be stable, we must have (x(t), y(t)) (x, y ) as t, or in other words, [ ] u(t) 0 as t. v(t) That implies both eigenvalues λ 1 and λ 2 have negative real parts. Therefore we can pinpoint exactly under which region in the Tr(A)-det(A) plane stability is achieved. Example: When det(a) < 0, we get two real eigenvalues, one positive and one negative, hence unstable. 7 / 27

8 Stability Region 8 / 27

9 Lotka-Volterra Equations x(t) : population of the prey (fish) y(t) : population of the predator (shark) dx = rx axy. dt dy = bxy ky. dt Or in terms of per captia rate: 1 dx x dt = r ay. 1 dy y dt = bx k. 9 / 27

10 Equilibrium Population f (x, y) = rx axy. g(x, y) = bxy ky. Equilibrium points: (0, 0) and (k/b, r/a). At (0, 0), [ ] fx f A = y = g x det(a) = rk < 0, unstable. At (k/b, r/a), g y [ ] fx f A = y = g x g y [ ] r 0 0 k [ 0 ] ak/b br/a 0 det(a) = (br/a)(ak/b) = kr > 0, Tr(A) = 0, center, unstable. 10 / 27

11 Theory Revisited Let s walk through the stability theory by looking at this particular example. We first do a perturbation around the equilibrium point: x(t) = x + u(t). y(t) = y + v(t). Deriving a system for u and v, we get du dt = dx dt = r(x + u) a(x + u)(y + v). dv dt = dy dt = b(x + u)(y + v) k(y + v). Ignoring the higher-order terms of u and v, du dt = (r ay )u ax v + rx ax y. dv dt = by u + (bx k)v + bx y ky. 11 / 27

12 At (0, 0) The system of equation is magically decoupled! du dt = ru. dv dt = kv. Solution is u(t) = u(0)e rt, v(t) = v(0)e kt. Therefore if we have a small population for both the prey and predator, the prey will flourish, whereas the predators will die of starvation. This shows that (0, 0) is unstable because a small perturbation will NOT lead the solution back to (0, 0). 12 / 27

13 At (k/b, r/a) This time we are not so lucky... du dt = ak b v. dv dt = br a u. But it is still possible to decouple. We can differentiate the first equation and substitute the second equation to get, A second order ODE for u. d 2 u dt 2 = ak b dv dt = kru. 13 / 27

14 General Solution The general solution for u(t) is then u(t) = Acos( krt) + Bsin( krt). From the first equation we know v(t) = b du ak dt = b r ( Asin( krt) Bcos( ) krt). a k The initial condition gives These are oscillatory solutions. A = u(0), B = v(0) a b k r. 14 / 27

15 Plot of u(t) and v(t) 15 / 27

16 Phase Plane To see what happens around the equilibrium points, we can draw the phase plane. dy dx = dy/dt dx/dt = y(bx k) x(r ay). 16 / 27

17 Average Population Since the solution around (k/b, r/a) is periodic, we can define the average population as x ave = 1 T t0 +T t 0 x(t)dt, y ave = 1 T t0 +T where T is the period, and t 0 is any time. t 0 y(t)dt, Volterra showed that these averages can be found by integrating the original system, t0 +T 1 dx t0 +T t 0 x dt dt = 1 t 0 x dx = ln x(t 0 + T ) ln x(t 0 ) = / 27

18 Average Population Continued For the right hand side, t0 +T t 0 t0 +T (r ay)dt = rt a y(t)dt = rt aty ave. t 0 Therefore y ave = r/a. Similarly, by integrating the second ODE, we find x ave = k/b. This is the equilibrium point, about which the solution is cycling. 18 / 27

19 Harvest Consider the harvesting of predator and prey We can rewrite it as dx dt = rx axy qe 1x. dy dt = bxy ky qe 2y. dx dt = (r qe 1)x axy. dy dt = bxy (k + qe 2)y. The natural growth rate of x is shifted by qe 1 while the natural death rate of y is shifted by qe 2. The new center becomes (k + qe 2 )/b, (r qe 1 )/a 19 / 27

20 Limited Resource for Prey When the resource is limited for prey, dx dt = rx(1 x K ) axy, dy = ky + bxy, dt where K is the carrying capacity of the prey in the absence of predator. In this case there are three equilibrium points (0, 0) (K, 0) ( k b, r a (1 k bk )) 20 / 27

21 Stability of Equilibria The stability matrix [ ] fx f y = At (0, 0), Unstable. At (K, 0), g x Stable or unstable? g y [ 2r K [ ] fx f y = g x g y [ ] fx f y = g x g y ] x + r ay ax by k + bx [ ] r 0 0 k [ r ak ] 0 k + bk 21 / 27

22 Predator Efficiency The parameter b/k measures the efficiency of the predator in utilizing food (prey) to reproduce itself. for b/k > 1/K (efficient predators, for example, reptiles), (K, 0) is unstable. for b/k < 1/K (inefficient predators, for example, mammals), (K, 0) is stable. The predator becomes extinct. Finally, at ( k b, r a (1 k bk )), [ ] [ fx f y = g x g y rk bk rb a (1 k ak b bk ) 0 This is stable if b/k > 1/K (efficient predators). The efficiency of the predator determines which equilibrium point is stable, or in other words, whether itself can survive or not. ] 22 / 27

23 Combat Model During World War I, Frederick William Lanchester, a British engineer in the Royal Air Force, discovered a way to model battle-field casualties using systems of differential equations. Suppose troops x and y are engaged in a fight. x(t): number of soldiers in force x. y(t): number of soldiers in force y. The CONCOM Model (conventional combat model) assumes dx dt = ax by + P(t), where a is the operational loss rate, b is the combat loss rate and P(t) is the reinforcement rate. 23 / 27

24 A Simplified CONCOM Model In the absence of operational loss and reinforcement, we have dx dt = by. dy dt = cx. Divide the two equations we get or Integrating gives us dy dx = cx by, bydy = cxdx. by(t) 2 cx(t) 2 = by 2 0 cx 2 0 = K(Constant). H(t) = by(t) 2 cx(t) 2 is called a conserved quantity. 24 / 27

25 Phase Plane We consider x wins if y vanishes first. Thus x wins if K < 0; y wins if K > 0; a stalemate occurs if K = / 27

26 K = 0 In this case we have or by 2 0 = cx 2 0, c b = ( y0 x 0 ) 2. That shows in order to stalemate an adversary twice as numerous, it does not suffice to be twice as effective; you must be 4 times as effective! This is known as Lanchester s square law. Question: Is dividing the army x into two equal forces and engage them separately with the enemy y a good strategy? Can you show this using the previous model? 26 / 27

27 Mixed Forces Suppose x is composed of units of two different effectiveness. x = x 1 + x 2. x 1 has effectiveness c 1, and x 2 has effectiveness c 2. And we assume y engages uniformly to both units. dx 1 dt = by x 1 x. dx 2 dt = by x 2 x. dy dt = c 1x 1 c 2 x 2. One can check that ( ) H(t) = by(t) 2 c1 x 1 (t) + c 2 x 2 (t) x(t) 2 x(t) is a conserved quantity for this system. So we can define the average effectiveness of x as c 1 x 1 (t) + c 2 x 2 (t). x(t) 27 / 27

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