These notes give a quick summary of the part of the theory of autonomous ordinary differential equations relevant to modeling zombie epidemics.

Transcription

1 NOTES ON AUTONOMOUS ORDINARY DIFFERENTIAL EQUATIONS MARCH 2017 These notes give a quick summary of the part of the theory of autonomous ordinary differential equations relevant to modeling zombie epidemics. 1. Autonomous linear differential equations, equilibria and stability Suppose that n 1. We are going to later multiply vectors of length n on the left by square matrices. So e ill consider vectors of length n to be column vectors x 1 x x 2. x n This ill lessen the need to take transposes of ro vectors as in the lectures in class. Let G : R n R n is a continuously differentiable vector valued function. Thus G assigns to each x R n a vector G(x) G 1 (x) G 2 (x) G n (x) and all the partial derivatives G i x j are continuous functions of x. Definition 1.1. An autonomous ordinary differential equation for a function f : R R n has the form df (1.1) (t) G(f(t)) and f(0) x in hich f : R R n is a function and x is an initial value for f. Here f 1 (t) f(t) f 2 (t) f n (t) for some functions f 1,..., f n from R to R and df (t) df 1 (t) df 2 (t) df n (t) The differential equation (1.1) is called autonomous because df depends only on hat f(t) is rather than depending on both f(t) and the independent variable t. If one thinks. 1

2 2 MARCH 2017 of t as time and f(t) as the state of a system at time t, then the rate of change df of the system depends only on its state f(t) at time t, not on hat time it is. Definition 1.2. The initial value x gives an equilibrium solution if f(t) x for all t is a solution of the differential equation (1.1). The folloing result is easy: Lemma 1.3. The vector x gives an equilibrium solution f(t) x for all t if and only if G(x) is the zero vector 0 (0,..., 0) transpose. Proof. If f(t) x for all t is a solution to the differential equation, e have df (1.2) (t) 0 G(f(t)) G(x) for all t. Conversely, if G(x) 0, and e let f(t) x for all t, then (1.2) holds, so f(t) solves the differential equation. Definition 1.4. An equilibrium solution f(t) x for all t is stable if there an ɛ > 0 such that the folloing is true for all x R n such that distance(x, x) < ɛ. Suppose f(x) is a solution of the differential equation ith initial condition f(0) x. Then d f (t) G( f(t)) f(t) x. t + 2. Linearizing the differential equation, and linear stability To stu stability near an equilibrium x, e use the Taylor expansion of G( x) for x near x. The Taylor expansion is G( x) G(x) + Jac(G)(x) ( x x) + Here Jac(G) is the n n matrix ( ) Gi higher order terms. x j 1 i,j n and Jac(G)(x) is the matrix of constants hich results from evaluating at x all the partial derivatives hich are the entries of Jac(G). The higher order terms in the expansion go to 0 more rapidly than any linear function of x x and x tends toard x. If f(t) ere a solution of the differential equation such that f(t) is alays close to x, then since G(x) 0 for an equilibrium value of x, e ould have G( f(t)) Jac(G)(x) ( f(t) x) + smaller order error. We use the first term on the right side of this expression to get a linear approximation to our original differential equation: Definition 2.1. The linear approximation to the differential equation d f (t) G( f(t)) near an equilibrium value x is d (2.3) f (t) A ( f(t) x) When A is the constant n n matrix A Jac(G)(x).

3 NOTES ON AUTONOMOUS ORDINARY DIFFERENTIAL EQUATIONS 3 Lemma 2.2. The initial value x is a equilibrium solution of (2.3), in the sense that defining f(t) x for all t is a solution of (2.3). Further, x is a stable equilibrium for (2.3) if and only if every solution y(t) of the differential equation (2.4) (t) Ay(t) has the property that t + y(t) 0 (0,..., 0) transpose, here A Jac(G)(x). Proof. Clearly f(t) x for all t is a solution of (2.3), since then d f (t) 0. Suppose first that all y(t) as in the Lemma have t + y(t) 0. By definition, x is a stable equilibrium solution of (2.3) if and only for all f(t) for hich (2.3) holds and f(0) x is sufficiently close to x, e have t + f(t) x. Set y(t) f(t) x. Then d f, so y(t) is a solution of (2.4). Thus f(t) (y(t) + x) 0 + x x t + t + and (2.3) has x as a stable equilibrium solution. Conversely, suppose that x is a stable equilibrium solution to (2.3), and that y(t) is a solution of (2.4). By definition, this means that there is an ɛ > 0 such that if distance(x, x) < ɛ, then every solution f of (2.3) such that f(0) x has the property that f(t) x. t + We can find a real constant c > 0 such that distance(0, c y(0)) < ɛ. Since y(t) is a solution of (2.4), e find that f(t) x + c y(t) is a solution of (2.3) since We have d f (t) c (t) c Ay(t) Ac y(t) A( f(t) x). distance( f(0), x) distance(x + c y(0), x) distance(cy(0), 0) < ɛ. Therefore t + f(t) x since x is a stable equilibrium, so y(t) ( f(t) x)/c 0. t + t + Definition 2.3. Let x be an equilibrium solution of the original differential equation (1.1). We ill say that x is a linearly stable equilibrium if the linearized differential equation (2.1) has x as a stable solution. By Lemma 2.2, this is equivalent to requiring that t + y(t) 0 for all solutions y(t) to hen A Jac(G)(x). (t) Ay(t) 3. Solutions of linear systems Suppose A is an n n matrix of complex numbers.

4 4 MARCH 2017 Theorem 3.1. The series of n n matrices e At (At) m converges for all values of t to an n n matrix of complex numbers. The unique solution y : R R n of the differential equation (3.5) is (t) Ay(t) and y(0) y(t) e At. This result does take some time to prove rigorously, so I ill not include a proof. The appearance of e At is justified by the folloing formal computation: (3.6) d eat d A m t m d A m t m A m mt m 1 A Am 1 t m 1 (m 1)! m1 m1 Ae At The folloing result is proved in linear algebra courses. Theorem 3.2. There is an invertible n n matrix B such that C BAB 1 has the upper triangular form λ 1 c 1,2 c 1,3 c 1,n 0 λ 2 c 2,3 c 2,n (3.7) C 0 0 λ 3 c 3,n λ n for some complex numbers λ 1,, λ n and c i,j ith i < j. The entries belo the diagonal of C are all 0. The numbers λ i for 1 i n may not all be distinct, and are called the eigenvalues of A. These λ i are the roots (counting mulitiplicities) of the characteristic polynomial c A (T ) det(t I n A) here T is an indeterminate and I n is the n n identity matrix. Observe no that if m 0, then C m (BAB 1 ) (BAB 1 ) (BAB 1 ) (BAB 1 ) BA m B 1

5 NOTES ON AUTONOMOUS ORDINARY DIFFERENTIAL EQUATIONS 5 since e can group terms in the product to take advantage of the fact that BB 1 I n. Thus e get ( ) Be At B 1 (At) m B B 1 BA m B 1 t m C m t m (3.8) e Ct Since C in (3.7) is upper triangular, the matrix C m is also upper triangular, ith diagonal terms given by λ m 1,..., λm n. Thus e Ct C m t m λ m 1 tm 0 λ m 2 tm 0 0 λ m 3 tm λ m n t m (3.9) e λ 1t 0 e λ 2t 0 0 e λ 3t e λnt here e don t need to calculate the entries above the diagonal. Suppose no that x is a linearly stable equilibrium solution of the original differential equation (1.1) in the sense of Definition 2.3. Let A Jac(G)(x). Then for all initial values y(0) of the differential equation (t) Ay(t) e should have t + y(t) 0. By Theorem 3.1, this is equivalent to t + eat 0. Taking to range over the vectors hich have exactly one component equal to 1 and all the others equal to 0, e ould be able to conclude that t + eat the zero matrix.

6 6 MARCH 2017 Then (3.8) ould give But no (3.9) ould sho t + ect B ( t + eat (3.10) t + eλ it 0 for i 1,..., n. Here if λ i a i + b i 1 e have So (3.10) is equivalent to ) B 1 the zero matrix. e λ it e a it (cos(b i t) + 1 sin(b i t)). (3.11) Re(λ i ) a i < 0 for i 1,..., n. With a little more ork, hich I on t include, one can sho that (3.11) is also sufficient for linear stability: Theorem 3.3. The differential equation (1.1) is linearly stable at an equilibrium x if and only if the eigenvalues λ 1,..., λ n of the Jacobian matrix A Jac(G)(x) at x have all negative real parts.