Calculus C (ordinary differential equations)

Transcription

1 Calculus C (ordinary differential equations) Lesson 1: Ordinary first order differential equations Separable differential equations Autonomous differential equations Logistic growth Frank Hansen Institute for Excellence in Higher Education Tohoku University

2 First order differential equations 1 / 11 Let F : D R be a function of two variables defined in an open domain D R 2. A solution to the first order ordinary differential equation (#) = F(t, x) is a differentiable function x : I R defined in an open interval I R such that (t, x(t)) D t I and F(t, x(t)) = The graph of a solution is called an integral curve. t I. G(x) = {(t, x(t)) t I}

3 Indefinite integrals as solutions 2 / 11 Consider the simple case where the function F is continuous but does not depend on the second variable. Then F is of the form F(t, x) = f (t) (t, x) D, where f is a continuous function defined in an open interval I. The differential equation (#) then takes the form = f (t), and the solutions are given by the indefinite integral x(t) = f (t) + C t I, where C is an arbitrary constant.

4 Separable differential equations 3 / 11 A differential equation of the form (#) is said to be separable. = F(t, x) = f (t)g(x) Suppose that (t, a) D for every t in an open interval I, and that g(a) = 0. Then obviously (#) has the constant solution x(t) = a t I. Suppose now x : I R is a solution to (#) satisfying g(x(t)) 0 t I. By inserting the explicit dependency of the variable t we may write x (t) = f (t) t I. g(x(t))

5 Solution using integration by substitution The left and write hand sides are now two identical functions of the variable t. In particular, they have the same indefinite integrals x (t) g(x(t)) = f (t) + C, where C is an arbitrary constant. Integration by substitution yields x (t) g(x(t)) = g(x) thus we obtain g(x) = f (t) + C t I. This expression may in some cases lead to a solution of (#). 4 / 11

6 Example of a separable differential equation I Consider the separable differential equation = F(x, t) = 2x 2 t (x, t) R 2. The constant function x(t) = 0 for t R is obviously a solution. Consider a solution t x(t) defined in an interval I, and suppose that x(t) 0 for t I. Then x (t) = 2t t I. x(t) 2 Integration by substitution then yields x 2 = x (t) x(t) 2 = 2t = t 2 + C. 5 / 11

7 Example of a separable differential equation II 6 / 11 By solving the first integral we then obtain or equivalently 1 x = t2 + C x = 1 t 2 + C and this function is well-defined for t 2 C. We may determine an integral curve going through the point (t, x) = (1, 1) by setting C = 2. It is given by x = 1 t 2 2 for 2 < t < 2. The same formula defines other solutions to the differential equation in the intervals < t < 2 and 2 < t <.

8 Autonomous differential equations 7 / 11 A differential equation written on the form (#) = F(x) F : I R is said to be autonomous. If F(a) = 0 then we realize that the constant function x(t) = a t I is a solution to (#). Solutions of this type are called stationary solutions. We shall later establish that the stationary solutions may be divided into two classes; called the stable and the unstable solutions. In practical applications we are often content with determining the stationary solutions which are stable.

9 Special autonomous differential equations I 8 / 11 We consider a differential equation of the form = A(x a)(x b), where A, a, b are constants and a b. The only constant solutions are given by x = a and x = b. Suppose x is a solution defined in an interval I such that x(t) a and x(t) b t I. For t I we then obtain 1 (#) (x a)(x b) = A = At + C 1 where C 1 is an arbitrary constant.

10 Special autonomous differential equations II 9 / 11 Since by an elementary calculation we have 1 (x a)(x b) = 1 b a ( 1 x b 1 ) x a 1 (x a)(x b) = 1 b a log x b x a. By inserting this expression in (#) we obtain 1 b a log x b x a = At + C 1 where C 1 is an arbitrary constant. It may be written on the form log x b x a = A(b a)t + C 2

11 Special autonomous differential equations III 10 / 11 where C 2 = C 1 (b a) R. Thus x b x a = CeA(b a)t where C = e C 2 R +. Suppose we are looking at an integral curve t x(t) not intersecting the constant solutions. Then x a and x b have constant signs, and we may write x b x a = CeA(b a)t where C R\{0}. If we solve this equation for x we obtain x = a + b a 1 Ce A(b a)t C R.

12 11 / 11 Logistic growth In particular, the differential equation for logistic growth 1 x = A(b x) A, b > 0, has the solution x = where x 0 = x(0) and 0 < x 0 < b. b 1 + (bx 1 0 1)e Abt b=2, A=