Math 285-3 ODE Notes Revised Spring 2007 E. Kosygina, C. Robinson, M. Stein Ordinary Differential Equations Definition. For a vector field F on Rn, the differential equation determined by the F is denoted by x0 = F(x).A solution to the differential equation is a differentiable path x(t) 0 0 in Rn such that x (t) = F(x(t)), i.e., the velocity vector x (t) of the path at time t is given by the vector field F at the point x(t). (In the calculus book by Colley, a solution is called a flow line. See p. 211.) We often specify the initial condition x0 at some time t0, i.e., we seek a solution x(t) such that x(t0) = x0. We shall concentrate on the following two questions: • How can we describe all possible solutions to a given differential equation? • How can we find a solution with given initial condition (t0, x0)? We will mainly consider linear differential equations of the form x0 = Ax, but will consider a few nonlinear examples. We will consider first n = 1 and then n ≥ 2. 1. Scalar Differential Equations When n = 1 and we have a vector field on the line, we shall denote it by a non-boldfaced F. Even though in this case F is just a function from R to R, we shall think of F(x) as a one-dimensional vector emanating from the point x on the line. Theorem 1. a. For a constant a ∈ R, all solutions y(t) to the linear differential equation y0 = ay are of the form y(t) = Ceat , where C is an arbitrary constant. a(t−t0) b. Given an initial condition (t0, y0), there is exactly one solution y(t) = y0e to 0 y = ay with y(t0) = y0. Proof. (a) It is very easy to check that y(t) of the form Ceat are solutions. (Do it!) We shall show that there are no other solutions. Let y(t) be an arbitrary solution of our equation. We compare this unknown solution y(t) to the known solution eat by defining z(t) = e−at y(t). Then z0(t) = −ae−at y(t) + e−at y0(t) = −ae−at y(t) + e−at ay(t) = 0, since y0(t) = ay(t). Therefore z(t) has to be a constant. Denoting the constant by C, we have that C = e−at y(t), or y(t) = Ceat for all t. (b) Suppose that we want to find a solution y(t) = Ceat that satisfies the initial condition at0 −at0 (t0, y0), i.e., y0 = y(t0) = Ce and C = y0e . Thus, our solution is −at0 at a(t−t0) y(t) = y0e e = y0e . Problem 1. Find the solution of the differential equation y0 = 2y, which satisfies the following initial conditions: a. y(0) = 0; b. y(0) = −1; c. y(2) = 3. Problem 2. Assume that a : R → R is a continuous function. Imitate the proof of Theorem 0 1 to show that the solution y(t) to the differential equation y = a(t)y with y(t0) = y0 is b(t) R t 0 given by y(t) = y0e , where b(t) = a(s) ds. Hint: b (t) = a(t). t0 1 2 Problem 3. Find the solution of the differential equation y0 = 2ty, which satisfies the initial condition y(0) = 1. Problem 4. Find the solution of the differential equation y0 = cos(t) y, which satisfies the − initial condition y( π/2) = −1. Theorem 2. Let a, g : R → R be continuous functions. Then the solution of the nonho- 0 mogeneous differential equation y = a(t)y + g(t) with y(t0) = y0 is given by Z t b(t) b(t) −b(s) (1) y(t) = y0 e + e e g(s) ds. t0 where b(t) = R t a(s) ds. t0 Proof. Let y(t) be the solution with y(t0) = y0. We use the the solution of the correspond- ing homogeneous equation found in Problem 2, to form what is called an integrating factor −b(t) −b(t) b(t0) e . We define z(t) = e y(t). Note that b(t0) = 0 so e = 1 and z(t0) = y0. Also, z0(t) = −e−b(t)b0(t)y(t) + e−b(t) y0(t) = −e−b(t)a(t)y(t) + e−b(t) [a(t)y(t) + g(t)] = e−b(t)g(t). Since g(t) and a(t) are given, and b(t) is determined by integration, we know the derivative of z(t). Integrating from t0 to t, Z t −b(s) z(t) − z(t0) = e g(s) ds and t0 Z t b(t) b(t) −b(s) y(t) = e z(t) = e y0 + e g(s) ds , t0 b(t) where we used the fact that z(t0) = y0. Notice that (i) y0 e is solution of the associated b(t) R t −b(s) homogeneous equation with y(t0) = y0 given in Problem 2 and (ii) e e g(s) ds is t0 a particular solution of the nonhomogeneous equation with initial condition (t0, 0). Example 1. An initial amount of money y0 is put in an account on which interest is compounded continuously at a rate of r > 0, for an effective annual rate of er . (For r = 0.05, the effective rate is 0.0513.) Assume that money is added continuously to the account at the rate of g. The differential equation governing the amount of money in the account is given by y0 = r y + g. The solution is determined as follows: a(t) = r, b(t) = ert , and Z t rt rt rs y(t) = y0 e + e e g ds 0 t rt rt h g rs = y0 e + e − e r 0 g = y ert + ert 1 − e rt 0 r h g i g = y + ert − . 0 r r 3 In the long term, the amount in the account approaches the amount of an initial deposit of g y + with no money added. 0 r Problem 5. Find the solution of the differential equation y0 = 2ty + t, which satisfies the condition y(0) = 0. Problem 6. For k = 2, 1, 0, 1, 2 find the solution (for t > 0) of the differential equation 0 y = y/t + t, which satisfies the initial condition y(1) = k. Graph these solutions in the (t, y) plane. What happens to these solutions when t → 0? Notice that at t = 0 the function a(t) = 1/t is undefined! Problem 7. Consider the nonhomogeneous linear equation (NH) y0 = a(t)y + g(t) with the associated homogeneous linear equation (H) y0 = a(t)y. a. If y p(t) is one (particular) solution of the nonhomogeneous equation (NH) and yh(t) is a solution of the homogeneous equation (H), show that y p(t) + yh(t) is a solution of the nonhomogeneous equation (NH). b. Assume that y p(t) is a particular solution of the nonhomogeneous equation (NH). Show that the general solution of the nonhomogeneous equation (NH) is y p(t) + Ceb(t) where b(t) is given as in Problem 2 and C is an arbitrary constant. Hint: For any solution y(t) of (NH), show that y(t) − y p(t) satisfies (H). Example 2 (Logistic equation). We consider one nonlinear differential equation, the so called logistic differential equation, y y0 = ry 1 − , K where K > 0 is a given constant. There are two constant solutions, y1(t) ≡ 0 and y2(t) ≡ K . To find solutions with y(0) 6= 0, K , this equation is solved by the method of separation of variables, that converts it into a problem of integrals, and then we use partial fractions: K y0 = r y (K − y) y0 y0 + = r. y K − y Integrating with respect to t, the term y0 dt changes it to and integral with respect to y: Z 1 Z 1 Z dy + dy = r dt y K − y ln(|y|) − ln(|K − y|) = rt + C1 |y| = C ert , where C = eC1 . |K − y| 2 2 Assuming 0 < y < K so we can drop the absolute value signs, we solve for y: rt rt y = C2 K e − C2e y rt rt (1 + C2e )y = C2 K e C K ert K y = 2 = , rt −rt (1 + C2e ) Ce + 1 4 1 where C = /C2. If y0 is the initial condition at t = 0, then some more algebra shows that (K − y0) C = /y0 so y K y(t; y ) = 0 . 0 −rt y0 + (K − y0)e It can be shown that this form of the solution is valid for any y0 and not just those with 0 < y0 < K . For the logistic equation, a solution y(t; y0) for an initial condition 0 < y0 < K has 0 0 y > 0 and it increases toward K . Also, for y0 > K , y < 0, and the solution y(t; y0) decrease toward K . So we can conclude, even without solving the differential equation, that for any y0 > 0 the solution y(t; y0) tends toward K as t goes to infinity. For this reason, K is called the carrying capacity. For y0 < 0, the denominator becomes 0 for = (K − y0) ∞ t1 ln /( y0) , and the solution goes to as t goes to t1. Remark. Notice that the solution y(t) of a linear differential equation or a nonhomoge- neous differential equation is equal to an expression found by means of integrals. A nonlinear equation of the form y0 = f (y) is solved by separation of variables and then integrals.