MATH 308 Differential Equations

Transcription

1 MATH 308 Differential Equations Summer, 2014, SET 5 JoungDong Kim Set 5: Section 3.1, 3.2 Chapter 3. Second Order Linear Equations. Section 3.1 Homogeneous Equations with Constant Coefficients. In this chapter, we will study second order linear ODE. In particular, we will focus on the solutions of second order linear ODE with constant coefficients. A second order linear ODE has the following form, If we devide P(t) on both sides, we obtain a more practical form, P(t)y +Q(t)y +R(t)y = G(t) (1) y +p(t)y +q(t)y = g(t), (2) where p(t) = Q(t) R(t) G(t), q(t) =,and g(t) = P(t) P(t) P(t). Definition 0.1 (Homogeneous). An equation is said to be homogeneous if the term g(t) in Eq. (2), or the term G(t) in Eq. (1) is zero for all t. Otherwise the equation is called nonhomogeneous. In this section, we study homogeneous equation with constant coefficients only. That is G(t) = 0, and P(t), Q(t), R(t) are constant in Eq. (1). Or we may write the equations as ay +by +cy = 0. (3) 1

2 Initial Value Problem To define an initial value problem for second order ODE, we need to specify two point values as follows; y(t 0 ) = y 0, y (t 0 ) = y 1. In general, an initial value problem of nth order ODE consists of a nth order equation with n initial conditions. Solve the Eq. (3). (Use Characteristic equation) 2

3 Ex1) Solve the initial value problem y +5y +6y = 0, y(0) = 0, y (0) = 1. 3

4 General Solution If the characteristic equation has two different real roots r 1, and r 2, then the general solution of Eq. (3) is y(t) = c 1 e r 1t +c 2 e r 2t. For initial value problem, we can use two initial conditions to determine the coefficients c 1 and c 2. Ex2) Solve the initial value problem 4y 8y +3y = 0, y(0) = 2, y (0) =

5 Behavior of solutions when t increases. Ex3) Find the general solution of the equation, y y = 0. Discuss the behavior of the solutions as t increases. 5

6 Section 3.2 Solutions of linear homogeneous equations: the Wronskian Let s talk about Existence and Uniqueness. And also if we can find more solutions, let s discuss how we can find the other solutions. Theorem 0.2 (Existence and Uniqueness theorem). Consider the initial value problem, y +p(t)y +q(t)y = g(t), y(t 0 ) = α, y (t 0 ) = β. If p,q, and g are continuous on an open interval I that contains the point t 0, then there is exactly one solution y = φ(t) solves the initial value problem, and it is defined throughout the whole interval I. Ex4) Find the longest interval in which the solution of the initial value problem, (t 2 3t)y +ty (t+3)y = 0, y(1) = 2, y (1) = 1. 6

7 Theorem 0.3 (Principle of Superposition). If y 1,y 2 are two solutions of a homogeneous ODE: y +p(t)y +q(t)y = 0, (4) then the linear combination c 1 y 1 +c 2 y 2 is also a solution for any values of the constant c 1 and c 2. Thm. 0.3 tells us that if we have two solutions, we can generate infinitely many solutions as called general solution. y = c 1 y 1 +c 2 y 2, 7

8 Theorem 0.4. If y 1,y 2 are two solutions of a homogeneous ODE; y +p(t)y +q(t) = 0, for any given initial condition; y(t 0 ) = α, y (t 0 ) = β, it is always possible to find the constant c 1,c 2 so that y = c 1 y 1 +c 2 y 2, solves the initial value problem if and only if the Wronskian: is NOT ZERO at t 0. W(y 1,y 2 ) := y 1 y 2 y 1 y 2 8

9 The next theorem justifies the term general solution we ve used already in the previous section. Theorem 0.5. Suppose that y 1 and y 2 are two solutions of the equation y +p(t)y +q(t)y = 0, then the family of solutions y = c 1 y 1 +c 2 y 2 with arbitrary coefficients c 1 and c 2 includes every solution of Eq. (4) if and only if there is a point t 0 where the Wronskian of y 1 and y 2 is not zero. If this is the case, we call y 1 and y 2 form a fundamental set of solutions of the equation. Ex5) Show that y 1 = t 1/2 and y 2 = t 1 form a fundamental set of solutions of 2ty +3ty y = 0. 9

10 In several cases we have been able to find a fundamental set of solutions, and therefore the general solution, of a given differential equation. However, this is often a difficult task, and the question arises as to whether a differential equation of the form of Eq. (4) always has a fundamental set of solutions. The following theorem provides an affirmative answer to this question. Theorem 0.6. Consider the differential equation, y +p(t)y +q(t) = 0, whose coefficients p and q are continuous on some open interval I. Choose some point t 0 in I. Let y 1 be the solution of Eq. (4) that satisfies the initial conditions, y(t 0 ) = 1, y (t 0 ) = 0, and let y 2 be the solution of Eq. (4) that satisfies the initial conditions y(t 0 ) = 0, y (t 0 ) = 1. Then y 1 and y 2 form a fundamental set of solution of Eq. (4). Summary Consider a general second order linear homogeneous differential equation, y +p(t)y +q(t)y = 0. To find all solutions of the equation, we need to find two specific solutions y 1 and y 2 such that their Wronskian W(y 1,y 2 ) = y 1 y 2 y 1 y 2 is none zero at some point t 0. Then all the solutions (the general solution) can be written as y = c 1 y 1 +c 2 y 2. y 1 and y 2 form a fundamental set of solutions. For an arbitrary initial condition, we can determine a unique pair c 1 and c 2 to solve the initial value problem. 10