MATH 312 Section 1.2: Initial Value Problems

Transcription

1 MATH 312 Section 1.2: Initial Value Problems Prof. Jonathan Duncan Walla Walla College Spring Quarter, 2007

2 Outline 1 Introduction to Initial Value Problems 2 Existence and Uniqueness 3 Conclusion

3 Families of Solutions As we saw in our examples last time, most particular solutions to a DE come from a family of solutions which contains one or more parameters. One way to eliminate these parameters is to specify initial conditions. Initial Value Problems An initial value problem (IVP) is a differential equation together with one or more initial conditions on the dependent variable which can be used to find a particular solution from a family of solutions. As we shall see in the next slide, the initial conditions we must specify vary depending on the order of the DE.

4 General Initial Value Problems We start with the simplest initial value problem, the first order IVP. First Order IVP A general first order IVP asks us to solve dy dx = f (x, y) for x I subject to the initial condition y(x 0 ) = y 0 for x 0 I. Second Order IVP A general second order IVP asks us to solve d2 y = f (x, y, y ) for dx 2 x I subject to the initial conditions y(x 0 ) = y 0 and y (x 0 ) = y 1 for x 0 I. nth Order IVP A general nth order IVP asks us to solve dn y dx n = f (x, y, y,..., y (n) ) for x I subject to the initial conditions y(x 0 ) = y 0, y (x 0 ) = y 1,..., y (n 1) (x 0 ) = y n 1 for x 0 I.

5 Initial Value Solutions Now that we ve defined an IVP, let s consider several examples. Solve 2y + y = 0 subject to y(0) = 4 and then solve it subject to y(2) = 1. Note: First verify that y = Ce x/2 is a family of solutions to this DE on (, ). Verify that y = C 1 x 1 + C 2 x + C 3 x ln x + 4x 2 is a family of solutions to x 3 d3 y + 2x 2 d2 y x dy dx 3 dx 2 dx + y = 12x 2 and find a particular solution subject to y(0) = 2, y (0) = 1, and y (0) = 0. Problem! We forgot to mention the interval of the solution, which in this case is I = (0, ) and x 0 = 0 is not in this I.

6 Questions of Existence The previous example brings up a fundamental question in solving IVPs. Existence Does the DE dy dx = f (x, y) have a solution? If it does, do any of these solutions meet the initial conditions (i.e. pass through the point (x 0, y 0 ) so that y(x 0 ) = y 0 is satisfied)? Revisiting the last example, the DE does have a family of solutions. However, none of them pass through the point (0, 2), or for that matter, none have a derivative y passing through (0, 1).

7 Questions of Uniqueness Another fundamental question we can ask about an IVP is if there is a solution, is it unique? Consider the following example. ( ) Find a solution to dy dx = 2 y 1 x subject to y(0) = 1, if it exists. on the interval I = (, ) Note that both y = 3x and y = 1 are solutions to the DE and satisfy the initial condition that y(0) = 1. Uniqueness Question The question we must now ask ourselves, is when can we be sure that there is exactly one solution curve to a given DE passing through the point (x 0, y 0 )?

8 The Existence and Uniqueness Theorem This question is answered by the following important theorem. Theorem 1.1 Let R be a rectangular region in the xy-plane defined by a x b and c y d that contains the point (x 0, y 0 ) in its interior. If f (x, y) and f y are both continuous on R, then there exists some interval I 0 given by (x 0 h, x 0 + h) for h > 0 contained in [a, b], and a unique function y(x) defined on I 0, that is a solution to the first order initial value problem. Applying this to the last example: ( ) y 1 f (x, y) = 2 x f y = 2 x So that both are discontinuous at x = 0 so the theorem fails, and we are not guaranteed a unique solution through (0, 1).

9 Applying the Theorem Let s apply this theorem to another example. Determine if the DE (4 y 2 )y = x 2 has a unique solution to the initial value problem subject to: 1 y(0) = 2 2 y(2) = 1 First, note that: so that the solutions are: f (x, y) = x 2 4 y 2 f y = 2yx 2 (4 y 2 ) 2 1 no, since both are discontinuous when y = 2 2 yes, since both are continuous when y = 1 and x = 2.

10 More about Existence-Uniqueness Theorem There are several important things which should be pointed out about this theorem. Notes about Theorem 1.1 If the conditions of the theorem are met, then a unique function is guaranteed only on some subinterval of [a, b]. The continuity of f (x, y) is what guarantees the existence of a solution. The continuity of f y is what guarantees the uniqueness of the solution. This theorem is not an if and only if statement. That is, even if the conditions of the theorem are not met, it is possible that there is a unique solution to the IVP.

11 Important Concepts Things to Remember from Section Finding Solutions to an IVP given a family of solutions 2 Stating the Existence and Uniqueness Theorem 3 Applying the Existence and Uniqueness Theorem