Review of Lecture 9 Existence and Uniqueness

Transcription

1 Review of Lecture 9 Existence and Uniqueness We consider y = f (x, y) with a given initial condition y(x 0 ) = y 0. There is a solution passing through (x 0, y 0 ). It is defined on some interval (a, b) with a < x 0 < b. The Picard iteration is the sequence y n+1 = x 0 f (x, y n (x)) dx The mapping from y n to y n+1 is a contraction mapping That, plus the bounded convergence theorem, is enough to prove the convergence.

2 Steps of the existence proof y(x) = lim n(x) n = lim n+1(x) n = lim = = = n x 0 x 0 x 0 x 0 f (x, y n (x)) dx lim f (x, y n(x)) dx n f (x, lim n y n(x)) dx f (x, y(x)) dx Hence y solves the integral equation, and hence the differential equation. Also y(0) = 0 so the initial condition is satisfied.

3 Initial value problems For a first order equation, we ask for a solution y such that y(t 0 ) is a specified value y 0.

4 Initial value problems For a first order equation, we ask for a solution y such that y(t 0 ) is a specified value y 0. We call the specification that y(t 0 ) = y 0 an initial condition.

5 Initial value problems For a first order equation, we ask for a solution y such that y(t 0 ) is a specified value y 0. We call the specification that y(t 0 ) = y 0 an initial condition. The general solution has one constant in it, that can be chosen to arrange that the initial condition is satisfied.

6 Initial value problems For a first order equation, we ask for a solution y such that y(t 0 ) is a specified value y 0. We call the specification that y(t 0 ) = y 0 an initial condition. The general solution has one constant in it, that can be chosen to arrange that the initial condition is satisfied. For a second order equation, there will be two constants of integration. So we get to specify two initial values: y(t 0 ) and y (t 0 ). An initial value problem is given by a second-order equation, together with specified initial values for y and y. Geometrically we ask for the solution to pass through a specified point with a specified slope. For a first order equation, we don t get to specify the slope. The equation determines it.

7 Existence for second order linear equations The basic existence theorem says that a second-order initial value problem has a solution defined in an interval about t 0. The interval is not just some interval, it is known. It extends as long as p, q, and g and their derivatives are defined and continuous, for the equation For the equation y + p(t)y + q(t)y = g(t) P(t)y + Q(t)y + R(t)y = G(t) it may also be stopped by the zeroes of P.

8 Homogeneity The equation P(t)y + Q(t)y + R(t)y = 0 is called homogeneous. What makes it homogeneous is that the right side is zero. The equation P(t)y + Q(t)y + R(t)y = G(t) is called non-homogeneous when G is not identically zero.

9 Homogeneity The equation P(t)y + Q(t)y + R(t)y = 0 is called homogeneous. What makes it homogeneous is that the right side is zero. The equation P(t)y + Q(t)y + R(t)y = G(t) is called non-homogeneous when G is not identically zero. For example y y = sin(t)

10 Homogeneity The equation P(t)y + Q(t)y + R(t)y = 0 is called homogeneous. What makes it homogeneous is that the right side is zero. The equation P(t)y + Q(t)y + R(t)y = G(t) is called non-homogeneous when G is not identically zero. For example y y = sin(t) is non-homogeneous and the corresponding homogeneous equation is y y = 0

11 Uniqueness for second order linear equations The solution to an initial-value problem is unique. If u and v are two solutions, then the difference y = u v satisfies a homogeneous equation with the initial values y 0 = 0 and y 0 = 0. So it suffices to prove uniqueness for the solution of homogeneous equations We won t prove this uniqueness today

12 Linear combinations A linear combination of u and v is a function of the form au + bv with a and b constant. For example, 3 sin t + 2t is a linear combination of sin t and t.

13 The superposition principle If u and v are two solutions of a homogeneous linear differential equation, then any linear combination y = au + bv is also a solution.

14 The superposition principle If u and v are two solutions of a homogeneous linear differential equation, then any linear combination y = au + bv is also a solution. P(t)y + Q(t)y + R(t)y = P(t)(au + bv) + Q(t)(au + bv) + R(t)(au + bv)

15 The superposition principle If u and v are two solutions of a homogeneous linear differential equation, then any linear combination y = au + bv is also a solution. P(t)y + Q(t)y + R(t)y = P(t)(au + bv) + Q(t)(au + bv) + R(t)(au + bv) = P(t)(au + bv ) + Q(t)(au + bv ) + R(t)(au + bv)

16 The superposition principle If u and v are two solutions of a homogeneous linear differential equation, then any linear combination y = au + bv is also a solution. P(t)y + Q(t)y + R(t)y = P(t)(au + bv) + Q(t)(au + bv) + R(t)(au + bv) = P(t)(au + bv ) + Q(t)(au + bv ) + R(t)(au + bv) = ap(t)u + aq(t)u + ar(t)u + bp(t)v + bq(t)v + br(t)v

17 The superposition principle If u and v are two solutions of a homogeneous linear differential equation, then any linear combination y = au + bv is also a solution. P(t)y + Q(t)y + R(t)y = P(t)(au + bv) + Q(t)(au + bv) + R(t)(au + bv) = P(t)(au + bv ) + Q(t)(au + bv ) + R(t)(au + bv) = ap(t)u + aq(t)u + ar(t)u + bp(t)v + bq(t)v + br(t)v = = 0

18 The superposition principle If u and v are two solutions of a homogeneous linear differential equation, then any linear combination y = au + bv is also a solution. P(t)y + Q(t)y + R(t)y = P(t)(au + bv) + Q(t)(au + bv) + R(t)(au + bv) = P(t)(au + bv ) + Q(t)(au + bv ) + R(t)(au + bv) = ap(t)u + aq(t)u + ar(t)u + bp(t)v + bq(t)v + br(t)v = = 0 It was important that the equation was homogeneous, otherwise the last step wouldn t work. This principle only works for linear homogeneous equations, and it s very very useful. That is one reason linear equations are so important.

19 An example Consider the equation y y = 0. It has two solutions, e t and e t. Hence y = Ae t + Be t is also a solution for every A and B.

20 An example Consider the equation y y = 0. It has two solutions, e t and e t. Hence y = Ae t + Be t is also a solution for every A and B. We can solve any initial-value problem with t 0 = 0 by choosing A and B correctly: y 0 = Ae 0 + Be 0 = A + B y 0 = Ae 0 Be 0 = A B Adding, we have 2A = y 0 + y 0, and subtracting, 2B = y 0 y 0.

21 We showed that any initial value problem for y y = 0 can be solved using a linear combination of e t and e t.

22 We showed that any initial value problem for y y = 0 can be solved using a linear combination of e t and e t. Therefore, by the uniqueness theorem, every solution is a linear combination of e t and e t, since if y is any solution, there is a linear combination of e t and e t solving the same initial value problem, so y must be that linear combination by uniqueness.

23 We showed that any initial value problem for y y = 0 can be solved using a linear combination of e t and e t. Therefore, by the uniqueness theorem, every solution is a linear combination of e t and e t, since if y is any solution, there is a linear combination of e t and e t solving the same initial value problem, so y must be that linear combination by uniqueness. We call the set {e t, e t } a fundamental set of solutions. We solve a homogeneous equation by finding a fundamental set of solutions.

24 Greek Lesson 3 Lambda is λ. Its use in the next few slides is traditional, although the textbook uses the Latin letter r instead. We already learned µ, which is mu.

25 The case of constant coefficients Consider the equation ay + by + cy = 0. Look for a solution y = e λt. Then y = λe λt y = λ 2 e λt ay + by + cy = e λt (aλ 2 + bλ + c) Hence we ll find a solution if and only if λ satisfies the characteristic equation aλ 2 + bλ + c = 0 or in other words is a root of the characteristic polynomial aλ 2 + bλ + c.

26 The characteristic polynomial ay + by + cy = 0. has characteristic polynomial aλ 2 + bλ + c. Since it s a quadratic polynomial, it always has complex roots. It might have two distinct real root. It might have one double root, which will be real (we assume a, b, c are real). It might have two non-real roots (which will be complex conjugates).

27 The case of two distinct real roots Let λ and µ be two unequal roots of the characteristic polynomial. Then e λt and e µt are a fundamental set of solutions. That is, every solution is a linear combination of these two. As above, it suffices (by uniqueness) to show that any initial value problem can be solved using a linear combination of these two solutions.

28 The case of two distinct real roots Let λ and µ be two unequal roots of the characteristic polynomial. Then e λt and e µt are a fundamental set of solutions. That is, every solution is a linear combination of these two. As above, it suffices (by uniqueness) to show that any initial value problem can be solved using a linear combination of these two solutions. If you just do the algebra, you find λ µ in the denominator. That s why we need λ µ for this to work. This algebra is on page 141, and I ll do it on the whiteboard in class.

29 The rest of class time was spent solving even-numbered homework problems.