First Order Initial Value Problems

Transcription

1 First Order Initial Value Problems A first order initial value problem is the problem of finding a function xt) which satisfies the conditions x = x,t) x ) = ξ 1) where the initial time,, is a given real number, the initial position, ξ IR d, is a given vector and : IR d IR IR d is a given function. We shall assume throughout these notes that is C. By definition, a solution to the initial value problem 1) on the interval I which may be open, closed or half open, but which, of course, contains ) is a differentiable function xt) which obeys x ) = ξ and xt) = xt),t ) for all t I Remark 1 The restriction to first order derivatives is not significant. Higher order systems can always be converted into first order systems, at the cost of introducing more variables. For example, substituting y = x z = x into the second order system x = x, x,t) converts it to the first order system d dt [ ] [ y = z ] z F y, z,t) Theorem 2 Let d IN, IR, ξ IR d and I be an interval in IR that contains. Assume that : IR d IR IR d is C. a) Regularity in t) A solution to the initial value problem 1) on I is C. b) Uniqueness) If xt) and yt) are both solutions to the initial value problem 1) on I, then xt) = yt) for all t I. Remark: For uniqueness, it is not sufficient to assume that is continuous. For example xt) = and xt) = 2 3 t) 3/2 both solve the initial value problem x) =, ẋ = 3 x. c) Local Existence) Le < ρ < and let C ρ ξ ) denote the closed ball in IR d with centre ξ and radius ρ. Then there exists a T > such that the initial value problem 1) has a c Joel Feldman. 27. All rights reserved. September 6, 27 First Order Initial Value Problems 1

2 solution xt) on the interval [ T, +T] with xt) C ρ ξ ) for all t [ T, +T]. Furthermore, if we fix any θ > and define M = sup { F x,t) x Cρ ξ ), t θ which is the maximum speed that x can have while t θ and x is in C ρ ξ )), then T min { θ, ρ M Remark: The bad news is that solutions of 1) need not exist for all t. For example, ẋ = x 2, x) = 1 has solution xt) = 1, which blows up at t = 1, despite the fact that t 1 the problem ẋ = x 2, x) = 1 shows no sign of pathology. Another example is ẋ = x 2 x) = 1 ẏ = x 2 cosx x 3 sinx y) = cos1 The solution is xt) = 1 t 1, yt) = 1 t 1 cos 1, which oscillates wildly while it is blowing t 1 up. The good news is that solutions, which we already know exist at least for times near, can only fail to globally exist i.e. exist for all t) under specific, known, circumstances, that can often be ruled out. d) Absence of a Global Solution) Let a < b with at least one of a, b finite. Let xt) be a solution of 1) on the interval a,b). Suppose that there does not exist a solution on any interval that properly contains a, b). Then b < = limsup xt) = t b a > = limsup xt) = t a+ Remark: It suffices to consider open intervals a, b), because given a solution on some nonopen interval I, we can always use part c) to extend it to a solution on an open interval that contains I. e) Global Existence) Suppose that a function H x) is conserved by all solutions of 1). In other words, suppose that if xt) obeys 1), then d dth xt)) =. Suppose further that lim H x) =. Then every solution of 1) on an interval properly contained in IR may x be extended to a global solution. f) Regularity with respect to initial conditions and other parameters) Let F x,t, α) be C. Let θ,ρ,a >. Then there is a T > such that i) For each τ, ζ and α obeying τ < θ, ξ ξ < ρ and α α < a, there is a solution xt; τ, ξ, α) to the initial value problem x = F x,t, α) xτ) = ξ on the interval [τ T,τ +T]. ii) xt; τ, ξ, α) is C as a function of all of its arguments in the region given in i). c Joel Feldman. 27. All rights reserved. September 6, 27 First Order Initial Value Problems 2

3 Proof: a) Let xt) be a solution to the initial value problem 1) on I. By hypothesis, it is differentiable on I. Hence it is continuous on I. As is continuous, xt) = xt),t) is a composition of continuous functions, and hence is continuous. That is, xt) is C 1 on I. So xt) = xt),t) is also C 1 with derivative d j=1 d F xt),t) dx j dt t)+ F t xt),t) = F F j xt),t)+ F t xt),t) j=1 which is again continuous. So xt) is C 2 on I. And so on. b) Let J be any closed finite interval containing and contained in I. We shall show that xt) = yt) for all t J. Since J is compact and xt) and yt) are continuous, both xt) and yt) are bounded for t J. Let S be a sphere in IR d that is sufficiently large that it contains xt) and yt) for all t J. Define and, for each T >, let Since x ) = y ), { d M S = sup i,j=1 F i x,t) x S, t J MT) = sup { xt) yt) t t T t ] xt) yt) = [ xτ) yτ) dτ [ ) )] = F xτ),τ F yτ),τ dτ Now, for any x, y S and τ J, x,τ ) y,τ ) = = d j,k=1 d dσf σ x+1 σ) y,τ ) dσ d ) x j y j ) xjf) σ x+1 σ) y,τ dσ j=1 x j y j xj F k σ x+1 σ) y,τ ) dσ x y M S dσ = M S x y 2) c Joel Feldman. 27. All rights reserved. September 6, 27 First Order Initial Value Problems 3

4 Hence, for all t J, t xt) yt) M S xτ) yτ) dτ If t J and t T, the same is true for all τ between and t so that t t xt) yt) M S xτ) yτ) dτ M S MT) dτ = t M S MT) TM S MT) Taking the supremum over t J with t T gives MT) TM S MT) If T 1, we have MT) 1 2 MT) which forces Mt) = for all t J with t T. 1 If isnot sufficiently big that [ 1, + 1 ] covers all ofj, repeat with replaced by ± 1 and then ±2 1, and so on. c) We shall construct a local solution to the integral equation xt) = ξ + xτ),τ ) dτ 3) which is equivalent to the initial value problem 1). To do so, we construct a sequence of approximate solutions by the algorithm x ) t) = ξ x 1) t) = ξ + x 2) t) = ξ +. x n) t) = ξ + x ) τ),τ ) dτ x 1) τ),τ ) dτ x n 1) τ),τ ) dτ 4) Ifthesequence x n) t)ofapproximatesolutionsconvergesuniformlyonsomeclosedinterval [ T, +T], then taking the limit of 4) as n shows that xt) = lim n xn) t) obeys 3) and hence 1). Define T = min { θ, ρ M { d L = sup i,j=1 F i x,t) x Cρ ξ ), t T µ k = sup{ x k) t) x k+1) t) t t T c Joel Feldman. 27. All rights reserved. September 6, 27 First Order Initial Value Problems 4

5 It suffices to prove that µ k <, since then we will have uniform convergence. k= We first verify that x n) t) remains in C ρ ξ ) for all n and t T. Since MT ρ, we have sup x n) t) t ξ sup F x n 1) τ),τ ) dτ t T t T T sup x n 1) τ),τ ) τ T TM ρ if x n 1) τ) C ρ ξ ) for all τ T and it follows, by induction on n, that x n) t) ξ ρ i.e. x n) t) C ρ ξ )) for all t T. Now x k) t) x k+1) t) = as in 2). Iterating this bound gives x k) t) x k+1) t) L 2 dτ 1. L k = L k L k [ F x k 1) τ),τ ) F x k) τ),τ )] dτ L x k 1) τ 1 ) x k) τ 1 ) dτ1 dτ 1 dτ 1 dτ 1 dτ 2 x k 2) τ 2 ) x k 1) τ 2 ) dτ 2 dτ 2 = L k ρ 1 k! t k dτ 2 τk 1 τk 1 τk 1 dτ k x ) τ k ) x 1) τ k ) dτ ξ k x 1) τ k ) dτ k ρ and hence µ k ρ 1 k! LT)k = ρe LT k= k= d) This is easy. If the solution stays bounded near one end of the interval, say on [b ε,b) with b <, we can use the local existence result of part c) to extend the solution past b simply by choosing very close to b. c Joel Feldman. 27. All rights reserved. September 6, 27 First Order Initial Value Problems 5

6 e) This is also easy. If h = H ξ ), then any solution xt) must remain in Since lim x X h = { x H x) = h H x) =, the region Xh is bounded. Apply part d). f) Outline of proof only.) The idea of the proof is the same as that for part c) local existence). For each fixed τ, ξ, α), the solution xt; τ, ξ, α) is the limit of the sequence defined recursively by x ) t; τ, ξ, α) = ξ x n) t; τ, ξ, α) = ξ + τ F x n 1) τ ; τ, ξ, α),τ, α ) dτ n = 1, 2, 3, 5) As in part a), it is obvious by induction on k that x k) t; τ, ξ, α) is C in all of its arguments. To prove that the limit as k is C, it suffices to prove that all derivatives of x k) converge uniformly as k. So it suffices to prove that for any possibly higher order) derivative D, µ D,k = sup D x k) D x k+1) obeys µ D,k <. But we can bound µ D,k by applying D to the recursion relation 5) k= and then using the method of part c). c Joel Feldman. 27. All rights reserved. September 6, 27 First Order Initial Value Problems 6