Chapter 19 Ordinary Dierential Equations 19.1 Classes of ODE In this chapter we will deal with the class of functions satisfying the following ordinary dierential equation {˙ (w)=i (w> { (w)) for almost all w 5 [w0>w0 + ] { (w0)={0 (19.1) i : R ×X $ X where i is a nonlinear function and X is a Banach space (any concrete space of functions). The Cauchy’s problem for (19.1) consists in re- solving of (19.1), or, in other words, in finding a function { (w) which satisfies (19.1). For the simplicity we also will use the following abbreviation: - ODE meaning an ordinary dierential equation, - DRHS meaning the discontinuous right-hand side. Usually the following three classes of ODE (19.1) are considered: 1) Regular ODE : 571 572 Chapter 19. Ordinary Dierential Equations i (w> {) is continuous in both variables. In this case { (w),satis- fying (19.1), should be continuous dierentiable, i.e., 1 { (w) 5 F [w0>w0 + ] (19.2) 2) ODE of the Caratheodory’s type: i (w> {) in (19.1) is measurable in w and continuous in {. 3) ODE with discontinuous right-hand side: i (w> {) in (19.1) is continuous in w and discontinuous in {.In fact, this type of ODE equation is related to the dierential inclusion: {˙ (w) 5 I (w> { (w)) (19.3) where I (w> {) is a set in R ×X. If this set for some pair (w> {) consists of one point, then I (w> {)=i (w> {). 19.2 Regular ODE 19.2.1 Theorems on existence Theorem based on the contraction principle Theorem 19.1 (on local existence and uniqueness) Let i (w> {) be a continuous in w on [w0>w0 + ], ( 0) and for any w 5 [w0>w0 + ] it satisfies the, so-called, local Lipschitz condition in {,thatis, there exists constant f> Oi A 0 such that ki (w> {)k f (19.4) ki (w> {1) i (w> {2)k Oi k{1 {2k for all w 5 [w0>w0 + ] and all {> {1>{2 5 Eu ({0) where Eu ({0):={{ 5 X|k{ {0k u} Then the Cauchy’s problem (19.1) has a unique solution on the time- interval [w0>w0 + 1],where © ª 31 1 ? min u@f> Oi > (19.5) 19.2. Regular ODE 573 Proof. 1) First, show that the Cauchy’s problem (19.1) is equivalent to finding the continuos solution to the following integral equation Zw { (w)={0 + i (v> { (v)) gv (19.6) v=w0 Indeed, if { (w) is a solution of (19.1), then, obviously, it is a dieren- tiable function on [w0>w0 + 1]. By integration of (19.1) on [w0>w0 + 1] we obtain (19.6). Inverse, suppose { (w) is continuos function satisfying (19.6). Then, by the assumption (19.4) of the theorem, it follows ki (v> { (v)) i (v0>{(v0))k = k[i (v> { (v)) i (v> { (v0))] + [i (v> { (v0)) i (v0>{(v0))]k ki (v> { (v)) i (v> { (v0))k + ki (v> { (v0)) i (v0>{(v0))k Oi k{ (v) { (v0)k + ki (v> { (v0)) i (v0>{(v0))k This implies that if v> v0 5 [w0>w0 + 1] and v $ v0, then the right hand-side of the last inequality tends to zero, and, hence, i (v> { (v)) is continuous at each point of the interval [w0>w0 + 1].And,moreover, we also obtain that { (w) is dierentiable on this interval, satisfies (19.1) and { (w0)={0. 2) Using this equivalence, let us introduce the Banach space F [w0>w0 + 1] of abstract continuous functions { (w) with values in X and with the norm k k k k { (w) F := max { (w) X (19.7) wM[w0>w0+1] Consider in F [w0>w0 + 1] the ball Eu ({0) and notice that the nonlinear operator x : F [w0>w0 + 1] $ F [w0>w0 + 1] defined by Zw x ({)={0 + i (v> { (v)) gv (19.8) v=w0 574 Chapter 19. Ordinary Dierential Equations transforms Eu ({0) in to Eu ({0) since ° ° ° Zw ° ° ° ° ° kx ({) {0k =max° i (v> { (v)) gv° wM[w0>w0+1] ° ° v=w0 Zw max ki (v> { (v))k gv 1f?u wM[w0>w0+1] v=w0 Moreover, the operator x is a contraction (see Definition 14.20) on Eu ({0). Indeed, by the local Lipschitz condition (19.4), it follows ° ° ° Zw ° ° ° ° ° kx ({1) x ({2)k =max° [i (v> {1 (v)) i (v> {2 (v))] gv° wM[w0>w0+1] ° ° v=w0 Zw max ki (v> {1 (v)) i (v> {2 (v))k gv wM[w0>w0+1] v=w0 k k k k 1Oi {1 {2 F = t {1 {2 F where t := 1Oi ? 1 for small enough u. Then, by Theorem (the contractionprinciple)14.17,weconcludethat(19.6)hasaunique solution { (w) 5 F [w0>w0 + 1]. Theorem is proven. Corollary 19.1 If in the conditions of Theorem 19.1 the Lipschitz condition (19.4) is fulfilled not locally, but globally, that is for all {1>{2 5 X (that corresponds with the case u = 4), then the Cauchy’s problem (19.1) has a unique solution for [w0>w0 + ] for any big enough. Proof. It directly follows from Theorem 19.1 if take u $4.But here we prefer to present also another proof based on another type of norm dierent from (19.7). Again, let us use the integral equivalent form (19.6). Introduce in the Banach space F [w0>w0 + 1] the following norm equivalent to (19.7): ° ° k k ° 3Oi w ° { (w) max := max h { (w) X (19.9) wM[w0>w0+] 19.2. Regular ODE 575 Then Zw k k 3Oi v Oi v k k x ({1) x ({2) Oi h h {1 (v) {2 (v) F = v=w0 Zw ¡ ¢ Oi v 3Oi v k k Oi h h {1 (v) {2 (v) F gv v=w0 Zw Oi v Oi h k{1 (w) {2 (w)kmax gv = v=w0 Zw ¡ ¢ Oi v Oi w Oi h gv k{1 (w) {2 (w)kmax = h 1 k{1 (w) {2 (w)kmax v=w0 Multiplying this inequality by h3Oi w and taking max we get wM[w0>w0+] ¡ ¢ 3Oi kx ({1) x ({2)kmax 1 h k{1 (w) {2 (w)kmax Since t := 1 h3Oi ? 1 we conclude that x is a contraction. Taking then big enough we obtain the result. Corollary is proven. Remark 19.1 Sure, the global Lipschitz condition (19.4) with u = 4 holds for very narrow class of functions which is known as the class of "quasi-linear" functions, that’s why Corollary 19.1 is too conservative. On the other hand, the conditions of Theorem 19.1 for finite (small enough) u?4 is not so restrictive valid for any function satisfying somewhat mild smoothness conditions. Remark 19.2 The main disadvantage of Theorem 19.1 is that the solution of the Cauchy’s problem (19.1) exists only on the interval [w0>w0 + 1] (where 1 satisfies (19.5)), but not at the complete interval [w0>w0 + ], that is very restrictive. For example, the Cauchy problem 2 {˙ (w)={ (w) >{(0) = 1 1 has the exact solution { (w)= that exists only on [0> 1) but not 1 w for all [0> 4). 576 Chapter 19. Ordinary Dierential Equations The theorem presented below gives a constructive (direct) method of finding a unique solution of the problem (19.1). It has several forms. Here we present the version of this result which does not use any Lipschitz conditions: neither local, no global. Theorem 19.2 (Picard-Lindelöf, 1890) Consider the Cauchy’s problem (19.1) where the function i (w> {) is continuous on { R1+q || | k k } V := (w> {) 5 w w0 w0, { {0 F u (19.10) C and the partial derivative i : V $ Rq is also continuous on V. C{ Define the sets ° ° ° ° ° C ° M := max ki (w> {)k, O := max ° i (w> {)° (19.11) (w>{)MV (w>{)MV C{ and choose the real number such that 0 ? u, M u, t := O ? 1 (19.12) Then 1) the Cauchy’s problem (19.1) has unique solution on V; 2) the sequence {{q (w)} of functions generated iteratively by Zw { +1 (w)={0 + i (v> { (v)) gv q q (19.13) v=w0 {0 (w)={0, q =0> 1> ===; w0 w + w0 to { (w) in the Banach space X with the norm (19.7) converges geometrically as k k q+1 k k {q+1 (w) { (w) F t {0 { (w) F (19.14) Proof. Consider the integral equation (19.6) and the integral op- erator x (19.8) given on V. So, (19.6) can be represented as x ({ (w)) = { (w) , { (w) 5 Eu ({0) 19.2. Regular ODE 577 where x : P $ X . For all w 5 [w0>w0 + ] we have ° ° ° Zw ° ° ° k k ° ° x ({ (w)) {0 =max° i (v> {q (v)) gv° wM[w0>w0+] ° ° v=w0 max (w w0)maxki (w> {)k M u wM[w0>w0+] (w>{)MV i.e., x (P) P. By the classical mean value theorem 16.5 ° ° ° ° k k ° C | ° k k i (w> {) i (w> |) = ° i (w> }) }M[{>|] ({ |)° O { | C{ and, hence, ° ° ° Zw ° ° ° ° ° kx ({ (w)) x (| (w))k =max° [i (v> { (v)) i (v> | (v))] gv° wM[w0>w0+] ° ° v=w0 k k k k O max { (w) | (w) = t { (w) | (w) F wM[w0>w0+] Applying now the contraction principle we obtain that (19.6) has a unique solution { 5 Eu ({0).Wealsohave Zw k k {q+1 (w) { (w) F =max [i (v> {q (v)) i (v> { (v))] gv wM[w0>w0+] v=w0 k k q+1 k k t {q (w) { (w) F t {0 { (w) F Theorem is proven.