COMPLETE RECUPERATION AFTER THE BLOW UP TIME FOR SEMILINEAR PROBLEMS

We consider explosive solutions y0(t), t ∈ [0, Ty0 ), of some ordinary differential equations P (Ty0 ) : dy dt (t) = f(y(t)), y(0) = y0, where f : Rd → Rd is a locally Lipschitz superlinear function and d ≥ 1. In this work we analyze the following question of controlability: given > 0, a continuous deformation y(t) de y0(t), built as a solution of the perturbed control problem obtained by replacing f(y(t)) by f(y(t)) +u(t), for a suitable control u, such that y(t) = y0(t) for any t ∈ [0, Ty0 − ] and such that y(t) also blows up in t = Ty0 but in such a way that y(t) could be extended beyond Ty0 as a function y ∈ Lloc(0,+∞ : R d)?

1. Introduction.We consider blowing-up solutions y 0 (t), t ∈ [0, T y0 ), of some ODEs It is well known that the solutions of P (f, y 0 ) develop blow-up processes in the sense that the maximal existence interval is of the form [0, T y0 ), for some finite time T y0 (i.e.there is a complete blow-up in the norm of y(t) after T y0 ).From the point of view of Control Theory, it is easy to see (by arguing as in [5]) that we can avoid the blow-up phenomenon by introducing a suitable control function u(t).To be more precise, for any small enough > 0 we can find a continuous deformation y(t) of the given trajectory, y 0 (t), built as solution of the control perturbed problem P (f, y 0 , u) = dy dt (t) = f (y(t)) + u(t) in R d , y(0) = y 0 , for a suitable control u ∈ L 1 loc (0, +∞ : R d ) and defined on the whole interval [0, +∞) such that y(t) = y 0 (t) for any t ∈ [0, T y0 − ].Indeed, fix any T e > T y0 − and let us consider w ∈ C 1 [0, +∞) such that w(t) = y 0 (t) for any t ∈ [0, T y0 − ] and w(t) = 0 for any t ∈ [T e , +∞).Then, defining u(t) = dw dt (t) − f (w(t)) if t ∈ (T y0 − , +∞) we get the required conditions and that, in fact, y(t) = w(t) = 0 for any t ∈ [T e , +∞).

ALFONSO C. CASAL, JES ÚS ILDEFONSO D ÍAZ AND JOS É MANUEL VEGAS
In this work our goal is completely different since we do not try to avoid the blow-up phenomenon but to control it in such a way that the solution let defined in the whole interval [0, +∞) at least as a L 1 loc (0, +∞ : R d ) function.We shall show that we can control the explosion by allowing a more singular class of controls.
Definition.We say that the trajectory y 0 (t) of problem P (f, y 0 ), with blow-up time T u 0 , has a controllable explosion if for any small enough > 0 we can find a continuous deformation, y(t), of the trajectory y 0 (t), built as solution of the control perturbed problem P (f, y 0 , u), for a suitable control u ∈ W −1,q loc (0, +∞ : R d ) [the dual space of W 1,q 0,loc (0, +∞ : R d )], for some q > 1, such that y(t) = y 0 (t) for any t ∈ [0, T y0 − ], y(t) also blows-up at t = T y0 (the controlled explosion) but y(t) can be extended beyond T y0 as a function y ∈ L 1 loc (0, +∞ : R d ).
Theorem 1. Assume f locally Lipschitz continuous and superlinear.Then, for any y 0 ∈ R d the blowing up trajectory y 0 (t) of the associated problem P (f : y 0 ) has a controlled explosion by means of the control problem P (f, y 0 , u).
Our main tools are the study of a suitable delayed feedback problems (in the spirit of a previous work by the authors [Casal,Díaz and Vegas [4]] and the application of a powerful nonlinear variation of constants formula.This type of formula was first established in the literature for nonlinear terms of class C 2 [Alekseev [2], Laksmikantham and Leela [6], ...].In this work we shall show that, as a matter of fact, the formula holds also for Lipschitz functions f (which at this stage can be assumed to be in fact globally Lipschitz) and with a very general perturbation term (which in fact can be even multivalued).For instance, given such a f and a family of maximal monotone operators β(t, y), on the space H = R d , with β(t,•) ∈ L 1 loc (0, +∞ : R d ), we consider the perturbed problem We know that once that f is globally Lipschitz function, the solutions of P (f, β, ξ) are well defined, as absolutely continuous functions on [0, T ], for any given T > 0 (this is an easy consequence of the general theory: see [3]).Now, we reformulate the trajectory y 0 (t) in more general terms (by modifying the initial time and the initial condition) as y 0 (t) = φ(t, t 0 , ξ) with φ(t, t 0 , ξ) the unique solution of the ODE We introduce the formal notation Φ(t, t 0 , ξ) = ∂ ξ φ(t, t 0 , ξ), where ∂ ξ denotes partial differentiation.Then we shall prove: Theorem 2. The flow map φ is Lipschitz continuous, Φ is absolutely continuous and the solution y(t) of the "perturbed problem" P * (f, β, ξ) has the integral representation where y 0 (t) = φ(t, t 0 , ξ) is the solution of the "unperturbed" problem P * (f, 0, ξ).
In the above formula we assumed, for simplicity, that β(t,•) is single-valued but a suitable similar expression can be stated if β(t,•) is multivalued.Applications of this arguments to parabolic partial differential equations (see Remark 3) will be presented elsewhere.
Theorem 3. Assume f ∈ C 2 and superlinear.Then, for any y 0 ∈ R d the blowing up trajectory y 0 (t) of the associated problem P (f : y 0 ) has a controlled explosion.
Proof.Step 1 (the strategy).Define τ = T y0 − .We make the change of variable t = t − τ and consider the delayed problem (where, for simplicity we denote again t by t, so that, for any −τ ≤ θ ≤ 0 we are identifying y 0 (θ) with y 0 (θ +T y0 − ), for some suitable functions B(t) and where g(r) is any C 2 function (for instance g(r) = r).Our goal is to show that we can chose the control term such that the solution of P (f, y 0 , B) is defined on the whole interval [0, τ ) and that u ∈ W −1,q (0, τ : R d ).Since y(t − τ ) = y 0 (t − τ ) for any t ∈ [0, T y0 − ], this will prove the result by iteration on the intervals τ < t < 2τ, ..., nτ < t < (n + 1)τ, n ∈ N.
Step 2 (choice of function B and reformulation as neutral equation).Given q > 1, a > 0 and α ∈ (0, 1 q ) and a continuous function m (to be taken, for instance, in order to have B(0) = 0) we define with t * = in this new time scale (i.e.t = T y 0 in the original time scale).We assume that t * ∈ (0, τ ), i.e. 2 < T y 0 .As in [4] we can reformulate P (f, y 0 , B) as the neutral problem Instead, we will change our strategy and apply a very useful, but little-known mathematical device: Alekseev's nonlinear variation of constants formula [2].We now briefly recall this result in a very simple setting (a more general statements will be obtained in the next section).
Remark 1 Notice that Φ(t, t 0 , ξ) satisfies Φ(t, t, ξ) = 1.Notice also that Alekseev's formula is usually stated under stronger regularity conditions on G, and for d ≥ 1. Alekseev's formula will be obtained in Theorem 2 in a much greater generality.

Continuation of
Step 2 Fortunately, we can consider the retarded term as an external "forcing" and by setting t 0 = 0, ξ = z(0) = y 0 (0), y(t) = φ(t, 0, ξ), we can write (formally): and integrate by parts: By the remark above, Φ(t, t, z(t)) = 1.On the other hand, as we saw before, for y 0 ∈ W 1,q (−τ, 0) and g ∈ C 1 the composite function s → g(y 0 (s − τ )) is also W 1,q (−τ, 0) and so is its product by the C 1 function Φ(t, s, z(s)).Therefore, its derivative belongs to L q (−τ, 0) and the indefinite integral, as in all the previous cases, is an absolutely continuous function.This means that the integration by parts is legitimate and we may state the following result, which is an extension of the previous ones.We may summarize the previous comments in the following way: The initial value problem and initial function y 0 in W 1,q (−τ, 0) has a precise integral sense in [0, τ ] by means of the neutral equivalent equation and its unique solution z admits the integral representation (where y(t) = φ(t, 0, y 0 (0)))).Then, for every ξ ∈ W 1,r (0, τ ) (where 1/q + 1/r = 1) the neutral Cauchy problem has a unique solution given by the identity (14).Therefore z ∈ L q (0, τ ) and z(t)− B(t)g(y 0 (t − τ )) is an absolutely continuous function and we may write symbolically z(t) = B(t)g(y 0 (t − τ )) + AC (15) where "AC" means "an absolutely continuous function".As a consequence, the singularities of the solution on [0, τ ] are also singularities of B. Thus, in particular, let t * = (notice that t * = T y 0 in the original scale of time), 0 < α < 1, let m be continuous on [0, τ ] and let Since the initial function y 0 satisfies y 0 (t * − τ ) = y 0 ( ) = 0, then t * is also a singularity of z (the controlled explosion) and strongly in L 2 (0, T : M d×d ).Indeed, Φ n (t, t 0 , ξ) is the solution of the problem where H n (t, t 0 , ξ) = ∂ x f n (φ n (t, t 0 , ξ)).But, we know that, if M is given by ( 19) then H n (t, t 0 , ξ) L ∞ (t0,T :M d×d ) ≤ M for any t 0 ∈ (0, T ) and for any ξ ∈ R d .
Remark 2. Notice that since our main interest is to study the asymptotic, near T y0 , we do not need to identify the limit matricial function Φ(t, s, y).This is a complicated task over the set of points y ∈ R d where f is not Frechet differentiable in y (see a nonlinear characterization in Mirica [7]).
once we assume p > 1, for suitable conditions on y 0 ∈ L 2 (Ω) and for an appropriate choice of the control function (taken as a suitable delayed feedback control) can be given in a similar way to the results presented in [4].By limitations in the length of this work, those results will be given elsewhere.