**HAL Id: hal-01955484**

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**Exact Controllability of nonlinear Heat equations in**

**spaces of analytic functions**

### Camille Laurent, Lionel Rosier

**To cite this version:**

Camille Laurent, Lionel Rosier. Exact Controllability of nonlinear Heat equations in spaces of analytic functions. 2018. �hal-01955484�

IN SPACES OF ANALYTIC FUNCTIONS

CAMILLE LAURENT AND LIONEL ROSIER

ABSTRACT. It is by now well known that the use of Carleman estimates allows to establish the
control-lability to trajectories of nonlinear parabolic equations. However, by this approach, it is not clear how to
decide whether a given function is indeed reachable. In this paper, we pursue the study of the reachable
states of parabolic equations based on a direct approach using control inputs in Gevrey spaces by
consider-ing a nonlinear heat equation in dimension one. The nonlinear part is assumed to be an analytic function of
*the spatial variable x, the unknown y, and its derivative*∂*xy*. By investigating carefully a nonlinear Cauchy
problem in the spatial variable and the relationship between the jet of space derivatives and the jet of time
derivatives, we derive an exact controllability result for small initial and final data that can be extended as
analytic functions on some ball of the complex plane.

2010 Mathematics Subject Classification: 35K40, 93B05

Keywords: Nonlinear heat equation; viscous Burgers’ equation; Allen-Cahn equation; exact control-lability; ill-posed problems; Gevrey class; reachable states.

1. INTRODUCTION

The null controllability of nonlinear parabolic equations is well understood since the nineties. It was derived in [6] in dimension one by solving some “ill-posed problem” with Cauchy data in some Gevrey spaces, and in [4, 5] in any dimension and for any control region by using some “parabolic Carleman estimates”.

The null controllability was actually extended to the controllability to trajectories in [5]. However, it is a quite hard task to decide whether a given state is the value at some time of a trajectory of the system without control (free evolution). In practice, the only known examples of such states are the steady states. As noticed in [16], in the linear case, the steady states are Gevrey functions of order 1/2 in x (and thus analytic over C) for which infinitely many traces vanish at the boundary, a fact which is also a very conservative condition leading to exclude e.g. all the nontrivial polynomial functions.

The recent paper [16] used the flatness approach and a Borel theorem to provide an explicit set of
*reachable states composed of states that can be extended as analytic functions on a ball B. It was also*
noticed in [16] that any reachable state could be extended as an analytic function on a square included in
*the ball B. We refer the reader to [1, 7] for new sets of reachable states for the linear 1D heat equation,*
*with control inputs chosen in L*2*(0, T ). We notice that the flatness approach applied to the control of*
PDEs, first developed in [11, 3, 18, 23], was revisited recently to recover the null controllability of (i) the
heat equation in cylinders [14]; (ii) a family of parabolic equations with unsmooth coefficients [15]; (iii)
the Schr¨odinger equation [17]; (iv) the Korteweg-de Vries equation with a control at the left endpoint

[13]. One of the main features of the flatness approach is that it provides control inputs developed as explicit series, which leads to very efficient numerical schemes.

The aim of the present paper is to extend the results of [16] to nonlinear parabolic equations. Roughly,
we shall prove that a reachable state for the linear heat equation is also reachable for the nonlinear one,
provided that its magnitude is not too large and its poles and those of the nonlinear term are sufficiently
*far from the origin. The method of proof is inspired by [6] where a Cauchy problem in the variable x*
*is investigated. The main novelty is that we prove an exact controllability result (and not only a null*
controllability result as in [6]), and we need to investigate the influence of the nonlinear terms on the
*jets of the time derivatives of two traces at x*= 0. Here, we do not use some series expansions of the
control inputs as in the flatness approach, but we still use some Borel theorem as in [21, 16]. It is unclear
whether the same results could be obtained by the classical approach using the exact controllability of
the linearized system and a fixed-point argument.

To be more precise, we are concerned with the exact controllability of the following nonlinear heat equation

∂*ty*= ∂*x*2*y+ f (x, y, ∂xy),* *x∈ [−1,1], t ∈ [0,T ],* (1.1)

*y _{(−1,t) = h}*

_{−1}

*(t),*

*t*(1.2)

_{∈ [0,T ],}*y(1,t) = h*1*(t),* *t∈ [0,T ],* (1.3)

*y(x, 0) = y*0_{(x),}_{x}_{∈ [−1,1],} _{(1.4)}

*where f : R*3_{→ R is analytic with respect to all its arguments in a neighborhood of (0,0,0). More}
precisely, we assume that

*f _{(x, 0, 0) = 0 ∀x ∈ (−4,4),}* (1.5)
and that

*f(x, y*0

*, y*1) =

### ∑

*(p,q,r)∈N*3

*ap,q,r(y*0)

*p(y*1)

*qxr*

*∀(x,y*0

*, y*1) ∈ (−4,4)3, (1.6) with

*|ap,q,r*| ≤

*M*

*b*

_{0}

*pbq*

_{1}

*br*2

*∀p,q,r ∈ N*(1.7)

for some constants

*M*> 0, *b*0*> 4, b*1*> 4, and b*2> 4. (1.8)

*Note that a*0,0,r*= 0 for all r ∈ N by (1.5). For p,q ∈ N let*

*Ap,q(x) =*

### ∑

*r*∈N

*ap,q,rxr*, *|x| < b*2.

We infer from (1.6) and (1.7) that

*f(x, y*0*, y*1) =

### ∑

*p*

_{, q ∈ N,}*p+ q > 0*

*Ap,q(x)(y*0)

*p(y*1)

*q*,

*|Ap,q(x)| ≤*

*M*

*b*

_{0}

*pbq*

_{1}1 1

_{−}

*|x|*2 ,

_{b}*2.*

_{|x| < b}*(1) the heat equation with an analytic potential:*

∂*ty*= ∂*x*2*y+ ϕ(x)y*

where*ϕ(x) = ∑r*≥0*arxr*, with*|ar| ≤ M/br*2*for all r∈ N and some constants M > 0, |b*2| > 4.

*(2) the Allen-Cahn equation*

∂*ty*= ∂*x*2*y+ y − y*3

*(3) the viscous Burgers’ equation*

∂*ty*= ∂*x*2*y− y∂xy.* (1.9)

Note that our controllability result is still valid when the nonlinear term* _{−y∂}xy*in (1.9) is replaced

by a term like*ϕ(x)yp*_{(∂}

*xy)q*with*ϕ as in (1), and p, q ∈ N.*

Because of the smoothing effect, the exact controllability result has to be stated in a space of analytic
*functions (see [16] for the linear heat equation). For given R> 1 and C > 0, we denote by RR,C*the set

R_{R}_{,C}_{:}_{= {y : [−1,1] → R; ∃(α}_{n}_{)}* _{n}*
≥0∈ RN, |α

*n| ≤ C*

*n!*

*Rn*

*∀n ≥ 0 and y(x) =*∞

### ∑

*n*=0 α

*n*

*xn*

*n!*

*∀x ∈ [−1,1]}.*

*We say that a function h*∞

_{∈ C}*([t*1

*,t*2

*]) is Gevrey of order s ≥ 0 on [t*1

*,t*2

*], and we write h ∈ Gs([t*1

*,t*2]), if

*there exist some positive constants M, R such that*

|∂*tph(t)| ≤ M*

*(p!)s*

*Rp* *∀t ∈ [t*1*,t*2*], ∀p ≥ 0.*

*Similarly, we say that a function y _{∈ C}*∞

*([x*1

*, x*2

*] × [t*1

*,t*2

*]) is Gevrey of order s*1

*in x and s*2

*in t, with*

*s*1*, s*2*≥ 0, and we write y ∈ Gs*1*,s*2*([x*1*, x*2*] × [t*1*,t*2*]), if there exist some positive constants M, R*1*, R*2such

that
|∂*p*1
*x* ∂
*p*2
*t* *y(x,t)| ≤ M*
*(p*1!)*s*1*(p*2!)*s*2
*Rp*1
1 *R*
*p*2
2
*∀(x,t) ∈ [x*1*, x*2*] × [t*1*,t*2*], ∀(p*1*, p*2) ∈ N2.

*The main result in this paper is the following exact controllability result.*
*Theorem 1.1. Let f= f (x, y*0*, y*1*) be as in (1.5)-(1.8) with b*2> ˆ*R*:= 4e*(2e)*

−1

*≈ 4.81. Let R > ˆR and T*>
*0. Then there exists some number ˆC> 0 such that for all y*0* _{, y}*1

_{∈ R}

*R*, ˆ*C, there exists h*−1*, h*1*∈ G*2*([0, T ])*

*such that the solution y of* *(1.1)-(1.4) is defined for all _{(x,t) ∈ [−1,1]×[0,T ] and satisfies y(x,T ) = y}*1

_{(x)}*for all x _{∈ [−1,1]. Furthermore, we have that y ∈ G}*1,2

_{([−1,1] × [0,T ]).}*A similar result with only one control can be derived assuming that f is odd w.r.t.* *(x, y*0). Consider

the control system

∂*ty*= ∂*x*2*y+ f (x, y, ∂xy),* *x∈ [0,1], t ∈ [0,T ],* (1.10)

*y(0,t) = 0,* *t _{∈ [0,T ],}* (1.11)

*y(1,t) = h(t),* *t _{∈ [0,T ],}* (1.12)

*y(x, 0) = y*0*(x),* *x*_{∈ [0,1].} (1.13)

*Corollary 1.2. Let f* *= f (x, y*0*, y*1*) be as in (1.5)-(1.8) with b*2> ˆ*R*:= 4e*(2e)*

−1

*≈ 4.81, and assume that*
*f _{(−x,−y}*0

*, y*1

*) = − f (x,y*0

*, y*1

*) ∀x ∈ [−1,1], ∀y*0

*, y*1∈ (−4,4). (1.14)

*Let R*> ˆ*R and T* *> 0. Then there exists some number ˆC> 0 such that for all y*0* _{, y}*1

_{∈ R}

*R*, ˆ*C* *with*

*(y*0* _{(−x),y}*1

*0*

_{(−x)) = (−y}*1*

_{(x), −y}*2*

_{(x)) for all x ∈ [−1,1], there exists h ∈ G}

_{([0, T ]) such that the }*so-lution y of* *(1.10)-(1.13) is defined for all _{(x,t) ∈ [−1,1]× ∈ [0,T ] and satisfies y(x,T ) = y}*1

*(x) for all*

*x*1,2

_{∈ [0,1]. Furthermore, we have that y ∈ G}

_{([0, 1] × [0,T ]).}*Corollary 1.2 can be applied e.g. to (i) the heat equation with an even analytic potential; (ii) the*
Allen-Cahn equation; (iii) the viscous Burgers’ equation.

The constant ˆ*R*:= 4e*(2e)*−1 is probably not optimal, but our main aim was to provide an explicit
(rea-sonable) constant. It is expected that the optimal constant is ˆ*R*:= 1, with a diamond-shaped domain of
analyticity as in [1] and [7] for the linear heat equation.

The paper is organized as follows. Section 2 is concerned with the wellposedness of the Cauchy
*problem in the x-variable (Theorem 2.1). The relationship between the jet of space derivatives and the jet*
of time derivatives at some point (jet analysis) for a solution of (1.1) is studied in Section 3. In particular,
we show that the nonlinear heat equation (1.1) can be (locally) solved forward and backward if the initial
*data y*0can be extended as an analytic function in some ball of C (Proposition 3.6). Finally, the proofs of

Theorem 1.1 and Corollary 1.2 are displayed in Section 4.

2. CAUCHY PROBLEM IN THE SPACE VARIABLE

*2.1. Statement of the global wellposedness result. Let f* *= f (x, y*0*, y*1) be as in (1.5)-(1.8). We are

*concerned with the wellposedness of the Cauchy problem in the variable x:*
∂2

*xy*= ∂*ty− f (x,y,∂xy),* *x∈ [−1,1], t ∈ [t*1*,t*2], (2.1)

*y(0,t) = g*0*(t),* *t∈ [t*1*,t*2], (2.2)

∂*xy(0,t) = g*1*(t),* *t∈ [t*1*,t*2], (2.3)

*for some given functions g*0*, g*1*∈ G*2*([t*1*,t*2]). The aim of this section is to prove the following result.

*Theorem 2.1. Let f* *= f (x, y*0*, y*1*) be as in (1.5)-(1.8). Let −∞ < t*1*< t*2*< +∞ and R > 4. Then there*

*exists some number C> 0 such that for all g*0*, g*1*∈ G*2*([t*1*,t*2*]) with*

*|g(n)i* *(t)| ≤ C*

*(n!)*2

*Rn* , *i= 0, 1, n ≥ 0, t ∈ [t*1*,t*2], (2.4)

*there exist some numbers R*1*, R*2 *with*4/e < R1*< R*2*and a solution y∈ G*1,2*([−1,1] × [t*1*,t*2*]) of *

*(2.1)-(2.3) satisfying for some constant M*> 0

|∂*p*1
*x* ∂
*p*2
*t* *y(x,t)| ≤ M*
*(p*1*+ 2p*2)!
*Rp*1
1 *R*
*2p*2
2
*∀(x,t) ∈ [−1,1] × [t*1*,t*2*], ∀(p*1*, p*2) ∈ N2. (2.5)

2.2. Abstract existence theorem. We consider a family of Banach spaces*(Xs*)*s*∈[0,1]satisfying for 0≤

*s*′_{≤ s ≤ 1}

*Xs⊂ Xs*′, (2.6)

*k f kXs′* *≤ k f kXs*; (2.7)

We are concerned with an abstract Cauchy problem

∂*xU(x) = G(x)U (x),* *−1 ≤ x ≤ 1,*

*U(0) = U*0,
*where X*0* _{∈ X}*1

*and G(x)*

*x*∈[−1,1]*is a familly of nonlinear operators with possible loss of derivatives.*

Our first result is a global wellposedness result.

*Theorem 2.2. For anyε ∈ (0,1/4), there exists a constant D > 0 such that for any family (G(x))x*∈[−1,1]

*of nonlinear maps from Xsto Xs*′ *for*0*≤ s*′*< s ≤ 1 satisfying*

*kG(x)UkXs′* ≤
ε
*s _{− s}*′

*kUkXs*(2.8)

*kG(x)U − G(x)V )kXs′*≤ ε

*s*′

_{− s}*kU −V kXs*(2.9)

*for*0* _{≤ s}*′

_{< s ≤ 1, x ∈ [−1,1] and U,V ∈ X}s*with*

*kUkXs*

*≤ D, kV kXs*

*≤ D, there exists η > 0 so that*

*for any U*0* _{∈ X}*1

*with*

*
U*0
_{
}

*X*1 *≤ η, there exists a solution U ∈ C([−1,1],Xs*0*) for some s*0*∈ (0,1) to the*

*integral equation*

*U(x) = U*0_{+}Z *x*
0

*G(τ)U (τ)dτ.*
*Moreover, we have the estimate*

*kU(x)kXs* *≤ C*1
1_{−} *λ |x|*
*a*∞*(1 − s)*
_{−1}_{
}
*
U*0
_{
}
*X*1, for 0*≤ s < 1, |x| <*
*a*∞
λ *(1 − s),*
*whereλ ∈ (0,1), a*∞*∈ (λ ,1) and C*1*> 0 are some constants. In particular, we have*

*kU(x)kXs≤ C*1
1_{−} _{a}_{∞}2
λ + 1
−1_{
}
*
U*0
_{
}
*X*1, for 0*≤ s ≤ s*0=
1
2(1 −
λ
*a*∞*), |x| ≤ 1.*

*If, in addition, we assume that*

*for all U*0*∈ XswithkU*0k*Xs≤ D, the map τ ∈ [−1,1] → G(τ)U*0*∈ Xs*′ *is continuous,* (2.10)

*then U is solution in the classical sense of*

∂*xU(x) = G(x)U (x),* *−1 ≤ x ≤ 1,*

*U(0) = U*0_{.} (2.11)

We prove first the existence of a solution on an interval_{[−(1 − δ ),1 − δ ], where δ ∈ (0,1). Next, we}
*use a scaling argument to obtain a solution for x*_{∈ [−1,1].}

*Proposition 2.3. For any _{ε ∈ (0,1/4), any δ ∈ (0,1) and any G and D as in (2.10)-(2.9), there exists}*

*some numbers a*∞

*> 1 − δ and η > 0 such that for any U*0

*∈ X*1

*with*

*
U*0
_{
}

*X*1 *≤ η, there exists a unique*

*solution for x _{∈ (−a}*∞

*, a*∞

*) to (2.11) in the space Y*∞

*(see below). Moreover, we have the estimate*

*kU(x)kXs≤ C*1
1_{−} *|x|*
*a*∞*(1 − s)*
−1_{
}
*
U*0
_{
}
*X*1, for 0*≤ s < 1, |x| < a*∞*(1 − s),*
*where C*1*> 0 is a constant.*

*Proof of Proposition 2.3.* We follow closely the proof of [8], taking care of the choice of the constants
and of the time of existence.

Consider a sequence of numbers*(ak*)*k*≥0 satisfying the following properties (the existence of such a

sequence is proved in Lemma 2.4, see below):
*(i) a*0= 1;

(ii) *(ak*)*k*≥0*is a decreasing sequence converging to a*∞> 1 − δ ;

(iii) ∑∞*i*=0 (4ε)
*i*

1−*ai+1 _{ai}* < +∞.

Next, we pickη small enough so that η ∑∞_{i}_{=0} (4ε)*i*

1−*ai+1 _{ai}*

*< D.*

*We define, for k _{∈ N∪{∞}, the (Banach) space Y}k= {U ∈*T0

*<*

_{≤s<1}C(−ak(1−s),ak(1−s),Xs); kUkYk∞} with the norm

*kUkYk*:= sup
*|x|<ak(1−s)*
0*≤s<1*
*kU(x)kXs*
1_{−} *|x|*
*ak(1 − s)*
*if k*_{∈ N,} (2.12)
*kUkY*∞ := sup
*|x|<a*∞*(1−s)*
0*≤s<1*
*kU(x)kXs*
1_{−} *|x|*
*a*∞*(1 − s)*
*if k*= ∞. (2.13)

Note that for* _{|x| < a}k(1 − s), 0 ≤ s < 1, we have that 1 −_{a}_{k(1−s)}|x|* ∈ (0,1]. Note also that we have

*Yk⊂ Yk*+1 and*kUkYk*+1*≤ kUkYk*, for the sequence*(ak*)*k*∈Nis decreasing.

We want to define a sequence*(Uk*)*k*≥0by the relations

*U*0= 0, *Uk*+1*= TUk* *for k*∈ N
where
*(TU )(x) = U*0+
Z *x*
0
*G(τ)U (τ)dτ.*
*Note that U*1*(x) = (TU*0*)(x) = U*0for*|x| < 1. Introduce*

*Vk*:= U*k*+1*−Uk*, *k*∈ N.

*We prove by induction on k*_{∈ N the following statements (that contain the fact that the sequence}
*(Uk*)*k*∈Nis indeed well defined):

λ*k*:*= kVk*k*Yk* ≤ (4ε)
*k*_{η,} _{(2.14)}
*kUk*+1*(x)kXs*≤
*k*

### ∑

*i*=0 λ

*i*1

_{−}

*ai*+1

*ai*

*≤ D for |x| < ak*+1

*(1 − s),*(2.15)

*so that G(x)Uk*+1*(x) is well defined in Xs*′ for*|x| ≤ a _{k}*

_{+1}

*(1 − s).*

*Let us first check that (2.14)-(2.15) are valid for k*= 0. For (2.14), we have that
λ0*= kU*1*−U*0k*Y*0 *≤ kU*

0

k*X*1 ≤ η.

For (2.15), we notice that

*kU*1k*Xs* *= kU*
0_{k}
*Xs* *≤ kU*
0_{k}
*X*1 ≤ η ≤
η
1* _{− a}*1

*≤ D.*

*Assume that (2.14)-(2.15) are true up to the rank k. Let us check that they are also true at the rank*
*k*+ 1.

*Take s and x (for simplicity, we assume x _{≥ 0) so that 0 ≤ x < a}k*+1

*(1 − s). For any 0 ≤ τ ≤ x, (2.15)*

gives max * _{kU}k*+1(τ)k

*s, kUk*(τ)k

*s*

*≤ D (recall ak*+1*≤ ak). In particular, we can apply (2.9) replacing s*′

*by s and s by s(τ) =*1_{2}1* _{+ s −}_{a}*τ

*k*+1 , obtaining

*kG(τ)Uk*+1

*(τ) − G(τ)Uk*(τ)k

*Xs*≤ ε

*s(τ) − skUk*+1(τ)

*−Uk*(τ)k

*Xs*(τ)

Note that we have indeed 0_{≤ s < s(τ) < 1. Next}

*kVk*+1*(x)kXs* *= k(TUk*+1*)(x) − (TUk)(x)kXs*
≤
Z *x*
0 *kG(τ)Uk*+1*(τ) − G(τ)Uk*(τ)k*Xs*
*dτ*
≤
Z *x*
0
ε
*s _{(τ) − s}kUk*+1(τ)

*−Uk*(τ)k

*Xs*(τ)

*dτ*

*≤ ε kVk*k

*Yk*Z

*x*0

*ak*+1

*s*1

_{(τ) − s}

_{− s(τ)}*ak*+1

*(1 − s(τ)) − τ*

*dτ*

*where we have used the fact that s(τ) satisfies τ < ak*+1*(1 − s(τ)) (for ak*+1*(1 − s(τ)) − τ =* 1_{2}*(ak*+1(1 −

*s _{) − τ) > 0) and 0 < s(τ) < 1, so that with (2.12)}*

*kVk*(τ)k

*Xs*(τ) ≤ 1

_{−}|τ|

*ak*+1

*(1 − s(τ))*−1

*kVk*k

*Yk*+1≤ 1

_{−}|τ|

*ak*+1

*(1 − s(τ))*−1

*kVk*k

*Yk*. (2.16)

*Let us go back to the estimate of the integral. To simplify the notations, we denote A= ak*+1*(1 − s)*

and recall 0* _{≤ τ ≤ x < A. We have}*
Z

*x*0

*ak*+1

*(1 − s(τ))*

*(s(τ) − s)(ak*+1

*(1 − s(τ)) − τ)*

*dτ*

*= 2ak*+1 Z

*x*0

*A*+ τ

*(A − τ)*2

*dτ*≤ Z

*x*0

*4ak*+1

*A*

*(A − τ)*2

*dτ = ak*+1

*4A*

*A*

_{− τ}

*x*0

*≤ ak*+1

*4A*

*A*· So, recalling

_{− x}

_{A}A_{−x}_{= 1 −}

*−1=1*

_{A}x_{−}

_{a}*|x|*

*k*+1

*(1−s)*−1 , we have obtained

*kVk*+1

*(x)ks*1

_{−}

*|x|*

*ak*+1

*(1 − s)*

*≤ 4ak*+1

*ε kVk*k

*.*

_{Y}_{k}So, we have proved that

*kVk*+1k*Yk*+1*≤ 4a*0*ε kVk*k*Yk*, (2.17)

and henceλ*k*+1*≤ 4a*0ελ*k. This yields (2.14) at rank k*+ 1.

*Let us proceed with the proof of (2.15) at rank k*+ 1.

*Since Uk*+2*= Uk*+1*+Vk*+1, we only need to prove*kVk*+1*(x)ks*≤

λ*k*+1

1−*ak+2 _{ak+1}* for

*|x| < ak*+2

*(1 − s). This is*

obtained by noticing that

*kVk*+1*(x)ks*≤
1_{−} *|x|*
*ak*+1*(1 − s)*
_{−1}
*kVk*+1k*Yk*+1≤
1_{−}*ak*+2
*ak*+1
_{−1}
λ*k*+1

since*|x| < ak*+2*(1 − s). The proof by induction of (2.14)-(2.15) is complete.*

We are now in a position to prove the existence of a solution to (2.11). Let us introduce the function
*U*∞:= lim*k*→+∞*Uk*= ∑+∞_{k}_{=0}*Vk. Note that the convergence of the series is normal in Y*∞. Indeed,

+∞

### ∑

*k*=0

*kVk*k

_{Y}_{∞}≤ +∞

### ∑

*k*=0

*kVk*k

*≤ +∞*

_{Y}_{k}### ∑

*k*=0 λ

*k*< +∞.

*Note also that (2.15) remains true for U*∞for*|x| < a*∞*(1 − s), so that G(τ)U*∞is well defined.

*Let us prove that U*∞*is indeed a solution of (2.11). Using the fact that Uk*+1*− TUk*= 0, we have

*U*∞*− (TU*∞*)(x) = U*∞*−Uk*+1+

Z *x*

0

*[G(τ)Uk(τ) − G(τ)U*∞*(τ)] dτ*

*where all the terms in the equation are in Y*∞. The same estimates as before yield for 0*≤ s < 1 and*

*|x| < a*∞*(1 − s)*

k
Z *x*

0

*[G(τ)Uk(τ) − G(τ)U*∞*(τ)] dτkXs* *≤ 4a*∞*εkUk−U*∞k*Y*∞

1_{−} *|x|*
*a*∞*(1 − s)*
−1
,
and hence

*kU*∞*− TU*∞k*Y*∞*≤ kU*∞*−Uk*+1k*Y*∞*+ 4a*∞*εkUk−U*∞k*Y*∞ → 0

*as k _{→ +∞. Thus U}*∞is a solution of (2.11).

*Let us prove the uniqueness of the solution of (2.11) in the same space Y*∞*. Assume that U and eU* are

*two solutions of (2.11) in Y*∞. Pick*λ ∈ (0,1), and let a*′0*= 1, a*′*k= λ akfor k*∈ N∗∪ {∞}. Notice that (i),

*(ii) and (iii) are still valid for the a*′* _{k}. Denote by Y_{k}*′

*, k*′

_{∈ N ∪ {∞}, the space associated with a}*. Note that*

_{k}*a*′

*∞*

_{k}< a*and hence Y*∞

*⊂ Yk*′

*for k*≫ 1. Then we have by the same computations as above that

*kU − eU*_{k}*Y _{k}*′ ≤ (4ε)

*kkU − eU*k

*Y*

_{0}′

which yields

*kU − eU*k*Y*′

∞≤ lim_{k}

→+∞*kU − eU*k*Yk*′ = 0.

*Thus U(x) = eU _{(x) for |x| < a}*′

_{∞}

*= λ a*∞, withλ as close to 1 as desired.

It remains to prove the existence of the sequence*(ak*)*k*≥0. This is done in the next lemma.

*Lemma 2.4. There exists a sequence(ak*)*k*∈N*satisfying (i), (ii) and (iii).*

*Proof.* *We denote C*0= 4ε < 1 and we require ∑∞*i*=0

*Ci*

0

1−*ai+1 _{ai}*

*< +∞. Picking a*0= 1 and γ > 0 small enough,

we define the sequence*(ak*)*k*_{∈N}by induction by setting

*ak*+1*= ak*
1_{−} γ
*(1 + k)*2
, *k*_{∈ N.}
The sequence *(ak*)*k*∈N is clearly decreasing, *C*

*i*
0
1−*ai+1 _{ai}* = γ−1

*(1 + i)*2

*0, and hence ∑∞*

_{C}i*i*=0

*Ci*0 1−

*ai+1*< +∞, for

_{ai}*C*0*< 1. Finally, bk= ln(ak*) converges to ∑∞*k*=0ln

1_{−}* _{(1+k)}*γ 2

≥ −2γ ∑∞*k*=0*(1+k)*1 2 forγ small enough. In

Let us complete the proof of Theorem 2.2 by using a scaling argument. Pick any numberλ ∈ (0,1) with ε λ < 1 4·

Let* _{δ = 1 − λ ∈ (0,1) and pick a}*∞

*∈ (λ ,1). Introduce the new variables ˜x := λ x ∈ [−λ ,λ ] = [−(1 −*

*δ ), 1 − δ ] for x ∈ [−1,1], and the new unknown*
˜

*U( ˜x) := U (x) = U (λ*−1*x).*˜
Then ˜*U* should solve

∂*x*˜*U*˜ = λ−1*G(λ*−1*x)U (λ*˜ −1*x) = ˜*˜ *G( ˜x) ˜U( ˜x),*
˜
*U(0) = U*0,
where
˜
*G( ˜x) :=*
λ−1* _{G(λ}*−1

_{x)}_{˜}

_{if ˜}

_{x}_{∈ [−λ ,λ ],}λ−1

*G(1)*if ˜

*x*

_{∈ [λ ,1],}λ−1

*G*

_{(−1)}if ˜

*x*

_{∈ [−1,−λ ].}Then

*G( ˜x)*˜

_{x}_{˜}

∈[−1,1]*is a family of nonlinear maps from Xsto Xs*′for 0*≤ s*′*< s ≤ 1 satisfying for 0 ≤ s*′<

*s _{≤ 1, ˜x ∈ [−1,1] and U,V ∈ X}s*with

*kUkXs*

*≤ D, kV kXs≤ D*

k ˜*G( ˜x)U kX _{s′}* ≤
˜
ε

*s*′

_{− s}*kUkXs*k ˜

*G*≤ ˜ ε

_{( ˜x)U − ˜}G_{( ˜x)V k}X_{s′}*s*′

_{− s}*kU −VkXs*

where ˜_{ε = ε/λ ∈ (0,1/4). Since (2.10)-(2.9) are satisfied, we infer from Proposition 2.3 the existence}
of a solution ˜*U*of

∂*x*˜*U*˜ = ˜*G( ˜x) ˜U(x),* *U*˜*(0) = U*0

*for x*_{∈ [−(1 − δ ),1 − δ ] = [−λ ,λ ], and satisfying}
˜

*U*_{∈} \

0*≤s<1*

*C _{(−a}*∞

*(1 − s),a*∞

*(1 − s),Xs*).

*Then the function U defined for x _{∈ [−1,1] by U(x) = ˜}U( ˜x) solves*
∂

*xU= G(x)U (x),*

*U(0) = U*0,

*for x _{∈ [−1,1], and it satisfies U ∈}*T

_{0}

_{≤s<1}C_{(−}

*a*∞

λ *(1 − s),*
*a*∞
λ*(1 − s),Xs*) and
*kU(x)kXs* *≤ C*1
1_{−} *λ |x|*
*a*∞*(1 − s)*
_{−1}_{
}
*
U*0
_{
}
*X*1, for 0*≤ s < 1, |x| <*
*a*∞
λ *(1 − s).*

The proof of Theorem 2.2 is complete.

2.4. Definitions. We define several spaces of Gevreyλ functions for λ > 1. For our application to the heat equation, we shall takeλ = 2, but for the moment we stay in the generality. Introduce

Γ_{λ}*(k) = 2*−5*(k!)*λ*(1 + k)*−2,

and letΓ denote the Gamma function of Euler. It is increasing on [2, +∞).

*We also introduce a variant of those functions with a parameter a _{∈ R (a is not necessarily an integer):}*
Γλ

*,a(k) = 2*−5

*(Γ(k + 1 − a))*λ

*(1 + k)*−2,

*for k∈ N s.t. k > |a| + 1,*(2.18)

Γλ*,a(k) = Γ*λ*(k),* *for k∈ N s.t. 0 ≤ k ≤ |a| + 1.* (2.19)

Clearly,Γ_{λ}_{,0}= Γ_{λ}*. Note that for k _{> |a| + 1, we have k + 1 − a ≥ 2, so we are in an interval where Γ is}*

*increasing. Thus we have for all k*

_{∈ N}

Γλ*,a(k) ≤ Γ*λ*(k),* *if a*≥ 0 (2.20)

Γλ*(k) ≤ Γ*λ*,a(k),* *if a*≤ 0. (2.21)

*The intermediate space will be the set of functions in C*∞*(K) (where K = [t*1*,t*2*] with −∞ < t*1*< t*2< ∞)

such that
*|u|L,a*:= sup
(
*u(k) _{(t)}*

_{}

*L|k−a|*Γ

_{λ}

*) < ∞.*

_{,a}(k),t ∈ K,k ∈ N*Note that for a _{= 0, we recover the spaces defined earlier in [24], and |u|}_{L}*

_{,0}

*.*

_{= |u|}_{L}*Definition1. Yamanaka [24] defined the norms*

*kukL*:= max
n
26* _{kuk}_{L}*∞

*, 23*

_{(K)}*L*−1

*u*′

_{}

*L*o ,

*and similarly we define for a*

_{∈ R}

*kukL,a*:= max
n
26* _{kuk}_{L}*∞

*, 23*

_{(K)}*L*−1

*u*′

_{}

*L,a*o .

*We denote by G*λ_{L}*(resp. G*λ* _{L}_{,a}) the (Banach) space of functions u_{∈ C}*∞

*< ∞ (resp.*

_{(K) such that kuk}_{L}*kukL,a*< ∞).

*The space G*λ* _{L}_{,a}* is supposed to “represent” the space of functions Gevrey

*λ with radius L*−1

_{with a}*derivatives. Roughly, we may think that u _{∈ G}*λ

*λ*

_{L}_{,a}if Dau_{∈ G}

_{L}, even if it is not completely true if a_{∈ N.}/

*Note that, as a direct consequence of (2.20)-(2.21), we have the embeddings G*λ

*λ*

_{L}_{,a}_{⊂ G}

_{L}*if a*

_{≥ 0,}

*G*λ

*λ*

_{L}_{⊂ G}

_{L}_{,a}if a_{≤ 0, together with the inequalities}

*kukL≤ max(L*
*a _{, L}_{−a}*

*) kukL,a* *if a*≥ 0, (2.22)

*kukL,a≤ max(La, L−a) kukL* *if a*≤ 0. (2.23)

*Furthermore, for any a _{∈ R and 0 < L < L}*′

*, we have the embedding G*λ

*λ*

_{L}_{,a}_{⊂ G}*′*

_{L}*with*

_{,a}*kukL*′* _{,a}≤ kuk_{L}_{,a}*. (2.24)

The following result [24, Theorem 5.4] will be used several times in the sequel.
*Lemma 2.5. (Algebra property) [24, Theorem 5.4]*

*kuvkL≤ kukLkvkL* *∀u,v ∈ G*

λ

2.5. Cost of derivation. The following result is a variant of Proposition 2.3 of Kawagishi-Yamanaka [9], where the spaces we consider contain some non-integer “derivatives”.

*Lemma 2.6 (Cost of derivatives for Gevrey spaces containing derivatives). Letλ > 0 and δ > 0. Let*
*q _{∈ N and a,b ∈ R with d = q − a + b > 0. Then there exists some number C = C(λ ,δ ,a,b,q) such that}*

*for all L> 0, all α > 1, and all u*λ

_{∈ G}

_{L}_{,a}, we have
*u(q)*
_{α}_{L}*,b*≤ *ChLi*
*C*_{+ (1 + δ )α}*b _{L}d*

*λ d*

*e*lnα λ

*d*!

*|u|L,a*

*and hence*

*u(q)*α

*L,b*≤

*ChLi*

*C*

_{+ (1 + δ )α}

*b*

_{L}d*λ d*

*e*lnα λ

*d*!

*kukL,a*.

*Proof.* The main tool will be the asymptotic of the Gamma function *Γ(x+d) _{Γ(x)}*

_{∼ x}d*as x*

_{→ +∞, which}follows at once from Stirling’s formula (see [22])

lim

*x*→+∞

*Γ(x + 1)*
*(x/e)x*√_{2πx}= 1.

In particular, for any_{δ > 0, there exists a number N = N(λ , δ , a, b, q) such that for all k ∈ N with k ≥ N,}

*Γ(k + 1 + q − a)*
*Γ(k + 1 − b)*

λ

*≤ (1 + δ )k*λ*d*.

*We can also assume that k≥ N implies k + q > |a| + 1, and k > |b| + 1, so that Γ*λ*,a(k + q) and Γ*λ*,b(k)*

are given by (2.18). Note that we always have * _{(1+k+q)}(1+k)*22

*≤ 1 if k ∈ N, for q ≥ 0.*

*For k _{≥ N, we have}*

*|u(k+q)(t)|*

*(αL)k*

_{−b}_{Γ}λ

*,b(k)*≤

*|u|L,aLk+q−a*Γλ

*,a(k + q)*

*(αL)k*

_{−b}_{Γ}λ

*,b(k)*≤

*|u|L,aL*

*d*α

*k−b*

*Γ(k + 1 + q − a)*

*Γ(k + 1 − b)*λ ≤ (1 + δ )

*|u|L,aL*

*d*α

*k−b*

*k*λ

*d*≤ (1 + δ )α

*b|u|L,aLd*sup

*t*≥0 (α

*−tt*λ

*d*) ≤ (1 + δ )α

*b|u|L,aLd*

_{λ d}*e*ln(α) λ

*d*,

where we used the fact that sup_{t}_{≥0}(α*−tt*λ*d*) = λ*d*
*e*ln(α)

λ*d*

, where* _{α > 1 and λ d > 0. If k ≤ N, we still}*
have

*|u(k+q)*

_{(t)|}*(αL)|k−b|*

_{Γ}

_{λ}

*≤*

_{,b}_{(k)}*|u|L,aL|k+q−a|−|k−b|*α

*|k−b|*Γ

_{λ}

*Γλ*

_{,a}(k + q)*,b(k)*·

This yields the result, for_{|k+q−a|−|k−b| ≤C(λ ,δ ,a,b,q), α}−|k−b|_{≤ 1, and}Γλ*,a(k+q)*

Γλ*,b(k)* *≤C(λ ,δ ,a,b,q)*

*for k _{≤ N(λ ,δ ,a,b,q).}*

The second statement follows by using the definition of_{k · k}α*L,b*and the estimate

*
u(q)*
*L*∞_{(K)}≤ L*|q−1−a|*_{Γ}
λ*,a(q − 1)*
*u*′_{}
*L,a≤ C(λ ,q,b,a)kukL,a*

*for q _{≥ 1, the case q = 0 being immediate.}*

2.6. Application to the semilinear heat equation. We aim to solve the system: ∂2

*xu*= ∂*tu− f (x,u,∂xu),* *x∈ [−1,1], t ∈ [0,T ],* (2.26)

*u(0,t) = u*0*(t),* *t∈ [0,T ],* (2.27)

∂*xu(0,t) = u*1*(t),* *t∈ [0,T ],* (2.28)

This is equivalent to solve the first order system

∂*xU* *= AU + F(x,U ),* (2.29)

*U(0) = U*0 (2.30)

*with U= (u, ∂xu), A =*

0 1
∂*t* 0
*, and F(x, (u*0*, u*1)) =
0
*− f (x,u*0*, u*1)
.
*Let L*> 0. We define the space X*L*:*= {U = (u*0*, u*1*) ∈ G*λ_{L}_{,}1

2*× G*

λ

*L*}, with

*kUk*X* _{L}= k(u*0

*, u*1)kX

*0k*

_{L}= ku

_{L}_{,1/2}

*+ ku*1k

*,*

_{L}where the norms are those defined in Definition 1 with*λ = 2. (Note that u*0 *is more regular than u*1of

one half derivative.) In particular, we have that

*kAUk*X* _{L}= ku*1k

_{L}_{,1/2}+ k∂

*tu*0k

*.*

_{L}*In the following result, L*1stands for the inverse of the radius of the initial datum.

*Theorem 2.7. Pick any L*1*< 1/4. Then there exists a number η > 0 such that for any U*0∈ X*L*1 *with*

*kU*0kX* _{L1}≤ η, there exists a solution to (2.29)-(2.30) for x ∈ [−1,1] in C([−1,1],XL*0

*) for some L*0

*> 0.*

*Proof of Theorem 2.7.* In order to apply Theorem 2.2, we introduce a scale of Banach spaces*(Xs*)*s*∈[0,1]

*as follows: for s*_{∈ [0,1], we set}
*kUkXs* *= e*
−τ*(1−s) _{k(u}*
0

*, u*1)kX

*−τ*

_{L}_{(s)}= e*(1−s)*

*ku*0k

*L(s),1/2+ ku*1k

*L(s)*, (2.31)

*L(s) = er(1−s)*1 (2.32)

_{L}*where r*= 2 and τ > 0 will be chosen thereafter. Note that (2.7) is satisfied because of (2.24) and the fact
*that L(s*′* _{) > L(s) for s}*′

_{< s. Actually, we have even that}*kUkX _{s′}*

*≤ e*−τ

*(s−s*

′_{)}

*kUkXs*. (2.33)

*Lemma 2.8 and Lemma 2.9 (see below) will allow us to select parameters so that G= A + F satisfies the*
assumptions of Theorem 2.2.

*Lemma 2.8. Let L*1*< 1/4. There exist τ*0*> 0 large enough and ε*0*< 1/4 such that we have the estimates*

*kAUkXs′* ≤

ε0

*s− s*′*kUkXs* *∀U ∈ Xs*,

*for all*τ ≥ τ0*and all s, s*′*with*0*≤ s*′*< s ≤ 1.*

*Proof.* By assumption, we have *L*

1/2 1

2 < 1/4 and we can pick δ > 0 so that

(1 + δ )*L*

1/2 1

2 < 1/4. (2.34)

Applying Lemma 2.6 with*λ = 2 and δ > 0 as in (2.34), and with q = 0, b = 1/2, a = 0 (respectively*
*q= 1, b = 0, a = 1/2), so that λ d = 1 in both cases, we obtain the existence of some number C = C*_{δ}> 0
such that
*kAUk*X_{α}_{L}*= ku*1kα*L*,1/2+ k∂*tu*0kα*L*
≤
*ChLiC*+1+ δ
*e*lnα*(αL)*
1/2* _{ku}*
1k

*L+ ku*0k

*L*,1/2 ≤

*C*+1+ δ

_{hLi}C*e*lnα

*(αL)*1/2

*kUk*X

*(2.35)*

_{L}uniformly for*α > 1 and L > 0. So, (2.35) applied with L = L(s), α =* *L _{L}(s_{(s)}*′)

*= er*′)

_{(s−s}_{> 1 becomes for}

*s*′*< s (we also use L*1*≤ C, for 0 < L*1< 1/4)

*kAUkX _{s′}*

*≤ e*−τ

*(s−s*′

_{)}

*CerC*+ (1 + δ )

*e*

*r*1−s′

_{2}

*1/2 1*

_{L}*er*′) !

_{(s − s}*kUkXs*≤

*Ce*−τ

*(s−s*′)

*erC+ (1 + δ )er*2

*L*1/2 1

*er(s − s*′

_{)}!

*kUkXs*≤

*e*−1

*τ(s − s*′

_{)}

*Ce*

*rC*

_{+ (1 + δ )}

*e*

*r*2

*L*1/2 1

*er*′

_{(s − s}_{)}!

*kUkXs*

where we have used

*e*−τ*(s−s*′)=*τ(s − s*′*)e*−

τ*(s−s*′_{)}

*τ(s − s*′_{)} ≤

*e*−1

*τ(s − s*′_{)} (2.36)

*and the fact that te−t* * _{≤ e}*−1

*for t*

_{≥ 0. Minimizing the constant in the r.h.s. leads to the choice r = 2.}*(Note that the initial space X*1= X

*L*1

*is independent on the choice of r.) We arrive to the estimate*

*kAUkXs′* ≤
*C*
τ + (1 + δ )
*L*1/2_{1}
2
!
1
*s _{− s}*′

*kUkXs*·

By (2.34), we can then pick τ0 large enough so that *C*_{τ} + (1 + δ )*L*

1/2 1

2 < 1/4 for τ ≥ τ0. The proof of

*Lemma 2.9. Let f be as in (1.5)-(1.8), and let F(x,U ) =*
0
*− f (x,u*0*, u*1)
*for x _{∈ [−1,1] and U =}*

*(u*0

*, u*1

*) ∈ L*∞

*(K)*2

*with*sup

*(ku*0k

*L*∞

_{(K)}, ku_{1}k

*∞*

_{L}

_{(K)}) < 4. Let r = 2, L_{1}

*> 0, and ε > 0. Then there exists*

τ0*> 0 (large enough) such that for τ ≥ τ*0*, there exists D> 0 (small enough) such that we have the*

*estimates*
*kF(x,U)kXs′* ≤
ε
*s _{− s}*′

*kUkXs*(2.37)

*kF(x,U) − F(x,V )kXs′*≤ ε

*s*′

_{− s}*kU −V kXs*(2.38)

*for*0* _{≤ s}*′

*0*

_{< s ≤ 1, and U = (u}*, u*1

*) ∈ Xs,V = (v*0

*, v*1

*) ∈ Xswith*

*kUkXs* *≤ D, kV kXs* *≤ D.* (2.39)

*Finally, for*0* _{≤ s ≤ 1 and U = (u}*0

*, u*1

*) ∈ Xs*

*withkUkXs*

*≤ D, the map x ∈ [−1,1] → F(x,U) ∈ Xs*

*is*

*continuous.*

*Proof.* *Since (2.37) follows from (2.38), for F _{(x, 0) = 0, it is sufficient to prove (2.38). Pick 0 ≤ s}*′

_{< s ≤}*1, D*satisfying (2.39). Then

_{> 0 and U,V ∈ X}s*kF(x,U) − F(x,V )kXs′* = k −
0
*f _{(x,U ) − f (x,V )}*
k

*Xs′*

*= e*−τ

*(1−s*′)

*0*

_{k f (x,u}*, u*1

*) − f (x,v*0

*, v*1)k

*L(s*′

_{)}

*≤ e*−τ

*(1−s*′)

### ∑

*p+q>0*

*kAp,q(x)[u*0

*pu*

*q*1

*− v*

*p*0

*v*

*q*1]k

*L(s*′

_{)}

*≤ e*−τ

*(1−s*′)

### ∑

*p+q>0*

*|Ap,q(x)| ku*0

*p− v*

*p*0k

*L(s*′

_{)}

*kuq*

_{1}k

*′*

_{L}_{(s}_{)}

*+kv*0

*p*k

*L(s*′

_{)}

*ku*

_{1}

*q− vq*

_{1}k

*′*

_{L}_{(s}_{)}

where we used the triangle inequality and Lemma 2.5. Note that, by (2.22), we have for a constant
*C= C(L*1*) ≥ 1 and any 0 ≤ s*′< 1
*ku*0k*L(s*′_{)}*+ ku*1k*L(s*′_{)}*≤ C ku*0k*L(s*′_{),1/2}*+ ku*1k*L(s*′_{)}*≤ Ce*τ*(1−s*
′_{)}
*kUkXs′* *≤ CDe*
τ_{,} _{(2.40)}
and similarly
*kv*0k*L(s*′_{)}*+ kv*1k*L(s*′_{)}*≤ CDe*τ.
Since, by Lemma 2.5,
*ku*0*p− v*
*p*
0k*L(s*′_{)} *= k(u*0*− v*0*)(up*0−1*+ u*
*p*_{−2}
0 *v*0*+ ··· + v*0*p*−1)k*L(s*′_{)}
*≤ ku*0*− v*0k*L(s*′_{)}
*ku*0k* _{L}p*−1

*′*

_{(s}_{)}

*+ ku*0k

*−2*

_{L}p*′*

_{(s}_{)}

*kv*0k

*L(s*′

_{)}

*+ ··· + kv*0k

*−1′*

_{L}p_{(s}_{)}

*≤ p(CDe*τ)

*p*−1

*0*

_{ku}*− v*0k

*L(s*′

_{)},

we infer that
*kF(x,U) − F(x,V )kXs′* *≤ e*
−τ*(1−s*′_{)}

### ∑

*p+q>0*

*|Ap,q(x)|p(CDe*τ)

*p+q−1ku*0

*− v*0k

*L(s*′

_{)}+

_{∑}

*p+q>0*

*|Ap,q(x)|q(CDe*τ)

*p+q−1ku*1

*− v*1k

*L(s*′

_{)}!

*=: e*−τ

*(1−s*′)

*(S*1

*+ S*2). (2.41)

*Let us estimate S*1*. Set M*′:*= M/(1 − b*−12 ). Since

*|Ap,q(x)| ≤*
*M*
*b*_{0}*pbq*_{1}(1 −
*|x|*
*b*2
)−1_{≤} *M*′
*b*_{0}*pbq*_{1} for*|x| ≤ 1,*
we have that
*S*1 ≤

### ∑

*p*>0

*M*′

*b*0

*p*

*CDe*τ

*b*0

*p*−1

### ∑

*q*≥0

*CDe*τ

*b*1

*q*

*ku*0

*− v*0k

*L(s*′

_{)}=

*M*′

*b*0

_{p}### ∑

_{≥0}

*(p + 1)*

*CDe*τ

*b*0

*p*(1 −

*CDe*τ

*b*1 )−1

*0*

_{ku}*− v*0k

*L(s*′

_{)}=

*M*′

*b*0(1 −

*CDe*τ

*b*0 )−2

_{(1 −}

*CDe*τ

*b*1 )−1

*0*

_{ku}*− v*0k

*L(s*′

_{)}≤

*8M*′ 0

_{b}*ku*0

*− v*0k

*L(s*′

_{)}

*≤ 2M*′

*ku*0

*− v*0k

*L(s*′

_{)}provided that

*D*

_{≤ min(}

*b*0

*e*−τ

*2C*,

*b*1

*e*−τ

*2C*)· (2.42)

Similarly, we can prove that

*S*2*≤ 2M*′*ku*1*− v*1k*L(s*′_{)}.

Therefore, using (2.33), (2.36) and (2.40), we infer that
*kF(x,U) − F(x,V )kX _{s′}*

*≤ 2M*′

*e*−τ

*(1−s*′

_{)}

*ku*0

*− v*0k

*L(s*′

_{)}

*+ ku*1

*− v*1k

*L(s*′

_{)}

*≤ 2M*′

*e*−τ

*(1−s*′)

*C*0

_{ku}*− v*0k

*′*

_{L}_{(s}_{),}1 2

*+ ku*1

*− v*1k

*L(s*′)

*≤ 2CM*′

*kU −V kX*

_{s′}*≤ 2CM*′

*e*−τ

*(s−s*′)

*≤*

_{kU −Vk}Xs*2CM*′ 1

_{e}*τ(s − s*′

_{)}

*kU −V kXs*·

To complete the proof of (2.38), it is sufficient to pick_{τ ≥ τ}0 withτ0*such that 2CM*′*/(eτ*0*) ≤ ε, and D*

For given 0*≤ s ≤ 1 and U = (u*0*, u*1*) ∈ Xs*with*kUkXs* *≤ D, let us prove that the map x ∈ [−1,1] →*

*F _{(x,U ) ∈ X}sis continuous. Pick any x, x*′

*∈ [−1,1]. From the mean value theorem, we have for r ∈ N*

that* _{|x}r_{− x}′r_{| ≤ r|x − x}*′

_{| for r ∈ N, and hence}*|Ap,q(x) − Ap,q(x*′*)| ≤ |x − x*′|

### ∑

*r*∈N

*rM*

*b*

_{0}

*pbq*

_{1}

*br*

_{2}=

*M*

*b*

_{0}

*pbq*

_{1}(1 − 1

*b*2 )−2

*′*

_{|x − x}_{|.}We infer that

*kF(x,U) − F(x*′

*,U )kXs*

*= e*− τ

*(1−s)*0

_{k f (x,u}*, u*1

*) − f (x*′

*, u*0

*, u*1)k

*L(s)*

*≤ e*−τ

*(1−s)*

### ∑

*p+q>0*

*|Ap,q(x) − Ap,q(x*′

*)|ku*

_{0}

*puq*

_{1}k

*L(s)*

*≤ e*−τ

*(1−s)M*

_{(1 −}1

*b*2 )−2

*′*

_{|x − x}_{|}

_{∑}

*p+q>0*

*0k*

_{ku}*L(s)*

*b*0

*p*

*ku*1k

*L(s)*

*b*1

*q*,

the last series being convergent for_{kUk}Xs*≤ D.*

We are in a position to prove Theorem 2.1.

*Proof of Theorem 2.1. Let f* *= f (x, y*0*, y*1*) be as in (1.5)-(1.8), −∞ < t*1*< t*2*< +∞ and R > 4. Pick*

*g*0*, g*1*∈ G*2*([t*1*,t*2*]) such that (2.4) holds. We will show that Theorem 2.7 can be applied provided that C*

*is small enough. Pick L*1*∈ (1/R,1/4). Let η = η(L*1*) > 0 be as in Theorem 2.7. Let U*0*= (u*0*, u*1) =

*(g*0*, g*1). We have to show that

*kU*0kX* _{L1}= kg*0k

_{L}_{1}

_{,}1

2*+ kg*1k*L*1≤ η

*for C small enough. It is sufficient to have*

*kg*0k*L*1,12 ≤
η
2, (2.43)
*kg*1k*L*1 ≤
η
2. (2.44)
Recall that
*kg*0k*L*112 = max
26* _{kg}*
0k

*L*∞

*([t*1

*,t*2]), 2 3

*−1 1 sup*

_{L}*t∈[t*1

*,t*2

*],k∈N*

*|g(k+1)*0

*(t)|*

*L|k−*1 2| 1 Γ2,1 2

*(k)* , (2.45)

*kg*1k

*L*1 = max 2 6

*1k*

_{kg}*L*∞

*1*

_{([t}*,t*2]), 2 3

*−1 1 sup*

_{L}*t∈[t*1

*,t*2

*],k∈N*

*|g(k+1)*1

*(t)|*

*Lk*

_{1}2−5

*(k!)*2

*−2 ! , (2.46) where Γ*

_{(1 + k)}_{2,}1 2

*(k) =*2−5

*Γ(k +*1

_{2})2

*(1 + k)*−2,

*if k*> 3/2, 2−5

*(k!)*2

*−2*

_{(1 + k)}_{if 0}

_{≤ k ≤ 3/2.}Then, if follows that (2.43) is satisfied provided that
*kg*0k*L*∞* _{([t}*
1

*,t*2]) ≤ 2 −7

_{η,}

_{(2.47)}

*kg(k+1)*0 k

*L*∞

*1*

_{([t}*,t*2]) ≤ 2 −4

*1*

_{ηL}*+|k−*1 2| 1 Γ2,1 2

*(k)*

*∀k ∈ N.*(2.48)

Since*Γ(k +*1_{2}* _{) ∼ Γ(k)k}*12

*∼ (k − 1)!k*1 2

*as k→ +∞, we have that Γ(k +*1 2) 2

*∼ (k!)*2

*of (2.48) is equivalent to 2−9*

_{/k. Thus, the r.h.s.}*ηLk*+12

1 *(k!)*2*k*−3 *as k→ +∞. Using (2.4) and the fact that L*1*> 1/R, we*

*have that (2.48) holds if C is small enough. The same is true for (2.47). Similarly, we see that (2.44) is*
satisfied provided that

*kg*1k*L*∞* _{([t}*
1

*,t*2]) ≤ 2 −7

_{η,}

_{(2.49)}

*kg(k+1)*1 k

*L*∞

*1*

_{([t}*,t*2]) ≤ 2 −9

*+1 1*

_{ηL}k*(k!)*2

*(k + 1)*−2

*∀k ∈ N.*(2.50)

*Again, (2.49) and (2.50) are satisfied if the constant C in (2.4) is small enough.*

*We infer from Theorem 2.7 the existence of a solution U* *= (y, ∂xy) ∈ C([−1,1],Xs*0*) for some s*0∈

*(0, 1) of (2.1)-(2.3). Let us check that y ∈ C*∞*([−1,1] × [t*1*,t*2]). To this end, we prove by induction on

*n*∈ N the following statement

*U _{∈ C}n_{([−1,1],C}k([t*1

*,t*2])2

*) ∀k ∈ N.*(2.51)

*The assertion (2.51) is true for n= 0, for Xs*0 *⊂ C*

*k _{([t}*

1*,t*2])2*for all k*∈ N. Assume (2.51) true for some

*n _{∈ N. Since A is a continuous linear map from X}sto Xs*′for 0

*≤ s*′

*< s ≤ 1, we have that*

*AU _{∈ C}n_{([−1,1],X}s) ⊂ Cn([−1,1],Ck([t*1

*,t*2])2

*) ∀s ∈ (0,s*0

*), ∀k ∈ N.*

*On the other hand, as f is analytic and hence of class C*∞*, we infer from (2.51) that F _{(x,U ) ∈ C}n*

_{([−1,1],}

*Ck*

_{([t}1*,t*2])2*) for all k ∈ N. Since ∂xU= AU + F(x,U ), we obtain that (2.51) is true with n replaced by*

*n _{+ 1. The proof of y ∈ C}*∞

*1*

_{([−1,1] × [t}*,t*2

*]) is complete. Finally, the proof of y ∈ G*1,2

*([−1,1] × [t*1

*,t*2]),

which uses some estimates of the next section, is given in appendix, with eventually a stronger smallness

assumption on the initial data.

3. CORRESPONDENCE BETWEEN THE SPACE DERIVATIVES AND THE TIME DERIVATIVES

We are concerned with the relationship between the time derivatives and the space derivatives of any solution of a general nonlinear heat equation

∂*ty*= ∂*x*2*y+ f (x, y, ∂xy),* (3.1)

*where f* *= f (x, y*0*, y*1*) is of class C*∞on R3.

*When f* = 0, then the jet (∂*n*

*xy(0, 0))n*≥0 is nothing but the reunion of the jets (∂*tny(0, 0))n*≥0 and

(∂*n*
*t* ∂*xy(0, 0))n*≥0, for
∂*n*
*t* *y*= ∂*x2ny,* *∀n ∈ N,* (3.2)
∂*n*
*t* ∂*xy*= ∂*x2n+1y,* *∀n ∈ N.* (3.3)

*When f is no longer assumed to be 0, then the relations (3.2)-(3.3) do not hold anymore. Nevertheless,*
there is still a one-to-one correspondence between the jet(∂*n*

*xy(0, 0))n*≥0 and the jets(∂*tny(0, 0))n*≥0and

(∂*n*

*t* ∂*xy(0, 0))n*≥0.

*Proposition 3.1. Let _{−∞ < t}*1

*≤ τ ≤ t*2

*< +∞. Assume that f ∈ C*∞(R3

*) and that y ∈ C*∞

*([−1,1]×[t*1

*,t*2])

*satisfies* *(3.1) on _{[−1,1] × [t}*1

*,t*2

*]. Then the determination of the jet (∂xny(0, τ))n*≥0

*is equivalent to the*

*determination of the jets*(∂*n*

*t* *y(0, τ))n*≥0 *and*(∂*tn*∂*xy(0, τ))n*≥0*.*

*Lemma 3.2. Let f _{∈C}*∞(R3

*∗*

_{) and n ∈ N}

_{. Then there exist two smooth functions H}_{n}_{= H}_{n}_{(x, y}_{0}

_{, y}_{1}

*−1)*

_{, ..., y}_{2n}*and ˜Hn*= ˜*Hn(x, y*0*, y*1*, ..., y2n) such that any solution y ∈ C*∞*([−1,1] × [t*1*,t*2*]) of (3.1) satisfies*

∂*n*

*t* *y* = ∂*x2ny+ Hn(x, y, ∂xy, ..., ∂x2n*−1*y)* *for(x,t) ∈ [−1,1] × [t*1*,t*2], (3.4)

∂*n*

*t* ∂*xy* = ∂*x2n+1y*+ ˜*Hn(x, y, ∂xy, ..., ∂x2ny)* *for(x,t) ∈ [−1,1] × [t*1*,t*2]. (3.5)

*Proof of Lemma 3.2.* *Assume first that n= 1. Then (3.4) holds with H*1*(x, y*0*, y*1*) = f (x, y*0*, y*1). Taking

*the derivative with respect to x in (3.1) yields*

∂*x*∂*ty*= ∂*x*3*y*+
*∂ f*
*∂ x(x, y, ∂xy) +*
*∂ f*
*∂ y*0
*(x, y, ∂xy)∂xy*+
*∂ f*
*∂ y*1
*(x, y, ∂xy)∂x*2*y,*

and hence (3.5) holds with ˜*H*1*(x, y*0*, y*1*, y*2) =∂_{∂}* _{x}f(x, y*0

*, y*1) +

_{∂}∂

_{y}f_{0}

*(x, y*0

*, y*1

*)y*1+

_{∂}∂

_{y}f_{1}

*(x, y*0

*, y*1

*)y*2.

*Assume now that (3.4) and (3.5) are satisfied at rank n*_{− 1, and let us prove that they are satisfied at}
*rank n. For (3.4), we notice that*

∂*n*
*t* *y* = ∂*t*(∂*tn*−1*y)*
= ∂*t*(∂*x2n*−2*y+ Hn*−1*(x, y, ∂xy, ..., ∂x2n*−3*y))*
= ∂*2n*−2
*x* ∂*ty*+
*2n*−3

### ∑

*k*=0

*∂ Hn*−1

*∂ yk*

*(x, y, ∂xy, ..., ∂x2n*−3

*y)∂t*∂

*xky*= ∂

*2n*−2

*x*(∂

*x*2

*y+ f (x, y, ∂xy)) +*

*2n*−3

### ∑

*k*=0

*∂ Hn*−1

*∂ yk*

*(x, y, ∂xy, ..., ∂x2n*−3

*y)∂xk*(∂

*x*2

*y+ f (x, y, ∂xy)) (3.6)*=: ∂

*−1*

_{x}2ny+ Hn(x, y, ∂xy, ..., ∂x2n*y)*

*for some smooth function Hn= Hn(x, y*0*, ..., y2n*−1). For (3.5), we notice that

∂*n*
*t* ∂*xy* = ∂*t*(∂*tn*−1∂*xy)*
= ∂*t*(∂*x2n*−1*y*+ ˜*Hn*−1*(x, y, ∂xy, ..., ∂x2n*−2*y))*
= ∂*2n*−1
*x* ∂*ty*+
*2n*−2

### ∑

*k*=0 ∂ ˜

*Hn*−1

*∂ yk*

*(x, y, ∂xy, ..., ∂x2n*−2

*y)∂t*∂

*xky*= ∂

*2n*−1

*x*(∂

*x*2

*y+ f (x, y, ∂xy)) +*

*2n*−2

### ∑

*k*=0 ∂ ˜

*Hn*−1

*∂ yk*

*(x, y, ∂xy, ..., ∂x2n*−2

*y)∂xk*(∂

*x*2

*y+ f (x, y, ∂xy))*=: ∂

*+ ˜*

_{x}2n+1y*Hn(x, y, ∂xy, ..., ∂x2ny)*

for some smooth function ˜*Hn*= ˜*Hn(x, y*0*, y*1*, ..., y2n*).

*Next, we relate the behaviour as n*_{→ +∞ of the jets (∂}* _{t}ny(0, τ))n*≥0 and(∂

*tn*∂

*xy(0, τ))n*≥0 to those of

the jet(∂*n*

*xy(0, τ))n*≥0. To do that, we assume that in (3.1) the nonlinear term reads

*f(x, y*0*, y*1) =

### ∑

*(p,q,r)∈N*3

*ap,q,r(y*0)*p(y*1)*qxr* *∀(x,y*0*, y*1) ∈ (−4,4)3, (3.7)

*where the coefficients ap,q,r*,*(p, q, r) ∈ N*3, satisfy (1.7)-(1.8).

*Proposition 3.3. Let _{−∞ < t}*1

*≤ τ ≤ t*2

*< +∞ and f = f (x, y*0

*, y*1

*) be as in (1.5)-(1.6) with the coefficients*

min( e*R, b*2*). Then there exists some number eC> 0 such that for any C ∈ (0, eC], , we can find a number*

*C*′*= C*′*(C, R, R*′*) > 0 with limC*→0+*C*′*(C, R, R*′*) = 0 such that for any function y ∈ C*∞*([−1,1] × [t*_{1}*,t*_{2}])

*satisfying(3.1) on _{[−1,1] × [t}*0

*,t*1

*] and*

*y(x, τ) = y*0

_{(x) =}_{∑}

∞
*n*=0

*an*

*xn*

*n!*,

*∀x ∈ [−1,1]*(3.8)

*for some y*0_{∈ R}* _{R}*˜

_{,C}, is such that|∂*k*
*x*∂*tny(0, τ)| ≤ C*′
*(2n + k)!*
*Rk _{R}′2n* ,

*∀k,n ∈ N.*(3.9)

*In particular, we have*|∂

*tny(0, τ)| ≤ C*′

*(2n)!*

*R′2n*,

*∀n ∈ N,*(3.10) |∂

*x*∂

*tny(0, τ)| ≤ C*′

*(2n + 1)!*

*RR′2n*,

*∀n ∈ N.*(3.11)

*Proof.* We know from Proposition 3.1 that the jets(∂*n*

*t* *y(0, τ))n*≥0 and(∂*tn*∂*xy(0, τ))n*≥0 are completely

determined by the jet(∂*n*

*xy(0, τ))n*_{≥0}*, that is by y*0*. A direct proof of estimates (3.10) and (3.11) (which*

follow at once from (3.9)) seems hard to be derived, whereas a proof of (3.9) can be obtained by induction
*on n. We shall need several lemmas.*

*Lemma 3.4. (see [10, Lemma A.1]) For all k _{, q ∈ N and a ∈ {0,...,k + q}, we have}*

### ∑

*j+ p = a*0

*≤ j ≤ k*0

*≤ p ≤ q*

*k*

*j*

*q*

*p*=

*k+ q*

*a*.

*The following Lemma gives the algebra property for the mixed Gevrey spaces G*1,2* _{([−1,1] × [t}*1

*,t*2]).

A slight modification of its proof actually yields Lemma 2.5, making the paper almost self-contained.
*Lemma 3.5. Let(x*0*,t*0*) ∈ [−1,1] × [t*1*,t*2*], R, R*′*∈ (0,+∞), q ∈ N, µ ∈ (q + 2,+∞), k*0*, n*0*∈ N, C*1*,C*2∈

*(0, +∞), and y*1*, y*2*∈ C*∞*([−1,1] × [t*1*,t*2*]) be such that*

|∂*xk*∂*tnyi(x*0*,t*0*)| ≤ Ci*
*(2n + k + q)!*
*Rk _{R}′2n_{(2n + k + 1)}*µ

*∀i = 1,2, ∀k ∈ {0,...,k*0

*}, ∀n ∈ {0,...,n*0}. (3.12)

*Then we have*|∂

*xk*∂

*tn(y*1

*y*2

*)(x*0

*,t*0

*)| ≤ Kq*,µ

*C*1

*C*2

*(2n + k + q)!*

*Rk*µ

_{R}′2n_{(2n + k + 1)}*∀k ∈ {0,...,k*0

*}, ∀n ∈ {0,...,n*0}, (3.13)

*where*

*Kq*,µ:= 2µ

*−q(1 + q)2q*

### ∑

*j*≥0

*i*

### ∑

≥0 1*(2i + j + 1)*µ

*−q*< ∞.

*Proof of Lemma 3.5:*Using*(2n + k + q)q _{≤ (1 + q)}q_{(1 + 2n + k)}q*

, we obtain

So, denoting*µ := µ − q > 2 and e*e *Ci*:= (1 + q)*qCi*, we have

|∂*k*

*x*∂*tnyi(x*0*,t*0)| ≤ e*Ci*

*(2n + k)!*

*Rk _{R}_{′2n}_{(2n + k + 1)}*eµ

*∀i = 1,2, ∀k ∈ {0,...,k*0

*}, ∀n ∈ {0,...,n*0}. (3.14)

From Leibniz’ rule, we have that
|∂*xk*∂*tn(y*1*y*2*)(x*0*,t*0)|
=
0*≤ j≤k*

### ∑

0*≤i≤n*

### ∑

*k*

*j*

*n*

*i*(∂

*∂*

_{x}j*1*

_{t}iy*)(x*0

*,t*0)(∂

*xk− j*∂

*tn−iy*2

*)(x*0

*,t*0) ≤

### ∑

0*≤ j≤k*0

*≤i≤n*

### ∑

*k*

*j*

*n*

*i*e

*C*1

*(2i + j)!*

*Rj*eµ

_{R}′2i_{(2i + j + 1)}*C*e2

*(2(n − i) + k − j)!*

*Rk− j*µe =

_{R}′2(n−i)_{(2(n − i) + k − j + 1)}*C*e1

*C*e2

*Rk*

_{R}′2n(2n + k)!### ∑

0*≤ j≤k*0

_{≤i≤n}### ∑

*k*

*j*

*n*

*i*

*2n+ k*

*2i+ j*−1

*(2i + j + 1)*eµ

*eµ | {z }*

_{(2(n − i) + k − j + 1)}*I*·

We infer from Lemma 3.4 that
*k*
*j*
*q*
*p*
≤
*k+ q*
*j+ p*
, for 0* _{≤ j ≤ k, 0 ≤ p ≤ q.}* (3.15)
This yields

*n*

*i*≤

*n*

*i*2 ≤

*2n*

*2i*,

and hence (using again (3.15))
*k*
*j*
*n*
*i*
≤
*k*
*j*
*2n*
*2i*
≤
*2n+ k*
*2i+ j*
·
*Finally, by convexity of x _{→ x}*µe, we have that

### ∑

0*0*

_{≤ j≤k}

_{≤i≤n}### ∑

*(2n + k + 1)*eµ

*(2i + j + 1)*eµ

*eµ =*

_{(2(n − i) + k − j + 1)}### ∑

0*0*

_{≤ j≤k}

_{≤i≤n}### ∑

1*2i+ j + 1*+ 1 2

*µe ≤ 2eµ−1*

_{(n − i) + k − j + 1}### ∑

0*≤ j≤k*0

_{≤i≤n}### ∑

_{1}

*(2i + j + 1)*µe+ 1

*(2(n − i) + k − j + 1)*eµ ≤ 2µe

### ∑

*j*

_{≥0}

*i*

### ∑

_{≥0}1

*(2i + j + 1)*µe < ∞,

where we used the fact that*µ = µ − q > 2.*e
It follows that
*I*≤ 2eµ

_{∑}

*j*≥0

*i*

### ∑

≥0 1*(2i + j + 1)*eµ ! 1

*(2n + k + 1)*eµ = 2 µ

*−q*

_{∑}

*j*≥0

*i*

### ∑

≥0 1*(2i + j + 1)*µ

*−q*!

*(2n + k + 1)q*

*(2n + k + 1)*µ,

and hence the proof of Lemma 3.5 is complete once we have noticed that *(2n + k)!(2n + k + 1)q*_{≤}

Let us go back to the proof of Proposition 3.3. Pick any numberµ > 3. We shall prove by induction
*on n*_{∈ N that}

|∂*xk*∂*tny(0, τ)| ≤ Cn*

*(2n + k)!*

*Rk _{R}′2n_{(2n + k + 1)}*µ,

*∀k ∈ N,*(3.16)

where 0*< Cn≤ C*′*< +∞. For n = 0, using the fact that R < eR, we have that*

|∂*xky(0, τ)| = |ak| ≤ C*
*k!*
e
*Rk* *≤ C* sup* _{p}*
∈N
(

*R*e

*R*)

*p*µ !

_{(p + 1)}*k!*

*Rk*µ

_{(k + 1)}*≤ C*0

*k!*

*Rk*µ provided that

_{(k + 1)}*C*≤ e

*C*= sup

*p*∈N (

*R*e

*R*)

*p*µ !

_{(p + 1)}_{−1}

*C*0. (3.17)

*Assume that (3.16) is satisfied at the rank n _{∈ N for some constant C}n*> 0. Then, by (1.1), (1.6), we have

that
|∂*k*
*x*∂*tn*+1*y*(0, τ)| = |∂*xk*∂*tn* ∂*x*2*y*+

### ∑

*p*

_{,q,r∈N}*ap,q,ryp*(∂

*xy)qxr*(0, τ)| =

_{|∂}

*∂*

_{x}k*n*

*t*∂

*x*2

*y*+

### ∑

*p,q∈N*

*Ap,q(x)yp*(∂

*xy)q*(0, τ)| ≤ |∂

*n*

*t*∂

*xk*+2

*y(0, τ)*| +

### ∑

*p*≥1 |∂

*k*

*x*∂

*tn*

*Ap*,0

*(x)yp*(0, τ)| +

_{∑}

*q*≥1

*p*

### ∑

≥0 |∂*k*

*x*∂

*tn*

*Ap,q(x)yp*(∂

*xy)q*(0, τ)|

*=: I*1

*+ I*2

*+ I*3. (3.18)

*(Note that the sum for I*2*is over p≥ 1, for A*0,0*(x) = 0.)*

*Since R*′*< R, we can pick some ε*_{∈ (0,1) such that}
*R*′2* _{≤ (1 − ε)R}*2.

*For I*1, we have that

*I*1*≤ Cn*

*(2n + k + 2)!*

*Rk*+2* _{R}′2n_{(2n + k + 3)}*µ

*≤ (1 − ε)Cn*

*(2n + k + 2)!*

*Rk _{R}′2n+2_{(2n + k + 3)}*µ· (3.19)

*Since Ap,q* *does not depend on t, we have that*∂*xk*∂*tnAp,k= 0 for n ≥ 1 and k ≥ 0. Next, for k ≥ 0, we*

have that
|∂*xkAp,q(0)| = k!|ap,q,k*| ≤
*k!*
*bk*_{2}
*M*
*b*_{0}*pbq*_{1} ≤
*C k!*
*(k + 1)*µ* _{R}k_{b}p*
0

*b*

*q*1

*for some constant C> 0 depending on R, b*2,*µ, for R < b*2.

Note that, still by (3.16), the function∂*xy*satisfies the estimate

|∂*xk*∂*tn*(∂*xy*)(0, τ)| ≤

*Cn*

*R*

*(2n + k + 1)!*

Usingµ − 1 > 2, we infer from iterated applications of Lemma 3.5 that
∂*k*
*x*∂*tn* *Ap*,0*yp*(0, τ) ≤
*CCnpKp(2n + k)!*
*Rk _{R}′2n_{(2n + k + 1)}*µ

*0 , (3.20) ∂*

_{b}p*k*

*x*∂

*tn*

*Ap,qyp*(∂

*xy)q*(0, τ) ≤

*CC*

*p+q*

*n*

*Kp+q(2n + k + 1)!*

*Rk*µ

_{R}′2n_{(2n + k + 1)}*0*

_{b}p*b*

*q*1

*Rq*

*∀q ≥ 1,*(3.21)

*where we denote K := max(K*0,µ*, K*1,µ). We infer from (3.20)-(3.21) that

*I*2 ≤

### ∑

*p*≥1

*CCnpKp(2n + k)!*

*Rk*µ

_{R}′2n_{(2n + k + 1)}*0 , (3.22)*

_{b}p*I*3 ≤

### ∑

*q*≥1

*p*

### ∑

≥0*CCnp+qKp+q(2n + k + 1)!*

*Rk*µ

_{R}_{′2n}_{(2n + k + 1)}*0*

_{b}p*b*

*q*1

*Rq*· (3.23)

Using (3.18)-(3.19) and (3.22)-(3.23), we see that the condition

|∂*xk*∂*tn*+1*y(0, τ)| ≤ Cn*+1

*(2n + k + 2)!*

*Rk _{R}′2n+2_{(2n + k + 3)}*µ,

*∀k ∈ N,*

is satisfied provided that

*(1 − ε)Cn*+

### ∑

*p*≥1

*CR*′2

*(2n + k + 1)(2n + k + 2)*

*2n+ k + 3*

*2n+ k + 1*µ (

*CnK*

*b*0 )

*p*+

_{∑}

*q*

_{≥1}

*p*

### ∑

_{≥0}

*CR*′2

*(2n + k + 2)*

_{2n}_{+ k + 3}*2n+ k + 1*µ (

*CnK*

*b*0 )

*p*(

*CnK*

*b*1

*R*)

*q*+1· (3.24)

_{≤ C}nPick a numberδ ∈ (0,1). Assume that

*Cn*≤ δ · min(

*b*0

*K*,
*b*1*R*

*K* ), (3.25)

*so that CnK/b*0*≤ δ and CnK/(b*1*R*) ≤ δ . Set

*Cn*+1 = λ*nCn*:= [(1 − ε) +
*K*
*b*0
*CR*′2
*(2n + 1)(2n + 2)*
3µ
1_{− δ} +
*K*
*b*1*R*
*CR*′2
*(2n + 2)*
3µ
(1 − δ )2*]Cn*·

*Then, with this choice of Cn*+1, (3.24) holds provided that (3.25) is satisfied. Next, one can pick some

*n*0*∈ N such that for n ≥ n*0, we have

λ*n*≤ 1.

*This yields Cn*+1 *≤ Cn* *for n≥ n*0*, provided that (3.25) holds for n= n*0*. To ensure (3.25) for n*=

0, 1, ..., n0*, it is sufficient to choose C*0small enough (or, equivalently, e*C*small enough) so that

*max C*0, λ0*C*0, λ1λ0*C*0, ..., λ*n*0−1···λ0*C*0

≤ δ · min(*b _{K}*0,

*b*1

*R*

*K*)· The proof by induction of (3.16) is achieved.

We can pick

*C*′*(C, R, R*′*) := max C*0, λ0*C*0, λ1λ0*C*0, ..., λ*n*0−1··· λ0*C*0

*with C*0*= C supp*∈N(*RR*e)

*p _{(p + 1)}*µ

*′*

_{, so that C}*′*

_{(C, R, R}

_{) → 0 as C → 0. The proof of Proposition 3.3 is}complete.

*Proposition 3.6. Let _{−∞ < t}*1

*≤ τ ≤ t*2

*< +∞ and f = f (x, y*0

*, y*1

*) be as in (1.5)-(1.6) with the coefficients*

*ap,q,r,* *(p, q, r) ∈ N*3*, satisfying* *(1.7)-(1.8). Assume in addition that b*2 > ˆ*R*:= 4e*(2e)*

−1

*≈ 4.81. Let*
˜

*R*> ˆ*R. Then there exists some number eC _{> 0 such that for any C ∈ (0, e}C] and any numbers R, R*′

*, L with*ˆ

*R< R*′*< R < min( ˜R, b*2*) and 4ee*

−1

*/R*′2*< L < 1/4, there exists a number C*′′*= C*′′*(C, R, R*′*, L) > 0 with*
lim*C*_{→0}+*C*′′*(C, R, R*′*, L) = 0 such that for any y*0∈ R_{R}_{˜}* _{,C}, we can pick a function y∈ G*1,2

*([−1,1] × [t*

_{1}

*,t*

_{2}])

*satisfying(3.1) for _{(x,t) ∈ [−1,1] × [t}*1

*,t*2

*] and*

*y(x, τ) = y*0*(x) =*
∞

### ∑

*n*=0

*an*

*xn*

*n!*,

*∀x ∈ [−1,1],*(3.26)

*and such that for all t _{∈ [t}*1

*,t*2]

|∂*tny(0,t)| ≤ C*′′*Ln(n!)*2, (3.27)

|∂*x*∂*tny(0,t)| ≤ C*′′*Ln(n!)*2. (3.28)

*Proof.* Let ˆ*R*:= 4e*(2e)*−1

, ˜*R*> ˆ*Rand R, R*′_{with ˆ}* _{R}_{< R}*′

_{< R < min( ˜}_{R, b}_{2}

_{). Pick e}

_{C,C as in Proposition 3.3,}*and pick any y*0_{∈ R}*R*˜*,C. If a function y as in Proposition 3.6 does exists, then both sequences of numbers*

*dn* := ∂*tny(0, τ),* *n*∈ N,

˜

*dn* := ∂*x*∂*tny(0, τ),* *n*∈ N

*can be computed inductively in terms of the coefficients an* = ∂*xny*0*(0), n ∈ N, according to *

Propo-sition 3.1. Furthermore, it follows from PropoPropo-sition 3.3 (see (3.10)-(3.11)) that we have for some
*C*′*= C*′*(C, R, R*′_{) > 0 and all n ∈ N}*|dn| ≤ C*′
*(2n)!*
*R′2n* ,
| ˜*dn| ≤ C*′
*(2n + 1)!*
*RR′2n* .

Note that both sequences*(dn*)*n*∈N and( ˜*dn*)*n*∈N(as above) can be defined in terms of the coefficients

*an’s, even if the existence of the function y is not yet established.*

*Let L*_{∈ (}*4e _{R}e−1*′2 ,14

*), R*′′∈ (

q

*4ee−1*

*L* *, R*′*) and M = M(R, R*′*, R*′′) > 0 such that

| ˜*dn| ≤ MC*′

*(2n)!*

*R′′2n*, *∀n ∈ N.*

*The following lemma is a particular case of [16, Proposition 3.6] (with ap= [2p(2p − 1)]*−1 *for p*≥ 1).

*Lemma 3.7. Let(dq*)*q*≥0*be a sequence of real numbers such that*

*|dq| ≤ CHk(2q)!* *∀q ≥ 0*

*for some H> 0 and C > 0. Then for all ˜H> ee*−1_{H there exists a function f}* _{∈ C}*∞

_{(R) such that}*f(q)(0) = dq* *∀q ≥ 0,* (3.29)