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HAL Id: hal-00807047

https://hal.archives-ouvertes.fr/hal-00807047

Submitted on 2 Apr 2013

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Remarks on the Cauchy problem for the

one-dimensional quadratic (fractional) heat equation

Luc Molinet, Slim Tayachi

To cite this version:

Luc Molinet, Slim Tayachi. Remarks on the Cauchy problem for the one-dimensional quadratic

(fractional) heat equation. Journal of Functional Analysis, Elsevier, 2015, 269, pp.2305-2327. �hal-

00807047�

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ONE-DIMENSIONAL QUADRATIC (FRACTIONAL) HEAT EQUATION

LUC MOLINET AND SLIM TAYACHI

Abstract.

We prove that the Cauchy problem associated with the one dimensional quadratic (fractional) heat equation:

ut

=

Dxu∓u2, t ∈

(0, T ), x

R or T , with 0

< α≤

1 is well-posed in

Hs

for

s ≥

max(−α, 1/2

2α) except in the case

α

= 1/2 where it is shown to be well-posed for

s > −1/2 and ill-posed for s

=

−1/2. As a

by-product we improve the known well-posedness results for the heat equation (α = 1) by reaching the end-point Sobolev index

s

=

−1. Finally, in the case 1/2< α≤

1, we also prove optimal results in the Besov spaces

B2s,q.

Keywords: Nonlinear heat equation, Fractional heat equation, Ill-posedness, Well- posedness, Sobolev spaces, Besov spaces.

2000 AMS Classification: 35K15, 35K55, 35K65, 35B40

1. Introduction and main results

The Cauchy problem for the quadratic fractional heat equation reads

(1.1) u

t

− D

x

u = ∓u

2

,

(1.2) u(0, ·) = u

0

,

where u = u(t, x) ∈ R , α ∈]0, 1], t ∈ (0, T ), T > 0, x ∈ R or T and D

x

is the Fourier multiplier by |ξ|

. In this paper, we consider actually the corresponding integral equation which is given by

(1.3) u(t) = S

α

(t)u

0

Z

t

0

S

α

(t − σ) u

2

(σ) dσ,

where S

α

(t) is the linear fractional heat semi-group and are interested in local well- posedness and ill-posedness results in the Besov spaces B

2s,q

(K) with s ∈ R , q ∈ [1, ∞[

and K = R or T .

1

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Let us recall that the Cauchy problem associated with the nonlinear heat equation in R

n

(1.4) u

t

− ∆u = ∓u

k

has been studied in many papers (see for instance [3, 4, 5, 6, 7, 9, 11, 12, 13, 14, 18, 19, 20, 21] and references therein). It is well-known that this equation is invariant by the space- time dilation symmetry u(t, x) 7→ u

λ

(t, x) = λ

k−12

u(λ

2

t, λx) and that the homogeneous Sobolev space ˙ H

n2k−12

is invariant by the associated space dilation symmetry ϕ(x) 7→

λ

k−12

ϕ(λx). The Cauchy problem (1.4) is known to be well-posed in H

s

for s ≥ s

c

=

n

2

k−12

except in the case (n, k) = (1, 2). Indeed, in this case the well-posedness is only known in H

s

for s > −1 and in [9] it is proven that the flow-map cannot be of class C

2

below H

−1

. Hence, this result is close to be optimal if one requires the smoothness of the flow-map. Recently, it was proven in [8] that the associated solution-map : u

0

7→ u cannot be even continuous in H

s

for s < −1. The first aim of this work is to push down the well-posedness result to the end point H

−1

. The second step is to extend these type of results for the one-dimensional quadratic fractional heat equation (1.1). Indeed we will derive optimal results for the Cauchy problem (1.1) in the scale of the Besov spaces B

2s,q

in the case

12

< α ≤ 1. In particular we will prove that the lowest reachable Sobolev index is −α that is strictly bigger then the critical Sobolev index for dilation symmetry that is 1/2 − 2α.

To reach the end-point index H

−α

we do not follow the classical method for parabolic equations (cf. [4, 12, 21]) that does not seem to be applicable here. We rather rely on an approach that was first introduced by Tataru [16] in the context of wave maps. Note that we mainly follow [10] where this method has been adapted for dispersive-dissipative equations. The fact that our equation is purely parabolic enables us to simplify the proof.

The optimality of our results follows from an approach first introduced by Bejenaru-Tao [1] for a one-dimensional quadratic Schr¨ odinger equation. This approach is based on a high-to low frequency cascade argument.

Finally we consider the case 0 < α ≤ 1/2. By classical parabolic methods we obtain the well-posedness in the Sobolev space

1

H

s

( R ), s ≥ 1/2 − 2α, unless α = 1/2. On the other hand, following a very nice result by Iwabuchi-Ogawa [8], we prove that (1.1) is ill-posed in H

−1/2

( R ) for α = 1/2. It is worth noticing that (1/2, −1/2) is the intersection of the straight borderlines for well-posedness that are s = −α and s = 1/2 − 2α.

1

Recall that 1/2

2α is the critical Sobolev index for dilation symmetry.

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Before stating our main result, let us give the precise definition of well-posedness we will use in this paper.

Definition 1.1. We will say that the Cauchy problem (1.1)-(1.2) is (locally) well-posed in some normed function space B if, for any initial data u

0

∈ B , there exist a radius R > 0, a time T > 0 and a unique solution u to (1.3), belonging to some space-time function space continuously embedded in C([0, T ]; B), such that for any t ∈ [0, T ] the map u

0

7→ u(t) is continuous from the ball of B centered at u

0

with radius R into B. A Cauchy problem will be said to be ill-posed if it is not well-posed.

Theorem 1. Let K = R or T and α ∈]1/2, 1]. The Cauchy problem (1.1) is locally well- posed in the Besov space B

2s,q

(K) if and only if (s, q) ∈ R × [1, +∞[ satisfies s > −α or s = −α and q ∈ [1, 2].

Remark 1.2. Our negative results can be stated more precisely in the following way : For any couple (s, q) ∈ R × [1, +∞[ satisfying, s < −α or s = −α and q > 2, there exists T > 0 such that the flow-map u

0

7→ u(t) is not continuous at the origin from B

2s,q

(K) into D

0

(K) for any t ∈]0, T [.

This paper is organized as follows. In the next section we define our resolution spaces in the case K = R . In Section 3 we derive the needed linear estimates on the free term and the retarded Duhamel operator and in Section 4 we prove our well-posedness result.

Section 5 is devoted to the non-continuity results for the same range of α. In Section 6 we complete the well-posedness results by considering the case 0 < α ≤ 1/2. First, by classical parabolic methods, we prove that we can reach the critical Sobolev index for dilation symmetry that is 1/2 − 2α unless α = 1/2. Then, following [8], we prove that (1.1) is ill-posed in H

−1/2

( R ) for α = 1/2. Finally we explain the needed adaptations in the periodic case K = T .

Throughout the paper, we will write f . g, whenever a constant C ≥ 1, only depending on parameters and not on t or x, exists such that f ≤ Cg. We write f ∼ g if f . g, and g . f. If C depends on parameters a, we write f .

a

g, instead.

2. Resolution Space We use the following definition for the Fourier transform

F (f )(ξ) = Z

R

f (x)e

−ixξ

dx, and the inverse Fourier transform is

F

−1

(f )(x) = 1 2π

Z

R

F(f )(ξ)e

ixξ

dξ,

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for f in S( R ), the Schwartz space of rapidly decreasing smooth functions, and by duality if f in S

0

( R ), the space of tempered distributions. We denote sometimes F(f) by ˆ f . The fractional power of the Laplacien can be defined by the Fourier transform: For α ∈ R ,

F (−∂

x2

)

α

f

(ξ) = |ξ|

F(f )(ξ).

Let s be a real number. The Sobolev space H

s

( R ) is defined by H

s

( R ) = {u ∈ S

0

( R ) |

Z

R

(1 + |ξ|

2

)

s

|F (u)(ξ)|

2

dξ < ∞}

where F (u) is the Fourier transform of u. The norm on H

s

( R ) is defined by

||u||

Hs(R)

= Z

R

(1 + |ξ|

2

)

s

|F (u)(ξ)|

2

1/2

.

We will need a Littlewood-Paley analysis. Let η ∈ C

0

( R ) be a non negative even function such that supp η ⊂ [−2, 2] and η ≡ 1 on [−1, 1]. We define ϕ(ξ) = η(ξ/2) − η(ξ) and the Fourier multipliers

F (∆

j

u)(ξ) = ϕ(2

−j

ξ)F u(ξ), j ≥ 0, and F(∆

−1

u)(ξ) = η(ξ)Fu(ξ) .

For any s ∈ R and q ≥ 1, the Besov space B

2s,q

( R ) is defined as the completion of S( R ) for the norm

kuk

Bs,q

2 (R)

= X

j≥−1

2

jsq

||∆

j

u||

qL2(R)

1/q

.

For s ∈ R , s

1

< s

2

, 1 ≤ q

1

≤ q

2

and q ≥ 1 we have the following embeddings B

2s,q1

, → B

2s,q2

and B

2s2,q

, → B

2s1,1

.

Moreover, It is well-known that the H

s

( R )-norm is equivalent to the B

2s,2

-norm so that H

s

( R ) = B

2s,2

( R ).

Finally, for 1 ≤ p ≤ ∞ we consider the space-time space ˜ L

p

( R ; B

s,q2

) equipped with the norm

kuk

˜

LptB2s,q

= h X

j≥−1

2

sjq

k∆

j

u(t)k

q

LptL2x

i

1/q

.

We are now able to define our resolution space. For T > 0 fixed, we consider the space X

α,Ts,q

= ˜ L

T

B

2s,q

∩ L ˜

2T

B

2s+α,q

equipped with the norm:

kuk

Xs,q

α,T

= hX

j

sup

t∈]0,T[

2

sjq

k∆

j

u(t)k

q

L2x

i

1/q

+ hX

j

2

j(s+α)q

k∆

j

uk

q

L2TL2x

i

1/q

.

Let us also consider the space Y

αs,q

:=

n

u ∈ L ˜

1

R

+

; B

s+2α,q2

( R )

and ∂

t

u ∈ L ˜

1

R

+

; B

s,q2

( R ) o

(6)

equipped with the norm:

kuk

Ys,q

α

= hX

j

2

(s+2α)jq

k∆

j

uk

q

L1tL2x

i

1/q

+ hX

j

2

sjq

k∆

j

u

t

k

q

L1tL2x

i

1/q

For T > 0, the restriction space Y

α,Ts,q

of Y

αs,q

is endowed with the usual norm kuk

Ys,q

α,T

= inf

v∈Y

{kvk

Ys,q

α

, v ≡ u on ]0, T [ } .

For T > 0 our resolution space will be E

α,Ts,q

= X

α,Ts,q

+ Y

α,Ts,q

endowed with the usual norm for a sum space :

kuk

Es,q

α,T

:= inf

u=v+w

(kvk

Xs,q

α,T

+ kwk

Ys,q

α,T

).

3. Linear estimates We first establish the following lemma.

Lemma 3.1. Let 0 < T ≤ 1 and ϕ ∈ B

2s,q

. Then we have

(3.1) kS

α

(t)ϕk

Xs,q

α,T

. kϕk

Bs,q

2

.

Proof. The standard smoothing effect of the (fractional) heat semi-group is not sufficient here since we have

kS

α

(t)ϕk

Bs+α,q

2

. t

12

kϕk

Bs,q

2

and the right hand side of this inequality is not square integrable near t = 0. Integrating by parts the linear fractional heat equation

(3.2) ∂

t

u − D

x

u = 0

on ]0, t[× R , t > 0, against u and using that u(0) = ϕ, we obtain Z

R

u

2

(t, x) dx + Z

t

0

Z

R

|D

αx

u(s, x)|

2

dx ds = Z

R

ϕ

2

(x) dx .

Using that for each j ∈ N , ∆

j

S

α

(t)D

sx

ϕ satisfies the linear fractional heat equation (3.2) with D

sx

ϕ as initial datum, powering in q/2 and then summing in j ≥ 0, we get for any T > 0,

X

j≥0

2

sjq

k∆

j

S

α

(t)ϕk

qL TL2x

1/q

+ X

j≥0

2

(s+α)jq

k∆

j

S

α

(t)ϕk

q

L2TL2x

1/q

. X

j≥0

2

sjq

k∆

j

ϕk

qL2

x

1/q

.

On the other hand, for j = −1 we write k∆

−1

S

α

(t)ϕk

L

TL2x

+ k∆

−1

S

α

(t)ϕk

L2

TL2x

≤ 2 T

1/2

k∆

−1

S

α

(t)ϕk

L

TL2x

≤ 2 T

1/2

k∆

−1

ϕk

L2 x

and the result follows.

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As a direct consequence we get the following estimate on the semi-group : Let 0 < T ≤ 1 and ϕ ∈ B

2s,q

then it holds

(3.3) kS

α

(t)ϕk

Es,q

α,T

. kS

α

(t)ϕk

Xs,q

α,T

. kϕk

Bs,q

2

.

Let us now define the operator L

α

by

(3.4) L

α

(f )(t, x) =

Z

t 0

S

α

(t − t

0

)f (t

0

)dt

0

. Then we have

Lemma 3.2. Let 0 < T ≤ 1 and f ∈ E

α,Ts,q

. Then we have

(3.5) kL

α

(f)k

Ys,q

α,T

. (1 + T )kf k

L˜1 TB2s,q

.

Proof. It suffices to prove the result for a time extension of L

α

(f ). More precisely, it suffices to prove that

kηL

α

(f )k

Ys,q

α,T

. kf k

L˜1 t>0B2s,q

,

for any f ∈ L ˜

1t>0

B

2s,q

supported in time in [0, 1] and where η ∈ C

0

( R ) is defined in Section 2 . Let u be the solution of the Cauchy problem

t

u − D

x

u = f, u(0) = 0.

It is easy to check that u = L

α

(f ). Multiplying this equation by u and integrating by parts, we get

1 2

d dt

Z

R

u

2

+ Z

R

(D

αx

u)

2

= Z

R

f u .

Applying this equality to localizing in frequencies equation and using Bernstein inequality and the Cauchy-Schwarz one, we get for any j ∈ N ,

1 2

d dt

Z

R

u

2j

+ 2

2αj

Z

R

u

2j

≤ Z

R

f

j2

1/2

Z

R

u

2j

1/2

. Here u

j

= ∆

j

u, f

j

= ∆

j

f . If R

R

u

2j

1/2

6= 0 we divide this last inequality by R

R

u

2j

1/2

to obtain

d dt

Z

R

u

2j

1/2

+ 2

2αj

Z

R

u

2j

1/2

≤ Z

R

f

j2

1/2

. On the other hand, the smoothness and non negativity of t 7→ ku

j

(t)k

2L2

x

forces

dtd

ku

j

(t)k

2L2

x

= 0 as soon as ku

j

(t)k

L2

x

= 0. This ensures that the above differential inequality is actually valid for all t > 0. Integrating this differential inequality in time we get for any j ∈ N ,

(3.6) 2

2αj

ku

j

k

L1

tL2x

. kf

j

k

L1 tL2x

.

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Now, in the case j = −1, we get in the same way k∆

−1

uk

L2

. k∆

−1

f k

L1

tL2x

. Integrating on [0, 2T ] this leads to kη∆

−1

uk

L1

tL2x

. T kf

j

k

L1 tL2x

.

Finally, in view of the linear fractional heat equation, the triangle inequality leads to (3.7) k∂

t

(ηu

j

)k

L1

tL2x

. kη∂

t

u

j

k

L1

tL2x

+ ku

j

k

L1

tL2x

. kf

j

k

L1 tL2x

.

Since u = L

α

(f ), summing in j ∈ N using Bernstein inequalities and recalling the expres- sion of the norm in Y

s,α

, we conclude that

(3.8) kηL

α

(f)k

Ys,q

α,T

. kf k

L˜1 tBs,qx

.

Lemma 3.3. Let 0 < T ≤ 1 and u ∈ Y

α,Ts,q

. Then it holds

(3.9) kuk

Xs,q

α,T

= kuk

L˜

TB2s,q

+ kuk

L˜2

TB2s+α,q

. kuk

Ys,q

α,T

. In particular, E

α,Ts,q

, → X

α,Ts,q

.

Proof. Again it suffices to prove this estimate for the non restriction spaces. Actually, by localizing in space frequencies it suffices to prove that for any function u ∈ L

1

( R

+

; L

2

( R )) with u

t

∈ L

1

( R

+

; L

2

( R )) it holds

(3.10) kuk

L

t L2x

. ku

t

k

L1

tL2x

and kuk

2L2

t>0L2x

. kuk

L1

t>0L2x

ku

t

k

L1 t>0L2x

.

Indeed, applying (3.10) to the space frequency localization u

j

of u, Bernstein’s inequalities lead to

2

js

ku

j

k

L

t L2x

. 2

js

k∂

t

u

j

k

L1

tL2x

and 2

jq(s+α)

ku

j

k

q

L2t>0L2x

. 2

jq(s+2α)/2

ku

j

k

q/2

L1t>0L2x

2

jqs/2

k∂

t

u

j

k

q/2

L1t>0L2x

, which yields the result by summing in j and applying Cauchy-Schwarz in j on the right-

hand member of the second inequalities.

Let us now prove (3.10). The first part is a direct consequence of the equality u(t) =

− R

t

u

t

(s)ds and Minkowsky integral inequality. To prove the second part we notice that u

2

(t) = −u(t) R

t

u

t

(s)ds so that we can write Z

0

Z

R

u

2

(t, x) dx dt = Z

0

Z

R

u(t, x) Z

t

0

u

t

(s, x) ds dx dt .

Z

R

Z

∞ 0

|u(t, x)| dt Z

0

|u

t

(t, x)| dt dx . k

Z

∞ 0

|u(t, ·)| dtk

L2 x

k

Z

∞ 0

|u

t

(t, ·)| dtk

L2 x

. kuk

L1

t>0L2x

ku

t

k

L1 t>0L2x

,

where we used Minkowsky integral inequality in the last step.

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4. Well-posedness for 1/2 < α ≤ 1 According to Lemma 3.2 we easily get for 0 < T ≤ 1,

kL

α

(u

2

)k

Es,q

α,T

. kL

α

(u

2

)k

Ys,q

α,T

. X

j

2

jsq

k∆

j

(u

2

)k

q

L1TL2x

1/q

. X

j

2

jq(s+1/2)

k∆

j

(u

2

)k

q

L1TL1x

1/q

. (4.1)

Now, by para-product decomposition we have k∆

j

(u

2

)k

L1

TL1x

. kuk

L2

T ,x

k∆

j

uk

L2

T ,x

+ X

|k−k0|≤3, k&j

k∆

k

uk

L2

T ,x

k∆

k0

uk

L2 T ,x

.

The contribution to (4.1) of the first term of the above right-hand side member can be estimated by

kuk

L2 T ,x

X

j

2

jq(s+1/2)

k∆

j

uk

q

L2TL2x

1/q

= kuk

L2 T ,x

kuk

˜

L2TBs+1/2,q2

which is acceptable as soon as α ≥ 1/2 and (s > −α or s = −α and 1 ≤ q ≤ 2 ). Indeed, this last condition ensures that kuk

L2

T ,x

. kuk

L˜2

TB2s+α,q

. For the second term, we notice that for α > 1/2, we can estimate its contribution by

kuk

L˜2

TB0,∞2

kuk

L˜2 TB2s+α,q

X

j

2

jq(1/2−α)

1/q

≤ C(α) kuk

L˜2

TB0,∞2

kuk

L˜2

TB2s+α,q

, where C(α) > 0 only depends on α > 1/2.

In view of Lemma 3.3, this proves that for α > 1/2, (4.2) kL

α

(u

2

)k

Es,q

α,T

. kuk

L˜2 T x

kuk

L˜2

TBs+α,q2

. kuk

E−α,2 α,T

kuk

Es,q

α,T

,

where the implicit constants only depends on α. In the same way, for any α ∈]1/2, 1], there exists C

α

> 0 such that

kL

α

(uv)k

Es,α

T

. kuk

L˜2 T x

kvk

L˜2

TB2s+α,q

+ kvk

L˜2 T x

kuk

L˜2 TB2s+α,q

≤ C

α

kuk

E−α,2

α,T

kvk

Es,q

α,T

+ kvk

E−α,2

α,T

kuk

Es,q

α,T

. (4.3)

Let us now fixed α ∈]1/2, 1]. (4.3) together with (3.3) lead to the existence of β > 0 such that for all u

0

∈ B

2−α,2

( R ) with

(4.4) ku

0

k

B−α,2

2 (R)

≤ β , the mapping

u 7→ S

α

(·)u

0

+ L

α

(u

2

)

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is a strict contraction in the ball of E

α,1−α,2

centered at the origin of radius (2C

α

)

−1

. Noticing that E

α,1s,q

, → E

α,1−α,2

as soon as

(4.5) (s ≥ −α and 1 ≤ q ≤ 2) or s > −α,

this ensures that the above mapping is also strictly contractive is a small ball of E

α,Ts,q

as soon as (4.5)-(4.4) are satisfied. Since S

α

is a continuous semi-group in B

2s,q

( R ) and accord- ing to Lemma 3.3, E

α,1s,q

, → L ˜

1

B

2s,q

, this leads to the well-posedness result in B

2s,q

( R ) under conditions (4.5) for initial data satisfying (4.4). The result for general initial data follows by a simple dilation argument. Indeed, the equation (1.1) is invariant under the dilation u(t, x) 7→ u

λ

(t, x) = λ

u(λ

t, λx) whereas kλ

u

0

(λ·)k

B−α,2

2 (R)

≤ λ

α−1/2

ku

0

k

B−α,2 2 (R)

→ 0 as λ & 0. Classical arguments then lead to the well-posedness result in B

2s,q

( R ) for arbitrary large initial data with a minimal time of existence T ∼ 1 + ku

0

k

B−α,2

2 (R)

2α−1

. Note that, the well-posedness being obtained by a fixed point argument, as a by-product we get that the solution-map : u

0

7→ u is real analytic from B

2s,q

( R ) into C [0, T ]; B

s,q2

( R )

.

5. Ill-posedness results for 1/2 < α ≤ 1 .

In this section we prove discontinuity results on the flow map u

0

7→ u(t) for any fixed t > 0 less than some T > 0. To clarified the presentation we separate the case s < −α and the case s = −α and q > 2.

5.1. The case s < −α. We take the counter example of [10] used for the KdV-Burgers equation.

We define the sequence of initial data {φ

N

}

N≥1

⊂ C

( R ) via its Fourier transform by

(5.6) φ ˆ

N

(ξ) = N

α

χ

IN

(ξ) + χ

IN

(−ξ) ,

where I

N

= [N, N + 2] and χ

IN

is the characteristic function of the interval I

N

, χ

IN

(ξ) =

1 if ξ ∈ I

N

, 0 if ξ 6∈ I

N

. That is

φ

N

(x) =

Nα π

sin(x) x

cos

(N + 1)x

if x 6= 0,

Nα

π

if x = 0.

Clearly φ

N

∈ C

0

( R ) :=

n

f ∈ C( R )| lim

|x|→∞

f(x) = 0 o

. For any (s, q) ∈ R × [1, +∞] we have kφ

N

k

Bs,q

2 (R)

∼ N

α+s

and thus kφ

N

k

B−α,q

2 (R)

∼ 1

whereas φ

N

→ 0 in B

2s,q

( R ), for s < −α.

(11)

Let us consider the following bilinear operator, closely related to second iteration of the Picard scheme,

A

2

(t, h, h) = 2 Z

t

0

S

α

(t − t

0

)[S

α

(t

0

)h]

2

dt

0

,

where S

α

is the semi-group of the linear heat equation. Let us denote by F

x

the partial Fourier transform with respect to x. Recall that

F

x

S

α

(t)ϕ

(ξ) = e

−t[ξ|

F

x

(ϕ)(ξ), ∀ ϕ ∈ S

0

( R ), and F

x

(f g) = F

x

(f ) ? F

x

(g), where ? is the convolution product.

It follows that

F

x

A

2

(t, φ

N

, φ

N

)

(ξ) = 2 Z

t

0

e

−(t−t0)|ξ|

Z

R

e

−t01|

φ ˆ

N

1

)e

−t0|ξ−ξ1|

φ ˆ

N

(ξ − ξ

1

)dξ

1

dt

0

= 2

Z

R

φ ˆ

N

1

) ˆ φ

N

(ξ − ξ

1

) Z

t

0

e

−(t−t0)|ξ|

e

−[|ξ1|+|ξ−ξ1|]t0

dt

0

1

= 2

Z

R

φ ˆ

N

1

) ˆ φ

N

(ξ − ξ

1

) e

−[|ξ1|+|ξ−ξ1|]t

− e

−|ξ|t

Θ

α

(ξ, ξ

1

)

1

, (5.7)

where

Θ

α

(ξ, ξ

1

) = |ξ|

− |ξ

1

|

− |ξ − ξ

1

|

. Note that the integrand is nonnegative. In particular, F

x

A

2

(t, φ

N

, φ

N

)

(ξ) = |F

x

A

2

(t, φ

N

, φ

N

) (ξ)|.

Let

K

1N

(ξ) = n

ξ

1

| (ξ − ξ

1

, ξ

1

) ∈ I

N

× I

N

or (ξ − ξ

1

, ξ

1

) ∈ I

−N

× I

−N

o

and

K

2N

(ξ) = n

ξ

1

| (ξ − ξ

1

, ξ

1

) ∈ I

N

× I

−N

or (ξ − ξ

1

, ξ

1

) ∈ I

−N

× I

N

o For any |ξ| ≤

12

, K

1N

(ξ) = ∅ and thus

F

x

A

2

(t, φ

N

, φ

N

)

(ξ) = 2 Z

K2N(ξ)

φ ˆ

N

1

) ˆ φ

N

(ξ − ξ

1

) e

−[|ξ1|+|ξ−ξ1|]t

− e

−|ξ|t

Θ

α

(ξ, ξ

1

)

1

.

On the other hand, for any (a, b) ∈ R

+

× R

one has obviously,

|a|

+ |b|

− |a + b|

≥ |a| ∧ |b|

.

Moreover, it is easy to check that |K

2N

(ξ)| ≥ 1 and that in K

2N

(ξ) it holds N

α

(ξ, ξ

1

)| ≤ 2(N + 2)

Hence, fixing t ∈]0, 1[, it holds F

x

A

2

(t, φ

N

, φ

N

)

(ξ) ≥ e

−t/2

N

1 − e

−Nt

2(N + 2)

≥ 1

4 e

−t/2

, ∀ξ ∈ [−1/2, 1/2],

(12)

for any N > 0 large enough. This ensures that for any fixed (s, q) ∈ R × [1, +∞] and any fixed t ∈]0, 1[,

(5.8) kA

2

(t, φ

N

, φ

N

)k

Bs,q

2

≥ 1

4 e

−t/2

for N > 0 large enough. Taking s < −α this proves the discontinuity of the map u

0

7→ u(t) in B

2s,q

. To prove the discontinuity with value in D

0

( R ), we proceed as follows. Let g ∈ S ( R ) be such that ˆ g is positive equal to 1 on [−1/4, 1/4] and supported in [−1/2, 1/2]. We obtain for N > 0 large enough,

| Z

R

A

2

(t, φ

N

, φ

N

)(x)g(x)dx| ≥ 1 8 e

−t/4

.

On the other hand the analytical well-posedness ensures that A

2

(t, φ

N

, φ

N

) is bounded in B

2−α,1

uniformly in N . Then, since D( R ) is dense in S( R ), there exists ϕ ∈ D( R ) such that (5.9)

Z

R

A

2

(t, φ

N

, φ

N

)(x)ϕ(x)dx ≥ 1

2

4

e

−t/4

.

This shows that A

2

(t, φ

N

; φ

N

) does not converge to 0 in D

0

( R ) and proves the discon- tinuity from B

2s,q

( R ), s < −α into D

0

( R ).

We now turn to prove the discontinuity of the flow-map u(t, ·) : B

2s,q

( R ) −→ B

2s,q

( R )

h 7−→ u(t, h) = S

α

(t)h + R

t

0

S

α

(t − σ) u

2

(σ) dσ.

By the theorem of well posedness, there exist T > 0 and

0

> 0 such that for any 0 < ≤

0

, khk

B−α,1

2

≤ 1 and 0 ≤ t ≤ T,

u(t, h) = S

α

(t)h +

X

k=2

k

A

k

(t, h

k

),

where h

k

= (h, · · · , h), h

k

7→ A

k

(t, h

k

) are k−linear continuous maps from (B

2−α,1

( R ))

k

into C([0, T ]; B

−α,12

( R )) and the series converges absolutely in C([0, T ]; B

2−α,1

( R )).

Hence

u(t, φ

N

) −

2

A

2

(t, φ

N

, φ

N

) = S

α

(t)φ

N

+

X

k=3

k

A

k

(t, φ

kN

).

Using the inequalities

kS

α

(t)φ

N

k

Bs,1

2 (R)

≤ kφ

N

k

Bs,1

2 (R)

≤ 2N

s+α

and

X

k=3

k

A

k

(t, φ

kN

)

B2−α,1(R)

0

3

X

k=3

k0

A

k

(t, φ

kN

)

B2−α,1(R)

≤ C

3

,

(13)

where C is a positive constant, we deduce that for s ≤ −α,

(5.10) sup

t∈[0,T]

ku(t, φ

N

) −

2

A

2

(t, φ

N

, φ

N

)k

Bs,1

2 (R)

≤ C

3

+ 2

0

N

s+α

. According to (5.8) this leads, for ≤

C−12e5−t/4

, to

ku(t, φ

N

)k

Bs,q

2 (R)

≥ C

0

2

/2 − 2

0

N

s+α

.

By letting N → ∞ we obtain the discontinuity result since u(t, 0) = 0 and φ

N

→ 0 in B

2s,q

( R ) for s < −α. The discontinuity of the flow-map from B

2s,q

( R ) into D

0

( R ) follows in the same way by combining (5.9) and (5.10).

5.2. The case s = −α and q > 2. This case is similar to the precedent except that we have to change a little the sequence of initial data. Here we take the same sequence as in the work of Iwabuchi and Ogawa [8]. For any N ≥ 10 we define

ψ

N

= N

12

X

N≤j≤2N

φ

2j

.

where φ

2j

is defined in (5.6).

Noticing that ∆

k

φ

2j

= δ

k,j

φ

2j

, we can easily check that kψ

N

k

B−α,q

2

∼ N

12+1q

. In particular, kψ

N

k

Bα,q

2

→ 0 for any q > 2 whereas kψ

N

k

B−α,2

2

= kψ

N

k

H−α

∼ 1. Since the equation is analytically well-posed in H

−α

( R ), in view of the preceding case, it suffices to prove that A

2

N

, ψ

N

, t) does not tend to 0 in D

0

. By the localization, it holds φ

2j

2j0

≡ 0 on ] − 1/2, 1/2[ as soon as j 6= j

0

≥ 10 and the same reasons as above lead to

F

x

A

2

(t, ψ

N

, ψ

N

)

(ξ) = N

12

X

N≤j≤2N

Z

K22j(ξ)

φ ˆ

2j

1

) ˆ φ

2j

(ξ − ξ

1

) e

−[|ξ1|+|ξ−ξ1|]t

− e

−|ξ|t

Θ

α

(ξ, ξ

1

)

1

≥ N

12

e

−t/2

N

12

N

1 − e

−Nt

2(N + 2)

≥ 1

4 e

−t/2

, ∀ξ ∈ [−1/2, 1/2], for any N > 0 large enough. This completes the proof of the ill-posedness results for 1/2 < α ≤ 1.

6. Further remarks

6.1. Wellposedness results in the case 0 < α ≤ 1/2. In this case we only consider

the well-posedness results in the Sobolev spaces H

s

( R ). We prove by standard parabolic

methods that one can reach the dilation critical Sobolev exponant s

c

= 1/2 − 2α except

in the case α = 1/2 where 1/2 − 2α = −α. See for instance [12], [4] or [21] for the same

kind of results in the case α = 1.

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Theorem 2. Let (α, s) ∈ R

2

be such that α ∈ (0, 1/2] and s ≥ 1/2 − 2α with s > −α.

Then the Cauchy problem (1.1) is locally well-posed in H

s

( R ).

Proof. The proof is done using a fixed point argument on a suitable metric space. The case s > 1/2 is trivial since H

s

( R ) is an algebra and the semi-group S

α

is contractive on H

s

( R ). One can thus simply perform a fixed point argument in C([0, T ]; H

s

( R )) on the Duhamel formula for a suitable T > 0 related to ku

0

k

Hs(R)

. The case s = 1/2 is also rather easy and is postponed at the end of the proof. So let us assume that

(6.1) 1/2 − 2α ≤ s < 1/2 if 0 < α < 1/2 and − 1/2 < s < 1/2 if α = 1/2 , that is α, s belong to the set

n

(α, s) ∈ R

2

| 0 < α ≤ 1/2, s ≥ 1/2 − 2α and s > −α o .

For s fixed as above we take 0 < s

0

< 1/2 such that 0 < s

0

− s < α and 2s

0

− 1

2 < s .

This is obviously possible for α = 1/2 since s > −1/2, and for 0 < α < 1/2 since s + 1/2 > s + α ≥ (1/2 − 2α) + α = 1/2 − α > 0. We first establish the existence and uniqueness of a solution of (1.3) in

X

M,T

:=

n

u ∈ C (0, T ], H

s0

( R )

| kuk

XT

:= sup

t∈(0,T]

t

s0

−s

ku(t)k

Hs0(R)

≤ M o

by proving that the mapping

Λ

u0

(u)(t) = S

α

(t)u

0

∓ Z

t

0

S

α

(t − σ) u

2

(σ) dσ, is a strict contraction in X

M,T

for suitable M > 0, T > 0.

From classical regularizing effects for the fractional heat equation it holds (6.2) kS

α

(t)f k

Hs2(R)

≤ Ct

s2−s1

kf k

Hs1(R)

, ∀ s

1

≤ s

2

, ∀ f ∈ H

s1

( R ).

Applying (6.2) with (s

1

, s

2

) = (s, s

0

), yields

(6.3) t

s0

−s

kS

α

(t)u

0

k

Hs0(R)

. ku

0

k

Hs(R)

. Now, according to [11], since 0 < s

0

< 1/2, it holds :

(6.4) kuvk

H2s012(R)

≤ Ckuk

Hs0(R)

kvk

Hs0(R)

,

(15)

where C is a positive constant. We thus obtain for any t > 0, t

s0−s

Z

t 0

S

α

(t − t

0

)u

2

(t

0

) dt

0

Hs0(R)

. t

s0−s

Z

t

0

S

α

(t − t

0

)u

2

(t

0

)

Hs0(R)

dt

0

. t

s0

−s

Z

t 0

(t − t

0

)

s0

−1/2

ku

2

(t

0

)k

H2s0−1/2(R)

dt

0

. t

s0

−s

Z

t

0

(t − t

0

)

s0

−1/2

ku(t

0

)k

2Hs0(R)

dt

0

. sup

τ∈]0,t[

τ

s0

−s

ku(τ )k

Hs0(R)

2

t

s−(1/2−2α)

Z

1 0

(1 − θ)

2s0

−1

θ

s−s0

α

. t

s−(1/2−2α)

kuk

2X

(6.5)

t

where in the last step we used that 0 < s

0

− s < α and that s

0

> 1/2 − 2α since s

0

> s ≥ 1/2 − 2α. In view of (6.5) we easily get for 0 < T < 1 and v

i

∈ X

T

, i = 1, 2, (6.6) kΛ

u0

(v

i

)k

XT

. kS

α

(·)u

0

k

XT

+ T

s−(1/2−2α)

kv

i

k

2X

T

and

(6.7) kΛ

u0

(v

1

− v

2

)k

XT

. T

s−(1/2−2α)

(kv

1

k

XT

+ kv

2

k

XT

)kv

1

− v

2

k

XT

.

Combining these estimates with (6.3) we infer that for s > 1/2 − 2α, Λ

u0

is a strict contraction on X

M,T

with M ∼ ku

0

k

Hs(R)

and T ∼ ku

0

k

−2α s−(1/2−2α)

Hs(R)

if s > 1/2 − 2α. This leads to the existence and uniqueness in X

T

for any u

0

∈ H

s

( R ). For s = 1/2 − 2α, Λ

u0

is also a strict contraction on X

M,T

with M ∼ ku

0

k

Hs(R)

and T ∼ 1 but only under a smallness assumption on ku

0

k

Hs(R)

. Hence, we get the existence in X

T

for any u

0

∈ H

s

( R ) with small initial data. Now to prove that the solution u belongs to C([0, T ]; H

s

( R )) we first notice that S

α

(u

0

) ∈ C( R

+

; H

s

( R )). Moreover , according to (6.2), we have

sup

t∈]0,T[

Z

t 0

S

α

(t − t

0

)(u

2

− v

2

)(t

0

) dt

0

Hs(R)

. sup

t∈]0,T[

Z

t 0

S

α

(t − t

0

)(u

2

− v

2

)(t

0

)

Hs(R)

dt

0

. sup

t∈]0,T[

Z

t 0

(t − t

0

)

min(0,(2s0

−1/2)−s

)

ku

2

− v

2

k

H2s0−1/2(R)

dt

0

. sup

t∈]0,T[

Z

t 0

(t − t

0

)

min(0,(2s0

−1/2)−s

)

ku − vk

Hs0(R)

ku + vk

Hs0(R)

dt

0

. ku + vk

XT

ku − vk

XT

T

min(1+

s−s0

α ,s−(1/2−2α) )

Z

1 0

(1 − θ)

min(0,

(2s0−1/2)−s

)

θ

s−s0

α

. T

min(1+s−sα0,s−(1/2−2α) )

ku + vk

XT

ku − vk

XT

(6.8)

(16)

where in the last step we used that 0 < s

0

− s < α and that

2s0−1/2−s

> −1 since 2s

0

− s > s ≥ 1/2 − 2α. This ensures that starting with a continuous function v ∈ C [0, T ]; H

s

( R )

∩ X

M,T

, the sequence of function constructed by the Picard sheme that converges to the solution in u ∈ X

T

is a Cauchy sequence in C [0, T ]; H

s

( R )

and thus u ∈ C [0, T ]; H

s

( R )

. The continuous dependence with respect to initial data in H

s

( R ) follows also easily from (6.8).

It remains to handle the case of arbitrary large initial data in H

sc

( R ) when s = s

c

= 1/2 −2α. We first notice that, according to (6.6)-(6.7), Λ

u0

is a strict contraction in X

M,T

as soon as M = 2kS

α

(·)u

0

k

XT

is small enough. Then, fixing u

0

∈ H

sc

( R ), by the density of H

s0

( R ) in H

sc

( R ) we infer that for any ε > 0 there exists u

0,ε

∈ H

s0

( R ) such that ku

0

− u

0,ε

k

Hsc(R)

< ε. Since u

0,ε

∈ H

s0

( R ) it holds kS

α

(·)u

0,ε

k

XT

≤ T

s0−s

ku

0,ε

k

Hs0(R)

. This leads to

kS

α

(·)u

0

k

XT

. T

s0

−s

ku

0,ε

k

Hs0(R)

+ ε .

Noticing that the right-hand side member of the above inequality can be made arbi- trary small by choosing suitable ε > 0 and T > 0, this proves the local existence in C [0, T ]; H

sc

( R )

∩ X

T

for arbitrary large initial data in H

sc

( R ). Note that here T > 0 does not depend only on ku

0

k

Hsc(R)

but on the Fourier profile of u

0

. The uniqueness holds in {f ∈ X

T

/ kf k

Xt

→ 0 as t & 0} . This completes the proof for (α, s) satisfying (6.1).

Finally for s = 1/2 we apply the fixed point argument in X ˜

M,T

:=

n

u ∈ C (0, T ], H

1+α2

( R )

| kuk

X˜

T

:= sup

t∈(0,T]

t

14

ku(t)k

H1+α2 (R)

≤ M o

.

Using that H

1+α2

( R ) is an algebra we easily get

t

14

Z

t 0

S

α

(t − t

0

)u

2

(t

0

) dt

0

H1+α2 (R)

. t

14

Z

t

0

ku

2

k

H1+α2 (R)

dt

0

. t

14

Z

t

0

kuk

2

H1+α2 (R)

dt

0

. t

34

kuk

2˜

Xt

Z

1 0

θ

−1/2

dθ . (6.9)

This gives the local existence and uniqueness in ˜ X

M,T

for M ∼ ku

0

k

H1/2(R)

and T ∼ ku

0

k

−4 3

H1/2(R)

. The fact that the solution u belongs to C [0, T ]; H

12

( R )

and the continuous

(17)

dependence with respect to initial data in H

12

( R ) follows by noticing that sup

t∈]0,T[

Z

t 0

S

α

(t − t

0

)(u

2

− v

2

)(t

0

) dt

0

H12(R)

. sup

t∈]0,T[

Z

t 0

kS

α

(t − t

0

)(u

2

− v

2

)(t

0

)k

H12(R)

dt

0

. sup

t∈]0,T[

Z

t 0

k(u

2

− v

2

)(t

0

)k

H12(R)

dt

0

. sup

t∈]0,T[

Z

t 0

ku(t

0

)

2

− v(t

0

)

2

k

H1+α2 (R)

dt

0

. sup

t∈]0,T[

Z

t 0

ku(t

0

) + v(t

0

)k

H1+α2 (R)

ku(t

0

) − v(t

0

)k

H1+α2 (R)

dt

0

. ku + vk

X˜

T

ku − vk

X˜

T

T

12

Z

1

0

θ

−1/2

dθ . (6.10)

6.2. Illposedness result for α = 1/2 and s = −1/2. Let us now prove an ill-posedness result at the crossing point (α, s) = (1/2, −1/2) of the two lines s = −α and s = 1/2 − 2α.

Recall that there exists T

0

> 0 and R

0

> 0 such that the solution-map u

0

7→ u associated with (1.1) for α = 1/2 is well-defined and continuous from the ball B (0, R

0

)

L2

of L

2

( R ) with values in C([0, T ]; L

2

( R )). The following norm inflation result clearly disproves the continuity of this solution map from B(0, R

0

)

L2

endowed with the H

−1/2

-topology with values in C([0, T ]; H

−1/2

), for any T ≤ T

0

.

Theorem 3. There exists a sequence T

N

& 0 and a sequence of initial data {φ

N

} ⊂ L

2

( R ) such that the sequence of emanating solutions {u

N

} of (1.1)is included in C([0, T

N

]; L

2

( R )) and satisfy

(6.11) kφ

N

k

H−1/2

→ 0 and ku

N

(T

N

)k

H−1/2

→ +∞ as N → ∞ .

We follow exactly the very nice proof of Iwabuchi-Ogawa [8] that proved the ill-posedness in H

−1

of the 2-D quadratic heat equation. Note that (1, −1) is the intersection of the two lines s = −α and s = 1 − 2α, this last line corresponding to the scaling critical Sobolev exponent in dimension 2. We need to introduce the rescaled modulation spaces (M

2,1

)

N

that are defined for any integer N ≥ 1 by

(M

2,1

)

N

:= n

u ∈ S

0

( R ) | kuk

(M2,1)N

< ∞ o where

kuk

(M2,1)N

:= X

k∈2NZ

kˆ uk

L2(k,k+2N)

.

Références

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