HAL Id: hal-00807047
https://hal.archives-ouvertes.fr/hal-00807047
Submitted on 2 Apr 2013
HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
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�
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=
Dx2αu∓u2, t ∈(0, T ), x
∈R or T , with 0
< α≤1 is well-posed in
Hsfor
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 aby-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
x2αu = ∓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
2αxis the Fourier multiplier by |ξ|
2α. In this paper, we consider actually the corresponding integral equation which is given by
(1.3) u(t) = S
α(t)u
0∓
Z
t0
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
Let us recall that the Cauchy problem associated with the nonlinear heat equation in R
n(1.4) u
t− ∆u = ∓u
khas 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−12u(λ
2t, λx) and that the homogeneous Sobolev space ˙ H
n2−k−12is invariant by the associated space dilation symmetry ϕ(x) 7→
λ
k−12ϕ(λx). The Cauchy problem (1.4) is known to be well-posed in H
sfor s ≥ s
c=
n
2
−
k−12except in the case (n, k) = (1, 2). Indeed, in this case the well-posedness is only known in H
sfor s > −1 and in [9] it is proven that the flow-map cannot be of class C
2below 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
07→ u cannot be even continuous in H
sfor 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,qin 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
−αwe 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
1H
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.
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
07→ u(t) is continuous from the ball of B centered at u
0with 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
07→ 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 .
ag, instead.
2. Resolution Space We use the following definition for the Fourier transform
F (f )(ξ) = Z
R
f (x)e
−ixξdx, and the inverse Fourier transform is
F
−1(f )(x) = 1 2π
Z
R
F(f )(ξ)e
ixξdξ,
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)
αf
(ξ) = |ξ|
2αF(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)|
2dξ < ∞}
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)|
2dξ
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 (∆
ju)(ξ) = ϕ(2
−jξ)F u(ξ), j ≥ 0, and F(∆
−1u)(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,q2 (R)
= X
j≥−1
2
jsq||∆
ju||
qL2(R) 1/q.
For s ∈ R , s
1< s
2, 1 ≤ q
1≤ q
2and q ≥ 1 we have the following embeddings B
2s,q1, → B
2s,q2and 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
sjqk∆
ju(t)k
qLptL2x
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
∞TB
2s,q∩ L ˜
2TB
2s+α,qequipped with the norm:
kuk
Xs,qα,T
= hX
j
sup
t∈]0,T[
2
sjqk∆
ju(t)k
qL2x
i
1/q+ hX
j
2
j(s+α)qk∆
juk
qL2TL2x
i
1/q.
Let us also consider the space Y
αs,q:=
n
u ∈ L ˜
1R
∗+; B
s+2α,q2( R )
and ∂
tu ∈ L ˜
1R
∗+; B
s,q2( R ) o
equipped with the norm:
kuk
Ys,qα
= hX
j
2
(s+2α)jqk∆
juk
qL1tL2x
i
1/q+ hX
j
2
sjqk∆
ju
tk
qL1tL2x
i
1/qFor T > 0, the restriction space Y
α,Ts,qof Y
αs,qis endowed with the usual norm kuk
Ys,qα,T
= inf
v∈Y
{kvk
Ys,qα
, v ≡ u on ]0, T [ } .
For T > 0 our resolution space will be E
α,Ts,q= X
α,Ts,q+ Y
α,Ts,qendowed with the usual norm for a sum space :
kuk
Es,qα,T
:= inf
u=v+w
(kvk
Xs,qα,T
+ kwk
Ys,qα,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,qα,T
. kϕk
Bs,q2
.
Proof. The standard smoothing effect of the (fractional) heat semi-group is not sufficient here since we have
kS
α(t)ϕk
Bs+α,q2
. t
−12kϕk
Bs,q2
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) ∂
tu − D
2αxu = 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
t0
Z
R
|D
αxu(s, x)|
2dx ds = Z
R
ϕ
2(x) dx .
Using that for each j ∈ N , ∆
jS
α(t)D
sxϕ satisfies the linear fractional heat equation (3.2) with D
sxϕ as initial datum, powering in q/2 and then summing in j ≥ 0, we get for any T > 0,
X
j≥0
2
sjqk∆
jS
α(t)ϕk
qL∞ TL2x 1/q+ X
j≥0
2
(s+α)jqk∆
jS
α(t)ϕk
qL2TL2x
1/q. X
j≥0
2
sjqk∆
jϕk
qL2x
1/q.
On the other hand, for j = −1 we write k∆
−1S
α(t)ϕk
L∞TL2x
+ k∆
−1S
α(t)ϕk
L2TL2x
≤ 2 T
1/2k∆
−1S
α(t)ϕk
L∞TL2x
≤ 2 T
1/2k∆
−1ϕk
L2 xand the result follows.
As a direct consequence we get the following estimate on the semi-group : Let 0 < T ≤ 1 and ϕ ∈ B
2s,qthen it holds
(3.3) kS
α(t)ϕk
Es,qα,T
. kS
α(t)ϕk
Xs,qα,T
. kϕk
Bs,q2
.
Let us now define the operator L
αby
(3.4) L
α(f )(t, x) =
Z
t 0S
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,qα,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,qα,T
. kf k
L˜1 t>0B2s,q,
for any f ∈ L ˜
1t>0B
2s,qsupported 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
∂
tu − D
2αxu = 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
αxu)
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
2αjZ
R
u
2j≤ Z
R
f
j21/2Z
R
u
2j1/2. Here u
j= ∆
ju, f
j= ∆
jf . If R
R
u
2j1/26= 0 we divide this last inequality by R
R
u
2j1/2to obtain
d dt
Z
R
u
2j1/2+ 2
2αjZ
R
u
2j1/2≤ Z
R
f
j21/2. On the other hand, the smoothness and non negativity of t 7→ ku
j(t)k
2L2x
forces
dtdku
j(t)k
2L2x
= 0 as soon as ku
j(t)k
L2x
= 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
2αjku
jk
L1tL2x
. kf
jk
L1 tL2x.
Now, in the case j = −1, we get in the same way k∆
−1uk
L2. k∆
−1f k
L1tL2x
. Integrating on [0, 2T ] this leads to kη∆
−1uk
L1tL2x
. T kf
jk
L1 tL2x.
Finally, in view of the linear fractional heat equation, the triangle inequality leads to (3.7) k∂
t(ηu
j)k
L1tL2x
. kη∂
tu
jk
L1tL2x
+ ku
jk
L1tL2x
. kf
jk
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,qα,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,qα,T
= kuk
L˜∞TB2s,q
+ kuk
L˜2TB2s+α,q
. kuk
Ys,qα,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
tk
L1tL2x
and kuk
2L2t>0L2x
. kuk
L1t>0L2x
ku
tk
L1 t>0L2x.
Indeed, applying (3.10) to the space frequency localization u
jof u, Bernstein’s inequalities lead to
2
jsku
jk
L∞t L2x
. 2
jsk∂
tu
jk
L1tL2x
and 2
jq(s+α)ku
jk
qL2t>0L2x
. 2
jq(s+2α)/2ku
jk
q/2L1t>0L2x
2
jqs/2k∂
tu
jk
q/2L1t>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
t0
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 xk
Z
∞ 0|u
t(t, ·)| dtk
L2 x. kuk
L1t>0L2x
ku
tk
L1 t>0L2x,
where we used Minkowsky integral inequality in the last step.
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,qα,T
. kL
α(u
2)k
Ys,qα,T
. X
j
2
jsqk∆
j(u
2)k
qL1TL2x
1/q. X
j
2
jq(s+1/2)k∆
j(u
2)k
qL1TL1x
1/q. (4.1)
Now, by para-product decomposition we have k∆
j(u
2)k
L1TL1x
. kuk
L2T ,x
k∆
juk
L2T ,x
+ X
|k−k0|≤3, k&j
k∆
kuk
L2T ,x
k∆
k0uk
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 ,xX
j
2
jq(s+1/2)k∆
juk
qL2TL2x
1/q= kuk
L2 T ,xkuk
˜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
L2T ,x
. kuk
L˜2TB2s+α,q
. For the second term, we notice that for α > 1/2, we can estimate its contribution by
kuk
L˜2TB0,∞2
kuk
L˜2 TB2s+α,qX
j
2
jq(1/2−α)1/q≤ C(α) kuk
L˜2TB0,∞2
kuk
L˜2TB2s+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,qα,T
. kuk
L˜2 T xkuk
L˜2TBs+α,q2
. kuk
E−α,2 α,Tkuk
Es,qα,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,αT
. kuk
L˜2 T xkvk
L˜2TB2s+α,q
+ kvk
L˜2 T xkuk
L˜2 TB2s+α,q≤ C
αkuk
E−α,2α,T
kvk
Es,qα,T
+ kvk
E−α,2α,T
kuk
Es,qα,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
0k
B−α,22 (R)
≤ β , the mapping
u 7→ S
α(·)u
0+ L
α(u
2)
is a strict contraction in the ball of E
α,1−α,2centered at the origin of radius (2C
α)
−1. Noticing that E
α,1s,q, → E
α,1−α,2as 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,qas soon as (4.5)-(4.4) are satisfied. Since S
αis a continuous semi-group in B
2s,q( R ) and accord- ing to Lemma 3.3, E
α,1s,q, → L ˜
∞1B
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
2αu(λ
2αt, λx) whereas kλ
2αu
0(λ·)k
B−α,22 (R)
≤ λ
α−1/2ku
0k
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
0k
B−α,22 (R)
−2α−14α. Note that, the well-posedness being obtained by a fixed point argument, as a by-product we get that the solution-map : u
07→ 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
07→ 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
αχ
IN(ξ) + χ
IN(−ξ) ,
where I
N= [N, N + 2] and χ
INis 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
Nk
Bs,q2 (R)
∼ N
α+sand thus kφ
Nk
B−α,q2 (R)
∼ 1
whereas φ
N→ 0 in B
2s,q( R ), for s < −α.
Let us consider the following bilinear operator, closely related to second iteration of the Picard scheme,
A
2(t, h, h) = 2 Z
t0
S
α(t − t
0)[S
α(t
0)h]
2dt
0,
where S
αis the semi-group of the linear heat equation. Let us denote by F
xthe partial Fourier transform with respect to x. Recall that
F
xS
α(t)ϕ
(ξ) = e
−t[ξ|2αF
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
xA
2(t, φ
N, φ
N)
(ξ) = 2 Z
t0
e
−(t−t0)|ξ|2αZ
R
e
−t0|ξ1|2αφ ˆ
N(ξ
1)e
−t0|ξ−ξ1|2αφ ˆ
N(ξ − ξ
1)dξ
1dt
0= 2
Z
R
φ ˆ
N(ξ
1) ˆ φ
N(ξ − ξ
1) Z
t0
e
−(t−t0)|ξ|2αe
−[|ξ1|2α+|ξ−ξ1|2α]t0dt
0dξ
1= 2
Z
R
φ ˆ
N(ξ
1) ˆ φ
N(ξ − ξ
1) e
−[|ξ1|2α+|ξ−ξ1|2α]t− e
−|ξ|2αtΘ
α(ξ, ξ
1)
dξ
1, (5.7)
where
Θ
α(ξ, ξ
1) = |ξ|
2α− |ξ
1|
2α− |ξ − ξ
1|
2α. Note that the integrand is nonnegative. In particular, F
xA
2(t, φ
N, φ
N)
(ξ) = |F
xA
2(t, φ
N, φ
N) (ξ)|.
Let
K
1N(ξ) = n
ξ
1| (ξ − ξ
1, ξ
1) ∈ I
N× I
Nor (ξ − ξ
1, ξ
1) ∈ I
−N× I
−No
and
K
2N(ξ) = n
ξ
1| (ξ − ξ
1, ξ
1) ∈ I
N× I
−Nor (ξ − ξ
1, ξ
1) ∈ I
−N× I
No For any |ξ| ≤
12, K
1N(ξ) = ∅ and thus
F
xA
2(t, φ
N, φ
N)
(ξ) = 2 Z
K2N(ξ)
φ ˆ
N(ξ
1) ˆ φ
N(ξ − ξ
1) e
−[|ξ1|2α+|ξ−ξ1|2α]t− e
−|ξ|2αtΘ
α(ξ, ξ
1)
dξ
1.
On the other hand, for any (a, b) ∈ R
+× R
−one has obviously,
|a|
2α+ |b|
2α− |a + b|
2α≥ |a| ∧ |b|
2α.
Moreover, it is easy to check that |K
2N(ξ)| ≥ 1 and that in K
2N(ξ) it holds N
2α≤
|Θ
α(ξ, ξ
1)| ≤ 2(N + 2)
2αHence, fixing t ∈]0, 1[, it holds F
xA
2(t, φ
N, φ
N)
(ξ) ≥ e
−t/2N
2α1 − e
−N2αt2(N + 2)
2α≥ 1
4 e
−t/2, ∀ξ ∈ [−1/2, 1/2],
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,q2
≥ 1
4 e
−t/2for N > 0 large enough. Taking s < −α this proves the discontinuity of the map u
07→ 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,1uniformly 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
4e
−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
t0
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,12
≤ 1 and 0 ≤ t ≤ T,
u(t, h) = S
α(t)h +
∞
X
k=2
kA
k(t, h
k),
where h
k= (h, · · · , h), h
k7→ A
k(t, h
k) are k−linear continuous maps from (B
2−α,1( R ))
kinto 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) −
2A
2(t, φ
N, φ
N) = S
α(t)φ
N+
∞
X
k=3
kA
k(t, φ
kN).
Using the inequalities
kS
α(t)φ
Nk
Bs,12 (R)
≤ kφ
Nk
Bs,12 (R)
≤ 2N
s+αand
∞
X
k=3
kA
k(t, φ
kN)
B2−α,1(R)≤
0 3∞
X
k=3
k0A
k(t, φ
kN)
B2−α,1(R)≤ C
3,
where C is a positive constant, we deduce that for s ≤ −α,
(5.10) sup
t∈[0,T]
ku(t, φ
N) −
2A
2(t, φ
N, φ
N)k
Bs,12 (R)
≤ C
3+ 2
0N
s+α. According to (5.8) this leads, for ≤
C−12e5−t/4, to
ku(t, φ
N)k
Bs,q2 (R)
≥ C
02/2 − 2
0N
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
−12X
N≤j≤2N
φ
2j.
where φ
2jis defined in (5.6).
Noticing that ∆
kφ
2j= δ
k,jφ
2j, we can easily check that kψ
Nk
B−α,q2
∼ N
−12+1q. In particular, kψ
Nk
Bα,q2
→ 0 for any q > 2 whereas kψ
Nk
B−α,22
= kψ
Nk
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(ψ
N, ψ
N, t) does not tend to 0 in D
0. By the localization, it holds φ
2j?φ
2j0≡ 0 on ] − 1/2, 1/2[ as soon as j 6= j
0≥ 10 and the same reasons as above lead to
F
xA
2(t, ψ
N, ψ
N)
(ξ) = N
−12X
N≤j≤2N
Z
K22j(ξ)
φ ˆ
2j(ξ
1) ˆ φ
2j(ξ − ξ
1) e
−[|ξ1|2α+|ξ−ξ1|2α]t− e
−|ξ|2αtΘ
α(ξ, ξ
1)
dξ
1≥ N
−12e
−t/2N
12N
2α1 − e
−N2αt2(N + 2)
2α≥ 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.
Theorem 2. Let (α, s) ∈ R
2be 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
αis 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
0k
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
2α
ku(t)k
Hs0(R)≤ M o
by proving that the mapping
Λ
u0(u)(t) = S
α(t)u
0∓ Z
t0
S
α(t − σ) u
2(σ) dσ, is a strict contraction in X
M,Tfor 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
−s22α−s1kf 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
2α
kS
α(t)u
0k
Hs0(R). ku
0k
Hs(R). Now, according to [11], since 0 < s
0< 1/2, it holds :
(6.4) kuvk
H2s0−12(R)
≤ Ckuk
Hs0(R)kvk
Hs0(R),
where C is a positive constant. We thus obtain for any t > 0, t
s02α−sZ
t 0S
α(t − t
0)u
2(t
0) dt
0Hs0(R)
. t
s02α−sZ
t0
S
α(t − t
0)u
2(t
0)
Hs0(R)dt
0. t
s0−s 2α
Z
t 0(t − t
0)
s0−1/2
2α
ku
2(t
0)k
H2s0−1/2(R)dt
0. t
s0−s 2α
Z
t0
(t − t
0)
s0−1/2
2α
ku(t
0)k
2Hs0(R)dt
0. sup
τ∈]0,t[
τ
s0−s
2α
ku(τ )k
Hs0(R) 2t
s−(1/2−2α) 2α
Z
1 0(1 − θ)
2s0−1 4α
θ
s−s0
α
dθ
. t
s−(1/2−2α)2αkuk
2X(6.5)
twhere 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
0k
XT+ T
s−(1/2−2α)2αkv
ik
2XT
and
(6.7) kΛ
u0(v
1− v
2)k
XT. T
s−(1/2−2α)2α(kv
1k
XT+ kv
2k
XT)kv
1− v
2k
XT.
Combining these estimates with (6.3) we infer that for s > 1/2 − 2α, Λ
u0is a strict contraction on X
M,Twith M ∼ ku
0k
Hs(R)and T ∼ ku
0k
−2α s−(1/2−2α)
Hs(R)
if s > 1/2 − 2α. This leads to the existence and uniqueness in X
Tfor any u
0∈ H
s( R ). For s = 1/2 − 2α, Λ
u0is also a strict contraction on X
M,Twith M ∼ ku
0k
Hs(R)and T ∼ 1 but only under a smallness assumption on ku
0k
Hs(R). Hence, we get the existence in X
Tfor 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 0S
α(t − t
0)(u
2− v
2)(t
0) dt
0Hs(R)
. sup
t∈]0,T[
Z
t 0S
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
2α )
ku
2− v
2k
H2s0−1/2(R)dt
0. sup
t∈]0,T[
Z
t 0(t − t
0)
min(0,(2s0−1/2)−s
2α )
ku − vk
Hs0(R)ku + vk
Hs0(R)dt
0. ku + vk
XTku − vk
XTT
min(1+s−s0
α ,s−(1/2−2α)2α )
Z
1 0(1 − θ)
min(0,(2s0−1/2)−s
2α )
θ
s−s0
α
dθ
. T
min(1+s−sα0,s−(1/2−2α)2α )ku + vk
XTku − vk
XT(6.8)
where in the last step we used that 0 < s
0− s < α and that
2s0−1/2−s2α> −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
Tis 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
u0is a strict contraction in X
M,Tas soon as M = 2kS
α(·)u
0k
XTis 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,εk
Hsc(R)< ε. Since u
0,ε∈ H
s0( R ) it holds kS
α(·)u
0,εk
XT≤ T
s02α−sku
0,εk
Hs0(R). This leads to
kS
α(·)u
0k
XT. T
s0−s
2α
ku
0,εk
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
Tfor arbitrary large initial data in H
sc( R ). Note that here T > 0 does not depend only on ku
0k
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
14ku(t)k
H1+α2 (R)
≤ M o
.
Using that H
1+α2( R ) is an algebra we easily get
t
14Z
t 0S
α(t − t
0)u
2(t
0) dt
0H1+α2 (R)
. t
14Z
t0
ku
2k
H1+α2 (R)
dt
0. t
14Z
t0
kuk
2H1+α2 (R)
dt
0. t
34kuk
2˜Xt
Z
1 0θ
−1/2dθ . (6.9)
This gives the local existence and uniqueness in ˜ X
M,Tfor M ∼ ku
0k
H1/2(R)and T ∼ ku
0k
−4 3
H1/2(R)
. The fact that the solution u belongs to C [0, T ]; H
12( R )
and the continuous
dependence with respect to initial data in H
12( R ) follows by noticing that sup
t∈]0,T[
Z
t 0S
α(t − t
0)(u
2− v
2)(t
0) dt
0H12(R)
. sup
t∈]0,T[
Z
t 0kS
α(t − t
0)(u
2− v
2)(t
0)k
H12(R)
dt
0. sup
t∈]0,T[
Z
t 0k(u
2− v
2)(t
0)k
H12(R)
dt
0. sup
t∈]0,T[
Z
t 0ku(t
0)
2− v(t
0)
2k
H1+α2 (R)
dt
0. sup
t∈]0,T[
Z
t 0ku(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
12Z
10
θ
−1/2dθ . (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
07→ u associated with (1.1) for α = 1/2 is well-defined and continuous from the ball B (0, R
0)
L2of 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)
L2endowed 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
Nk
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
−1of 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)
Nthat 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