Well-posedness and regularity of the generalized Burgers equation in periodic Gevrey spaces

Accepted Manuscript Well-posedness and regularity of the generalized Burgers equation in periodic Gevrey spaces

John Holmes

PII: DOI: Reference:

S0022-247X(17)30399-2 http://dx.doi.org/10.1016/j.jmaa.2017.04.045 YJMAA 21332

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Journal of Mathematical Analysis and Applications

Received date:

1 June 2016

Please cite this article in press as: J. Holmes, Well-posedness and regularity of the generalized Burgers equation in periodic Gevrey spaces, J. Math. Anal. Appl. (2017), http://dx.doi.org/10.1016/j.jmaa.2017.04.045

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Well-posedness and regularity of the generalized Burgers equation in periodic Gevrey spaces John Holmes Department of Mathematics, The Ohio State University, Columbus, OH 43210

Abstract We consider the generalized Burgers equation in a class of Gevrey functions Gσ,δ,s (T). We show that the generalized Burgers equation is well–posed in this space. Furthermore, we show that the solution is Gevrey-σ in the spacial variable and Gevrey-2σ in the time variable. Keywords: Burgers equation; Gevrey regularity; Initial value problem; Multilinear estimates; Sobolev spaces; Uniform radius of analyticity 2010 MSC: Primary: 35Q53

1. Introduction For k = 1, 2, 3, . . . we consider the initial-value problem for the generalized Burgers equation (k-gBurgers)  ∂t u = ∂x2 u + uk ∂x u, x ∈ T, t ∈ R (1.1) u(x, 0) = u0 (x), and study its well–posedness and the regularity properties of the solution when the initial data belong to a class of analytic Gevrey spaces. These spaces are defined by the norm  1/σ f 2G˙ σ,δ,s (T) = |n|2s e2δ|n| |f(n)|2 < ∞ (1.2) n=0

for any s ∈ R, δ > 0 and σ ≥ 1. A function in G˙ 1,δ,s (T) is a real analytic function which has a holomorphic extension on a symmetric strip of width 2δ in the complex plane. For an analytic periodic function ϕ, there exists δ0 > 0 such that ϕ ∈ G˙ 1,δ0 ,s (T). Thus, well-posedness in these spaces shows that the width of the strip of analyticity does not collapse during the time interval of existence. It is clear that the dominant weighting of these L2 1/σ spaces is by the exponential term, i.e. e2δ|n| . We include the Sobolev weight, |n|2s , in the definition of these spaces simply to simplify the multilinear estimates we will need. Therefore, in our proofs, we will fix the index s = sk = 12 − k1 to simplify our computations. Using these spaces, our main result about the well-posedness of k-gBurgers in analytic Gevrey spaces reads as follows. Theorem 1. Let σ ≥ 1, δ > 0 and sk = 12 − k1 . Given small initial data, u0 (x), in the Gevrey spaces G˙ σ,δ,sk (T), for any T > 0 the initial-value problem for the k-gBurgers equation (1.1) is well-posed in the space C([0, T ]; G˙ σ,δ,sk ). Preprint submitted to Elsevier

April 26, 2017

The Cauchy problem for the k-gBurgers equation (1.1) with initial data in homogeneous Sobolev spaces, ˙ p,s (R), has been studied by Bekiranov [B] in 1996. In particular, well-posedness was established for W 1 < p < ∞ and s ≥ sp,k = p1 − k1 . Although Bekiranov provided details for the proof on the real line, the proof holds on the circle with some modification. Our work is an expansion upon this result, and since we are working in classical function spaces, we assume s = sk for simplicity. [B] provided two different proofs of local well-posedness. The first proof employed the methodology introduced by Kato and Fujita [KF] which relied on Lp Sobolev spaces in the space variable with a special weight in the time variable. The second proof used the methodology based on Lqt Lpx estimates, introduced by Giga [G]. In particular, our result uses the multilinear estimates provided in the first proof. Also, expanding upon these results, Zahrouni [Z] proved well-posedness of k-gBurgers in Besov spaces. In Ferrari and Titi [FTi], the authors study the initial value problem for the following generalization of Burgers equation ∂t u = ∂x2 u + G(u, ∂x u) in multiple dimensions and where G(x, y) is an entire function in both variables. They assume the initial data is of a Sobolev class and show that the corresponding solution, u(x, t), is analytic in a strip of width 2t in the complex plane in the spacial variable. They accomplish this via a Galerkin approximation argument in non-homogeneous Gevrey spaces similar to the ones we defined above. The study of well-posedness for the k-gBurgers equation for rough initial data is motivated by the work of Dix [D] who studied the viscous Burgers equation (corresponding to k = 1). He showed the viscous Burgers equation is well posed in H s (= W 2,s ) for s > − 12 and ill-posed below this index by finding infinitely many solutions. These solutions were constructed by using the Cole–Hopf transformation [C, Ho], which may not generalize for k > 1. However, this argument was generalized in [B] for initial data in the Sobolev space W p,s . Related to the k-gBurgers equation is the following semilinear parabolic equation ∂t u − ∂x2 u + P (D)F (u) = 0,

(1.3)

where P (D) is a pseudo-differential operator of order d for d ∈ [0, 2), and F (u) is a non-linear function which behaves like |u|α for α > 1. This equation has been studied by numerous authors over the past several decades. In particular Giga [G], Henry [H], Ribaud [R1, R2], Taylor [Ta] and Weissler [W] contributed to the study of well-posedness in inhomogeneous Sobolev spaces H p,s . Well-posedness for the 2−d initial-value problem (1.3), both local and global, has been proved above the critical value sc = p1 − α−1 . In addition, ill-posedness results below the critical index were completed in 2002 by Molinet, Ribaud and Youssfi [MRY] for the case p = 2, by showing the initial-value problem cannot be solved using iterative methods. This index, for the k-gBurgers equation corresponds to the scaling index by which we mean, if u(x, t) is a solution to the Cauchy problem for the generalized Burgers equation, with initial data u0 (x), then uλ (x, t) = λ1/k u(λx, λ2 t),

(1.4)

is a solution to the Cauchy problem with initial data uλ,0 = λ1/k u0 (λx).

(1.5)

Moreover, the norm of uλ,0 is invariant under the H˙ sk (T) norm. A closely related equation is the so called fractal Burgers equation ∂t u − (∂x2 )α u + u∂x (u) = 0. (1.6) 2

This equation has been shown to be well posed in homogeneous Sobolev spaces H s for s > 0 and ill-posed below this index by Xu [X], extending results on well-posedness and traveling wave solutions by Biler, Funaki and Woyczynski [BFW]. Our proof of Theorem 1 is similar to the methodology for proving well-posedness in analytic Gevrey spaces for the generalized Korteweg-de Vries (k-gKdV) equation ∂t u + ∂x3 u + uk ∂x u = 0,

(1.7)

developed in [HHP2]. Similar theorems using these spaces were done by Gruji´c and Kalisch [GK1, GK2] in the non-periodic case, and by Hannah, Himonas and Petronilho [HHP1], Himonas and Petronilho [HP2] in the periodic case. Using the spaces G1,δ,s introduced by Foias and Temam [FT], they showed that for initial data that are analytic in a symmetric strip of width 2δ in the complex plane, there exists a time T such that the k-gKdV solution is analytic in the same strip for all t ∈ [0, T ]. Also, similar results for a modified KdV equation, with higher order dispersion, but with k = 1 non-linearity can be found in Gorski, Himonas, Holliman and Petronilho [GHHP]. For other related results see [BGK, GH, HP1] and the references contained within. Our second result, which is partially motivated by Tahara [T], is about regularity properties of k-gBurgers solutions when the initial data belong to analytic Gevrey spaces. A periodic function, u(x), is Gevrey-α, denoted Gα (T), if there exists a constant C > 0 such that sup |∂x u(x)| ≤ C +1 (!)α ,  = 0, 1, . . . . x∈T

A function of two variables, u(x, t), is said to be in the non-isotropic space Gα,β (T, [0, T ]), if there exists a constant C > 0 such that sup (x,t)∈T×(0,T )

|∂tj ∂x u(x, t)| ≤ C j++1 (!)α (j!)β , for (j, ) ∈ Z+ × Z+ .

Note that, if a function, u(x, t) ∈ Gα,β (T, [0, T ]), then for each t fixed, u(·, t) ∈ Gα (T) and for each x fixed, u(x, ·) ∈ Gβ ([0, T ]), however, the converse need not be true. We state the second main result as follows. Theorem 2. For σ ≥ 1 and δ > 0, if the initial data of the k-gBurgers Cauchy problem (1.1) are in G˙ σ,δ,sk (T), then the solution, u(x, t) ∈ Gσ,2σ (T, [0, T ]). In particular, this implies the solution is Gevrey-σ in the spatial variable and Gevrey-2σ in the time variable. Furthermore, the result is sharp; when k is odd, there exists real valued u0 (x) ∈ Gσ,δ,sk (T) for any σ ≥ 1 and δ > 0, such that the solution u(x, t) is not in the space Gσ,r (T, [0, T ]) for any 1 ≤ r < 2σ. Lysik [L1] showed a similar result for the related equation ∂t u = ∂x2 u + u2 . In particular, Lysik constructed an example which showed that the solution was not analytic in the time variable for t near 0. In [L2], Lysik showed for a large class of Burgers type equations, that if the solution was analytic in the spacial variable on a domain Ω, it was Gevrey-2 in the time variable. This proof relied upon constructing the formal power series solution to the equation, and then relied upon some combinatorial estimates to estimate the time variable regularity. Since we are only considering the periodic case, will adopt the notation Lp = Lp (T), H˙ s = H˙ s (T), etc, wherever it is convenient and conserves space. We begin in Section 2 with the proof of Theorem 1. In section 3, we will show the time regularity of the solution claimed in Theorem 2, and in section 4 we complete the proof of Theorem 2 by showing the sharpness of the regularity in the time variable. 3

˙ σ,δ,s 2. Well-posedness in G In this section, we prove that the k-gBurgers equation is well posed in a class of analytic Gevrey spaces used in [GK2] and [HP2] for the generalized KdV equation. Also note that we assume mean zero data, which is a conserved property in the solution. Thus, we do not include the point n = 0 in summations in the phase space to ensure that there are no times in which we divide by zero. We begin by defining the spaces used in this work. Let δ > 0, σ ≥ 1 and 1 ≤ p ≤ ∞. Let f be a test function, we define the  = eδ|n|1/σ |f(n)|, and we define the space Lp , as the set sub-linear operator A applied to f by Af σ,δ   ˙ Af Lp < ∞ . (2.1) Lpσ,δ = f ∈ D : f Lpσ,δ = Also, for α > 0, we define the following spaces that involve both variables (x, t)  C

α

((0, T ); Lpσ,δ )

=

f∈

C((0, T ); Lpσ,δ )



: sup t Af Lp < ∞ , α

(2.2)

t∈(0,T )

where C((0, T ); X) denotes the space of continuous functions from the time interval (0, T ) into the Banach space X, and using this space, we define the Banach space

C0α ((0, T ); Lpσ,δ ) = f ∈ C α ((0, T ); Lpσ,δ ) : lim tα f Lp = 0 . (2.3) + σ,δ

t→0

For sk =

1 2

− k1 , for any lifespan T > 0 and for γk =

we define

k+2 , 4k(k + 1)

(2.4)

  2(k+1) Y σ,δ,sk = f ∈ C([0, T ]; G˙ σ,δ,sk ) ∩ C0γk ((0, T ); Lσ,δ ) : f Y σ,δ,s < ∞ ,

(2.5)

where ˙ sup f G˙ σ,δ,sk + sup tγk f L2(k+1) . f Y σ,δ,sk = t∈(0,T )

t∈(0,T )

σ,δ

The Y σ,δ,sk spaces are an extension of the Y sk (= Y σ,0,sk ) used in [B]. We now state the main result in this section, from which Theorem 1 will follow. Theorem 3. Suppose that δ > 0, σ ≥ 1, sk = 12 − k1 and u0 ∈ G˙ σ,δ,sk . Then, provided u0 G˙ σ,δ,sk is sufficiently small, for any T > 0, there exists a unique solution u of the k-gBurgers initial value problem satisfying 2(k+1) u ∈ C([0, T ]; G˙ σ,δ,sk ) ∩ C0γk ((0, T ); Lσ,δ ) = Y σ,δ,sk .

(2.6)

Moreover, the map u0 → u(t) from G˙ σ,δ,sk into Y σ,δ,sk is Lipschitz. Proof of Theorem 3. Defining the operator 2 . Φu(x, t) = e−D t u0 (x) +



t

e−D

0

2

(t−τ )

1 ∂x (uk+1 )(x, τ )dτ, k+1

(2.7)

we see that solving (1.1) amounts to finding a fixed point for the operator Φ in an appropriate space. We will show this space is the Y σ,δ,sk space, and we begin by showing that it is a subspace of C([0, T ]; G˙ σ,δ,sk ). 4

Lemma 1. For all σ ≥ 1 and δ > 0, Y σ,δ,sk → C([0, T ]; G˙ σ,δ,sk ). Proof. This follows directly from the definition of the Y σ,δ,s space. In fact, f Y σ,δ,sk = sup f G˙ σ,δ,sk + sup tγk f L2(k+1) t∈(0,T )

σ,δ

t∈(0,T )

≥ sup f G˙ σ,δ,sk = f C([0,T ],G˙ σ,δ,sk ) . t∈(0,T )

Also we will use the multilinear estimates found in the following lemma. Lemma 2. Let g1 , . . . , gk+1 ∈ Y sk (T) and 0 ≤ t < T < ∞. Then the following inequalities hold  t −D2 (t−τ ) ∂x (g1 · · · gk+1 )(τ ) s dτ  sup tγk (k+1) g1 L2(k+1) · · · gk+1 L2(k+1) , e ˙ 0

t

γk

H

k

 t −D2 (t−τ ) ∂x (g1 · · · gk+1 )(τ ) e

L2(k+1)

0

(2.8)

t∈(0,T )

dτ  sup tγk (k+1) g1 L2(k+1) · · · gk+1 L2(k+1) .

(2.9)

t∈(0,T )

Proof. Defining the Riesz potential Dx = (−∂x2 )1/2 (equivalently, Dx f = F −1 [|n|f(n)] for any test function f ) we have ˙ Ik =

 t  t −D2 (t−τ ) sk −D2 (t−τ ) ∂x (g1 · · · gk+1 )(τ ) s dτ = e ∂ (g · · · g )(τ ) e D x 1 k+1 x ˙

Since the product e−D tive to the first term

k

H

0

2

(t−τ )

L2

0

dτ.

∂x (g1 · · · gk+1 ) is a convolution with the heat kernel, we may move the derivaIk ≤

 t 2 ∂x Dxsk e−D (t−τ ) (g1 · · · gk+1 )(τ )

L2

0

dτ,

We shall need the following heat kernel estimate. Lemma 3. For 0 ≤ α, and p = 2 = q or 0 < α ≤ 1 and 1 ≤ p < q ≤ ∞. Then there exists a positive constant c = c(p, q, α) such that 2

Dxα e−D t f Lq ≤ ct−r f Lp , where r = 12 ( p1 − 1q ) + α2 . Proof. The proof is contained in the appendix. Now we apply the heat kernel estimate, Lemma 3, with p = 2, q = 2 and α = sk + 1 ≥ 1/2  Ik ≤

t 0

1

(t − τ )− 2 (sk +1) g1 · · · gk+1 (τ )L2 dτ.

We apply the generalized H¨older inequality to obtain  t 1 (t − τ )− 2 (sk +1) g1 (τ )L2(k+1) · · · gk+1 (τ )L2(k+1) dτ. Ik ≤ 0

5

(2.10)

Now we multiply by 1 = τ γk τ −γk to obtain  t 1 (t − τ )− 2 (sk +1) τ −γk τ γk g1 (τ )L2(k+1) · · · gk+1 (τ )L2(k+1) dτ, Ik ≤ 0

Taking supreme in the time variable, we obtain  Ik ≤ sup t

γk

t∈(0,T )

Since 12 ≤ inequality.

3 4

−

1 2k

=

1 2 (sk

g1 (t)L2(k+1) · · · gk+1 (t)L2(k+1)

t 0

1

(t − τ )− 2 (sk +1) τ −γk dτ.

+ 1) ≤ 3/4 the remaining integral is bounded using the following calculus

Lemma 4. Let 0 < α, β < 1 and θ = α + β − 1, such that 0 ≤ θ ≤ 1. Then we have the following bound  t (t − y)−β y −α dy ≤ cα,β t−θ . B(α, β, t)= ˙ 0

Applying the calculus lemma, we have  t 1 (t − τ )− 2 (sk +1) τ −γk dτ ≤ ck t−θ < ∞, 0

where θ = γk +

1 2 (sk

+ 1) − 1 is responsible for our choice of γk . This gives us 1

Ik ≤ ck sup t1− 2 (sk +1) g1 (t)L2(k+1) · · · gk+1 (t)L2(k+1) , t∈(0,T )

and the exponent 1 1 1 − (sk + 1) = 1 − 2 2 as desired. Now we consider Jk =t ˙ Again, the operators e−D

2

γk

3 1 − 2 k

 =1−

3k − 2 k+2 = = γk (k + 1), 4k 4k

 t −D2 (t−τ ) ∂x (g1 · · · gk+1 )(τ ) e

L2(k+1)

0

dτ.

(t−τ )

and ∂x commute  t 2 Jk = tγk ∂x e−D (t−τ ) (g1 · · · gk+1 )(τ )

L2(k+1)

0

dτ.

We apply Lemma 3 with p = 2, q = 2(k + 1) and α = 1 to obtain  t 1 1 1 1 J k ≤ c k t γk (t − τ )− 2 ( 2 − 2(k+1) )− 2 (g1 · · · gk+1 )(τ )L2 dτ. 0

We apply the generalized H¨older inequality with  J k ≤ c k t γk We now multiply by 1 = τ  J k ≤ c k t γk

t 0

t 0

1

1

1

1 2

=

1 2(k+1)

+ ··· +

1 2(k+1)

1

(t − τ )− 2 ( 2 − 2(k+1) )− 2 g1 (τ )L2(k+1) · · · gk+1 (τ )L2(k+1) dτ.

γk (k+1)−γk γk −γk (k+1)

τ

, and let −rk = − 12 ( 12 −

1 2(k+1) )

−

1 2

(t − τ )−rk τ γk −γk (k+1) τ γk (k+1)−γk g1 (τ )L2(k+1) · · · gk+1 (τ )L2(k+1) dτ, 6

and then taking supreme of the g’s and the τ γk (k+1)−γk we obtain  t (t − τ )−rk τ γk −γk (k+1) dτ sup tγk (k+1) g1 (t)L2(k+1) · · · gk+1 (t)L2(k+1) . J k ≤ ck t∈(0,T )

0

Noticing the remaining integral is again a generalized Beta distribution, we may apply Lemma 4, to obtain Jk ≤ ck t−θ sup tγk (k+1) g1 (t)L2(k+1) · · · gk+1 (t)L2(k+1) , t∈(0,T )

where

1 θ = rk − γk + γk (k + 1) − 1 = 2

1 1 − 2 2(k + 1)



k+2 1 + − 1 = 0, 2 4(k + 1)

+

and so the remaining terms are what we desired. Combining the estimates of Ik and Jk , the proof of the lemma is complete. 2(k+1) In order to take advantage of the estimate in Lemma 2, we notice that for u(·, t) ∈ G˙ σ,δ,sk ∩ Lσ,δ , Au(·, t)∈˙ H˙ sk ∩ L2(k+1) and in particular we have

AuH˙ sk = uG˙ σ,δ,sk , and AuLp = uLpσ,δ .

(2.11)

Moreover, multiplying the second equation by tγ , add the above two equations together and then taking the supremum over t ∈ (0, T ) we get sup AuH˙ sk + sup tγ AuLp = sup uG˙ σ,δ,sk + sup tγ uLpσ,δ ⇒ AuY sk = uY σ,δ,sk .

t∈(0,T )

t∈(0,T )

t∈(0,T )

t∈(0,T )

The following proposition states the desired multilinear estimate. Proposition 1. Let σ ≥ 1, δ > 0, sk = 12 − k1 and T > 0. Then there is a constant C > 0 such that for all g1 , . . . , gk+1 ∈ Y σ,δ,sk we have  t −D 2 (t−τ ) e ∂ (g · · · g )(τ )dτ ≤ Cg1 Y σ,δ,sk · · · gk+1 Y σ,δ,sk . (2.12) x 1 k+1 0

Y σ,δ,sk

Proof. We will show the claim by computing the two portions of the norm individually. Part 1: supt∈(0,T )  · G˙ σ,δ,sk . We apply Minkowski’s inequality, and then open the norm to find  t −D 2 (t−τ ) e ∂ (g · · · g )(τ )dτ P1 = sup x 1 k+1 ˙ σ,δ,s t∈(0,T )

0

G

k

1  t   2 2  sk +1 −n2 (t−τ ) δ|n|1/σ  ≤ sup e e | g1  · · ·  gk+1 | (n, τ ) dτ. |n| t∈(0,T )

0

(2.13)

n∈Z

The convolution is defined as | g1  · · ·  gk+1 |(n, τ ) = |



g1 (n − n1 , τ ) g2 (n1 − n2 , τ ) · · · gk+1 (nk , τ )|.

(2.14)

n1 ,··· ,nk

Using eδ|n| eδ|n|

1/σ

1/σ

1/σ

≤ eδ|n−n1 |

1/σ

· · · eδ|nk |

| g1  · · ·  gk+1 |(n, τ ) ≤

, from (2.14) we get  1/σ 1/σ |eδ|n−n1 | g1 (n − n1 , τ )| · · · |eδ|nk | gk+1 (nk , τ )|.

n1 ,··· ,nk

7

(2.15)

Now, use the definition of A and convolution to conclude   1/σ 1  · · ·  Ag  eδ|n| | g1  · · ·  gk+1 |(n, τ ) ≤ Ag k+1 (n, τ ).

(2.16)

We substitute the above expression, inequality (2.16), back into (2.13) which yields P1 ≤ sup

 t 

t∈(0,T )

0

|n|

2sk

 2    −n2 (t−τ )    Ag1  · · ·  Ag ne k+1 (n, τ )

 12 dτ.

n∈Z

Hence, P1 ≤ sup

t∈(0,T )

 t −D2 (t−τ ) ∂x (Ag1 · · · Agk+1 )(τ ) s dτ. e ˙ H

0

k

Now we can apply Lemma 2 to the right hand side of the above to conclude P1 ≤ C sup tγk (k+1) Ag1 L2(k+1) · · · Agk+1 L2(k+1) , t∈(0,T )

which completes the first part after applying relation (2.11). Part 2: supt∈(0,T ) tγk  · L2(k+1) . For the second estimate, we apply Minkowski’s inequality and use the 2(k+1)

definition of the Lσ,δ

σ,δ

space to obtain

 t −D 2 (t−τ ) P2 = sup t e ∂x (g1 · · · gk+1 )(τ )dτ 2(k+1) t∈(0,T ) 0 Lσ,δ  t 1  inx −n2 (t−τ ) δ|n|1/σ γk ≤ sup t e e ne | g1  · · ·  gk+1 |(n, τ ) 2π t∈(0,T ) 0 γk

n∈Z

dτ.

(2.17)

L2(k+1)

We use inequality (2.15), relation (2.17) and the definition of A to conclude the following inequality P2 ≤ sup t

γk

t∈(0,T )

 t −D2 (t−τ ) ∂x (Ag1 · · · Agk+1 )(τ ) e

L2(k+1)

0

dτ.

Now we use Lemma 2 P2 ≤ C sup tγk (k+1) Ag1 L2(k+1) · · · Agk+1 L2(k+1) , t∈(0,T )

and applying relation (2.11), we complete the proof of the second estimate. Combining the estimates from Part 1 and Part 2 completes the proof of the proposition. Lemma 5. Let σ ≥ 1, δ > 0, sk =

1 2

ΦuY σ,δ,sk

1 k

and T > 0. For all u ∈ Y σ,δ,sk and u0 ∈ G˙ σ,δ,sk we have 

1 uk+1 . (2.18) ≤C + u  σ,δ,s ˙ 0 G k Y σ,δ,sk k+1

−

Proof. In light of the previous proposition, it remains to show 2

2

2

e−D t u0 Y σ,δ,sk = sup e−D t u0 G˙ σ,δ,sk + sup tγk e−D t u0 L2(k+1) ≤ Cu0 G˙ σ,δ,sk . t∈(0,T )

t∈(0,T )

8

σ,δ

We bound the first part using H¨older’s inequality as follows 2

2

sup e−D t u0 G˙ σ,δ,sk = sup |n|sk e−n t eδ|n|

t∈(0,T )

t∈(0,T )

1/σ

| u0 |2 (Z)

2

≤ sup e−n t L∞ |n|sk eδ|n|

1/σ

t∈(0,T )

| u0 |2 (Z) = u0 G˙ σ,δ,sk . 2(k+1)

The second part we estimate in two cases. First, when sk < 0, recalling the Lσ,δ 2 2 tγk e−D t u0 L2(k+1) = tγk e−D t Au0 2(k+1) . σ,δ

norm we have

L

Applying the operator Dx−sk Dxsk we have

2 2 tγk e−D t u0 L2(k+1) = tγk Dx−sk e−D t Dxsk Au0 σ,δ

L2(k+1)

.

We now apply the estimate found in Lemma 3 with f = Dxsk Au0 , q = 2(k + 1) and r = 1/4 − 1/(4(k + 1)) − sk /2 which yields the inequality 2

tγk e−D t u0 L2(k+1) ≤ ctγk t−r Dxsk Au0 L2 = Ctγk −r u0 G˙ σ,δ,sk . σ,δ

When we consider the exponent of t, we find for k = 1, 2 γk − r =

1 1 1 1 1 k+2 + ( − )− + = 0, 4k(k + 1) 2 2 k 4 4(k + 1)

and therefore the result holds. Now we estimate when sk > 0. In this case, we apply 1 = Dx−βk Dxβk , for βk = k/(2(k + 1)) so that 2 2 tγk e−D t u0 L2(k+1) = tγk Dx−βk Dxβk e−D t Au0 2(k+1) . σ,δ

L

We shall next use the following result which can be found in [S]. Lemma 6 (Hardy-Littlewood-Sobolev). Suppose that 0 < α < 1, 1 < p < 1/α and q satisfies 1 1 = − α. q p If f ∈ Lp , then Dx−α f ∈ Lq (recall Dx = (−∂x2 )1/2 is the Riesz potential) and there exists a constant C = C(n, α, p) such that Dx−α f Lq ≤ Cf Lp . Applying the Hardy-Littlewood-Sobolev lemma, with p = 2, q = 2(k + 1) and α = βk we obtain 2 2 tγk e−D t u0 L2(k+1) ≤ Ctγk Dxβk e−D t Au0 . 2 σ,δ

L

Next, we write βk = βk − sk + sk and estimate the above as follows 2 2 tγk e−D t u0 L2(k+1) ≤ Ctγk Dxβk −sk Dxsk e−D t Au0 σ,δ

L2

so that we can now we apply Lemma 3, with βk − sk ≥ 0, to obtain 2

tγk e−D t u0 L2(k+1) ≤ Ctγk −βk /2+sk /2 Au0 H˙ sk , σ,δ

9

,

and we can compute γk −

sk k+2 k 1 1 βk + = − + − = 0, 2 2 4k(k + 1) 4(k + 1) 4 2k

which yields the result 2

tγk e−D t u0 L2(k+1) ≤ C u0 G˙ σ,δ,sk . σ,δ

The following proposition summarizes our results and completes the proof to Theorem 3 under a smallness assumption. Proposition 2. Let σ ≥ 1, δ > 0 and sk = condition

1 2

− k1 . For initial data u0 ∈ G˙ σ,δ,sk satisfying the smallness 2k+1 (k + 1) − 1

u0 G˙ σ,δ,sk ≤

2

k+1 k

C

k+1 k

(k + 1)

k+1 k

,

(2.19)

if we choose the ball B(0, R)={u ˙ ∈ Y σ,δ,sk : uY σ,δ,sk ≤ R} with radius R = 2−1/k C −1/k then Φ : B(0, R) → B(0, R) and is a contraction. Proof. Applying Lemma 5 we get 



1 1 k+1 k+1 uY σ,δ,sk + u0 G˙ σ,δ,sk ≤ C R ΦuY σ,δ,sk ≤ C + u0 G˙ σ,δ,sk . k+1 k+1 which we require is less than R. Solving the resulting inequality for u0 G˙ σ,δ,sk we obtain ΦuY σ,δ,sk ≤ R ⇐⇒ u0 G˙ σ,δ,sk ≤

1 R − Rk+1 . C k+1

To show Φ is a contraction, notice  t  t 2 1 1 −D 2 (t−τ ) k+1 Φu − Φv = e ∂x u dτ − e−D (t−τ ) ∂x v k+1 dτ k+1 0 k+1 0  t   2 1 = e−D (t−τ ) ∂x uk+1 − v k+1 dτ k+1 0 ⎛ ⎞  t k  2 1 = e−D (t−τ ) ∂x ⎝ uk−j v j (u − v)⎠ dτ. k+1 0 j=0 By applying the derivative to each term in the summation separately, we have Φu − Φv =

k  j=0

1 k+1



t

e−D

2

0

(t−τ )

  ∂x uk−j v j (u − v) dτ.

Therefore, applying Proposition 1 to each term in (2.20) we get C  uk−j vjY σ,δ,sk u − vY σ,δ,sk . Y σ,δ,sk k + 1 j=0 k

Φu − ΦvY σ,δ,sk ≤

Using the estimate vY σ,δ,s , uY σ,δ,sk ≤ R, C  k−j j R R u − vY σ,δ,sk = CRk u − vY σ,δ,sk . k + 1 j=0 k

Φu − ΦvY σ,δ,sk ≤

10

(2.20)

To apply the contraction mapping theorem, we require CRk < 1. Using the definition of R we have Φu − ΦvY σ,δ,sk ≤

1 u − vY σ,δ,sk , 2

(2.21)

which completes the proof.

3. Regularity of the solution to Burgers equation in both variables Here we will show that given G˙ σ,δ,sk (δ > 0) initial data, one can obtain at least Gevrey-2σ regularity in the time variable and Gevrey-σ in the spatial variable. In particular, we will show that for every x ∈ T, for every 0 < t < T , and for all non-negative integers  and j, the following estimate holds sup (x,t)∈T×(0,T )

|∂tj ∂x u(x, t)| ≤ C j++1 (j!)2σ (!)σ

(3.1)

for some C > 0. Notice, G˙ σ,δ,sk is a collection of H˙ sk functions whose smoothness was measured in the phase space by 1/σ the additional weighting eδ|n| . For δ > 0, this regularity may also be expressed in the base space in a more classical sense, through the growth rate of the derivatives under the L∞ norm. Lemma 7. If u(x, t) ∈ C((0, T ); G˙ σ,δ,s ), for σ ≥ 1, δ > 0 and s ∈ R, then following estimate holds for  ∈ N+ ,

  ∂x u(x, t) ∞ ≤ M 1 (!)σ , (3.2) L C0 for some positive constants M and C0 . Before we continue our proof, let us give some notation introduced by Alinhac and M´etivier [AM] and used in [HHP1]. For q ∈ N+ , σ ≥ 1, ε > 0 and c > C40 > 0 set mq =

c(q!)σ c , M0 = max{1, } and Mq = ε1−q mq . (q + 1)2 8

The following inequalities can be found in [HHP1]  k   k  m mk− ≤mk , M Mk− ≤ εMk , Mj ≤ εMj+1 for j ≥ 2.   0≤≤k

(3.3)

(3.4)

0<
Also, given 0 < C0 < 1 there exists 0 < ε0 such that for any 0 < ε ≤ ε0 we have 1 C0j+1

j!σ ≤ Mj , for j ≥ 2.

(3.5)

To prove the claimed regularity in t, we will need the following result. Lemma 8. Given , k ∈ {0, 1, 2, . . . } we have   k     k p=0 q=0

p

q

M−p+2(k−q) Mp+1+2q

m   m Mr Mm−r ≤ r r=1

where m =  + 2k + 1, and Mj , j = 0, 1, . . . , m, are defined in (3.3). 11

(3.6)

Proof. For non-negative integers  and k, let m =  + 2k + 1. We begin the estimate by changing the order of summation   k  k         k  k (3.7) M−p+2(k−q) Mp+1+2q = M−p+2(k−q) Mp+1+2q . p q p q p=0 q=0 q=0 p=0 Making the change of variables r = p + 2q + 1 we obtain     k +2q+1 k       k  k M−p+2(k−q) Mp+1+2q = M−r+2q+1+2(k−q) Mr−2q+2q p q r − 2q − 1 q p=0 q=0 q=0 r=2q+1 =

k +2q+1   q=0 r=2q+1

Using the convention that for a, b ∈ N,

a −b

= 0 and

 a

  k     k p=0 q=0

p

q

M−p+2(k−q) Mp+1+2q ≤

0

  k Mm−r Mr . q

 r − 2q − 1 =1=

It now remains to show that

k  q=0

a

k  m  q=0 r=1

(3.8)

a

, this is bounded by   k Mm−r Mr . q

 r − 2q − 1

(3.9)

   k m ≤ . q r

 r − 2q − 1

We first show the first few cases, to illustrate the idea, then we proceed for general k. Also we will assume without loss of generality that r − 2q − 1 ≥ 0, since the terms in which this fails are zero. When k = 1, we have      1   1   +2 = + = , r − 2q − 1 q r−1 r−2−1 r−1 q=0 where in the last equality we used

   a a a+1 + = . b b+1 b+1

(3.10)

  a a+1 ≤ , b b+1

(3.11)

Next, using

we obtain

1  q=0

 r − 2q − 1

    1 +3 m ≤ = . q r r

Similarly, when k = 2 we have 2  q=0

 r − 2q − 1

      2   2  = + + . q r−1 r−3 1 r−5

Using equality 3.10 we have 2  q=0

 r − 2q − 1

     2 +4  2 ≤ + . q r−1 r−3 1 12

Next using the inequality

we obtain

   a c a+c ≤ , b d b+d 2  q=0

 r − 2q − 1

(3.12)

    2 +4 +2 ≤ + . q r−1 r−2

Using inequality 3.11 we have       2   2 +4 +4 +5 m ≤ + ≤ = . r − 2q − 1 q r − 1 r r r q=0 where in the last inequality we again used property 3.10. We will now proceed by analyzing the general case    

    k   k  k  k  k = + ··· + + . r − 2q − 1 q r − 1 0 r − 2k + 1 k − 1 r − 2k − 1 k q=0 Using equality 3.10 on the first and last terms, and then inequality 3.12 on the intermediate terms we obtain     

  k   k  + 2k  k  k = + + ··· + r − 2q − 1 q r−1 r−3 1 r − 2k + 1 k − 1 q=0

   

  + 2k +k +k +k +k ≤ + + + + ··· + . r−1 r−2 r−3 r−4 r−k Now using equality 3.10 and inequality 3.11 iteratively on the k − 2 remaining terms we obtain     k   k  + 2k  + 2k − 2 ≤ + . r − 2q − 1 q r−1 r−2 q=0 Applying the bound in inequality 3.11 again, we find       k   k  + 2k  + 2k  + 2k + 1 m ≤ + = = , r − 2q − 1 q r − 1 r r r q=0 which complete the proof of the lemma. Using this lemma, we will show the following result concerning the regularity in both derivatives, given Gevrey-σ regularity in the time variable. Lemma 9. Let u(x, t) be a solution to the Cauchy problem (1.1). If u(x, t) satisfies inequality

 1  ∂x u(t, ·)L∞ ≤ M !σ (3.13) C0   for some C0 > 0 and M ≥ max 2, cC8 0 , cC4 2 and σ ≥ 1, then there exists an ε0 > 0 s.t. for any 0 0 < ε < ε0 , |∂tj ∂x u(x, t)| ≤ M kj+1 M+2j , j ∈ {0, 1, 2, ...},  ∈ {0, 1, 2, ...} for all (x, t) ∈ T × (0, T ). 13

(3.14)

Proof. We first verify that the spacial derivatives satisfy the estimate. Thus, let j = 0. For  = 0 it follows from equation (3.13) and the definition of M0 that |u(x, t)| ≤ M ≤ M M0 , ∀(x, t) ∈ T × (0, T ).   For  = 1 we have |∂x u(x, t)| ≤ M C10 (1!)σ < M 4c = M M1 , ∀(x, t) ∈ T × (0, T ). For  ≥ 2 we have   |∂x u(x, t)| ≤ M C10 (!)σ ≤ M M , ∀(x, t) ∈ T × (0, T ), by applying assumption (3.13) and property (3.5). Thus we have the inequality is satisfied for all  if j = 0. We will now proceed by induction on j. Therefore, assume (3.14) is true for 0 ≤ q ≤ j and  ∈ {0, 1, . . . }, and we will consider the j + 1 derivative in t. Notice |∂tj+1 ∂x u| = |∂tj ∂x (uk ∂x u) + ∂tj ∂x+2 u| ≤ |∂tj ∂x (uk ∂x u)| + |∂tj ∂x+2 u|. We estimate the second term as |∂tj ∂x+2 u| ≤ M kj+1 M+2+2j = M kj+1 M+2(j+1) ,

(3.15)

by applying the inductive hypothesis. We expand the first term using Leibniz rule ∂tj ∂x (uk ∂x u)

=

j  j1 

p1    

jk−1

···

···

jk =0 p1 =0 p2 =0

j1 =0 j2 =0



pk−1 pk =0

   

  p1 pk−1 j j1 jk−1 j−j1 −p1  ... ... ∂t × ∂x u · · · ∂tjk ∂xpk +1 u, p2 pk j2 jk j1 p1 and applying the assumption to each term to yields |∂tj ∂x (uk ∂x u)| ≤

j  j1 =0

×M

  

jk−1

···

jk =0 p1 =0

k(j−j1 )+1

     pk−1 j jk−1 ... ... pk jk p1 j1 p =0 pk−1

···

k

M(−p1 )+2(j−j1 ) M k(j1 −j2 )+1 M(p1 −p2 )+2(j1 −j2 ) · · ·

× M k(jk−1 −jk )+1 M(pk−1 −pk )+2(jk−1 −jk ) M kjk +1 M(pk +1)+2jk .

(3.16)

Next, using Lemma 8 with p = pk ,  = pk−1 q = jk , k = jk−1 , m = pk−1 + 2jk−1 + 1 we obtain m    pk−1  jk−1   m M(pk−1 −pk )+2(jk−1 −jk ) Mpk +1+2jk ≤ Mr Mm−r . pk jk r p =0 j =0 r=1 pk−1 jk−1

k

(3.17)

k

Using the second inequality of (3.4) we obtain m   m r=1

r

Mr Mm−r = 2Mm M0 +

m−1  r=1

 m Mr Mm−r ≤ (M0 + ε)Mpk−1 +2jk−1 +1 . r

Similarly, applying Lemma 8 with p = pk−1 ,  = pk−2 q = jk−1 , k = jk−2 , m = pk−2 + 2jk−2 + 1, followed by the second inequality of (3.4) we find  pk−2  jk−2  M(pk−2 −pk−1 )+2(jk−2 −jk−1 ) Mpk−1 +1+2jk−1 ≤ (M0 + ε)Mpk−2 +2jk−2 +1 . pk−1 jk−1 =0 j =0



pk−2 pk−1

jk−2

k−1

Continue this way to obtain all such inequalities, and substituting them into (3.16) yields |∂tj ∂x (uk ∂x u)| ≤ M kj+k+1 (M0 + ε)k M+2j+1 . 14

Now use the third inequality of (3.4) |∂tj ∂x (uk ∂x u)| ≤ M k(j+1)+1 (M0 + ε)k εM+2(j+1) . Let ε ≤ ε0 =

1 2(M0 +1)k

< 1 so that (M0 + ε)k ε ≤ (M0 + 1)k ε ≤ (M0 + 1)k

1 1 = . 2(M0 + 1)k 2

Therefore, we may conclude that |∂tj ∂x (uk ∂x u)| ≤

1 k(j+1)+1 M M+2(j+1) . 2

Combining this estimate with the first estimate (3.15) yields |∂tj+1 ∂x u| ≤

1 k(j+1)+1 M M+2(j+1) + M kj+1 M+2(j+1) ≤ M k(j+1)+1 M+2(j+1) , 2

which is precisely the claim. Before we continue, we state the following two inequalities which will be needed shortly. Lemma 10. If a, b ∈ N+ , then (a + b)! ≤ e2a+2b (a!)(b!) and (2a)! ≤ (4)a (a!)2 . Using the previous two lemmas, the we can now complete the proof of Gσ,2σ (T, [0, T ]) regularity. Theorem 4. For initial data u0 ∈ G˙ σ,δ,sk (σ ≥ 1, δ > 0, sk = 12 − k1 ), for every (x, t) ∈ T × (0, T ), and for all non-negative integers j and , there exists a constant C > 0 such that |∂tj ∂x u(x, t)| ≤ C j++1 (j!)2σ (!)σ .

(3.18)

Proof. Applying Lemma 9 for j and  ∈ Z+ yields |∂tj ∂x u(x, t)| ≤ M kj+1 M+2j = M M jk ε1−2j−

c(2j + )!σ ≤ M εc (2j +  + 1)2

Mk ε2

j

1 (2j + )!σ . ε

Next, we apply the first estimate in Lemma 10 to estimate (2j + )!σ :

|∂tj ∂x u(x, t)|

≤ M εc

Mk ε2

j

1 4j 2 e e ((2j)!)σ (!)σ . ε

We estimate the factorial (2j)! using the second estimate in Lemma 10 to obtain

|∂tj ∂x u(x, t)|

≤ M εc

M k e4 ε2

j

e2 ε



(4σ )j (j!)2σ (!)σ

≤ LC1j C2 (j!)2σ (!)σ , where L = M εc, C1 = M k e4 4σ /ε2 and C2 = e2 /ε. Combining constants yields precisely our definition of Gσ,2σ (T, [0, T ]), and therefore, our proof is complete.

15

4. Sharp regularity of the solution to Burgers equation In this section, we assume k is odd and are in the periodic case. Notice that if w solves the initial value problem  1 (wx )k+1 wt = wxx + k+1 , (4.1) w(x, 0) = w0 (x) d then u = ∂x w solves the k-gBurgers Cauchy problem (1.1) with initial data u0 (x) = dx w0 (x). We will σ,r show that w(x, t) need not be G (T, [0, T ]), 1 ≤ r < 2σ, and use this result to imply u(x, t) is not either. The methodology of proof is to analyze the solution near t = 0, however, we notice that time derivatives do not exist at t = 0. Thus, we take right hand derivatives in the time variable when we evaluate derivatives at t = 0, since these derivatives exist for 0 < t < T . We give the details for initial data in the space G˙ 1,1/2,sk (T), the proofs for the general case are similar.

Theorem 5. Suppose that for k = 1, 5, 9, . . .



w0 (x) = Re

∞ 

 e

−n inx

e

,

(4.2)

n=1

and for k = 3, 7, 11, . . .



∞ 

w0 (x) = −Re

 e

−n inx

e

.

(4.3)

n=1

Then a solution to the Cauchy problem (4.1) is not G1,r (T, [0, T ]) for 1 ≤ r < 2. Proof. We provide the details for the case k = 1, 5, 9, . . . . Since the above formulas are written as the Fourier coefficients, it is easy to check using the formula given in equation (4.2), that the initial data is G˙ 1,1/2,s for any s ∈ R. In fact we have    1/σ n2s e2δn e−2n ≤ n2s e−2n+n ≤ n2s e−n < ∞. n≥1

n≥1

n≥1

To estimate the derivatives, we will use the following formula for the solution w(x, t) of the initial value problem (4.1) Lemma 11. If w is a solution to (4.1), then the following formula holds for x ∈ T and t ∈ [0, T ], where we take left and right hand derivatives at the endpoints ∂tj w = ∂x2j w +

j 



α

Cαq ∂xα1 w · · · ∂x qk+1 w,

q=1 |α|+q−qk=2j

for j ∈ {1, 2, 3, . . . }, Cαq ≥ 0 and αμ ≥ 1. By using the fact that

dj dxj w0 (0)

= Re(ij )Aj , where Aj =

∂tj w(0, 0) = Re(i2j )A2j +

j 



∞ n=1

e−n nj and last lemma we obtain

Cαq Aα1 . . . Aαqk+1 Re(iα1 ) . . . Re(iαqk+1 )

q=1 |α|+q−qk=2j

= (−1)j A2j +

j 



Cαq Aα1 . . . Aαqk+1 (−1)

q=1 |α|+q−qk=2j

16

|α| 2

.

(4.4)

Using the fact |α| + q − qk = 2j we have ∂tj w(0, 0) = (−1)j A2j +

j 



Cαq Aα1 . . . Aαqk+1 (−1)

2j+qk−q 2

q=1 |α|+q−qk=2j



= (−1)j A2j +

j 



Cαq Aα1 . . . Aαqk+1 (−1)

q(k−1) 2

 ,

(4.5)

q=1 |α|+q−qk=2j

and therefore, by using the fact that in this case k − 1 is a multiple of four we have |∂tj w(0, 0)| ≥ A2j =

∞ 

e−n n2j ≥ e−2j (2j)2j = C j (2j)2j ≥ C j (j)2j ≥ C j (j!)2 ,

n=1

which shows the solution w(0, ·) ∈ Gr (R) for any 1 ≤ r < 2. We show the previous estimates force sharpness of G1,2 (T, [0, T ]) regularity in time of u. Corollary 1. For odd values of k, the solution u(x, t) may fail to be G1,r (T, [0, T ]) for any 1 ≤ r < 2 when it’s initial data are G˙ 1,1/2,sk (T). Proof. Suppose by contradiction there exists a C > 0 and 1 ≤ r < 2 such that for all (x, t) ∈ T × (0, T ) |∂tj ∂x u(x, t)| ≤ C j++1 (j!)r (!), for (j, ) ∈ Z+ × Z+ .

(4.6)

Then, taking right hand derivatives at t = 0, and considering  = 0 and j ∈ Z+ we have by the definition of w(x, t) |∂tj ∂x w(0, 0)| = |∂tj u(0, 0)| ≤ C j+1 (j!)r ,

(4.7)

and taking  = 1 and j ∈ Z+ we have |∂tj ∂x2 w(0, 0)| = |∂tj ∂x u(0, 0)| ≤ C j+1 (j!)r .

(4.8)

From the equation for w and the initial data stated in the previous lemma, we have for all j ∈ Z+ ((j + 1)!)2 ≤ |∂tj+1 w(0, 0)| = |∂tj ∂x2 w(0, 0) +

1 ∂ j (∂x w(0, 0))k+1 (0, 0)|. k+1 t

(4.9)

Applying the triangle inequality, and estimates (4.7) and (4.8) we have ((j + 1)!)2 ≤ |∂tj+1 w(0, 0)| ≤ 2C j+1 (j!)r .

(4.10)

A contradiction. Thus, u(x, t) may not be better than G1,2 (T, [0, T ]). 1/σ

Using the example provided in [GHHP], i.e., replacing the weight e−n with e−2δn , and the above methodology, the analogous result holds for initial data in the space Gσ,δ,sk when σ > 1, δ > 0. Corollary 2. Let σ ≥ 1 and δ > 0 and suppose 1 ≤ r < 2σ. For k = 1, 5, 9, . . . and for k = 3, 7, 11, . . . , consider the G˙ σ,δ,sk (T) initial data ∞ ∞     −2δn1/σ inx −2δn1/σ inx e e e e w0 (x) = Re , and w0 (x) = −Re . (4.11) n=1

n=1

Then for x fixed, the solution, w(x, ·) to the Cauchy problem (4.1) is not Gr in the time variable. Furthermore, this implies the solution u(x, t) is not Gσ,r (T, [0, T ]). 17

Proof. The proof of this is similar to the case σ = 1. Finally, we note that one can obtain optimality of the 2σ regularity in the time variable if we admit complex valued G˙ σ,δ,sk (T) initial data for all values of k ∈ Z+ . Corollary 3. Suppose for k = 1, 2, . . . , σ ≥ 1 and δ > 0, the initial data for the k-gBurgers equation is given by u0 (x) = (i)

1/k

∞ 

e−2δn

1/σ

einx .

(4.12)

n=1

Then for x fixed, a solution of the k-gBurgers equation, u(x, ·) is not in Gevrey-r in the time variable for any 1 ≤ r < 2σ. Proof. In this case, the following formula holds for the initial data, u0 (x) ∂xj u0 (x) = (i)

1/k

∞ 

(in) e−n einx . j

(4.13)

n=1

Next, observe the solution u(x, t) satisfies the following estimate. Lemma 12. For a solution u(x, t) of the initial value problem (1.1), the following formula holds: ∂tj u = ∂x2j u +

j 



α

Cαq ∂xα1 u · · · ∂x qk+1 u

(4.14)

q=1 |α|+q=2j

for j ∈ {1, 2, 3, . . . }, Cαq ≥ 0 and αμ ≥ 0. Using the above formula for our particular initial data, the relation |α| + q = 2j, and identity (4.14) we get ∂tj u(0, 0)

= (i)

1 2j+ k

∞ 

2j

(n) e

−n

+ (i)

1 2j−q+ k +q

n=1

 1 2j+ k

= (i)

∞ 

2j



q=1 |α|+q=2j

 (n) e−n + B

j 

Cαq

∞ 

(n)

α1

e−n · · ·

n=1

∞ 

(n)

αqk+1

e−n

n=1

,

n=1

where B > 0. Taking the absolute value of both sides, we get ∞      j 2j 2j (n) e−n ≥ (2j) e−2j ≥ C j (j)2j ≥ C j (j!)2 , ∂t u(0, 0) ≥ n=1

which shows that u(x, t) cannot be Gevrey-r for any 1 ≤ r < 2 in time near zero.

Appendix A. Proof of Lemma 3 For completeness, we provide a proof of Lemma 3.

18

(4.15)

Proof. The case p = 2 = q follows from Parseval’s lemma, and the case p = 1, q = ∞ is a simple calculation. For the remaining p–q estimate, we break the estimate into cases based on α. First we do the case α = 1, then we do the intermediate cases. 2

Case α = 1. Let Γ(x, t)  f (x) = F −1 (e−D t f ), then Γ(x, t) = √

1  −(x+2πn)2 /4t e . 4πt n∈Z

(A.1)

Using the convolution theorem for the Fourier transform, we have 2

∂x e−D t f Lq = ∂x Γ(x, t)  f (x)Lq , Applying Young’s convolution inequality, with 1 < p < q < ∞ and r =

1

1 1 q − p +1

, we obtain

2

∂x e−D t f Lq ≤ ∂x Γ(x, t)Lr f Lp . It is easy to calculate the following two estimates ∂x Γ(x, t)L1 ≤ π −1/2 t−1/2 , ∂x Γ(x, t)L∞ ≤ (e

−1/2

+ 1)t

−1

(A.2)

,

(A.3)

from which we can conclude, by the interpolation of Lp spaces θ

∂x Γ(x, t)Lr  t−θ/2 (t−1 )1−θ = t 2 −1 , where

(A.4)

θ 1−θ 1 1 = + =⇒ θ = , r 1 ∞ r

which implies, 1

∂x Γ(x, t)Lr ≤ c˜t 2r −1 . Finally, using

1 r

=

1 q

−

1 p

(A.5)

+ 1, we have 1

1

1

1

∂x Γ(x, t)Lr ≤ c˜t− 2 ( p − q )− 2 ,

(A.6)

which completes the result in the case α = 1. Case α ∈ (0, 1). We have 2

2

Dxα e−D t f Lq = Dxα−1 Dx e−D t f Lq . Since −1 < α − 1 < 0, we may apply the Hardy-Littlewood-Sobolev (HLS) lemma and we calculate q p = q(1−α)+1 . 2

2

Dxα e−D t f Lq ≤ CDx e−D t f Lp , where C = C(α, p, q). We now apply Young’s inequality for convolutions, where the fact that Dx e

2

−D t

f = Dx Γ(x, t)  f (x) to find 2

Dxα e−D t f Lq ≤ Dx Γ(x, t)Lr f Lp . 19

1 p

+1 =

1 r

+

1 p ,

and use

We solve for r in terms of q and p to find pp = r=  p − p + pp

qp q(1−α)+1 p (q(1−α)+1) q q(1−α)+1 − q(1−α)+1

+

qp q(1−α)+1

=

qp q(1−α)+1 2p q−p qα+p −q q(1−α)+1

=

qp 2p q

−

p qα

+ p − q

,

and we find that the requirement ( 1 ≤ r, p ) to apply Young’s inequality, is satisfied when 1 1 q 1 = 2 − α + −  ≤ 1 =⇒ p ≤ = p, r q p q − αq + 1 which by assumption (from applying HLS) is strictly less then q. Now we estimate Dx Γ(x, t)Lr using inequality (A.5) to find 1 Dx Γ(x, t)Lr ≤ ct 2r −1 , and we compute 2p q − p qα + p − q α −p qα + p − q 1 1 1 1 −1= −1= =− (  − )− .  2r 2qp 2qp 2 p q 2 Thus, 2

Dxα e−D t f Lq ≤ ct

− 12 ( p1 − q1 )− α 2

f Lp ,

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