Remarks on a weighted energy estimate and its application to nonlinear wave equations in one space dimension

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ScienceDirect J. Differential Equations 256 (2014) 389–406 www.elsevier.com/locate/jde

Remarks on a weighted energy estimate and its application to nonlinear wave equations in one space dimension Makoto Nakamura Faculty of Science, Kojirakawa-machi 1-4-12, Yamagata 990-8560, Japan Received 24 May 2013; revised 17 August 2013 Available online 7 October 2013

Abstract A weighted energy estimate with tangential derivatives on the light cone is applied for the Cauchy problem of semilinear wave equations with the null conditions in one space dimension. The well-posedness and lifespan of the solutions are considered based on the vector field method. © 2013 Elsevier Inc. All rights reserved. MSC: 35L70 Keywords: Nonlinear wave equations; Null conditions; Weighted energy estimates

1. Introduction Let c > 0, T > 0. Let us consider the Cauchy problem of linear wave equations  2  ∂t − c2 ∂x2 u(t, x) = f (t, x) for (t, x) ∈ [0, T ) × R, u(0, ·) = u0 (·), ∂t u(0, ·) = u1 (·),

(1.1)

where u is the unknown function, f is the inhomogeneous term, u0 and u1 are initial data. The standard energy estimates show the inequality E-mail address: [email protected]. 0022-0396/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2013.09.005

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1 sup 0tT 2 1  2





2 2  ∂t u(t, x) + c2 ∂x u(t, x) dx



R



 2 2 u1 (x) + c2 ∂x u0 (x) dx +

R

T 

  ∂t u(t, x) · f (t, x) dx dt =: E,

(1.2)

0 R

where we have put the right hand side as E. We put r := |x|, ∂r := (x/|x|)∂x , Dc := ∂t + c∂r which denotes the tangential derivative on the light cone {(t, x) ∈ (0, ∞) × R: ct = |x|}. For κ  −1, we define a weight function W (cT , κ) by  W (cT , κ) :=

{1 − (1 + cT )−κ }/κ log(1 + cT )

if κ > 0 or − 1  κ < 0, if κ = 0.

(1.3)

We prepare the following weighted energy estimate. Lemma 1.1. The solution of (1.1) satisfies the estimate c sup κ∈R 12W (cT , κ)

T  0 R

(Dc u(t, x))2 dx dt  E. (1 + |ct − r|)1+κ

(1.4)

The weighted energy estimate (1.4) has been shown by Lindblad and Rodnianski [24, p. 76, Corollary 8.2] and Alinhac [2, Theorem 1] for three space dimensions (i.e. x ∈ R3 ) with κ > 0, and it plays an important role to control the nonlinear terms which satisfy the null conditions since it enables us to obtain the decay estimates for waves near the light cone where the singularity propagates. Remark 1.2. When we consider the application of Lemma 1.1, the simple bounds W (cT , κ)  1/κ if κ > 0, W (cT , κ)  (1 + cT )|κ| /|κ| if −1  κ < 0 are useful (see the proof of Theorem 1.3, below). Let us consider the Cauchy problem of nonlinear wave equations as the application of Lemma 1.1. 

 ∂t2 − c2 ∂x2 u(t, x) = f (∂t u, ∂x u)(t, x) u(0, ·) = u0 (·), ∂t u(0, ·) = u1 (·),

for (t, x) ∈ [0, T ) × R,

(1.5)

where f (∂t u, ∂x u) denotes the nonlinear term dependent on ∂t u and ∂x u. We use the vector fields ∂t ,

∂x ,

x Ωc := ct∂x + ∂t , c

L := t∂t + r∂r ,

Γc := (∂t , ∂x , Ωc , L),

(1.6)

where r := |x|. We put c := ∂t2 − c2 ∂x2 and note the commuting properties ∂t Ωc = Ωc ∂t + c∂x , ∂x Ωc = Ωc ∂x +∂t /c, ∂t L = (L+1)∂t , ∂x L = (L+1)∂x , LΩc = Ωc L, c ∂t,x = ∂t,x c , c Ωc = Ωc c , c L = (L + 2)c . We denote the size of initial data by

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ε = ε(u0 , u1 )

    := u0 H 2 (R) + u1 H 1 (R) + x∂x u0 (x)L2 (R) + x∂x2 u0 (x)L2 (R)        + xu1 (x)L2 (R) + x∂x u1 (x)L2 (R) +  1 + |x| f (u1 , ∂x u0 )(x)L2 (R) ,

(1.7)

which implies the bound   Γ α u(0, ·) c

|α|1

L2 (R)

+

  ∂t Γ α u(0, ·) c

|α|1

L2 (R)

+

  ∂x Γ α u(0, ·) c

|α|1

L2 (R)

 Cε (1.8)

for some constant C > 0. We define the function space X(T , κ) by    

X(T , κ) := u ∈ C [0, T ), H 2 (R) ∩ C 1 [0, T ), H 1 (R) : |u|X(T ,κ) < ∞ ,

(1.9)

where  |u|X(T ,κ) := max

  ∂t Γ α u

L∞ ((0,T ),L2 (R))

c

|α|1





|α|1

1 W (cT , κ)

T  0 R

,

  ∂x Γ α u

L∞ ((0,T ),L2 (R))

c

|α|1

(Dc Γcα u(t, x))2 dx dt (1 + |ct − r|)1+κ

,

1/2  .

(1.10)

For integer p with 0  p < ∞, let Np denote the set of homogeneous polynomials of order p on R2 . For example, 1 ∈ N0 , t, x ∈ N1 , t 2 , tx, x 2 ∈ N2 and their linear combinations for (t, x) ∈ R2 . Let χ = χ(x) ∈ C ∞ (R) be any fixed function which satisfies χ = 0 near the origin and χ = 1 away from the origin. The following theorem shows how the tangent derivative affects the existence time of the solutions of (1.5), although the cut-off function χ appears technically to remove the discontinuity of Dc at the origin. Theorem 1.3. Let p be an integer with 2  p < ∞. (1) Let h ∈ Np−1 , and let f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)χ(x). Put κ := p − 2. Then for any data (u0 , u1 ) with ε = ε(u0 , u1 ) < ∞, there exist T > 0 and a unique solution u of (1.5) in X(T , κ), where T can be taken as any number with T  C/ε 2(p−1)

if p  3,

T log(1 + cT )  C/ε2

if p = 2

(1.11)

for some constant C > 0 which is independent of u0 , u1 and ε. (2) Let p = 2 or p = 3. Let h ∈ Np−2 , and let f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)2 χ(x). Put κ := (p − 3)/2. Then for any data (u0 , u1 ) with ε = ε(u0 , u1 ) < ∞, there exist T > 0 and a unique solution u of (1.5) in X(T , κ), where T can be taken as any number with   T  C exp C  /ε 2

if p = 3,

T  C/ε2

if p = 2

for some constants C > 0 and C  > 0 which are independent of u0 , u1 and ε.

(1.12)

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(3) Let p  4. Let h ∈ Np−2 , and let f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)2 χ(x). Put κ := (p − 3)/2. Then for any data (u0 , u1 ) with ε = ε(u0 , u1 ) sufficiently small, there exists a unique solution u of (1.5) in X(∞, κ). Namely, we obtain time global solutions for small data. We are able to show the continuous dependence of the solutions on the initial data, and the asymptotic behavior of the global solutions as follows. Proposition 1.4. Let u be the solution obtained by Theorem 1.3. (1) Let v be the solution of (1.5) with u0 and u1 replaced by v0 and v1 , which is obtained by Theorem 1.3. If    ∂t Γ α (u − v)(0, ·) c

|α|1

L2 (R)

 

+ ∂x Γcα (u − v)(0, ·)L2 (R)

(1.13)

tends to 0, then |u − v|X(T ,κ) tends to 0, where T and κ are the numbers shown in Theorem 1.3. Moreover, if u0 − v0 L2 (R) tends to 0, then sup0t


 lim ∂t Γcα u(t, ·) − uα (t, ·) L2 (R) + ∂x Γcα u(t, ·) − uα (t, ·) L2 (R) = 0 (1.14)

t→∞

for any α with |α|  1, where uα denotes the free solution of (∂t2 − c2 ∂x2 )uα = 0 with uα (0, ·) = uα0 and ∂t uα (0, ·) = uα1 . We apply Theorem 1.3 for nonlinear terms which satisfy the null conditions. We note the equation (∂t u)2 − c2 (∂x u)2 = (∂t + c∂x )u(∂t − c∂x )u,

(1.15)

and Dc = ∂t + c∂x for x > 0 and Dc = ∂t − c∂x for x < 0. So that, (∂t u)2 − c2 (∂x u)2 behaves like ∂t,x uDc u, similarly, {(∂t u)2 − c2 (∂x u)2 }2 behaves like (∂t,x u)2 (Dc u)2 , to which we apply Theorem 1.3. We obtain the following corollary. Corollary 1.5. The following results hold. (1) Let p  2 be an integer. Let h ∈ Np−2 , and let f (∂t u, ∂x u) = h(∂t u, ∂x u){(∂t u)2 − c2 (∂x u)2 }. Then for any data (u0 , u1 ) with ε = ε(u0 , u1 ) < ∞, there exist T > 0 and a unique solution u of (1.5) in X(T , p − 2), where T can be taken as any number with (1.11) for some constant C > 0 which is independent of u0 , u1 and ε. (2) Let p  4 be an integer. Let h ∈ Np−4 , and let f (∂t u, ∂x u) = h(∂t u, ∂x u){(∂t u)2 − c2 (∂x u)2 }2 . Then for any data (u0 , u1 ) with ε = ε(u0 , u1 ) sufficiently small, there exists a unique solution u of (1.5) in X(∞, (p − 3)/2). Namely, we obtain time global solutions for small data.

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The results in Corollary 1.5 are also valid for the system of nonlinear wave equations. We give a typical example as follows. Let us consider the Cauchy problem ⎧  ∂t2 − c2 ∂x2 u(t, x) = f (∂t,x u, ∂t,x v)(t, x) for (t, x) ∈ [0, T ) × R, ⎪ ⎪ ⎪  ⎨ 2 ∂t − c2 ∂x2 v(t, x) = g(∂t,x u, ∂t,x v)(t, x) for (t, x) ∈ [0, T ) × R, ⎪ u(0, ·) = u0 (·), ∂t u(0, ·) = u1 (·), ⎪ ⎪ ⎩ v(0, ·) = v0 (·), ∂t v(0, ·) = v1 (·),

(1.16)

where ∂t,x denotes ∂t and ∂x , f (∂t,x u, ∂t,x v) and g(∂t,x u, ∂t,x v) denote f (∂t u, ∂x u, ∂t v, ∂x v) and g(∂t u, ∂x u, ∂t v, ∂x v) precisely. Let Np denote the set of homogeneous polynomials of p order on R4 in the following corollary. We put ε = ε(u0 , u1 , v0 , v1 ) =

   w0 H 2 (R) + w1 H 1 (R) + x∂x w0 (x)L2 (R) w=u,v

     

+ x∂x2 w0 (x)L2 (R) + xw1 (x)L2 (R) + x∂x w1 (x)L2 (R)    +  1 + |x| f (u1 , ∂x u0 , v1 , ∂x v0 )(x)L2 (R)    +  1 + |x| g(u1 , ∂x u0 , v1 , ∂x v0 )(x)L2 (R) . (1.17)

Corollary 1.6. The following results hold. (1) Let p  2 be an integer. Let {hj }1j 6 ⊂ Np−2 , and let

f (∂t,x u, ∂t,x v) = h1 (∂t,x u, ∂t,x v) (∂t u)2 − c2 (∂x u)2

+ h2 (∂t,x u, ∂t,x v) (∂t v)2 − c2 (∂x v)2 + h3 (∂t,x u, ∂t,x v)(∂t u∂x v − ∂x u∂t v),

g(∂t,x u, ∂t,x v) = h4 (∂t,x u, ∂t,x v) (∂t u)2 − c2 (∂x u)2

+ h5 (∂t,x u, ∂t,x v) (∂t v)2 − c2 (∂x v)2 + h6 (∂t,x u, ∂t,x v)(∂t u∂x v − ∂x u∂t v).

(1.18)

(1.19)

Then for any data (u0 , u1 ) and (v0 , v1 ) with ε(u0 , u1 , v0 , v1 ) < ∞, there exist T > 0 and a unique solution (u, v) of (1.16) in X(T , p − 2) ⊕ X(T , p − 2), where T can be taken as any number with (1.11) for some constant C > 0 which is independent of (u0 , u1 ), (v0 , v1 ) and ε. (2) Let p  4 be an integer. Let {hj }1j 6 ⊂ Np−4 , and let

2 f (∂t,x u, ∂t,x v) = h1 (∂t,x u, ∂t,x v) (∂t u)2 − c2 (∂x u)2

2 + h2 (∂t,x u, ∂t,x v) (∂t v)2 − c2 (∂x v)2 + h3 (∂t,x u, ∂t,x v)(∂t u∂x v − ∂x u∂t v)2 .

(1.20)

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2 g(∂t,x u, ∂t,x v) = h4 (∂t,x u, ∂t,x v) (∂t u)2 − c2 (∂x u)2

2 + h5 (∂t,x u, ∂t,x v) (∂t v)2 − c2 (∂x v)2 + h6 (∂t,x u, ∂t,x v)(∂t u∂x v − ∂x u∂t v)2 .

(1.21)

Then for any data (u0 , u1 ) and (v0 , v1 ) with ε(u0 , u1 , v0 , v1 ) sufficiently small, there exists a unique solution (u, v) of (1.16) in X(∞, (p − 3)/2) ⊕ X(∞, (p − 3)/2). Namely, we obtain time global solutions for small data. Remark 1.7. Instead of (1.20) and (1.21), we are able to consider the following nonlinear terms. Put F1 := (∂t u)2 − c2 (∂x u)2 ,

F2 := (∂t v)2 − c2 (∂x v)2 ,

F3 := ∂t u∂x v − ∂x u∂t v. (1.22)

Let p  4. Let {h1j k }1j,k3 ⊂ Np−4 , {h2j k }1j,k3 ⊂ Np−4 , and let f (∂t,x u, ∂t,x v) =



h1j k (∂t,x u, ∂t,x v)Fj Fk ,

1j,k3

g(∂t,x u, ∂t,x v) =



h2j k (∂t,x u, ∂t,x v)Fj Fk .

(1.23)

1j,k3

Then the result of (2) in Corollary 1.6 holds. We have considered the system of two equations in Corollary 1.6, we are also able to consider the system of any number of equations. Remark 1.8. In Corollaries 1.5 and 1.6, the continuous dependence of the solutions on the initial data, and the asymptotics of global solutions to free solutions hold analogously to Proposition 1.4. To introduce the known results, let us consider the Cauchy problem for n  1 spatial variables 

 ∂t2 − c2 u(t, x) = f (∂t u, ∇u)(t, x) u(0, ·) = εu0 (·), ∂t u(0, ·) = εu1 (·),

for (t, x) ∈ [0, T ) × Rn ,

(1.24)

 where := nj=1 ∂j2 , ∇ := (∂1 , . . . , ∂n ), u0 , u1 ∈ C0∞ (Rn ) and ε > 0. For integer p  2, let Np denote the set of homogeneous polynomials on R1+n . For f ∈ Np , the existence time T has been estimated from below as follows. When n = 1, T  C/εp−1 for p  2. When n = 2, T  C/ε2 for p = 2, T  C exp(C  /ε 2 ) for p = 3, T = ∞ for p  4. When n = 3, T  C exp(C  /ε) for p = 2, T = ∞ for p  3. When n  4, T = ∞ for p  2 (see [16]). Here, C > 0 and C  > 0 are some constants independent of ε. These lower bounds have been shown to be sharp by Lax [19] (see also [11] and the introduction in [12]) for n = 1 and p = 2, by John [12] and Zhou [30] for n = 2, 3 and p = 2, by Kong [18] for n = 1 and p  2, by Zhou [30] for n  1 and even p  2, by Zhou and Han [31] for n = 2 and p = 3. Especially, the existence time has been estimated by C/ε p−1 for p  2 when n = 1, however, our Corollary 1.5 improves the existence time to (1.11) for p  2, and to ∞ for p  4 when f satisfies the null conditions. We refer to the historical notes in [21, p. 70] by Li and Chen, and [31]. When n  2, we also refer to the introduction by Takamura and Wakasa [27] and Zhou and Han [32] for the

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estimates of the lifespan of solutions of nonlinear wave equations, especially for the power type nonlinear terms (∂t2 − )u = |u|p for p > 1. When n = 1, small global solutions and blow-up phenomena for the quasilinear hyperbolic system ∂t u + A(u)∂x u = 0 were shown by Lax in [20]. Lindblad has referred to global solutions for quasilinear equation (∂t2 − ∂x2 )u = u(∂t − ∂x )2 u in [22]. A space–time L2 -estimate for nonlinear terms which satisfy null conditions was shown by Bournaveas in [3] and it was applied for global solutions of Dirac Klein–Gordon equations. The propagation of singularities of time local solutions for a system of semilinear wave equations satisfying the null condition was studied by Ito and Kato in [10], where the estimate for the existence time T was not considered. The global solutions of energy conservative weak solutions for a system of variational wave equations modeling liquid crystals have been shown by Zhang and solutions of wave maps u : R × R → M of the form (∂t2 − ∂x2 )ui +  Zheng ini[28]. Thej global k j k 1j,kI Γj k (u)(∂t u ∂t u − ∂x u ∂x u ) = 0 for 1  i  I , where M is I dimensional complete Riemannian manifold and Γjik denotes the Christoffel symbols for M, have been shown by Gu [7] (see also [5,15,29]). When n = 2, the small global solutions under the null conditions have been shown by Godin [6], Hoshiga [9], and Katayama [13,14]. See also Alinhac [1]. When n = 3, the small global solutions under the null conditions have been shown by Christodoulou [4] and Klainerman [17]. See also [25] by Sideris and Tu for the system of wave equations. Lindblad and Rodnianski have pointed out in [24] that the type of weighted energy estimates (1.4) give a much simplified proof for the small global solutions for (1.24) when f satisfies the null conditions when n = 3. In this paper, we apply their idea to the case of one space dimension, and obtain time global solutions under the null conditions. Our proofs for the theorem and corollaries are simple and fundamental. We put ∇t,x := (∂t , ∂x ), r := (1+ r 2 )1/2 for r ∈ R. The notation a  b denotes the inequality a  Cb for some constant C > 0 which is not essential in the argument. 2. Proof of Lemma 1.1 The proof of Lemma 1.1 for κ  0 has been essentially given in [23, Lemma A.1]. We show the proof with more rigorous expression of the weight function W (cT , κ) as defined in this paper. By the multiplication of ∂t u to the both sides of (∂t2 − c2 ∂x2 )u = f , we have ∂t e0 (u) − c2 ∂x (∂t u∂x u) = (∂t u)f,

(2.1)

where we have put e0 (u) := {(∂t u)2 + c2 (∂x u)2 }/2. Integrating by t and x, we have the standard energy estimate  e0 (t, x) dx 

sup 0tT

T 



R

e0 (0, x) dx + R

  (∂t u)f  dx dt =: J1 .

(2.2)

0 R

For −∞ < q  T , we put



B0 (q) := (0, x): |x|  −cq , BT (q) := (T , x): |x|  c(T − q) ,

K(q) := (t, x): 0  t  T , |x|  c(t − q) ,

C(q) := (t, x): 0  t  T , |x| = c(t − q) .

(2.3) (2.4) (2.5)

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Integrating the both sides of (2.1) on K(q) and applying the divergence theorem, we have c J2 := √ 2 1 + c2 

 (Dc u)2 dσ C(q)



e0 (u)(T , x) dx −

= BT (q)

 e0 (u)(0, x) dx −

B0 (q)

(∂t u)f dx dt,

(2.6)

K(q)

where we have used (e0 (u), −c2 ∂t u∂x u) · (−c, x/|x|) = −c(Dc u)2 /2 on C(q). So that, by (2.2), we have J2  2J1 . Especially, we have T −T

J2 dq  (1 + c|q|)1+κ

T

−T

2J1 dq (1 + c|q|)1+κ

(2.7)

for any κ ∈ R, which shows T



0 tr/c−T

 log(1 + cT ) (Dc u)2 8J1 dx dt  · 1+κ {1 − (1 + cT )−κ }/κ c (1 + |ct − r|)

if κ = 0, if κ = 0.

(2.8)

T  On the other hand, 0 tr/c−T (Dc u)2 /(1 + |ct − r|)1+κ dx dt is bounded by 4J1 T / (1 + cT )1+κ due to (Dc u)2  4e0 (u) and (2.2). Therefore, we have obtained T  0 R

(Dc u(t, x))2 dx dt (1 + |ct − r|)1+κ 

 4J1 T /(1 + cT )1+κ +

8J1 c−1 log(1 + cT ) 8J1 {1 − (1 + cT )−κ }/cκ

if κ = 0, if κ = 0.

(2.9)

Since log(1 + cT )  cT /(1 + cT ) and {1 − (1 + cT )−κ }/κ  cT /(1 + cT )1+κ for κ  −1, we obtain the required result. 3. Proof of Theorem 1.3 We prove Theorem 1.3 in this section. First, we collect several facts. The tangential derivative Dc has commuting properties ∂t Dc = Dc ∂t ,

for x = 0, and we have the bound

∂x Dc = Dc ∂x , Dc L = (L + 1)Dc ,   x Dc Dc Ωc = Ωc + |x|

(3.1)

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      Γ α χ(x)Dc u   Dc Γ α u. c

(3.2)

c

|α|1

|α|1

We use the following Klainerman–Sobolev estimate. Lemma 3.1. (See [8, p. 118, Proposition 6.5.1], [26, p. 43, Theorem 1.3].) For any fixed c > 0, the following estimate holds for any u.    α  Γ u(t, ·) 2 , ct − r1/2 u(t, x)  c L (R)

(3.3)

|α|1

where α denotes the multi index. By this lemma and the bound |Γc ct − rl |  ct − rl for any l ∈ R, we have the bound         χ(x)Dc u(t, x)  ct − |x| −l−1/2  ct − |y| l Dc Γ α u(t, y) 2 c L (R) y

|α|1

(3.4)

for any function u and l ∈ R. 3.1. Continuity argument We adopt the continuity argument to show that T can be taken as (1.11) and (1.12). We prepare the following proposition. Proposition 3.2. Let p  2 be an integer. Let f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)χ(x) for h ∈ Np−1 , or f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)2 χ(x) for h ∈ Np−2 . Put  κ :=

p−2 (p − 3)/2

if h ∈ Np−1 , if h ∈ Np−2 .

(3.5)

Let T > 0, and let u ∈ C([0, T ), H 2 (Rn )) be the solution of (1.5). We define ε by (1.7). Assume that u satisfies     1/2  ct − |x| ∇t,x u(t, x), |u|X(T ,κ)  A0 ε |u|Y := max sup (3.6) (t,x)∈[0,T )×R

for some constant A0 > 0. Then the estimates sup

 1/2  ct − |x| ∇t,x u(t, x)  C0 |u|X(T ,κ)



(3.7)

(t,x)∈[0,T )×R

and  |u|X(T ,κ)  C0 ε + C0 (A0 ε) W (cT , κ) p

1/2

·

T 1/2 W (cT , κ)1/2

hold for some constant C0 > 0 which is independent of A0 , ε and T .

if h ∈ Np−1 , if h ∈ Np−2

(3.8)

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Proof. The estimate (3.7) follows from (3.3) and the definition (1.10). We put γ := (κ + 1)/2. By the energy estimates (1.2) and Lemma 1.1, we have |u|X(T ,κ)  C0 ε + C0

  Γ α c u c

(3.9)

L1 ((0,T ),L2 (R))

|α|1

for some constant C0 > 0 independent of A0 , ε and T , where we have used that Γc is commutable with c for the last term. First, let us consider the case f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)χ(x) for h ∈ Np−1 . We have     α  Γ α c u  Γ h(∂t u, ∂x u)(Dc u)χ(x)  c c |α|1

|α|1

 |∂u|p−2

       Γ α ∂u Dc u + |∂u|p−1 Dc Γ α u, c

c

|α|1

(3.10)

|α|1

where we have used (3.2) for the last inequality. By the assumption (3.6), we have      Γ α c u  (A0 ε)p−2 Γ α ∂uct − r−γ Dc Γ α u 2 c c c L (R)

|α|1

|α|1

  Dc Γ α u,

+ (A0 ε)p−1 ct − r−γ

(3.11)

c

|α|1

where we have used (3.3), (3.4) with l = −γ , and (p − 1)/2 = γ due to κ = p − 2. Taking L1 ((0, T ), L2 (R)) norm of both sides, we have   Γ α c u c

L1 ((0,T ),L2 (R))

|α|1

 (A0 ε)p−2 T 1/2

  Γ α ∂u c

|α|1

+ (A0 ε)p−1 T 1/2

L∞ ((0,T ),L2 (R))

  ct − r−γ Dc Γ α u c

  ct − r−γ Dc Γ α u c

|α|1

L2 ((0,T )×R)

L2 ((0,T )×R)

(3.12)

,

which shows   Γ α c u c

|α|1

L1 ((0,T ),L2 (R))

 C0 (A0 ε)p T 1/2 W (cT , κ)1/2

(3.13)

for some constant C0 > 0 which is independent of A0 , ε and T . We have obtained (3.8) as required. The case f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)2 χ(x) for h ∈ Np−2 follows similarly. We have       Γ α c u  |∂u|p−3 Γ α ∂u(Dc u)2 + |∂u|p−2 |Dc u| Dc Γ α u, c c c |α|1

which yields

|α|1

|α|1

(3.14)

M. Nakamura / J. Differential Equations 256 (2014) 389–406

399

     Γ α c u  (A0 ε)p−3 Γ α ∂uct − r−γ Dc Γ α u2 2 c

c

|α|1

c

|α|1

+ (A0 ε)p−2 ct − r−γ

L (R)

    Dc Γ α u ct − r−γ Dc Γ α u c

|α|1

c

|α|1

L2 (R)

.

(3.15) Taking L1 ((0, T ), L2 (R)) norm of both sides, we have   Γ α c u c

L1 ((0,T ),L2 (R))

 (A0 ε)p−3

  Γ α ∂u

|α|1

c

|α|1

+ (A0 ε)p−2

L∞ ((0,T ),L2 (R))

  ct − r−γ Dc Γ α u2 2 c

  ct − r−γ Dc Γ α u2 2 c

|α|1

L ((0,T )×R)

L ((0,T )×R)

(3.16)

,

which shows   Γ α c u c

|α|1

L1 ((0,T ),L2 (R))

 C0 (A0 ε)p W (cT , κ)

for some constant C0 > 0 independent of A0 , ε and T . We obtain (3.8) as required.

(3.17) 2

3.2. Proof of Theorem 1.3 We now prove Theorem 1.3. We use the following result for time local solutions whose proof is given in Appendix A. For T > 0, we define Z(T ) := {u: uZ(T ) < ∞}, where uZ(T ) :=

   t−1 Γ α u c

|α|1

L∞ ((0,T ),L2 (R))

  + ∂t Γcα uL∞ ((0,T ),L2 (R))

 

+ c∂x Γcα uL∞ ((0,T ),L2 (R)) .

(3.18)

Theorem 3.3. Let p be an integer with 2  p < ∞. Let h and f be given as those in (1) or (2) in Theorem 1.3, and let us consider the Cauchy problem (1.5). If   B = B(u0 , u1 , f ) := u0 L2 (R) + (1 + c)∂x u0 L2 (R) + c∂x2 u0 L2 (R)   + x∂x u0 L2 (R) + (1 + x∂x )∂x u0 L2 (R) + u1 L2 (R) + (1 + c)∂x u1 L2 (R)   + xu1 L2 (R) /c + (1 + x∂x )u1 L2 (R)   + c2 ∂x2 u0 + f (u1 , ∂x u0 )L2 (R)   + c∂x u0 + cx∂x2 u0 + xf (u1 , ∂x u0 )/cL2 (R)

(3.19)

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is finite, then there exists T =  T (B) > 0 such that there exists a unique time local solution u of (1.5) in Z(T ) with u ∈ j =0,1 C j ([0, T ), H 2−j (R)) and Γcα u ∈ C([0, T ), H 1 (R)) ∩ C 1 ([0, T ), L2 (R)) for any α with |α| = 1. Remark 3.4. The condition that B < ∞ is equivalent to    Γ α u(0, ·) c

|α|1

L2 (R)

   

  +  ∂t Γcα u (0, ·)L2 (R) + c ∂x Γcα u (0, ·)L2 (R) < ∞. (3.20)

We note that if uZ(T ) is shown bounded on the time interval [0, T ), then the solution u can be prolonged beyond T by Theorem 3.3 and it does not blow up at T . In the following, let us show that uZ(T ) is bounded on [0, T ) for T which satisfies (1.11) or (1.12). Then we know that u exists on [0, T ) at least. We note that the time local solution given by Theorem 3.3 satisfies the assumption (3.6) for some A0 > 0 and T > 0 by Lemma 3.1 and the definition of X(T , κ). Let us consider the case (1) in the theorem, namely, f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)χ(x) for h ∈ Np−1 . Since C0 is independent of A0 , ε and T in Proposition 3.2, we take A0 , ε and T such that 4C0 (1 + C0 )  A0 ,

C0 (A0 ε)p W (cT , κ)1/2 T 1/2  A0 ε/4(1 + C0 ).

(3.21)

Then (3.8) shows |u|X(T ,κ)  A0 ε/2(1 + C0 ), so that, (3.7) shows sup

 1/2  ct − |x| ∇t,x u(t, x)  A0 ε/2.



(3.22)

(t,x)∈[0,T )×R

Therefore, we obtain |u|Y  A0 ε/2. Namely, starting from the assumption (3.6), we obtain |u|Y  A0 ε/2 by these A0 , ε and T . This result shows that |u|Y is bounded on [0, T ). By Remark 1.2, we note that the second inequality in (3.21) is satisfied if  A0 ε/4(1 + C0 )  C0 (A0 ε) T p

1/2

·

(p − 2)−1/2 {log(1 + cT )}1/2

if p  3, if p = 2.

(3.23)

Since T can be taken arbitrary with this inequality, T can be estimated as (1.11) for some constant C > 0 which is independent of ε. Let us prove the results (2) and (3) in the theorem. We take A0 , ε and T such that 4C0 (1 + C0 )  A0 ,

C0 (A0 ε)p W (cT , κ)  A0 ε/4(1 + C0 ).

(3.24)

Then the same argument in the proof of (1) shows that |u|Y is bounded on [0, T ). Here, we note that the second inequality in (3.24) is satisfied if ⎧ ⎨ 2/(p − 3) A0 ε/4(1 + C0 )  C0 (A0 ε)p · log(1 + cT ) ⎩ 2(1 + cT )1/2

if p  4, if p = 3, ifp = 2.

(3.25)

Since T can be taken arbitrary with this inequality, T can be estimated as (1.12) for some constants C > 0 and C  > 0 which are independent of ε if p = 2 or p = 3. And if p  4 and ε > 0

M. Nakamura / J. Differential Equations 256 (2014) 389–406

401

is sufficiently small, then we can take any T > 0, namely, we obtain global solutions for small data. 4. Proof of Proposition 1.4 We define κ by (3.5). (1) Since u and v are the solutions obtained by the use of Proposition 3.2, they satisfy maxu,v |w|X(T ,κ)  B0 maxw=u,v ε(w0 , w1 ) for some constant B0 > 0. We put ε˜ := maxw=u,v ε(w0 , w1 ). By the energy estimate (1.2) and Lemma 1.1, we have |u − v|X(T ,κ) 

   ∂t Γ α (u − v)(0, ·) c

|α|1

+

L2 (R)

 

+ c∂x Γcα (u − v)(0, ·)L2 (R)

  c Γ α (u − v) c

L1 ((0,T ),L2 (R))

|α|1

.

(4.1)

By the similar argument to derive (3.13) and (3.17), we have   c Γ α (u − v) c

L1 ((0,T ),L2 (R))

|α|1

 (B0 ε˜ )p−1 T 1/2 W (cT , κ)1/2 |u − v|X(T ,κ)

(4.2)

for the case f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)χ(x) with h ∈ Np−1 , and   c Γ α (u − v) c

L1 ((0,T ),L2 (R))

|α|1

 (B0 ε˜ )p−1 W (cT , κ)|u − v|X(T ,κ)

(4.3)

for the case f (∂t u, ∂x u) = h(∂t u, ∂x u)(Dc u)2 χ(x) with h ∈ Np−2 . For both cases, since the coefficients (B0 ε˜ )p−1 T 1/2 W (cT , κ)1/2 and (B0 ε˜ )p−1 W (cT , κ) are small numbers by the condition on T in the proof Theorem 1.3, we have |u − v|X(T ,κ) 

   ∂t Γ α (u − v)(0, ·) c

|α|1

L2

 

+ c∂x Γcα (u − v)(0, ·)L2 ,

(4.4)

which yields the required result on |u − v|X(T ,κ) . The last result follows from the elementary inequality t (u − v)(t, ·) = (u − v)(0, ·) +

∂t (u − v)(s, ·) ds 0

for t ∈ (0, T ), and ∂t (u − v)L∞ ((0,T ),L2 (R))  |u − v|X(T ,κ) . √ (2) We put ω := − . For any α with |α|  1, we use the identity    sin tcω  Γcα u(t, ·) = (cos tcω) Γcα u (0, ·) + ∂t Γcα u (0, ·) cω

(4.5)

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M. Nakamura / J. Differential Equations 256 (2014) 389–406

t +

 sin(t − s)cω  c Γcα u (s, ·) ds cω

0

= (cos tcω)uα0 +

sin tcω α u1 − cω

∞

 sin(t − s)cω  c Γcα u (s, ·) ds, cω

(4.6)

t

where we have put uα0

  := Γcα u (0, ·) −

∞

 sin scω  c Γcα u (s, ·) ds, cω

0

uα1

  := ∂t Γcα u (0, ·) +

∞

  cos scω c Γcα u (s, ·) ds.

(4.7)

0

We put the free solution uα (t, ·) := (cos tcω)uα0 +

sin tcω α cω u1 .

Then we have

   α

 Γ u(t, ·) − uα (t, ·) ˙ 1 + ∂t Γ α u(t, ·) − uα (t, ·)  2 c c H (R) L (R)   α    c Γc u L1 ((t,∞),L2 (R)) .

(4.8)

By the definition of Γc , it follows that    uα  |α|1

0 H˙ 1 (R)

   

c Γ α u 1 + uα1 L2 (R)  B + , c L ((0,∞),L2 (R))

(4.9)

|α|1

where   B := u0 H˙ 1 (R) + x∂x2 u0 L2 (R) + u1 L2 (R)     + x∂x u1 L2 (R) + xf (u1 , ∂x u0 )L2 (R) .

(4.10)

 We note that our solution u satisfies |α|1 Γcα c uL1 ((0,∞),L2 (R)) < ∞ as we have shown in (3.17). So that, the left hand side of (4.8) tends to 0 as t → ∞, and uα0 ∈ H˙ 1 (R), uα1 ∈ L2 (R) for |α|  1 as required. 5. Proofs of Corollaries 1.5 and 1.6 We use (∂t u)2 − c2 (∂x u)2 = (∂t + c∂x )u(∂t − c∂x )u,

(5.1)

and the fact Dc = ∂t + c∂x for x > 0 and Dc = ∂t − c∂x for x < 0 to obtain     (∂t u)2 − c2 (∂x u)2   |∂t u| + |∂x u| |Dc u|.

(5.2)

M. Nakamura / J. Differential Equations 256 (2014) 389–406

403

We are able to show (3.10) and (3.14). Then the proof of Corollary 1.5 follows from the proof of Theorem 1.3. By ∂t u∂x v − ∂x u∂t v = Dc u∂x v − ∂x uDc v,

(5.3)

   |∂t u∂x v − ∂x u∂t v|  |∂x v| + |∂x u| |Dc u| + |Dc v| .

(5.4)

we have

By this bound and (5.2), the proof of Corollary 1.6 follows from the proof of Theorem 1.3 analogously. Acknowledgments The author is thankful to the anonymous referee for valuable suggestions to improve the paper. Appendix A √ For the completeness of the paper, we prove Theorem 3.3 in this section. We put ω := − . For any fixed (u0 , u1 ) with B(u0 , u1 , f ) < ∞, we define u2 (t, ·) = (cos tcω)u0 + (cω)−1 (sin tcω)u1 , namely the free solution, and we define {un }n3 inductively by t un+1 (t, ·) := u2 (t, ·) +

sin(t − s)cω f (∂t un , ∂x un )(s, ·) ds cω

(A.1)

0

for n  2. For any α with |α|  1, by the energy estimate, we have     ∂t Γ α un+1  ∞ + c∂x Γcα un+1 L∞ ((0,T ),L2 (R)) c L ((0,T ),L2 (R))         ∂t Γcα un+1 (0, ·)L2 (R) + c ∂x Γcα un+1 (0, ·)L2 (R)   + c Γcα un+1 L1 ((0,T ),L2 (R)) .

(A.2)

The first two terms are bounded by B, while the last term is estimated by   c Γ α un+1  c

p−1

L1 ((0,T ),L2 (R))

 T ∂t,x un L∞ ((0,T ),H 1 (R))

  Γ β ∂t,x un  c

|β|1

L∞ ((0,T ),L2 (R))

p

 T un Z(T ) .

(A.3)

We use Γcα un+1 (t, ·) = (Γcα un+1 )(0, ·) +  −1 α  t Γ un+1 (t, ·) c

t

α 0 ∂t Γc un+1 (s, ·) ds

L∞ ((0,T ),L2 (R))

p

So that, we have un+1 Z(T )  B + T un Z(T ) .

to obtain

     Γcα un+1 (0, ·)L2 (R)   + ∂t Γcα un+1 L∞ ((0,T ),L2 (R)) .

(A.4)

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M. Nakamura / J. Differential Equations 256 (2014) 389–406

Similarly, we have     ∂t Γ α (un+1 − un ) ∞ + c∂x Γcα (un+1 − un )L∞ ((0,T ),L2 (R)) c L ((0,T ),L2 (R))    c Γcα (un+1 − un )L1 ((0,T ),L2 (R)) ,

(A.5)

where we note that un+1 and un have same data and f (∂t un (0, ·), ∂x un (0, ·)) = f (∂t un−1 (0, ·), ∂x un−1 (0, ·)). By simple calculation, we have   c Γ α (un+1 − un ) c

L1 ((0,T ),L2 (R))

T

p−1

max

w=un ,un−1

wZ(T ) un − un−1 Z(T )

(A.6)

and  −1 α  t Γ (un+1 − un )(t, ·) c

L∞ ((0,T ),L2 (R))

   ∂t Γcα (un+1 − un )L∞ ((0,T ),L2 (R)) .

(A.7)

So that, we have un+1 − un Z(T )  T

max

w=un ,un−1

p−1

wZ(T ) un − un−1 Z(T ) .

(A.8)

Therefore, we have the limit u of {un }n1 in Z(T ) if T = T (B) > 0 is sufficiently small since Z(T ) is a Banach space. And u satisfies t u(t, ·) = u2 (t, ·) +

sin(t − s)cω f (∂t u, ∂x u)(s, ·) ds, cω

(A.9)

0

which shows that Γcα u ∈ C([0, T ), H 1 (R)) ∩ C 1 ([0, T ), L2 (R)) for any α with |α|  1 by the bound {cω}−1 (sin tcω)ψL2 (R)  tψH −1 (R) . Next, we show the uniqueness  of the solution. Let v be another solution of the Cauchy problem (1.5) such that v ∈ j =0,1 C j ([0, T ), H 2−j (R)) and Γcα v ∈ C([0, T ), H 1 (R)) ∩ C 1 ([0, T ), L2 (R)) for any α with |α| = 1. We define t0 := inf{t ∈ [0, T ): u(t, ·) = v(t, ·)} and show that t0 < T yields a contradiction. Let t0 < T . By the continuity of u and v, we have (u − v)(t0 , ·) = ∂t (u − v)(t0 , ·) = 0. Similarly to (A.8), we have u − vZ((t0 ,t0 +ε)) :=

   t − t0 −1 Γ α (u − v)(t, x) c

|α|1

2 L∞ t ((t0 ,t0 +ε),Lx (R))

  + ∂t Γcα (u − v)L∞ ((t ,t +ε),L2 (R)) 0 0  

α   + c ∂x Γc (u − v) L∞ ((t ,t +ε),L2 (R)) 0 0



p−1 Cε max wZ((t0 ,t0 +ε)) u − vZ((t0 ,t0 +ε)) w=u,v

(A.10)

for sufficiently small ε > 0. Since vZ((t0 ,t0 +ε)) < ∞, we obtain u − vZ((t0 ,t0 +ε)) = 0 for sufficiently small ε > 0, which shows u = v on [t0 , t0 + ε) and a contradiction to the definition of t0 .

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