On the Cauchy problem of a weakly dissipative μ HS equation
aa r X i v : . [ m a t h . A P ] S e p On the Cauchy problem of a weakly dissipative µ HS equation
Jingjing Liu ∗ Zhaoyang Yin † Department of Mathematics, Sun Yat-sen University,510275 Guangzhou, China
Abstract
In this paper, we study the Cauchy problem of a weakly dissipative µ HS equation. Wefirst establish the local well-posedness for the weakly dissipative µ HS equation by Kato’ssemigroup theory. Then, we derive the precise blow-up scenario for strong solutions to theequation. Moreover, we present some blow-up results for strong solutions to the equationand give the blowup rate of strong solutions to the equation when blowup occurs. Finally,we give two global existence results to the equation.
Keywords : A weakly dissipative µ HS equation, blow-up scenario, blow-up, strong solutions,global existence.
Recently, the µ -Hunter-Saxton (also called µ -Camassa-Holm) equation µ ( u ) t − u txx = − µ ( u ) u x + 2 u x u xx + uu xxx , which is originally derived and studied in [37] attracts a lot of attention. Here u ( t, x ) is a time-dependent function on the unit circle S = R / Z and µ ( u ) = R S udx denotes its mean. In [37],the authors show that if interactions of rotators and an external magnetic field is allowed, thenthe µ HS equation can be viewed as a natural generalization of the rotator equation. Moreover,the µ HS equation describes the geodesic flow on D s ( S ) with the right-invariant metric given atthe identity by the inner product [37]( u, v ) = µ ( u ) µ ( v ) + Z S u x v x dx. In [37, 43], the authors showed that the µ HS equation admits both periodic one-peakon solutionand the multi-peakons. Moreover, in [29, 31], the authors also discussed the µ HS equation.One of the closest relatives of the µ HS equation is the Camassa-Holm equation u t − u txx + 3 uu x = 2 u x u xx + uu xxx , ∗ e-mail: [email protected] † e-mail: [email protected] µ HSequation is the Hunter-Saxton equation [32] u txx + 2 u x u xx + uu xxx = 0 , which is an asymptotic equation for rotators in liquid crystals and modeling the propagationof weakly nonlinear orientation waves in a massive nematic liquid crystal. The orientation ofthe molecules is described by the field of unit vectors (cos u ( t, x ) , sin u ( t, x )) [56]. The Hunter-Saxton equation also arises in a different physical context as the high-frequency limit [23, 33]of the Camassa-Holm equation. Similar to the Camassa-Holm equation, the Hunter-Saxtonequation also has a bi-Hamiltonian structure [34, 46] and is completely integrable [1, 33].The initial value problem of the Hunter-Saxton equation also has been studied extensively,c.f.[3, 32, 42, 56]In general, it is difficult to avoid energy dissipation mechanisms in a real world. So, it isreasonable to study the model with energy dissipation. In [30] and [47], the authors discussedthe energy dissipative KdV equation from different aspects. Weakly dissipative CH equationand weakly dissipative DP equation have been studied in [51] and [27, 50, 54] respectively.Recently, some results for a weakly dissipative µ DP equation are proved in [38]. It is worthyto note that the local well-posedness result in [38] is obtained by using a method based on ageometric argument.In this paper, we will discuss the Cauchy problem of the following weakly dissipative µ HSequation: y t + uy x + 2 u x y + λy = 0 , t > , x ∈ R ,y = µ ( u ) − u xx , t > , x ∈ R ,u (0 , x ) = u ( x ) , x ∈ R ,u ( t, x + 1) = u ( t, x ) , t ≥ , x ∈ R , (1.1)or in the equivalent form: µ ( u ) t − u txx + 2 µ ( u ) u x − u x u xx − uu xxx + λ ( µ ( u ) − u xx ) = 0 , t > , x ∈ R ,u (0 , x ) = u ( x ) , x ∈ R ,u ( t, x + 1) = u ( t, x ) , t ≥ , x ∈ R . (1.2)2ere the constant λ is assumed to be positive and the term λy = λ ( µ ( u ) − u xx ) models energydissipation.The paper is organized as follows. In Section 2, we establish the local well-posedness of theinitial value problem associated with the equation (1.1). In Section 3, we derive the preciseblow-up scenario. In Section 4, we present some explosion criteria of strong solutions to theequation (1.1) with general initial data and give the blowup rate of strong solutions to theequation when blowup occurs. In Section 5, we give two new global existence results of strongsolutions to the equation (1.1). Notation
Given a Banach space Z , we denote its norm by k·k Z . Since all space of functionsare over S = R / Z , for simplicity, we drop S in our notations if there is no ambiguity. We let[ A, B ] denote the commutator of linear operator A and B . For convenience, we let ( ·|· ) s × r and( ·|· ) s denote the inner products of H s × H r , s, r ∈ R + and H s , s ∈ R + , respectively. In this section, we will establish the local well-posedness for the Cauchy problem of the equation(1.1) in H s , s > , by applying Kato’s theory.For convenience, we state here Kato’s theory in the form suitable for our purpose. Considerthe abstract quasi-linear equation: dvdt + A ( v ) v = f ( v ) , t > , v (0) = v . (2.1)Let X and Y be Hilbert spaces such that Y is continuously and densely embedded in X and let Q : Y → X be a topological isomorphism. Let L ( Y, X ) denote the space of all boundedlinear operator from Y to X ( L ( X ), if X = Y .). Assume that:(i) A ( y ) ∈ L ( Y, X ) for y ∈ Y with k ( A ( y ) − A ( z )) w k X ≤ µ k y − z k X k w k Y , y, z, w ∈ Y, and A ( y ) ∈ G ( X, , β ), (i.e. A ( y ) is quasi-m-accretive), uniformly on bounded sets in Y .(ii) QA ( y ) Q − = A ( y ) + B ( y ), where B ( y ) ∈ L ( X ) is bounded, uniformly on bounded sets in Y . Moreover, k ( B ( y ) − B ( z )) w k X ≤ µ k y − z k Y k w k X , y, z ∈ Y, w ∈ X. (iii) f : Y → Y and extends also to a map from X to X . f is bounded on bounded sets in Y ,and k f ( y ) − f ( z ) k Y ≤ µ k y − z k Y , y, z ∈ Y, k f ( y ) − f ( z ) k X ≤ µ k y − z k X , y, z ∈ Y. Here µ , µ , µ and µ depend only on max {k y k Y , k z k Y } . Theorem 2.1 [35] Assume that (i), (ii) and (iii) hold. Given v ∈ Y , there exist a maximal T > depending only on k v k Y and a unique solution v to Eq.(2.1) such that v = v ( · , v ) ∈ C ([0 , T ); Y ) ∩ C ([0 , T ); X ) . Moreover, the map v → v ( · , v ) is continuous from Y to C ([0 , T ); Y ) ∩ C ([0 , T ); X ) .
3n one hand, with y = µ ( u ) − u xx , the first equation in (1.2) takes the form of a quasi-linearevolution equation of hyperbolic type: u t + uu x = − ∂ x A − (2 µ ( u ) u + 12 u x ) − λu, (2.2)where A = µ − ∂ x is an isomorphism between H s and H s − with the inverse v = A − w givenexplicitly by [26, 37] v ( x ) = ( x − x µ ( w ) + ( x −
12 ) Z Z y w ( s ) dsdy (2.3) − Z x Z y w ( s ) dsdy + Z Z y Z s w ( r ) drdsdy. Since A − and ∂ x commute, the following identities hold A − ∂ x w ( x ) = ( x −
12 ) Z w ( x ) dx − Z x w ( y ) dy + Z Z x w ( y ) dydx, (2.4)and A − ∂ x w ( x ) = − w ( x ) + Z w ( x ) dx. (2.5)On the other hand, integrating both sides of the first equation in (1.2) with respect to x on S ,we obtain ddt µ ( u ) = − λµ ( u ) , it follows that µ ( u ) = µ ( u ) e − λt := µ e − λt , (2.6)where µ := µ ( u ) = Z S u ( x ) dx. Combining (2.2) and (2.6), the equation (1.2) can be rewrite as u t + uu x = − ∂ x A − (2 µ e − λt u + u x ) − λu, t > , x ∈ R ,u (0 , x ) = u ( x ) , x ∈ R ,u ( t, x + 1) = u ( t, x ) , t ≥ , x ∈ R . (2.7)The remainder of this section is devoted to the local well-posedness result. Firstly, we willgive a useful lemma. Lemma 2.1 [35] Let r,t be real numbers such that − r < t ≤ r . Then k f g k H t ≤ c k f k H r k g k H t , if r > , k f g k H t + r − ≤ c k f k H r k g k H t , if r < , where c is a positive constant depending on r, t. heorem 2.2 Given u ∈ H s , s > , then there exists a maximal T = T ( u ) > , and aunique solution u to (2.7) (or (1.1)) such that u = u ( · , u ) ∈ C ([0 , T ); H s ) ∩ C ([0 , T ); H s − ) . Moreover, the solution depends continuously on the initial data, i.e., the mapping u → u ( · , u ) : H s → C ([0 , T ); H s ) ∩ C ([0 , T ); H s − ) is continuous. Proof
For u ∈ H s , s > , we define the operator A ( u ) = u∂ x . Similar to Lemma 2.6 in[55], we have that A ( u ) belongs to G ( H s − , , β ), that is, − A ( u ) generates a C -semigroup T ( t )on H s − and k T ( t ) k L ( H s − ) ≤ e tβ for all t ≥
0. Analogous to Lemma 2.7 in [55], we get that A ( u ) ∈ L ( H s , H s − ) and k ( A ( z ) − A ( y )) w k H s − ≤ µ k z − y k H s − k w k H s , for all z, y, w ∈ H s .Let Q = Λ = (1 − ∂ x ) . Define B ( z ) = QA ( z ) Q − − A ( z ) for z ∈ H s , s > . Similar toLemma 2.8 in [55], we deduce that B ( z ) ∈ L ( H s − ) and k ( B ( z ) − B ( y )) w k H s − ≤ µ k z − y k H s k w k H s − , for all z, y ∈ H s and w ∈ H s − . Where µ , µ are positive constants.Set f ( u ) = − ∂ x ( µ − ∂ x ) − (2 µ e − λt u + 12 u x ) − λu. Let y, z ∈ H s , s > . Since H s − is an Banach algebra, it follows that k f ( y ) − f ( z ) k H s ≤ k − ∂ x ( µ − ∂ x ) − (2 µ e − λt ( y − z ) + 12 ( y x − z x )) k H s + λ k y − z k H s ≤ k µ e − λt ( y − z ) + 12 ( y x + z x )( y x − z x ) k H s − + λ k y − z k H s ≤ | µ |k y − z k H s − + 12 k y x + z x k H s − k y x − z x k H s − + λ k y − z k H s ≤ (2 | µ | + k y k H s + k z k H s + λ ) k y − z k H s . Furthermore, taking z = 0 in the above inequality, we obtain that f is bounded on boundedset in H s . Moreover, k f ( y ) − f ( z ) k H s − ≤ k − ∂ x ( µ − ∂ x ) − (2 µ e − λt ( y − z ) + 12 ( y x − z x )) k H s − + λ k y − z k H s − ≤ k µ e − λt ( y − z ) + 12 ( y x + z x )( y x − z x ) k H s − + λ k y − z k H s − ≤ | µ |k y − z k H s − + c k y x + z x k H s − k y x − z x k H s − + λ k y − z k H s − ≤ (2 | µ | + c ( k y k H s + k z k H s ) + λ ) k y − z k H s − , here we applied Lemma 2.1 with r = s − t = s − . Set Y = H s , X = H s − . It is obvious that Q is an isomorphism of Y onto X . Applying Theorem 2.1, we obtain the local well-posednessof Eq.(1.1) in H s , s > , and u ∈ C ([0 , T ); H s ) ∩ C ([0 , T ); H s − ). This completes the proofof Theorem 2.2. (cid:3) Remark 2.1
Similar to the proof of Theorem 2.3 in [55], we have that the maximal time ofexistence
T > in Theorem 2.2 is independent of the Sobolev index s > . The precise blow-up scenario
In this section, we present the precise blow-up scenario for strong solutions to the equation(1.1). We first recall the following lemmas.
Lemma 3.1 [36] If r > , then H r ∩ L ∞ is an algebra. Moreover k f g k H r ≤ c ( k f k L ∞ k g k H r + k f k H r k g k L ∞ ) , where c is a constant depending only on r. Lemma 3.2 [36] If r > , then k [Λ r , f ] g k L ≤ c ( k ∂ x f k L ∞ k Λ r − g k L + k Λ r f k L k g k L ∞ ) , where c is a constant depending only on r. Lemma 3.3 [11, 29] If f ∈ H ( S ) is such that R S f ( x ) dx = 0 , then we have max x ∈ S f ( x ) ≤ Z S f x ( x ) dx. Next we prove the following useful result on global existence of solutions to (1.1).
Theorem 3.1
Let u ∈ H s , s > , be given and assume that T is the maximal existence timeof the corresponding solution u to (2.7) with the initial data u . If there exists M > suchthat k u x ( t, · ) k L ∞ ≤ M, t ∈ [0 , T ) , then the H s -norm of u ( t, · ) does not blow up on [0,T). Proof
Let u be the solution to (2.7) with the initial data u ∈ H s , s > , and let T be themaximal existence time of the corresponding solution u , which is guaranteed by Theorem 2.2.Throughout this proof, c > s .Applying the operator Λ s to the first equation in (2.7), multiplying by Λ s u , and integratingover S , we obtain ddt k u k H s = − uu x , u ) s − u, ∂ x ( µ − ∂ x ) − (2 µ e − λt u + 12 u x )) s − λ ( u, u ) s . (3.1)Let us estimate the first term of the right hand side of (3.1). | ( uu x , u ) s | = | (Λ s ( u∂ x u ) , Λ s u ) | = | ([Λ s , u ] ∂ x u, Λ s u ) + ( u Λ s ∂ x u, Λ s u ) |≤ k [Λ s , u ] ∂ x u k L k Λ s u k L + 12 | ( u x Λ s u, Λ s u ) |≤ ( c k u x k L ∞ + 12 k u x k L ∞ ) k u k H s ≤ c k u x k L ∞ k u k H s , (3.2)6here we used Lemma 3.2 with r = s . Furthermore, we estimate the second term of the righthand side of (3.1) in the following way: | ( u, ∂ x ( µ − ∂ x ) − (2 µ e − λt u + 12 u x )) s |≤ k ∂ x ( µ − ∂ x ) − (2 µ e − λt u + 12 u x ) k H s k u k H s ≤ k µ e − λt u + 12 u x k H s − k u k H s ≤ c ( | µ |k u k H s + k u x k L ∞ k u x k H s − ) k u k H s ≤ c ( | µ | + k u x k L ∞ ) k u k H s , (3.3)where we used Lemma 3.1 with r = s − . Combining (3.2) and (3.3) with (3.1), we get ddt k u k H s ≤ c ( | µ | + k u x k L ∞ + 2 λ ) k u k H s . An application of Gronwall’s inequality and the assumption of the theorem yield k u k H s ≤ e c ( | µ | + M +2 λ ) t k u k H s . This completes the proof of the theorem. (cid:3)
The following result describes the precise blow-up scenario for sufficiently regular solutionsto the equation (1.1).
Theorem 3.2
Let u ∈ H s , s > be given and let T be the maximal existence time of thecorresponding solution u to (2.7) with the initial data u . Then the corresponding solutionblows up in finite time if and only if lim inf t → T { inf x ∈ S u x ( t, x ) } = −∞ . Proof
Applying a simple density argument, Remark 2.1 implies that we only need to considerthe case s = 3 . Multiplying the first equation in (1.1) by y and integrating over S with respectto x yield ddt Z S y dx = 2 Z S y ( − uy x − u x y − λy ) dx = − Z S uyy x dx − Z S u x y dx − λ Z S y dx = − Z S u x y dx − λ Z S y dx. So, if there is a constant
M > u x ( t, x ) ≥ − M, ∀ ( t, x ) ∈ [0 , T ) × S , then ddt Z S y dx ≤ (3 M − λ ) Z S y dx. Z S y dx ≤ e (3 M − λ ) Z S y (0 , x ) dx. Note that Z S y dx = µ ( u ) + Z S u xx dx ≥ k u xx k L . Since u x ∈ H ⊂ H and R S u x dx = 0 , Lemma 3.3 implies that k u x k L ∞ ≤ √ k u xx k L ≤ e M − λ k y (0 , x ) k L . Theorem 3.1 ensures that the solution u does not blow up in finite time.On the other hand, by Sobolev’s imbedding theorem it is clear that iflim inf t → T { inf x ∈ S u x ( t, x ) } = −∞ , then T < ∞ . This completes the proof of the theorem. (cid:3) In this section, we discuss the blow-up phenomena of the equation (1.1) and prove that thereexist strong solutions to (1.1) which do not exist globally in time.Firstly, for u ∈ H s , s > , we will give some useful estimates for the corresponding solution u. By the first equation of (1.2) and (2.6), a direct computation implies that ddt Z S u x dx = 2 Z S u ( − u txx ) dx = 2 Z S u ( − µ ( u ) t − µ ( u ) u x + 2 u x u xx + uu xxx − λµ ( u ) + λu xx ) dx = − µ ( u ) t µ ( u ) − λ ( µ ( u )) − λ Z S u x dx = − λ Z S u x dx. It follows that Z S u x dx = Z S u ,x dx · e − λt := µ e − λt , (4.1)where µ = (cid:0)R S u ,x dx (cid:1) . Note that R S ( u ( t, x ) − µ ( u )) dx = µ ( u ) − µ ( u ) = 0 . By Lemma 3.3,we find that max x ∈ S [ u ( t, x ) − µ ( u )] ≤ Z S u x ( t, x ) dx ≤ µ . So we have k u ( t, · ) k L ∞ ≤ | µ | + √ µ . (4.2)8 emma 4.1 [14] Let t > and v ∈ C ([0 , t ); H ( R )) . Then for every t ∈ [0 , t ) there existsat least one point ξ ( t ) ∈ R with m ( t ) := inf x ∈ R { v x ( t, x ) } = v x ( t, ξ ( t )) , and the function m is almost everywhere differentiable on (0 , t ) with ddt m ( t ) = v tx ( t, ξ ( t )) a.e. on (0 , t ) . Theorem 4.1
Let u ∈ H s , s > , u c for ∀ c ∈ R and T be the maximal time of thesolution u to (1.1) with the initial data u . If u satisfies the following condition Z S u ,x dx < − λµ − µ q λ µ + 2 K, where K = 6 | µ | µ ( | µ | + √ µ ) , then the corresponding solution to (1.1) blows up in finitetime. Proof
As mentioned earlier, here we only need to show that the above theorem holds for s = 3.Differentiating the first equation of Eq.(2.7) with respect to x , we have u tx = − u x − uu xx + 2 µ e − λt u − λu x − µ e − λt − µ e − λt (4.3)Then, it follows that ddt Z S u x dx = Z S u x u xt dx =3 Z S u x ( − u x − uu xx + 2 µ e − λt u − λu x − µ e − λt − µ e − λt ) dx ≤ − Z S u x dx − Z S uu x u xx dx + 6 µ e − λt Z S uu x dx − λ Z S u x dx = − Z S u x dx + 6 µ e − λt Z S uu x dx − λ Z S u x dx ≤ − Z S u x dx − λ Z S u x dx + 6 | µ | µ ( | µ | + √ µ ):= − Z S u x dx − λ Z S u x dx + K. Using the following inequality (cid:12)(cid:12)(cid:12)(cid:12)Z S u x dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18)Z S u x dx (cid:19) (cid:18)Z S u x dx (cid:19) ≤ (cid:18)Z S u x dx (cid:19) µ , and letting m ( t ) = Z S u x dx,
9e have ddt m ( t ) ≤ − µ m ( t ) − λm ( t ) + K = − µ (cid:18) m ( t ) + 3 λµ + µ q λ µ + 2 K (cid:19) (cid:18) m ( t ) + 3 λµ − µ q λ µ + 2 K (cid:19) . Note that if m (0) < − λµ − µ p λ µ + 2 K then m ( t ) < − λµ − µ p λ µ + 2 K for all t ∈ [0 , T ). From the above inequality we obtain m (0) + A + Bm (0) + A − B e Bµ t − ≤ Bm ( t ) + A − B ≤ A = 3 λµ , B = µ p λ µ + 2 K. Since 0 < m (0)+ A + Bm (0)+ A − B <
1, then there exists0 < T ≤ µ p λ µ + 2 K ln m (0) + 3 λµ − µ p λ µ + 2 Km (0) + 3 λµ + µ p λ µ + 2 K , such that lim t → T m ( t ) = −∞ . On the other hand, Z S u x dx ≥ inf x ∈ S u x ( t, x ) Z S u x dx = inf x ∈ S u x ( t, x ) · µ e − λt . Applying Theorem 3.2, the solution u blows up in finite time. (cid:3) Theorem 4.2
Let u ∈ H s , s > , and T be the maximal time of the solution u to (1.1) withthe initial data u . If inf x ∈ S u ′ ( x ) < − λ − p λ + 2 K, with K = 2 | µ | ( | µ | + √ µ ) , then the corresponding solution to (1.1) blows up in finite time. Proof
As mentioned earlier, here we only need to show that the above theorem holds for s = 3.Define now m ( t ) := min x ∈ S [ u x ( t, x )] , t ∈ [0 , T )and let ξ ( t ) ∈ S be a point where this minimum is attained by using Lemma 4.1. It followsthat m ( t ) = u x ( t, ξ ( t )) . Clearly u xx ( t, ξ ( t )) = 0 since u ( t, · ) ∈ H ( S ) ⊂ C ( S ) . Evaluating (4.3) at ( t, ξ ( t )), we obtain dm ( t ) dt = − m ( t ) + 2 µ e − λt u ( t, ξ ( t )) − λm ( t ) − µ e − λt − µ e − λt ≤ − m ( t ) − λm ( t ) + 2 | µ | ( | µ | + √ µ ):= − m ( t ) − λm ( t ) + K = −
12 ( m ( t ) + λ + p λ + 2 K )( m ( t ) + λ − p λ + 2 K ) . m (0) < − λ − √ λ + 2 K then m ( t ) < − λ − √ λ + 2 K for all t ∈ [0 , T ). From theabove inequality we obtain m (0) + λ + √ λ + 2 Km (0) + λ − √ λ + 2 K e √ λ +2 Kt − ≤ √ λ + 2 Km ( t ) + λ − √ λ + 2 K ≤ < m (0)+ λ + √ λ +2 Km (0)+ λ −√ λ +2 K <
1, then there exists0 < T ≤ √ λ + 2 K ln m (0) + λ − √ λ + 2 Km (0) + λ + √ λ + 2 K , such that lim t → T m ( t ) = −∞ . Theorem 3.2 implies the solution u blows up in finite time. (cid:3) Theorem 4.3
Let u ∈ H s , s > , and T be the maximal time of the solution u to (1.1) withthe initial data u . If u ( x ) is odd satisfies u ′ (0) < − λ, then the corresponding solution to(1.1) blows up in finite time. Proof
As mentioned earlier, here we only need to show that the above theorem holds for s = 3. By µ ( − u ( t, − x )) = − µ ( u ( t, x )) , we have (1.2) is invariant under the transformation( u, x ) → ( − u, − x ) . Thus we deduce that if u ( x ) is odd, then u ( t, x ) is odd with respect to x for any t ∈ [0 , T ) . By continuity with respect to x of u and u xx , we have u ( t,
0) = u xx ( t,
0) = 0 , ∀ t ∈ [0 , T ) . Evaluating (4.3) at ( t,
0) and letting h ( t ) = u x ( t, , we obtain dh ( t ) dt = − h ( t ) − λh ( t ) − µ e − λt − µ e − λt ≤ − h ( t ) − λh ( t )= − h ( t )( h ( t ) + 2 λ ) . Note that if h (0) < − λ then h ( t ) < − λ for all t ∈ [0 , T ). From the above inequality weobtain (cid:18) λh (0) (cid:19) e λt − ≤ λh ( t ) ≤ . Since h (0) h (0)+2 λ > , then there exists0 < T ≤ λ ln h (0) h (0) + 2 λ , such that lim t → T h ( t ) = −∞ . Theorem 3.2 implies the solution u blows up in finite time. (cid:3) Consequently, we will discuss the blow-up rate for the wave-breaking solutions to Eq.(1.1).The following result implies that the blow-up rate of strong solutions to weakly dissipative µ -HS equation is not affected by the weakly dissipative term even though the occurrence ofblow-up of strong solutions to Eq. (1.1) is affected by the dissipative parameter, see Theorem4.1-4.3. 11 heorem 4.4 Let u ∈ H s , s > , and T be the maximal time of the solution u to (1.1) withthe initial data u . If T is finite, we obtain lim t → T ( T − t ) min x ∈ S u x ( t, x ) = − . Proof
From the proof of Theorem 4.2, with m ( t ) := min x ∈ S [ u x ( t, x )] , t ∈ [0 , T ) , we have (cid:12)(cid:12)(cid:12)(cid:12) m ′ ( t ) + 12 m ( t ) + λm ( t ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) µ e − λt u ( t, ξ ( t )) − µ e − λt − µ e − λt (cid:12)(cid:12)(cid:12)(cid:12) ≤ | µ | ( | µ | + √ µ ) + 2 µ + 12 µ := K It follows that − K ≤ m ′ ( t ) + 12 m ( t ) + λm ( t ) ≤ K a.e. on (0 , T ) . (4.4)Thus, − K − λ ≤ m ′ ( t ) + 12 ( m ( t ) + λ ) ≤ K + 12 λ a.e. on (0 , T ) . (4.5)Let ε ∈ (0 , ). Since lim inf t → T m ( t ) = −∞ by Theorem 3.2, there is some t ∈ (0 , T ) with m ( t ) + λ < m ( t ) + λ ) > ε ( K + λ ). Let us first prove that( m ( t ) + λ ) > ε ( K + 12 λ ) , t ∈ [ t , T ) . (4.6)Since m is locally Lipschitz (it belongs to W , ∞ loc ( R ) by Lemma 4.1) there is some δ > m ( t ) + λ ) > ε ( K + 12 λ ) , t ∈ [ t , t + δ ) . Pick δ > δ < T − t we would have ( m ( t + δ )+ λ ) = ε ( K + λ )while m ′ ( t ) ≤ −
12 ( m ( t ) + λ ) + K + 12 λ < ( ε −
12 )( m ( t ) + λ ) < a.e. on ( t , t + δ ) . Being locally Lipschitz, the function m is absolutely continuous and therefore we would obtainby integrating the previous relation on [ t , t + δ ] that m ( t + δ ) + λ ≤ m ( t ) + λ < , which on its turn would yield( m ( t + δ ) + λ ) ≥ ( m ( t ) + λ ) > ε ( K + 12 λ ) . The obtained contradiction completes the proof of the relation (4.6).A combination of (4.5) and (4.6) enables us to infer12 + ε ≥ − m ′ ( t )( m ( t ) + λ ) ≥ − ε a.e. on (0 , T ) . (4.7)12ince m ( t ) + λ is locally Lipschitz on [0 , T ) and (4.6) holds, it is easy to check that m ( t )+ λ islocally Lipschitz on ( t , T ) . Differentiating the relation ( m ( t ) + λ ) · m ( t )+ λ = 1 , t ∈ ( t , T ) , weget (cid:18) m ( t ) + λ (cid:19) ′ = − m ′ ( t )( m ( t ) + λ ) a.e. on ( t , T ) , with m ( t )+ λ absolutely continuous on ( t , T ) . For t ∈ ( t , T ). Integrating (4.7) on ( t, T ) toobtain ( 12 + ε )( T − t ) ≥ − m ( t ) + λ ≥ ( 12 − ε )( T − t ) , t ∈ ( t , T ) , that is, 1 + ε ≤ − ( m ( t ) + λ )( T − t ) ≤ − ε , t ∈ ( t , T ) . By the arbitrariness of ε ∈ (0 , ) , we havelim t → T ( m ( t ) + λ )( T − t ) = − , so the statement of Theorem 4.4 follows. (cid:3) In this section, we will present some global existence results. Firstly, we give a useful lemma.Given u ∈ H s with s > . Theorem 2.2 ensures the existence of a maximal T > u to (2.7) such that u = u ( · , u ) ∈ C ([0 , T ); H s ) ∩ C ([0 , T ); H s − ) . Consider now the following initial value problem ( q t = u ( t, q ) , t ∈ [0 , T ) ,q (0 , x ) = x, x ∈ R . (5.1) Lemma 5.1
Let u ∈ H s with s > , T > be the maximal existence time. Then Eq.(5.1)has a unique solution q ∈ C ([0 , T ) × R ; R ) and the map q ( t, · ) is an increasing diffeomorphismof R with q x ( t, x ) = exp (cid:18)Z t u x ( s, q ( s, x )) ds (cid:19) > , ( t, x ) ∈ [0 , T ) × R . Moreover, with y = µ ( u ) − u xx , we have y ( t, q ( t, x )) q x = y ( x ) e − λt . Proof
The proof of the first conclusion is similar to the proof of Lemma 4.1 in [57], so we omitit here. By the first equation in (1.1) and the equation (5.1), we have ddt y ( t, q ( t, x )) q x = ( y t + y x q t ) q x + y · q x q xt = ( y t + uy x ) q x + 2 yu x q x = ( y t + uy x + 2 yu x ) q x = − λyq x .
13t follows that y ( t, q ( t, x )) q x = y ( x ) e − λt . (cid:3) Theorem 5.1 If y = µ − u ,xx ∈ H does not change sign, then the corresponding solution u of the initial value u exists globally in time. Proof
Note that given t ∈ [0 , T ) , there is a ξ ( t ) ∈ S such that u x ( t, ξ ( t )) = 0 by the periodicityof u to x -variable. If y ≥ , then Lemma 5.1 implies that y ≥ . For x ∈ [ ξ ( t ) , ξ ( t ) + 1] , wehave − u x ( t, x ) = − Z xξ ( t ) ∂ x u ( t, x ) dx = Z xξ ( t ) ( y − µ ( u )) dx = Z xξ ( t ) ydx − µ ( u )( x − ξ ( t )) ≤ Z S ydx − µ ( u )( x − ξ ( t )) = µ ( u )(1 − x + ξ ( t )) ≤ | µ | . It follows that u x ( t, x ) ≥ −| µ | . On the other hand, if y ≤ , then Lemma 5.1 implies that y ≤ . Therefore, for x ∈ [ ξ ( t ) , ξ ( t ) + 1] , we have − u x ( t, x ) = − Z xξ ( t ) ∂ x u ( t, x ) dx = Z xξ ( t ) ( y − µ ( u )) dx = Z xξ ( t ) ydx − µ ( u )( x − ξ ( t )) ≤ − µ ( u )( x − ξ ( t )) ≤ | µ | . It follows that u x ( t, x ) ≥ −| µ | . This completes the proof by using Theorem 3.3. (cid:3)
Corollary 5.1
If the initial value u ∈ H such that k ∂ x u k L ≤ √ | µ | , then the corresponding solution u of u exists globally in time. Proof
Note that R S ∂ x u dx = 0 , Lemma 3.3 implies that k ∂ x u k L ∞ ≤ √ k ∂ x u k L . If µ ≥
0, then y = µ − ∂ x u ≥ µ − √ k ∂ x u k L ≥ µ − | µ | = 0 . If µ ≤
0, then y = µ − ∂ x u ≤ µ + k ∂ x u k L ∞ ≤ µ + √ k ∂ x u k L ≤ µ + | µ | = 0This completes the proof by using Theorem 5.1. (cid:3) Acknowledgments
This work was partially supported by NNSFC (No. 10971235), RFDP(No. 200805580014), NCET (No. 08-0579) and the key project of Sun Yat-sen University. Theauthors thank the referees for useful comments and suggestions.14 eferences [1] R. Beals, D. Sattinger and J. Szmigielski, Inverse scattering solutions of the Hunter-Saxton equa-tions,
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