Global well-posedness for the defocusing, quintic nonlinear Schrödinger equation in one dimension
aa r X i v : . [ m a t h . A P ] O c t Global well-posedness for the defocusing, quinticnonlinear Schr¨odinger equation in one dimension
Ben DodsonJune 26, 2018
Abstract:
In this paper, we prove global well-posedness for low regularitydata for the one dimensional quintic defocusing nonlinear Schr¨odinger equa-tion. We show that a unique solution exists for u ∈ H s ( R ), s > . Thisimproves the result in [13], which proved global well-posedness for s > .The main new argument is that we obtain almost Morawetz estimates withimproved error. In this paper we study the initial value problem for the quintic, defocusing,nonlinear Schr¨odinger equation in one dimension, iu t + ∆ u = | u | u,u (0 , x ) = u ∈ H s ( R ) . (1.1)This is an L -critical equation. By the results of [4], this equation has alocal solution on some [0 , T ], T ( k u k H s ( R ) ) >
0, when s >
0. If a solutionto (1.1) fails to be global and only exists on [0 , T ∗ ), T ∗ < ∞ , thenlim t ր T ∗ k u ( t ) k H s ( R ) = ∞ . (1.2)[3] proved M ( u ( t )) = Z | u ( t, x ) | dx = M ( u (0)) , (1.3) E ( u ( t )) = 12 Z |∇ u ( t, x ) | dx + 16 Z | u ( t, x ) | dx = E ( u (0)) , (1.4)1re conserved, giving global well-posedness for u ∈ H ( R ). The regularitynecessary for global well-posedness has since been loweredto s > /
3, (see[13]). In this paper we will prove
Theorem 1.1 ( ) is globally well-posed for all u ∈ H s ( R ) , s > .Moreover, sup t ∈ [0 ,T ] k u ( t ) k H s ( R ) . (1 + T ) − s ) s s − + . (1.5)[13] used the I-method, a method that we will utilize in this paper as well.The I-method was first introduced for the defocusing, cubic initial valueproblem (see [7]).In §
2, we will start with some preliminary information, including the Strichartzestimates, Littlewood-Paley theory, a description of the I-method, and a lo-cal well-posedness result. In §
3, an energy increment will be obtained. In §
4, the almost Morawetz estimates will be proved. In §
5, we will prove thetheorem.
The proof of local well-posedness makes use of the Strichartz estimates.
Theorem 2.1
A pair ( p, q ) is called an admissible pair if p + q = . If ( p, q ) and (˜ p, ˜ q ) are admissible pairs and and u ( t, x ) solves iu t + ∆ u = F ( t ) ,u (0 , x ) = u , (2.1) then k u ( t, x ) k L pt L qx ( J × R ) . k u k L ( R ) + k F ( t ) k L ˜ p ′ t L ˜ q ′ x ( J × R ) . (2.2) Proof:
See [23]. p ′ denotes the Lebesgue exponent pp − .The Strichartz space will be defined by the norm k u k S ( J × R ) = sup ( p,q ) admissible k u k L pt L qx ( J × R ) . (2.3)2he space N ( J × R ) is the dual space to S ( J × R ). See [23] for moredetails.We will also make use of the Littlewood-Paley decomposition. Suppose φ ( x )is a smooth function, φ ( x ) = (cid:26) , | x | ≤ / , | x | >
1. (2.4) F ( P ≤ N u ) = φ ( ξN )ˆ u ( ξ ) , F ( P >N u ) = (1 − φ ( ξN ))ˆ u ( ξ ) , F ( P N u ) = P ≤ N u − P ≤ N u. (2.5)For convenience, let u N = P N u , similarly for u ≤ N , u >N .The I - operator is a Fourier multiplier, I N : H s ( R ) → H ( R ) , (2.6) d I N f ( ξ ) = m N ( ξ ) ˆ f ( ξ ) , (2.7) m N ( ξ ) = ( , | ξ | ≤ N ;( N | ξ | ) s − , | ξ | > N . (2.8) k Iu k H ( R ) . N − s k u k H s ( R ) , k u k H s ( R ) . k Iu k H ( R ) , (2.9)therefore, controlling E ( Iu ( t )) gives control of k u ( t ) k H s ( R ) . For the rest ofthe paper, If denotes I N f , and the presence of an N is implied. Lemma 2.2
Let I be a compact time interval, t ∈ I , N > , and suppose u , u are two solutions to ( ) such that u j ( t ) has Fourier support in theregion {| ξ j | ≤ N } for j = 1 , . Suppose also that the Fourier supports of u , u are separated by at least ≥ cN . Then for any q > , k u u k L qt,x ( I × R ) . N − /q k u k S ∗ ( I × R ) k u k S ∗ ( I × R ) , (2.10) where u k S ∗ ( I × R ) = k u k L ( R ) + k ( i∂ t + ∆) u k L / t,x ( I × R ) . (2.11) Proof:
See [22].To this end, let q = 2 + δ , p = − δ .In proving theorem 1.1, we will make use of a linear-nonlinear decomposition.See [19] for the linear-nonlinear decomposition for the defocusing, semilinearwave equation, [14] for the linear-nonlinear decomposition used for the threedimensional cubic defocusing nonlinear Schr¨odinger equation. Theorem 2.3 If k∇ Iu k L ( R ) ≤ and for some ǫ > sufficiently small, k Iu k L / t L x ( J × R ) ≤ ǫ, (2.13) then kh∇i Iu k S ( J × R ) . . (2.14) Moreover, the solution has the form e it ∆ u + u nl ( t ) , (2.15) k P >cN h∇i Iu nl k S ( J × R ) . N / − . (2.16) Proof:
The solution obeys the Duhamel formula, Iu ( t, x ) = e it ∆ u + Z t e i ( t − τ )∆ I ( | u ( τ ) | u ( τ )) dτ. (2.17)By the Strichartz estimates, k (1 − I ) u k L / t L x ( J × R ) . N kh∇i Iu k S ( J × R ) , (2.18)4 h∇i Iu k S ( J × R ) . kh∇i Iu k L ( R ) + kh∇i Iu k S ( J × R ) k u k L / t L x ( J × R ) (2.19) . kh∇i Iu k S ( J × R ) . kh∇i Iu k L ( R ) + kh∇i Iu k S ( J × R ) ( ǫ + kh∇i Iu k S ( J × R ) N ) . (2.20)Therefore, by the continuity method, kh∇i Iu k S ( J × R ) . . (2.21)This takes care of (2.14). Next, we remark that this also proves kh∇i I ( | u | u ) k L / t,x ( J × R ) . kh∇i Iu k S ( J × R ) . . (2.22)To estimate the nonlinearity, ∇ Iu nl ( t ) = Z t ∇ e i ( t − τ )∆ I ( | u ( τ ) | u ( τ )) dτ. (2.23) I ( | u ( τ ) | u ( τ )) = I ( | u ≤ cN ( τ ) | u ≤ cN ( τ )) + I ( O (( u ≤ cN ( τ )) ( u > cN ( τ ))))+ I ( O (( u > cN ( τ ))( u > cN ( τ )) u ( τ ) )) . (2.24)The first term, I ( | u ≤ cN ( τ ) | u ≤ cN ( τ )) is supported on | ξ | ≤ cN . To estimatethe second term, use the bilinear estimates, k∇ I ( O ( u > cN ( u ≤ cN ) )) k L t L x ( J × R ) . k ( ∇ Iu > cN )( u ≤ cN ) k L qt,x k u ≤ cN k L pt,x k u ≤ cN k L t L ∞ x . N / − . Finally, k∇ I ( O (( u > cN ( τ ))( u > cN ( τ )) u ( τ ) )) k L t L x ( J × R ) . k∇ Iu k L t L ∞ x ( J × R ) k u > cN k L / t L x ( J × R ) k u k L / t L x ( J × R ) . ǫ N kh∇i Iu k S ( J × R ) . N . k u > cN k L / t L x ( J × R ) . X cN ≤ N j k P N j u k L / t L x ( J × R ) . kh∇i Iu k S ( J × R ) X cN ≤ N j N sj N − s . N kh∇i Iu k S ( J × R ) . (2.25) (cid:3) In this section we prove almost conservation of the modified energy E ( Iu ( t )). Theorem 3.1
Let E ( Iu ( t )) = 12 Z |∇ Iu ( t, x ) | dx + 16 Z | Iu ( t, x ) | dx. (3.1) If J is an interval where a solution u ( t, x ) of ( ) exists, k u k L / t L x ( J × R ) ≤ ǫ , E ( Iu ) ≤ , then sup t ,t ∈ J | E ( Iu ( t )) − E ( Iu ( t )) | . N / − k P >cN ∇ Iu k L t L ∞ x ( J × R ) + 1 N − , (3.2) where c > is some small constant.Proof: ddt E ( Iu ( t )) = Re Z ( Iu t ( t, x ))[ I ( | u ( t, x ) | u ( t, x )) − | Iu ( t, x ) | Iu ( t, x )] dx. (3.3)Taking the Fourier transform, let Σ = { ξ + ... + ξ = 0 } , dξ is the Lebesguemeasure on the hyperplane, using the fact that Iu t = iI ∆ u − iI ( | u | u ) , (3.4) ddt E ( Iu ( t )) = − Re Z Σ ( i | ξ | c Iu ( t, ξ ))[1 − m ( ξ + ... + ξ ) m ( ξ ) m ( ξ ) · · · m ( ξ ) ] × c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) dξ (3.5)6 Re Z Σ ( i \ I ( | u | u )( t, ξ ))[1 − m ( ξ + ... + ξ ) m ( ξ ) m ( ξ ) · · · m ( ξ ) ] × c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) dξ. (3.6)We will estimate (3.5) and (3.6) separately by making a Littlewood-Paleydecomposition and consider several cases separately. Without loss of gener-ality let N ≥ N ≥ N ≥ N ≥ N . The term ( ) : When estimating this term, we will frequently use the bilinear estimate(2.10).
Case 1, N << N : In this case, m ( ξ i ) ≡
1, so1 − m ( ξ + ... + ξ ) m ( ξ ) · · · m ( ξ ) ≡ . (3.7) Case 2, N & N >> N : By the fundamental theorem of calculus, | − m ( ξ + ... + ξ ) m ( ξ ) · · · m ( ξ ) | . N N . (3.8)Recall that q = 2 + δ , p = − q . N ∼ N , so X N . N ∼ N X N ≤ N ≤ N ≤ N <
1, so1 − m ( ξ + ... + ξ ) m ( ξ ) · · · m ( ξ ) ≡ . (3.14) Case 2, N & N >> N : In this case, apply the fundamental theorem ofcalculus, | − m ( ξ + ... + ξ ) m ( ξ ) · · · m ( ξ ) | . N N . (3.15) X N . N ∼ N X N ≤ N ≤ N ≤ N < Theorem 4.1 Let u be the solution to the nonlinear Schr¨odinger equationin one dimension, iu t + ∆ u = | u | u. (4.1) Then k Iu k L t,x ([0 ,T ] × R ) . k Iu k L ∞ t ˙ H x ([0 ,T ] × R ) k u k L ( R ) + X J k N − kh∇i Iu k S ( J k × R ) , (4.2) where [0 , T ] = ∪ k J k .Proof: We start with the case I = 1. We will use the method found in [13]and [6]. Let ω ( t, z ) : R × R → C , ω ( t, z ) = u ( t, x ) u ( t, x ) u ( t, x ) u ( t, x ) , (4.3)where u ( t, x ) is a solution to (1.1). Then ω ( t, z ) obeys the equation iω t + ∆ z ω = ( X i =1 | u ( t, x i ) | ) ω ( t, z ) = N . (4.4)Next, define the interaction Morawetz quantity, M a ( t ) = 2 Z R ∂ j a ( z ) Im ( ω ( t, z ) ∂ j ω ( t, z )) dz, (4.5)following the convention that repeated indices are summed. Let T j ( t, z ) = 2 Im ( ω ( t, z ) ∂ j ω ( t, z )) , (4.6) L jk ( t, z ) = − ∂ jk ( | ω ( t, z ) | ) + 4 Re ( ∂ j ω∂ k ω )( t, z ) , (4.7) ∂ t T j + ∂ k L jk = 2 {N , ω } jp . (4.8) Z T Z R ∂ t ∂ j a ( z ) Im ( ω ( t, z ) ∂ j ω ( t, z )) dzdt (4.9)= Z T Z R ∂ j a ( z ) ∂ jkk ( | ω ( t, z ) | ) dzdt (4.10)13 Z T Z R ∂ j a ( z ) ∂ k Re ( ∂ j ω∂ k ω )( t, z ) dzdt (4.11)+ 2 Z T Z a j ( z ) {N , ω } jp dzdt. (4.12)Now, evaluate each term separately. Make a change of variables, y = Az ,where A = 12 − − − − − − (4.13)is an orthonormal matrix with inverse A − = 12 − 11 1 − − − − − . (4.14)In the new variables, let a ( y ) = ( y + y + y ) / , − ∆∆ a ( y ) = 4 πδ ( y , y , y ) .A − y = 12 y y y y . (4.15)Therefore, integrating (4.10) by parts,(4.10) = Z T Z ( − ∆∆ a ( y )) | ω ( t, z ) | dzdt = 8 π Z T Z R | u ( t, x ) | dxdt. (4.16) ∂ jk a ( z ) is a positive semidefinite matrix, so integrating (4.11) by parts,4 Z T Z R ( ∂ jk a ( z )) Re ( ∂ j ω∂ k ω )( t, z ) ≥ . (4.17)Finally, for (4.12), 14 N , ω } jp ( t, z ) = − | ω ( t, z ) | ∂ j ( X i =1 | u ( t, x i ) | ) . (4.18) | ω ( t, z ) | ∂ j ( X i =1 | u ( t, x i ) | ) = 23 X j =1 ∂ j ( | u ( t, x j ) | ( | Y i = j | u ( t, x i ) | )) , so integrating (4.12) by parts, − Z T Z a j ( z ) {N , ω } jp ( t, z ) dzdt = 43 Z T Z a jj ( z ) | ω ( t, z ) | | u ( t, x j ) | dzdt ≥ . (4.19)Therefore, Z T Z R | u ( t, x ) | dxdt . | Z T Z ∂ t ( a j ( z ) Im [ ω ( t, z ) ∂ j ω ( t, z )] dzdt | . k u k L ∞ t ˙ H / ([0 ,T ] × R ) k u k L ( R ) . (4.20)Next, we prove an almost Morawetz estimate (see [5], [11], [15] for the twodimensional case; [5], [12], [6] for discussion of the one dimensional case).If u ( t, x ) solves (4.1), then Iu ( t, x ) solves iIu t ( t, x ) + ∆ Iu ( t, x ) = I ( | u ( t, x ) | u ( t, x )) = N . (4.21)Split the nonlinearity into ”good” and ”bad” pieces, N = N g + N b . N g = X i =1 | Iu ( t, x i ) | Iu ( t, x i ) Y j = i Iu ( t, x j ) , (4.22) N b = X i =1 [ I ( | u ( t, x i ) | u ( t, x i )) − | Iu ( t, x i ) | ( Iu ( t, x i ))] Y j = i Iu ( t, x j ) . (4.23)Let ω ( t, z ) = Iu ( t, x ) Iu ( t, x ) Iu ( t, x ) Iu ( t, x ), performing the same anal-ysis will split Z T Z R ∂ t a j ( z ) T j ( t, z ) dzdt, N = N g , thenthe previous analysis would carry over identically. Indeed, Z T Z R ( − ∆∆ a ( z )) | ω ( t, z ) | dzdt = 8 π Z T Z | Iu ( t, x ) | dxdt. (4.24)4 Z T Z R ( ∂ jk a ( z )) Re ( ∂ j ω∂ k ω )( t, z ) dzdt ≥ . (4.25)2 Z T Z a j ( z ) {N g , ω } jp ( t, z ) dzdt = 43 Z T Z a jj ( z ) | ω ( t, z ) | | Iu ( t, x j ) | dz ≥ . (4.26)Therefore, Z T Z | u ( t, x ) | dxdt . k Iu k L ∞ t ˙ H x ([0 ,T ] × R ) k u k L ( R ) + | Z T Z a j ( z ) {N b , ω } jp ( t, z ) dzdt | . (4.27)To analyze the remainder N b , first consider a term of the form Z J k Z a j ( z ) N b ( t, z ) ∂ j ω ( t, z ) dzdt. (4.28)Recall (4.23), without loss of generality let i = 1 and estimate Z J k Z a j ( z )[ I ( | u ( t, x ) | u ( t, x )) − | Iu ( t, x ) | Iu ( t, x )] Iu ( t, x ) Iu ( t, x ) Iu ( t, x ) × ∂ j ( Iu ( t, x ) Iu ( t, x ) Iu ( t, x ) Iu ( t, x )) dxdt. (4.29)Because a j ( z ) ∈ L ∞ ,(4.28) . k I ( | u ( t, x ) | u ( t, x )) −| Iu ( t, x ) | Iu ( t, x ) k L t L x ( J k × R ) kh∇i Iu k L ∞ t L x ( R ) . (4.30)To evaluate I ( | u ( t, x ) | u ( t, x )) − | Iu ( t, x ) | Iu ( t, x ) , make a Littlewood - Paley partition of unity. Let16 ( t, ξ ) = Z ξ = ξ + ... + ξ [1 − m ( ξ + ... + ξ ) m ( ξ ) ...m ( ξ ) ] c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) c Iu ( t, ξ ) . (4.31)Without loss of generality, let N ≥ N ≥ N ≥ N ≥ N . Consider severalcases separately. Case 1: N << N In this case, | − m ( ξ + ... + ξ ) m ( ξ ) ...m ( ξ ) | ≡ . Case 2: N & N >> N In this case, by the fundamental theorem ofcalculus, | − m ( ξ + ... + ξ ) m ( ξ ) ...m ( ξ ) | . N N . (4.28) . X N . N X N ≤ N ≤ N ≤ N < 1, using (4.15), Z ∞−∞ Z y + y + y ≤ y + y + y | Iu ( t, y + y + y − y | | Iu ( t, y + y − y + y | ×| Iu ( t, y − y + y + y | | Iu ( t, y − y − y − y | dy dy dy dy . ( Z r r dr ) k Iu k L ∞ t L x ( J k × R ) . kh∇i Iu k S ( J k × R ) , so k a jj ( y ) ω ( t, y ) k L ∞ t L y ( J k × R ) . kh∇i Iu k S ( J k × R ) . 18n the other hand, when y + y + y ≥ ∂ jj a ( z ) is bounded and k Iu ( t, x ) Iu ( t, x ) Iu ( t, x ) Iu ( t, x ) k L t L x ( J k × R ) . kh∇i Iu k S ( J k × R ) . (4.36)Since k I ( | u ( t, x ) | u ( t, x )) −| Iu ( t, x ) | Iu ( t, x ) k L t L x ( J × R ) . N − kh∇i Iu k S ( J k × R ) , theorem 4.1 is proved. Theorem 5.1 ( ) is globally well-posed for u ∈ H s ( R ) , s > .Proof: Z |∇ Iu ( x ) | dx . N − s ) k u k H s ( R ) . Z | Iu ( x ) | dx . N − s k u k H s ( R ) . (5.1)If u ( t, x ) solves (1.1) on [0 , T ], then rescaling,1 λ / u ( tλ , xλ ) (5.2)solves (1.1) on [0 , λ T ], we will call the rescaled solution u λ ( t, x ). Choose λ ∼ N (1 − s ) /s so that E ( Iu λ (0)) = 1 / 2. Let W = { t : E ( Iu λ ( t )) ≤ } . (5.3)W is closed by the dominated convergence theorem and nonempty since0 ∈ W . To prove W = [0 , λ T ], it suffices to prove W is open in [0 , λ T ].Suppose W = [0 , T ], then by continuity there exists δ > E ( Iu λ ( t )) ≤ , T + δ ]. Lemma 5.2 When s > / , k Iu λ k L t,x ([0 ,T + δ ] × R ) ≤ C , (5.4) for some C ( T ) . roof: Let τ = sup { T ∗ ∈ [0 , T + δ ] : k Iu λ k L t,x ([0 ,T ∗ ] × R ) ≤ C } . If τ < T + δ ,there exists δ ′ > k Iu λ k L t,x ([0 ,τ + δ ′ ] × R ) ≤ C . (5.5)Recall E ( Iu λ ( t )) ≤ , τ + δ ′ ], k Iu λ k L / t L x ([0 ,τ + δ ′ ] × R ) ≤ ( λ T ) / k Iu λ k L t,x ([0 ,τ + δ ′ ] × R ) ≤ ( λ T ) / (2 C ) / . (5.6)Partition [0 , τ + δ ′ ] into ( λ T ) / (2 C ) / ǫ / subintervals such that k Iu λ k L / t L x ([0 ,τ + δ ′ ] × R ) ≤ ǫ (5.7)on each subinterval. Then apply the almost Morawetz estimate, k Iu λ k L t,x ([0 ,τ + δ ′ ] × R ) ≤ C k u k L ( R ) k Iu λ k L ∞ t ˙ H ([0 ,τ + δ ′ ] × R ) + X k kh∇i Iu k S ( J k × R ) N − ≤ C + C ( λ T ) / (2 C ) / ǫ / N − ≤ C + C N − s )3 s T / (2 C ) / ǫ / N − ≤ C , (5.8)when N is sufficiently large, as long as ( − ss ) < 2, or s > / 4. This provesthe lemma. (cid:3) Returning to the theorem, k Iu λ k L t,x ([0 ,T + δ ] × R ) ≤ C , k Iu λ k L / t L x ([0 ,T + δ ] × R ) . λ / T / . (5.9)Partition [0 , T + δ ] into . ( λ T ) / subintervals. We will call these the littleintervals. Take the union of the first N / − little subintervals, and call thisbig interval J = ∪ N / − l =1 J ,l . Take the union of the next N / − subintervals and call this big interval J ,and so on. 200 , T + δ ] = ( λ T / N / − [ k =1 J k = ( λ T / N / − [ k =1 N / − [ l =1 J k,l . (5.10) Lemma 5.3 Suppose E ( Iu ( t )) ≤ on J k . sup t ,t ∈ J k | E ( Iu ( t )) − E ( Iu ( t )) | . N / − . (5.11) Proof: By theorem (3.1),sup t ,t ∈ J k,l | E ( Iu ( t )) − E ( Iu ( t )) | . N / − k P >cN ∇ Iu k L t L ∞ x ( J k,l × R ) + 1 N − (5.12) N / X l =1 k P >cN ∇ Iu k L t L ∞ x ( J k,l × R ) . N / k P >cN ∇ Iu k L t L ∞ x ( J k × R ) . Let J k = [ a, b ], J k,m = [ a m , b m ], a m +1 = b m , a = a , b N / − = b . On eachlittle subinterval, perform the linear-nonlinear decomposition in theorem 2.3.The solution on [ a m , b m ] is of the form e i ( t − a m )∆ u ( a m ) + u nlm ( t ) . By induction, e i ( t − a m )∆ u ( a m ) = e it ∆ u + m − X j =1 e i ( t − a j )∆ u nlj ( b j ) . (5.13) k∇ e i ( t − a )∆ Iu ( a ) k L t L ∞ x ( J k × R ) . . (5.14) N / − X m =1 k∇ e i ( t − a )∆ Iu ( b m ) nl k L t L ∞ x ( J k × R ) . . (5.15) N / − X l =1 k∇ Iu nl ( t ) k L t L ∞ x ( J k,l × R ) . N / − . (5.16)21herefore, k∇ P >cN Iu k L t L ∞ x ( J k × R ) . . Plugging this back in to (5.11), N / − X l =1 sup t ,t ∈ J k,l | E ( Iu ( t )) − E ( Iu ( t )) | . 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