The initial-boundary value problem for the 1D nonlinear Schrödinger ...
The initial-boundary value problem for the 1D nonlinear Schr¨odinger equation on the half-line Justin Holmer
Department of Mathematics, University of California Berkeley, CA 94720 1 2
Abstract We prove, by adapting the method of Colliander-Kenig [9], local wellposedness of the initial-boundary value problem for the one-dimensional nonlinear Schr¨odinger equation i∂t u + ∂x2 u + λu|u|α−1 = 0 on the half-line under low boundary regularity assumptions.
1 2
AMS subject code: 35Q55 The content of this article appears as part of the author’s Ph.D. thesis at the University
of Chicago.
1
1
Introduction
We consider the initial-boundary value problem on the right half-line for the one-dimensional nonlinear Schr¨odinger (1D NLS) equation i∂t u + ∂x2 u + λu|u|α−1 = 0 for (x, t) ∈ (0, +∞) × (0, T ) u(0, t) = f (t) for t ∈ (0, T ) u(x, 0) = φ(x) for x ∈ (0, +∞)
(1)
where λ ∈ C. ˙ On R, we define the homogeneous L2 -based Sobolev spaces H˙ s = H(R) by 2 ˆ the norm kφkH˙ s = k|ξ|s φ(ξ)k L2ξ and the L -based inhomogeneous Sobolev spaces 2 1/2 ˆ H s = H s (R) by the norm kφkH s = khξis φ(ξ)k . In L2ξ , where hξi = (1 + |ξ| )
addition, we shall need L2 -based inhomogeneous Sobolev spaces on the half-line R+ = (0, +∞), which we denote H s (R+ ). These are defined, for s ≥ 0, as: ˜ φ ∈ H s (R+ ) if ∃ φ˜ ∈ H s (R) such that φ(x) = φ(x) for a.e. x > 0; in this ˜ H s (R) . We also similarly define, for s ≥ 0, case we set kφkH s (R+ ) = inf φ˜ kφk ˜ φ ∈ H s (0, L) if ∃ φ˜ ∈ H s (R) such that φ(x) = φ(x) a.e. on (0, L); in this case ˜ Hs . we set kφkH s (0,L) = inf φ˜ kφk The local smoothing inequality of [15] for the 1D Schr¨odinger group is 2
keit∂x φk
2s+1 4
˙ L∞ x Ht
≤ ckφkH˙ s
This inequality is sharp in the sense that
2s+1 4
cannot be replaced by any
higher number. We are thus motivated to consider initial-boundary data pairs (φ(x), f (t)) ∈ H s (R+ x)×H
2s+1 4
(R+ t ) and inclined to consider this configuration
optimal in the scale of L2 -based Sobolev spaces. Note that the trace map φ → φ(0) is well-defined on H s (R+ ) when s > 12 . 2
Thus, if s > 12 , then
2s+1 4
>
1 2
and both φ(0) and f (0) are well-defined quantities.
Since φ(0) and f (0) are both meant to represent u(0, 0), they must agree. Therefore, we consider (1) for 0 ≤ s < φ ∈ H s (R+ ), f ∈ H
2s+1 4
3 2
in the setting
(R+ ), and if
1 2
< s < 32 , φ(0) = f (0)
(2)
The solutions we construct shall have the following characteristics. Definition 1. u(x, t) will be called a distributional solution of (1), (2) on [0, T ∗ ) with strong traces if (a) u belongs to a space X with the property that u ∈ X implies u|u|α−1 is defined as a distribution. (b) u(x, t) satisfies the equation (1) in the sense of distributions on the set (x, t) ∈ (0, +∞) × (0, T ∗ ). (c) Space traces: ∀ T < T ∗ , we have u ∈ C([0, T ]; Hxs ) and u(·, 0) = φ in H s (R+ ). (d) Time traces: ∀ T < T ∗ , we have u ∈ C(Rx ; H in H
2s+1 4
2s+1 4
(0, T )) and u(0, ·) = f
(0, T ).
For the purposes of uniqueness in the high regularity setting s > 12 , we can consider a weaker notion of solution. Definition 2. u(x, t) will be called a distributional solution of (1), (2) on [0, T ∗ ) with weak traces if it satisfies conditions (a), (b) of Definition 1 and (c) One-sided space traces: ∀ T < T ∗ , we have u ∈ C([0, T ]; H s (R+ x )) and u(·, 0) = φ in H s (R+ ). 3
(d) Boundary values: ∀ T < T ∗ , we have lim ku(x, ·) − f k
H
x↓0
2s+1 4 (0,T )
= 0.
So that we may, at a later time, properly address the matter of uniqueness in the low regularity s
12 , u is a distributional solution of (1), (2) with weak traces if and only if it is a mild solution; in this case u is unique. Our main result is the following existence statement. Theorem 1. (a) Subcritical: Suppose 0 ≤ s < 12 , and 2 ≤ α < 4
5−2s 1−2s
or 1 2
< s < 32 , and 2 ≤ α < ∞
Then ∃ T ∗ > 0 and u that is both a mild solution and a distributional solution with strong traces of (1),(2) on [0, T ∗ ). If T ∗ < ∞, then limt↑T ∗ ku(·, t)kHxs = ∞. Also, ∀ T < T ∗ , ∃ δ0 = δ0 (s, T, φ, f ) > 0 such that if 0 < δ ≤ δ0 and kφ − φ1 kH s (R+ ) + kf − f1 k
H
2s+1 4 (R+ )
< δ then there is a solution u1 (as
above) on [0, T ], corresponding to (φ1 , f1 ), such that ku − u1 kC([0,T ]; Hxs ) + ku − u1 k
C(Rx ; H
2s+1 4 (0,T ))
≤ cδ, with c = c(s, T, f, φ).
(b) Critical: Suppose 0 ≤ s
0 maximal and u
that is both a mild solution and a distributional solution with strong traces of (1),(2) on [0, T ∗ ). Also, ∃ T = T (s, φ, f ) < T ∗ and ∃ δ0 = δ0 (s, φ, f ) > 0 such that if 0 < δ ≤ δ0 and kφ − φ1 kH s (R+ ) + kf − f1 k
H
2s+1 4 (R+ )
< δ then
there is a solution u1 (as above) on [0, T ], corresponding to (φ1 , f1 ), such that ku − u1 kC([0,T ]; Hxs ) + ku − u1 k
C(Rx ; H
2s+1 4 (0,T ))
≤ cδ, with c = c(s, f, φ).
Note that in (b), we may not have blow-up in the norm ku(·, t)k as t ↑ T ∗ . The proof of Theorem 1 involves the introduction of a boundary forcing operator analogous to that introduced by [9] in their treatment of the generalized Korteweg de-Vries equation (gKdV) on the half-line, and incorporates the techniques of the standard proof of local well-posedness of the corresponding initial-value problem based on the Strichartz estimates (see [7]). One could also consider the left half-line problem i∂t u + ∂x2 u + λu|u|α−1 = 0 for (x, t) ∈ (−∞, 0) × (0, T ) u(0, t) = f (t) for t ∈ (0, T ) u(x, 0) = φ(x) for x ∈ (−∞, 0) 5
although this is actually identical to the right half-line problem (1) by the transformation u(x, t) → u(−x, t). We plan, in a future publication, to examine the initial-boundary value problem for the line-segment i∂t u + ∂x2 u + λu|u|α−1 = 0 for (x, t) ∈ (0, L) × (0, T ) u(0, t) = f1 (t) for t ∈ (0, T ) for t ∈ (0, T )
u(L, t) = f2 (t) u(x, 0) = φ(x)
for x ∈ (0, L)
and consider global existence questions for the half-line and line-segment problems. We now briefly mention some earlier work and alternate perspectives on this problem and related problems. The main new feature of our work is the low regularity requirements for φ and f . Under higher regularity assumptions, more general results are already available. [18] considered a bounded or unbounded domain Ω ⊂ Rn with smooth boundary ∂Ω, and proved global existence of solutions to i∂t u + ∆u + λu|u|α−1 = 0 for (x, t) ∈ Ω × (0, T ) u(x, t) = f (x, t) for x ∈ ∂Ω u(x, 0) = φ(x) for x ∈ Ω
(3)
where f ∈ C 3 (∂Ω) is compactly supported, φ ∈ H 1 (Ω), and λ < 0. This solution is obtained as a limit of solutions to approximate problems after several a priori identities have been established. Earlier, [6] and [5] had obtained solutions to (1) for α > 3, λ < 0 and α = 3, λ ∈ R for φ ∈ H 2 (R+ ) and f ∈ C 2 (0, T ), using
6
semigroup techniques and a priori estimates. The problem (3) with f = 0 had been considered previously ([4] [22] [20] [21] [23]). [10] in the integrable case α = 3, λ = ±2 with φ Schwartz and f sufficiently smooth, obtained a solution to (1) by reformulating the problem as a 2×2 matrix Riemann-Hilbert problem. In this setting, [3] obtain an explicit representation for ∂x u(0, t). Outline: In §2, we discuss some notation, introduce function spaces and recall some needed properties of these function spaces. In §3, we review the definition and basic properties of the Riemann-Liouville fractional integral. In §4, 5, we state the needed estimates for the group and inhomogeneous solution operator. In §6, we define the boundary forcing operator, adapted from [9], and prove the needed estimates for it. In §7, we prove Theorem 1. In §8, we prove Prop. 1.
2
Notations and some function space properties
Let χS denote the characteristic function for the set S. We shall write LqT to R ˆ mean Lq ([0, T ]). Set φ(ξ) = x e−ixξ φ(x) dx. Define (τ − i0)α as the limit, in the sense of distributions, of (τ + iγ)−α as γ ↑ 0. Let hξis = (1 + |ξ|2 )s/2 . Let s f (ξ) = |ξ|s fˆ(ξ). The homogeneous L2 -based Sobolev spaces are H d ˙ s (R) = D
(−∂ 2 )−s/2 L2 (R) and the inhomogeneous L2 -based H s (R) = (1 − ∂ 2 )−s/2 L2 (R). We also set, for 1 ≤ p ≤ ∞, W s,p = (I − ∂ 2 )−s/2 Lp . We use the notation H s to mean H s (R) (and not H s (R+ ) or H0s (R+ )). The trace operator φ 7→ φ(0) is defined for φ ∈ H s (R) when s > 21 . For s ≥ 0, define φ ∈ H s (R+ ) if ∃ φ˜ ∈ H s (R) 7
˜ H s (R) . ˜ such that φ(x) = φ(x) for x > 0; in this case we set kφkH s (R+ ) = inf φ˜ kφk ˜ For s ≥ 0, define φ ∈ H0s (R+ ) if, when φ(x) is extended to φ(x) on R by setting ˜ ˜ H s (R) . φ(x) = 0 for x < 0, then φ˜ ∈ H s (R); in this case we set kφkH0s (R+ ) = kφk Define φ ∈ C0∞ (R+ ) if φ ∈ C ∞ (R) with supp φ ⊂ [0, +∞) (so that, in particular, ∞ φ and all of its derivatives vanish at 0), and C0,c (R+ ) as those members of ∞ C0∞ (R+ ) with compact support. We remark that C0,c (R+ ) is dense in H0s (R+ )
for all s ∈ R. We shall take a fixed θ ∈ Cc∞ (R) such that θ(t) = 1 on [−1, 1] and supp θ ⊂ [−2, 2]. Denote by θT (t) = θ(tT −1 ). Lemma 1 ([9] Lemma 2.8). If 0 ≤ α < 12 , then kθT hkH α ≤ chT iα khkH˙ α , where c = c(α, θ). Lemma 2 ([13] Lemma 3.5). If − 21 < α < 12 , then kχ(0,+∞) f kH α ≤ ckf kH α , where c = c(α). Lemma 3 ([9] Prop. 2.4, [13] Lemma 3.7, 3.8). If
1 2
< α
0, i.e.
α−1 t+ ,f Γ(α)
1 = Γ(α)
Z
+∞
tα−1 f (t) dt
0
Integration by parts gives, for Re α > 0, that α+k−1 α−1 t+ t+ k = ∂t Γ(α) Γ(α + k) for all k ∈ N. This formula can be used to extend the definition (in the sense of distributions) of
tα−1 + Γ(α)
to all α ∈ C. In particular, we obtain tα−1 + = δ0 (t) Γ(α) α=0 9
A change of contour calculation shows that α−1 b 1 t+ (τ ) = e− 2 πiα (τ − i0)−α Γ(α) where (τ − i0)−α is the distributional limit. If f ∈ C0∞ (R+ ), we define tα−1 Iα f = + ∗ f Γ(α) Thus, when Re α > 0, Z t 1 Iα f (t) = (t − s)α−1 f (s) ds Γ(α) 0 Rt and I0 f = f , I1 f (t) = 0 f (s) ds, and I−1 f = f 0 . Also Iα Iβ = Iα+β , which follows from the Fourier transform formula. For further details on the distribution tα−1 + , Γ(α)
see [11].
Lemma 7. If h ∈ C0∞ (R+ ), then Iα h ∈ C0∞ (R+ ), for all α ∈ C. Lemma 8 ([12]). If 0 ≤ α < +∞ and s ∈ R, then kI−α hkH0s (R+ ) ≤ ckhkH0s+α (R+ ) Lemma 9 ([12]). If 0 ≤ α < +∞, s ∈ R, µ ∈ C0∞ (R) kµIα hkH0s (R+ ) ≤ ckhkH0s−α (R+ ) where c = c(µ).
4
Estimates for the group
Set it∂x2
e
1 φ(x) = 2π
Z
2 ˆ dξ eixξ e−itξ φ(ξ)
ξ
10
(4)
so that
(i∂t + ∂x2 )eit∂x2 φ = 0 for (x, t) ∈ R × R eit∂x2 φ(x) = φ(x) for x ∈ R t=0
Lemma 10. Let s ∈ R. If φ ∈ H s (R), then 2
(a) Space traces: keit∂x φ(x)kC(Rt ; Hxs ) ≤ ckφkH s . 2
(b) Time traces: kθT (t)eit∂x φ(x)k
2s+1 4 )
C(Rx ; Ht
1 q
(c) Mixed-norm: If 2 ≤ q, r ≤ ∞ and
+
≤ chT i1/4 kφkH s . 1 2r
2
= 14 , then keit∂x φ(x)kLqt Wxs,r ≤
ckφkH s . Proof. (a) is clear from (4). (b) was obtained in [15]. (c) was obtained by [19] (see also [14]).
5
Estimates for the Duhamel inhomogeneous solution operator
Let Z Dw(x, t) = −i
t
0
2
ei(t−t )∂x w(x, t0 ) dt0
0
Then
(i∂t + ∂x2 )Dw(x, t) = w(x, t) for (x, t) ∈ R × R Dw(x, 0) = 0
for x ∈ R
Lemma 11. Suppose 2 ≤ q, r ≤ ∞ and
1 q
+
1 2r
= 14 , then
(a) Space Traces: If s ∈ R, then kDwkC(Rt ;Hxs ) ≤ ckwkLq0 W s,r0 . t
(b) Time Traces: If − 32 < s < 12 , then kθT (t)Dw(x, t)k 11
x
2s+1 4 )
C(Rx ; Ht
≤ chT i1/4 kwkLq0 W s,r0 . t
x
(c) Mixed-norm: If s ∈ R, then kDwkLqt Wxs,r ≤ ckwkLq0 W s,r0 . x
t
Proof. (a) and (c) are due to [19] (see also [14]). We now prove (b), following the techniques of Theorem 2.3 in [16]. We use the representation Z i +∞ 0 2 Dw(x, t) = − (sgn t0 )ei(t−t )∂x w(x, t0 ) dt0 2 −∞ Z Z 1 1 ˆ τ) itτ ixξ w(ξ, + lim e e dξ dτ →0+ 2π |τ +ξ 2 |> 2πi τ τ + ξ2 = I + II and Term II can also be written Z 1 II = eitτ [m(·, τ ) ∗ wˆ t (·, τ )](x) dτ 2π τ where w ˆ t (·, τ ) denotes the Fourier transform of w(x, t) in the t-variable alone and exp(−|x||τ |1/2 ) 1 sin(|x||τ |1/2 ) 1 + χ (τ ) m(x, τ ) = − χ(0,+∞) (τ ) (−∞,0) 2 |τ |1/2 2 |τ |1/2 First we treat Term I for all s and all admissible pairs q, r. Pairing Term I with f (x, t) such that kf k
− 2s+1 4
L1x Ht
≤ 1, we are left to show that
Z
(sgn t0 )e−it0 ∂x2 w(x, t0 ) dt0
0
t
Hxs
≤ ckwkLq0 W s,r0 t
x
and
Z
θT (t)e−it∂x2 f (x, t) dt
Hx−s
t
≤ ckf k
− 2s+1 4
L1x Ht
The first of these follows from the proof of (a), while the second is obtained by duality and Lemma 10(b). We address Term II separately for r0 = 2, q 0 = 1, and r0 = 1, q 0 = 43 ; the intermediate cases follow by interpolation. For the case 12
r0 = 2, q 0 = 1, we use the first representation of Term II with Lemma 1, the change of variable η = −ξ 2 , and L2 -boundedness of the Hilbert transform on A2 -weighted spaces, to obtain Z kθT (t)(Term II)k
2s+1 4
s
≤c
2
1/2
|ξ| |w(ξ, ˆ τ )| dξ
Ht
ξ
Z
s
≤c
2 !1/2 |wˆ (ξ, t)| dt dξ
Z
x
|ξ| ξ
t
where w ˆ x (ξ, t) denotes the Fourier transform in the x-variable alone. Complete the bound by applying Minkowskii’s integral inequality and the Placherel theorem. The validity of this step is restricted to − 23 < s < 12 . We shall only prove the r0 = 1, q 0 =
4 3
case for s = 0. Note that by the
second representation for Term II, k(Term II)kL∞ H 1/4 is x
Z Z
−1/2
|τ | τ
t
Z
m(x − y, τ )wˆ (y, τ ) dy
y
t
|τ |−1/2 m(x − z, τ )wˆ t (z, τ ) dz dτ
z
which is equivalent to Z K(y, s, z, t)w(y, s)w(z, t) dy ds dz dt y,s,z,t
where Z K(y, s, z, t) =
|τ |1/2 e−i(s−t)τ m(x − y, τ )m(x − z, τ ) dτ
τ
From the definition of m, we see that |K(y, s, z, t)| ≤ c|s − t|−1/2 . We conclude by applying the theorem on fractional integration (see Theorem 1 of Chapter V in [17]).
13
6
Estimates for the Duhamel boundary forcing operator
For f ∈ C0∞ (R+ ), define the boundary forcing operator Z t 0 2 i 14 π Lf (x, t) = 2e ei(t−t )∂x δ0 (x)I−1/2 f (t0 ) dt0 Z t0 1 ix2 0 −1/2 =√ (t − t ) exp I−1/2 f (t0 ) dt0 4(t − t0 ) π 0
(5) (6)
The equivalence of the two definitions is evident from the formula −i π sgn t 2 b e 4 ix 1 2 √ exp (ξ) = e−itξ 1/2 4t 2 π |t| From these two definitions, we see that 3 (i∂t + ∂x2 )Lf (x, t) = 2ei 4 π δ0 (x)I−1/2 f (t) for (x, t) ∈ R × R Lf (x, 0) = 0 for x ∈ R Lf (0, t) = f (t) for t ∈ R We now establish some continuity properties of Lf (x, t) when f is suitably nice. ∞ Lemma 12. Let f ∈ C0,c (R+ ).
(a) For fixed t, Lf (x, t) is continuous in x for all x ∈ R and ∂x Lf (x, t) is continuous in x for x 6= 0 with 1
lim ∂x Lf (x, t) = e− 4 πi I−1/2 f (t) x↑0
1
lim ∂x Lf (x, t) = −e− 4 πi I−1/2 f (t) x↓0
(7) (b) ∀ k = 0, 1, 2, . . . and for fixed x, ∂tk Lf (x, t) is continuous in t for all t ∈ R.
14
We also have the pointwise estimates, for k = 0, 1, 2, . . ., on [0, T ], |∂tk Lf (x, t)| + |∂x Lf (x, t)| ≤ chxi−N where c = c(f, N, k, T ). Proof. Let us denote “integration by parts” by IBP. It is clear from (6) and dominated convergence that, for fixed t, Lf (x, t) is continuous in x, and for 1
fixed x, Lf (x, t) is continuous in t. Let h = 2ei 4 π I−1/2 f ∈ C0∞ (R+ ) (by Lemma Rt 0 7) and φ(ξ, t) = 0 e−i(t−t )ξ h(t0 ) dt0 . By IBP in t0 , |∂ξk φ(ξ, t)| ≤ chξi−k−1 , where c = c(h, k, T ), and thus |∂ξk φ(ξ 2 , t)| ≤ chξi−k−2
(8)
We have Z Lf (x, t) =
eixξ φ(ξ 2 , t) dξ
(9)
ξ 0
2
and by IBP in ξ and (8), we have |Lf (x, t)| ≤ chxi−N . By ∂t [ei(t−t )∂x δ0 (x)] = 0
2
−∂t0 [ei(t−t )∂x δ0 (x)] and IBP in t0 in (5), ∂t Lf = L∂t f , and thus, for fixed x, 0
2
∂tk Lf (x, t) is continuous in t and |∂tk Lf (x, t)| ≤ chxi−N . By ∂x2 [ei(t−t )∂x δ0 (x)] = 0
3
2
i∂t0 [ei(t−t )∂x δ0 (x)] and IBP in t0 in (5), ∂x2 Lf (x, t) = 2ei 4 π δ0 (x)I−1/2 f (t) − iL(∂t f )(x, t). Hence i 43 π
∂x Lf (x, t) = e
Z
x
(sgn x)I−1/2 f (t) − i
L(∂t f )(x0 , t) dx0 + c(t)
x0 =0
Since all terms except c(t) are odd in x, we must have c(t) = 0. From this we obtain (7), and the bound |∂x Lf (x, t)| ≤ c. From (9), IBP in ξ and (8), we obtain that |∂x Lf (x, t)| ≤ c|x|−N . Combining the two previous bounds, we have |∂x Lf (x, t)| ≤ chxi−N . Now we provide an alternate representation of Lf (x, t). 15
∞ (R+ ). Then Lemma 13. Suppose f ∈ C0,c Z 1 1/2 Lf (x, t) = eitτ e−|x|(τ −i0) fˆ(τ ) dτ 2π τ
(10)
where 1
(τ − i0) 2 = χ(0,+∞) (τ )|τ |1/2 − iχ(−∞,0) (τ )|τ |1/2 Proof. It suffices to verify that (a) On [0, T ], |Lf (x, t)| + |∂t Lf (x, t)| ≤ chxi−N , with c = c(f, N, T ). (b) Lf (x, 0) = 0 3
(c) (i∂t + ∂x2 )Lf (x, t) = 2δ0 (x)e 4 πi I−1/2 f (t) (a) is integration by parts in τ in (10) using −2(τ − i0)1/2 |x|−1 ∂τ [e−|x|(τ −i0) −|x|(τ −i0)1/2
e
1/2
]=
∞ . To show (b), note that since f ∈ C0,c (R+ ), fˆ(τ ) extends to an
analytic function on Im τ < 0 satisfying |fˆ(τ )| ≤ chτ i−k with c = c(f, k), and thus 1 lim Lf (x, 0) = 2π γ↑0 Since |e−|x|τ
1/2
Z
e−|x|τ
1/2
fˆ(τ ) dτ
(11)
Im τ =γ
| ≤ 1 for Im τ < 0, by Cauchy’s theorem, (11) = 0. (c) is a direct
computation from (10). Denote the operator defined by (10) as L2 f (x, t) and the one given by (5)-(6) as L1 f (x, t). Setting w = L1 f − L2 f , we have w(x, 0) = 0 and (i∂t + ∂x2 )w = 0. R Compute ∂t x |w|2 dx = 0, which yields w = 0, to complete the proof. Lemma 14. Suppose q, r ≥ 2 and
1 q
+
1 2r
= 14 .
(a) Space traces: If − 12 < s < 32 , then kθT (t)Lf (x, t)kC(Rt ;Hxs ) ≤ chT i1/4 kf k
16
2s+1 4 (R+ )
H0
.
(b) Time traces: If s ∈ R, then kLf k
2s+1 4 (R+ )) t
≤ ckf k
2s+1 4 (R+ )
.
H0
C(Rx ;H0
(c) Mixed norm: If 0 ≤ s ≤ 1, r 6= ∞, we have kLf kLqt Wxs,r ≤ ckf k
2s+1 4 (R+ )
.
H0
∞ (R+ ). Proof. By density, it suffices to establish these facts for f ∈ C0,c
By pairing (a) with φ(x) such that kφkH −s ≤ 1, we see that it suffices to show Z
t
t0 =0
0 2 f (t0 )θT (t)ei(t−t )∂x φ x=0 dt0 ≤ chT i1/4 kf k
H
2s+1 4
But LHS ≤ kχ(−∞,t) f (t0 )k
0
2
kθT (t)ei(t−t )∂x φ(x)k
2s+1 Ht0 4
−2s−1 4
≤ RHS
Ht0
by Lemmas 10(b) and 2. To establish the continuity statement, write θT (t2 )Lf (x, t2 )− Rt θT (t1 )Lf (x, t1 ) = t12 ∂t [θ(t)Lf (x, t)] dt. By ∂t L = L∂t and the bound just derived, we have kθT (t2 )Lf (x, t2 ) − θT (t1 )Lf (x, t1 )k ≤ c|t2 − t1 |kf k
2s+5 4
.
H0
(b) is immediate from Lemma 13, except that we should confirm that (under ∞ (R+ ), that ∂tk Lf (x, 0) = 0 for all k = 0, 1, 2, . . .. This, the assumption f ∈ C0,c
however, follows from ∂t L = L∂t . The continuity statement follows by using Rx Lf (x2 , t) − Lf (x1 , t) = x12 ∂x Lf (x, t) dx. From Lemma 13, we have Z 1 1/2 − 41 πi ∂x Lf (x, t) = e (sgn x) eitτ e−|x|(τ −i0) [I−1/2 f ]b(τ ) dτ 2π τ and thus kLf (x2 , t) − Lf (x1 , t)k
2s+1 4 (R+ )
H0
≤ c|x2 − x1 |kf k
2s+3 4
H0
To prove (c), it suffices to establish ≤ ckf kH˙ 1/4 kLf (x, t)kL4t L∞ x
(12)
k∂x Lf (x, t)kL4t L∞ ≤ ckf kH˙ 3/4 x
(13)
and
17
Indeed, the proof of (a) in the case s = 1 shows 2 ≤ ckf k ˙ 1/4 kLf (x, t)kL∞ H t Lx
2 ≤ ckf k ˙ 3/4 k∂x Lf (x, t)kL∞ H t Lx
Interpolate (12) with the first inequality and (13) with the second inequality to obtain kLf (x, t)kLqt Lrx ≤ ckf kH˙ 1/4
k∂x Lf (x, t)kLqt Lrx ≤ ckf kH˙ 3/4
for admissible q, r. This implies kLf (x, t)kLqt Wxs,r ≤ ckf k
H
2s+1 4
,
r 6= ∞
for s = 0 and s = 1. Now interpolate over s between these two endpoints to obtain the result as stated. 4/3
By pairing LHS of (12) against w(x, t) ∈ Lt L1x , we see that it suffices to show
Z Z
1/2 itτ −|x|(τ −i0)
e e w(x, t) dx dt
˙ −1/4 x
Writing out the Z
L2τ
x,t,y,s
t
H
norm, we see that it suffices to show
K(x, t, y, s) w(x, t)w(y, s) dx dt dy ds ≤ ckwkL4/3 L1 t
x
where Z K(x, t, y, s) =
|τ |−1/2 ei(t−s)τ e−|x|(τ −i0)
1/2
e−|y|(τ +i0)
1/2
dτ
τ
By a change of contour calculation, it follows that |K(x, y, t, s)| ≤ c|t − s|−1/2 , and hence (12) follows by the theorem on fractional integration. For (13), the kernel is instead Z K(x, t, y, s) = (sgn x)(sgn y)
|τ |−1/2 ei(t−s)τ e−|x|(τ −i0)
τ
and hence the estimation of |K| is identical. 18
1/2
e−|y|(τ +i0)
1/2
dτ
7
Existence: Proof of Theorem 1
First we prove the subcritical assertion (a) in the case 0 ≤ s < 12 . Select an ˜ H s ≤ 2kφkH s (R+ ) . Set r = extension φ˜ ∈ H s of φ such that kφk q=
4(α+1) . (α−1)(1−2s)
α+1 1+(α−1)s
and
2 This is an admissible pair with r ≥ 2 and q ≥ 2( 1−2s + 1). Set 2s+1 4
Z = C(Rt ; Hxs ) ∩ C(Rx ; Ht
) ∩ Lqt Wxs,r
Take w ∈ Z. By the chain rule (Lemma 5), for α ≥ 1 (see below for details) kDs (|w|α−1 w)kLq0
r0 4T Lx
for some σ > 0. 2 f (t) − θ2T (t)eit∂x φ˜
≤ cT σ kwkαLq
(14)
r,s 4T Wx
Note that by Lemmas 10(b), 11(b), 2, if w ∈ Z, then 2s+1 4
x=0
∈ H0
2s+1 4
α−1 (R+ )(0, t) ∈ H0 t ) and θ2T (t)D(w|w|
(R+ t ),
and the evaluation at x = 0 in these statements is understood in the sense of 2s+1 4
C(Rx ; Ht
). Let
2 2 Λw(t) = θT (t)eit∂x φ˜ + θT (t)L(f − θ2T ei·∂x φ˜ x=0 )(t)
(15)
− λθT (t)D(w|w|α−1 )(t) + λθT (t)L(θ2T D(w|w|α−1 ) x=0 )(t) so that, on [0, T ], (i∂t + ∂x2 )Λw = −λw|w|α−1 for x 6= 0 in the sense of distributions. By Lemmas 10, 11, 14 and (14), kΛwkZ ≤ ckφkH s (R+ ) + ckf k
H
2s+1 4 (R+ )
+ cT σ kwkαZ
(16)
In the sense of C(Rt ; Hxs ), we have Λw x, 0) = φ(x) on R, and in the sense of 2s+1 4
C(Rx ; Ht
), we have Λw(0, t) = f (t) on [0, T ]. We therefore look to solve
Λw = w for some selection of T . By the chain rule and product rule (see below for details), for α ≥ 2, α−1 kΛw1 − Λw2 kZ ≤ cT σ (kw1 kα−1 ZT + kw2 kZ )kw1 − w2 kZ
19
(17)
Now choose T small in terms of kφkH s (R+ ) and kf k
H
2s+1 4 (R+ )
, so that, by (16)
and (17), Λ is a contraction, which yields a unique fixed point u, which on [0, T ] solves the integral equation 2 2 u(t) = eit∂x φ˜ + L(f − ei·∂x φ˜ x=0 )
(18)
− λD(u|u|α−1 ) + λL(D(u|u|α−1 ) x=0 ) Let S be the set of all times T > 0 for which (1) ∃ u ∈ Z such that u solves (18) on [0, T ] and (2) for each pair u1 , u2 ∈ Z, such that u1 solves (18) on [0, T1 ] with T1 ≤ T and u2 solves (18) on [0, T2 ] with T2 ≤ T , we have u1 = u2 on [0, min(T1 , T2 )]. We claim that T as given in the above contraction argument is in S. We need only show condition (2). But the integral equation (18) has a unique solution by the contraction argument in the space LqTm Wxs,r , where Tm = min(T1 , T2 ), by Lemmas 10(c), 11(c), 14(c) and the fact that χ[0,Tm ] Lg = χ[0,Tm ] L(θTm g), χ[0,Tm ] Dw = χ[0,Tm ] Dχ[0,Tm ] w. Let T ∗ = sup S. Define u∗ on [0, T ∗ ) by setting, for t < T ∗ , u∗ (t) = u(t) for some u ∈ Z whose existence is given by condition (1); this is well-defined by condition (2). Suppose T ∗ < ∞ and limt↑T ∗ ku(·, t)kH s (R+ ) 6= ∞. Then ∃ a and a sequence tn → T ∗ such that ku∗ (tn )kH s (R+ ) ≤ a. By the above existence argument applied at time tn for n sufficiently large, we obtain a contradiction, as follows. We shall select T = tn for n sufficiently large in a moment. We have, by assumption, u1 ∈ Z solving the integral equation 2 2 u1 (t) = eit∂x φ˜ + L(f − ei·∂x φ˜ x=0 )
(19)
− λD(u1 |u1 |α−1 ) + λL(D(u1 |u1 |α−1 ) x=0 ) on [0, T ]. Apply the above existence argument to obtain u2 ∈ Z solving, on 20
[T, T + δ], the integral equation 2 2 u2 (t) = ei(t−T )∂x u(T ) + LT (f − ei(·−T )∂x u(T ) x=0 ) − λDT (u2 |u2 |α−1 ) + λLT (DT (u2 |u2 |α−1 ) x=0 )
(20)
where LT g(t) = (g(· + T ))(t − T ) Since δ = δ(a, kf k
H
2s+1 4 (R+ )
DT v(t) = D(v(· + T ))(t − T )
), we can select n sufficiently large so that T + δ =
tn + δ > T ∗ . Now we show that we can concatenate these two integral equations. Define u(t) = u1 (t) for −∞ < t ≤ T and u(t) = u2 (t) for T ≤ t < +∞. Then clearly u ∈ Lqt Wxr,s ∩ C(Rt ; Hxs ). Evaluate (19) at t = T , substitute into (20), and apply the two identities 2 2 Lg(t) = ei(t−T )∂x Lg(T ) − LT (g − ei(·−T )∂x Lg(T ) x=0 )(t) for t ≥ T i(t−T )∂x2
Dv(t) = e
T
Dv(T ) + D v(t)
(21)
for all t
2 with v(t) = −λu|u|α−1 (t) and g(t) = f (t)−eit∂x φ˜ x=0 −Dv(0, t) on 0 ≤ t ≤ T +δ. This establishes that u solves, on [0, T + δ], the integral equation (18). Next, we show that u ∈ C(Rx ; H ψ(t) = 1 for
T 2
2s+1 4
). Let ψ ∈ C ∞ such that ψ(t) = 0 for t ≤ 0,
≤ t ≤ T + 2δ , ψ(t) = 0 for t > T + δ. It is clear from the definition 2s+1 4
of u that (1 − ψ)u ∈ C(Rx ; Ht
). Since by (18)
2 2 ψ(t)u(t) = ψ(t)eit∂x φ˜ + ψ(t)L(f − θ2(T +δ) ei·∂x φ˜ x=0 )(t)
− λψ(t)D(u|u|α−1 )(t) + λψ(t)L(θ2(T +δ) D(u|u|α−1 ) x=0 ) 2s+1 4
by Lemmas 10(b), 11(b), and 14(b) we have ψu ∈ C(Rx ; Ht
).
Next, we need to verify condition (2) in the definition of f . Now suppose u is a solution on [0, Tu ] with Tu ≤ T + δ, and v is a solution on [0, Tv ] with 21
Tv ≤ T + δ, and suppose min(Tu , Tv ) ≥ T ∗ . Then u(t) = v(t) for all t ≤ T (since T ∈ S). Then, again by (21), u solves (20) with u2 replaced by u, and v solves (20) with u2 replaced by v (u(T ) = v(T )). By uniqueness of the fixed point to (20) in Lq[T,T +δ] Wxs,r , we get that u(t) = v(t) on [T, T + δ]. We have thus established that sup S ≥ T + δ > T ∗ , which is a contradiction, so in fact limt↑T ∗ ku(·, t)kH s (R+ ) = ∞ if T ∗ < ∞. Now we move on the continuity claim. Suppose (φ, f ) gives a solution u of (18) on [0, T ∗ ), and consider (φ1 , f1 ) with kφ−φ1 kH s (R+ ) +kf −f1 k
H
2s+1 4 (R+ )
< δ.
Fix T < T ∗ . Let u1 be the solution corresponding to (φ1 , f1 ) on [0, T1 ], where T1 is the first time t such that ku1 kLq[0,t] Wxs,r = 2kukLqT Wxs,r . We claim that T1 > T provided we take δ sufficiently small. Indeed, taking the difference of the two integral equations, we find, for t ≤ min(T1 , T ) ku − u1 kLq[0,t] Wxs,r ≤ cδ + c(kukLqT Wxs,r + ku1 kLqT
1
Wxs,r )ku
− u1 kLq1
[0,t]
Wxs,r
where q1 < q, and c depends only upon operator norms. This gives, by Lemma 4, ku − u1 kLq[0,t] Wxs,r ≤ cδ
(22)
where now c depends on f , φ, and T . Now if T1 < T , then take t = T1 in (22) and δ sufficiently small to obtain a contradiction. The inequality (22) plus estimates on the difference of the integral equations for u and u1 also shows ku − u1 kC([0,T ]; Hxs ) + ku − u1 k
2s+1 4 (0,T ))
C(Rx ; H
≤ cδ
Now we remark on the proof in the subcritical case (a) for 2s+1 4
Z = C(Rt ; Hxs ) ∩ C(Rx ; Ht
22
)
1 2
< s < 32 . Let
Set r = 2, q = ∞ in the remainder of the argument above. Do note, how 2s+1 2 ever, that to show f (t) − θ2T (t)eit∂x φ˜ x=0 ∈ H 4 (R+ t ), we need to appeal to the compatibility condition f (0) = φ(0) and Lemma 3. Also, by Lemma 3, 2s+1 4
θ2T (t)D(w|w|α−1 )(0, t) ∈ H0
(R+ t )
Now we discuss the critical case (b). Let Z = Lqt Wxs,r with r = q=
4(α+1) . (α−1)(1−2s)
α+1 1+(α−1)s
and
The integral equation is
2 2 Λw(t) = θT (t)eit∂x φ˜ + θT (t)L(f − θ2T ei·∂x φ˜ x=0 )(t)
(23)
− λθT (t)D(w|w|α−1 )(t) + λθT (t)L(θ2T D(w|w|α−1 ) x=0 )(t) ˜ Lq W s,r → 0 as T ↓ 0 and kθT (t)L(f − Now, because q 6= ∞, kθT (t)eit∂x φk x t 2 θ2T ei·∂x φ˜ x=0 )(t)kLqt Wxr,s → 0 as T ↓ 0. Therefore, ∃ T > 0 such that 2
2 ˜ Lq W s,r + kθT (t)L(f − θ2T ei·∂x2 φ˜ )(t)kLq W s,r < δ kθT (t)eit∂x φk x x x=0 t t which gives kΛwkZ ≤ δ + ckwkαZ
(24)
For δ sufficiently small, there will be a fixed point in the space { w ∈ Z | kwkZ < 2δ }. From Λu = u, (23) and Lemmas 10(a), 11(a), 14(a), we can recover the bounds in C(Rt ; Hxs ), and by Lemmas 10(b), 11(b), 14(b), we can recover the 2s+1 4
bounds in C(Rx ; Ht
). Let T ∗ be the supremum of all existence times with a
uniqueness stipulation, as before. We are not able to show the blowup statement in this case. Moreover, we also can only establish the continuity assertion for some T < T ∗ .
23
7.1
Notes on applying the chain and product rule
We shall apply the chain rule (Lemma 5) with w : R → C and F : C → C given by F (w) = |w|α−1 w, for α ≥ 1. Then α−3 2 α−1 α−3 (α − 1)|w| (Re w) + |w| (α − 1)|w| (Re w)(Im w) F 0 (w) = α−3 α−3 2 α−1 (α − 1)|w| (Re w)(Im w) (α − 1)|w| (Im w) + |w| and consequently each component of F 0 (w) is bounded by |w|α−1 . Thus kDs |w|α−1 wkLrx0 ≤ cαk|w|α−1 kLrx00 kDs wkLrx where
1 r00
=
1 r0
− 1r = 1 − 2r and
1 so that (α−1)r 00 =
1 r
− s and
1 q 00
=
1 (α−1)q 00
1 q0
− 1q = 1 − 2q . Since r, q have been selected
> 1q , we have
kDs |w|α−1 wkLrx0 ≤ ckDs wkαLrx To handle differences, for w0 , w1 : R → C, set wθ = θw1 + (1 − θ)w0 . Then Z 1 α−1 α−1 |w1 | w1 − |w0 | w0 = (α − 1)|wθ |α−3 wθ (wθ ◦ (w1 − w0 ))+ |wθ |α−1 (w1 − w0 ) θ=0
where z1 ◦ z2 = (Re z1 )(Re z2 ) + (Im z1 )(Im z2 ). To this, apply Ds , and invoke the product rule (Lemma 6) and the chain rule (Lemma 5).
8
Uniqueness: Proof of Prop. 1
We shall begin by establishing uniqueness of a distributional solution with weak traces for the linear problem for s ≥ 0. Given two solutions u1 , u2 , consider the difference v = u1 − u2 . We are thus assuming v ∈ C([0, T ∗ ); L2 (R+ )) with v(x, 0) = 0 24
(25)
and lim kv(x, ·)kL2(0,T ) = 0
x→0+
(26)
Take T < T ∗ . Let θ(t) be a nonnegative smooth function supported on [−2, −1] R with θ = 1. Let θδ (t) = δ −1 θ(δ −1 t). For δ, > 0, let ZZ vδ, (x, t) = v(y, s)θδ (x − y)θ (t − s) dy ds (27) which defines, in the sense of distributions, vδ, (x, t) a smooth function on −δ < x < +∞, − < t < T − 2. Owing to the assumption (25) we can write Z Z v(y, s)θδ (x − y) dy ds vδ, (x, t) = θ (t − s) y
s
where the integrals are defined in the usual sense. From this it follows that kvδ, (·, t)kL2 (R+x ) ≤
sup t+≤s≤t+2
kv(·, s)kL2 (R+x )
Owing to the assumption (26), ∃ L > 0 such that sup0<x≤2L kv(x, ·)kL2(0,T ) ≤ 1. It follows that, for x + 2δ < 2L, (27) can be written Z Z v(y, s)θ (t − s) ds dy vδ, (x, t) = θδ (x − y) s
y
where the integrals are understood in the usual sense, and we also have kvδ, (x, ·)kL2 (0,T ) ≤
sup
kv(y, ·)kL2 (,T +2)
(28)
x+δ
Department of Mathematics, University of California Berkeley, CA 94720 1 2
Abstract We prove, by adapting the method of Colliander-Kenig [9], local wellposedness of the initial-boundary value problem for the one-dimensional nonlinear Schr¨odinger equation i∂t u + ∂x2 u + λu|u|α−1 = 0 on the half-line under low boundary regularity assumptions.
1 2
AMS subject code: 35Q55 The content of this article appears as part of the author’s Ph.D. thesis at the University
of Chicago.
1
1
Introduction
We consider the initial-boundary value problem on the right half-line for the one-dimensional nonlinear Schr¨odinger (1D NLS) equation i∂t u + ∂x2 u + λu|u|α−1 = 0 for (x, t) ∈ (0, +∞) × (0, T ) u(0, t) = f (t) for t ∈ (0, T ) u(x, 0) = φ(x) for x ∈ (0, +∞)
(1)
where λ ∈ C. ˙ On R, we define the homogeneous L2 -based Sobolev spaces H˙ s = H(R) by 2 ˆ the norm kφkH˙ s = k|ξ|s φ(ξ)k L2ξ and the L -based inhomogeneous Sobolev spaces 2 1/2 ˆ H s = H s (R) by the norm kφkH s = khξis φ(ξ)k . In L2ξ , where hξi = (1 + |ξ| )
addition, we shall need L2 -based inhomogeneous Sobolev spaces on the half-line R+ = (0, +∞), which we denote H s (R+ ). These are defined, for s ≥ 0, as: ˜ φ ∈ H s (R+ ) if ∃ φ˜ ∈ H s (R) such that φ(x) = φ(x) for a.e. x > 0; in this ˜ H s (R) . We also similarly define, for s ≥ 0, case we set kφkH s (R+ ) = inf φ˜ kφk ˜ φ ∈ H s (0, L) if ∃ φ˜ ∈ H s (R) such that φ(x) = φ(x) a.e. on (0, L); in this case ˜ Hs . we set kφkH s (0,L) = inf φ˜ kφk The local smoothing inequality of [15] for the 1D Schr¨odinger group is 2
keit∂x φk
2s+1 4
˙ L∞ x Ht
≤ ckφkH˙ s
This inequality is sharp in the sense that
2s+1 4
cannot be replaced by any
higher number. We are thus motivated to consider initial-boundary data pairs (φ(x), f (t)) ∈ H s (R+ x)×H
2s+1 4
(R+ t ) and inclined to consider this configuration
optimal in the scale of L2 -based Sobolev spaces. Note that the trace map φ → φ(0) is well-defined on H s (R+ ) when s > 12 . 2
Thus, if s > 12 , then
2s+1 4
>
1 2
and both φ(0) and f (0) are well-defined quantities.
Since φ(0) and f (0) are both meant to represent u(0, 0), they must agree. Therefore, we consider (1) for 0 ≤ s < φ ∈ H s (R+ ), f ∈ H
2s+1 4
3 2
in the setting
(R+ ), and if
1 2
< s < 32 , φ(0) = f (0)
(2)
The solutions we construct shall have the following characteristics. Definition 1. u(x, t) will be called a distributional solution of (1), (2) on [0, T ∗ ) with strong traces if (a) u belongs to a space X with the property that u ∈ X implies u|u|α−1 is defined as a distribution. (b) u(x, t) satisfies the equation (1) in the sense of distributions on the set (x, t) ∈ (0, +∞) × (0, T ∗ ). (c) Space traces: ∀ T < T ∗ , we have u ∈ C([0, T ]; Hxs ) and u(·, 0) = φ in H s (R+ ). (d) Time traces: ∀ T < T ∗ , we have u ∈ C(Rx ; H in H
2s+1 4
2s+1 4
(0, T )) and u(0, ·) = f
(0, T ).
For the purposes of uniqueness in the high regularity setting s > 12 , we can consider a weaker notion of solution. Definition 2. u(x, t) will be called a distributional solution of (1), (2) on [0, T ∗ ) with weak traces if it satisfies conditions (a), (b) of Definition 1 and (c) One-sided space traces: ∀ T < T ∗ , we have u ∈ C([0, T ]; H s (R+ x )) and u(·, 0) = φ in H s (R+ ). 3
(d) Boundary values: ∀ T < T ∗ , we have lim ku(x, ·) − f k
H
x↓0
2s+1 4 (0,T )
= 0.
So that we may, at a later time, properly address the matter of uniqueness in the low regularity s
12 , u is a distributional solution of (1), (2) with weak traces if and only if it is a mild solution; in this case u is unique. Our main result is the following existence statement. Theorem 1. (a) Subcritical: Suppose 0 ≤ s < 12 , and 2 ≤ α < 4
5−2s 1−2s
or 1 2
< s < 32 , and 2 ≤ α < ∞
Then ∃ T ∗ > 0 and u that is both a mild solution and a distributional solution with strong traces of (1),(2) on [0, T ∗ ). If T ∗ < ∞, then limt↑T ∗ ku(·, t)kHxs = ∞. Also, ∀ T < T ∗ , ∃ δ0 = δ0 (s, T, φ, f ) > 0 such that if 0 < δ ≤ δ0 and kφ − φ1 kH s (R+ ) + kf − f1 k
H
2s+1 4 (R+ )
< δ then there is a solution u1 (as
above) on [0, T ], corresponding to (φ1 , f1 ), such that ku − u1 kC([0,T ]; Hxs ) + ku − u1 k
C(Rx ; H
2s+1 4 (0,T ))
≤ cδ, with c = c(s, T, f, φ).
(b) Critical: Suppose 0 ≤ s
0 maximal and u
that is both a mild solution and a distributional solution with strong traces of (1),(2) on [0, T ∗ ). Also, ∃ T = T (s, φ, f ) < T ∗ and ∃ δ0 = δ0 (s, φ, f ) > 0 such that if 0 < δ ≤ δ0 and kφ − φ1 kH s (R+ ) + kf − f1 k
H
2s+1 4 (R+ )
< δ then
there is a solution u1 (as above) on [0, T ], corresponding to (φ1 , f1 ), such that ku − u1 kC([0,T ]; Hxs ) + ku − u1 k
C(Rx ; H
2s+1 4 (0,T ))
≤ cδ, with c = c(s, f, φ).
Note that in (b), we may not have blow-up in the norm ku(·, t)k as t ↑ T ∗ . The proof of Theorem 1 involves the introduction of a boundary forcing operator analogous to that introduced by [9] in their treatment of the generalized Korteweg de-Vries equation (gKdV) on the half-line, and incorporates the techniques of the standard proof of local well-posedness of the corresponding initial-value problem based on the Strichartz estimates (see [7]). One could also consider the left half-line problem i∂t u + ∂x2 u + λu|u|α−1 = 0 for (x, t) ∈ (−∞, 0) × (0, T ) u(0, t) = f (t) for t ∈ (0, T ) u(x, 0) = φ(x) for x ∈ (−∞, 0) 5
although this is actually identical to the right half-line problem (1) by the transformation u(x, t) → u(−x, t). We plan, in a future publication, to examine the initial-boundary value problem for the line-segment i∂t u + ∂x2 u + λu|u|α−1 = 0 for (x, t) ∈ (0, L) × (0, T ) u(0, t) = f1 (t) for t ∈ (0, T ) for t ∈ (0, T )
u(L, t) = f2 (t) u(x, 0) = φ(x)
for x ∈ (0, L)
and consider global existence questions for the half-line and line-segment problems. We now briefly mention some earlier work and alternate perspectives on this problem and related problems. The main new feature of our work is the low regularity requirements for φ and f . Under higher regularity assumptions, more general results are already available. [18] considered a bounded or unbounded domain Ω ⊂ Rn with smooth boundary ∂Ω, and proved global existence of solutions to i∂t u + ∆u + λu|u|α−1 = 0 for (x, t) ∈ Ω × (0, T ) u(x, t) = f (x, t) for x ∈ ∂Ω u(x, 0) = φ(x) for x ∈ Ω
(3)
where f ∈ C 3 (∂Ω) is compactly supported, φ ∈ H 1 (Ω), and λ < 0. This solution is obtained as a limit of solutions to approximate problems after several a priori identities have been established. Earlier, [6] and [5] had obtained solutions to (1) for α > 3, λ < 0 and α = 3, λ ∈ R for φ ∈ H 2 (R+ ) and f ∈ C 2 (0, T ), using
6
semigroup techniques and a priori estimates. The problem (3) with f = 0 had been considered previously ([4] [22] [20] [21] [23]). [10] in the integrable case α = 3, λ = ±2 with φ Schwartz and f sufficiently smooth, obtained a solution to (1) by reformulating the problem as a 2×2 matrix Riemann-Hilbert problem. In this setting, [3] obtain an explicit representation for ∂x u(0, t). Outline: In §2, we discuss some notation, introduce function spaces and recall some needed properties of these function spaces. In §3, we review the definition and basic properties of the Riemann-Liouville fractional integral. In §4, 5, we state the needed estimates for the group and inhomogeneous solution operator. In §6, we define the boundary forcing operator, adapted from [9], and prove the needed estimates for it. In §7, we prove Theorem 1. In §8, we prove Prop. 1.
2
Notations and some function space properties
Let χS denote the characteristic function for the set S. We shall write LqT to R ˆ mean Lq ([0, T ]). Set φ(ξ) = x e−ixξ φ(x) dx. Define (τ − i0)α as the limit, in the sense of distributions, of (τ + iγ)−α as γ ↑ 0. Let hξis = (1 + |ξ|2 )s/2 . Let s f (ξ) = |ξ|s fˆ(ξ). The homogeneous L2 -based Sobolev spaces are H d ˙ s (R) = D
(−∂ 2 )−s/2 L2 (R) and the inhomogeneous L2 -based H s (R) = (1 − ∂ 2 )−s/2 L2 (R). We also set, for 1 ≤ p ≤ ∞, W s,p = (I − ∂ 2 )−s/2 Lp . We use the notation H s to mean H s (R) (and not H s (R+ ) or H0s (R+ )). The trace operator φ 7→ φ(0) is defined for φ ∈ H s (R) when s > 21 . For s ≥ 0, define φ ∈ H s (R+ ) if ∃ φ˜ ∈ H s (R) 7
˜ H s (R) . ˜ such that φ(x) = φ(x) for x > 0; in this case we set kφkH s (R+ ) = inf φ˜ kφk ˜ For s ≥ 0, define φ ∈ H0s (R+ ) if, when φ(x) is extended to φ(x) on R by setting ˜ ˜ H s (R) . φ(x) = 0 for x < 0, then φ˜ ∈ H s (R); in this case we set kφkH0s (R+ ) = kφk Define φ ∈ C0∞ (R+ ) if φ ∈ C ∞ (R) with supp φ ⊂ [0, +∞) (so that, in particular, ∞ φ and all of its derivatives vanish at 0), and C0,c (R+ ) as those members of ∞ C0∞ (R+ ) with compact support. We remark that C0,c (R+ ) is dense in H0s (R+ )
for all s ∈ R. We shall take a fixed θ ∈ Cc∞ (R) such that θ(t) = 1 on [−1, 1] and supp θ ⊂ [−2, 2]. Denote by θT (t) = θ(tT −1 ). Lemma 1 ([9] Lemma 2.8). If 0 ≤ α < 12 , then kθT hkH α ≤ chT iα khkH˙ α , where c = c(α, θ). Lemma 2 ([13] Lemma 3.5). If − 21 < α < 12 , then kχ(0,+∞) f kH α ≤ ckf kH α , where c = c(α). Lemma 3 ([9] Prop. 2.4, [13] Lemma 3.7, 3.8). If
1 2
< α
0, i.e.
α−1 t+ ,f Γ(α)
1 = Γ(α)
Z
+∞
tα−1 f (t) dt
0
Integration by parts gives, for Re α > 0, that α+k−1 α−1 t+ t+ k = ∂t Γ(α) Γ(α + k) for all k ∈ N. This formula can be used to extend the definition (in the sense of distributions) of
tα−1 + Γ(α)
to all α ∈ C. In particular, we obtain tα−1 + = δ0 (t) Γ(α) α=0 9
A change of contour calculation shows that α−1 b 1 t+ (τ ) = e− 2 πiα (τ − i0)−α Γ(α) where (τ − i0)−α is the distributional limit. If f ∈ C0∞ (R+ ), we define tα−1 Iα f = + ∗ f Γ(α) Thus, when Re α > 0, Z t 1 Iα f (t) = (t − s)α−1 f (s) ds Γ(α) 0 Rt and I0 f = f , I1 f (t) = 0 f (s) ds, and I−1 f = f 0 . Also Iα Iβ = Iα+β , which follows from the Fourier transform formula. For further details on the distribution tα−1 + , Γ(α)
see [11].
Lemma 7. If h ∈ C0∞ (R+ ), then Iα h ∈ C0∞ (R+ ), for all α ∈ C. Lemma 8 ([12]). If 0 ≤ α < +∞ and s ∈ R, then kI−α hkH0s (R+ ) ≤ ckhkH0s+α (R+ ) Lemma 9 ([12]). If 0 ≤ α < +∞, s ∈ R, µ ∈ C0∞ (R) kµIα hkH0s (R+ ) ≤ ckhkH0s−α (R+ ) where c = c(µ).
4
Estimates for the group
Set it∂x2
e
1 φ(x) = 2π
Z
2 ˆ dξ eixξ e−itξ φ(ξ)
ξ
10
(4)
so that
(i∂t + ∂x2 )eit∂x2 φ = 0 for (x, t) ∈ R × R eit∂x2 φ(x) = φ(x) for x ∈ R t=0
Lemma 10. Let s ∈ R. If φ ∈ H s (R), then 2
(a) Space traces: keit∂x φ(x)kC(Rt ; Hxs ) ≤ ckφkH s . 2
(b) Time traces: kθT (t)eit∂x φ(x)k
2s+1 4 )
C(Rx ; Ht
1 q
(c) Mixed-norm: If 2 ≤ q, r ≤ ∞ and
+
≤ chT i1/4 kφkH s . 1 2r
2
= 14 , then keit∂x φ(x)kLqt Wxs,r ≤
ckφkH s . Proof. (a) is clear from (4). (b) was obtained in [15]. (c) was obtained by [19] (see also [14]).
5
Estimates for the Duhamel inhomogeneous solution operator
Let Z Dw(x, t) = −i
t
0
2
ei(t−t )∂x w(x, t0 ) dt0
0
Then
(i∂t + ∂x2 )Dw(x, t) = w(x, t) for (x, t) ∈ R × R Dw(x, 0) = 0
for x ∈ R
Lemma 11. Suppose 2 ≤ q, r ≤ ∞ and
1 q
+
1 2r
= 14 , then
(a) Space Traces: If s ∈ R, then kDwkC(Rt ;Hxs ) ≤ ckwkLq0 W s,r0 . t
(b) Time Traces: If − 32 < s < 12 , then kθT (t)Dw(x, t)k 11
x
2s+1 4 )
C(Rx ; Ht
≤ chT i1/4 kwkLq0 W s,r0 . t
x
(c) Mixed-norm: If s ∈ R, then kDwkLqt Wxs,r ≤ ckwkLq0 W s,r0 . x
t
Proof. (a) and (c) are due to [19] (see also [14]). We now prove (b), following the techniques of Theorem 2.3 in [16]. We use the representation Z i +∞ 0 2 Dw(x, t) = − (sgn t0 )ei(t−t )∂x w(x, t0 ) dt0 2 −∞ Z Z 1 1 ˆ τ) itτ ixξ w(ξ, + lim e e dξ dτ →0+ 2π |τ +ξ 2 |> 2πi τ τ + ξ2 = I + II and Term II can also be written Z 1 II = eitτ [m(·, τ ) ∗ wˆ t (·, τ )](x) dτ 2π τ where w ˆ t (·, τ ) denotes the Fourier transform of w(x, t) in the t-variable alone and exp(−|x||τ |1/2 ) 1 sin(|x||τ |1/2 ) 1 + χ (τ ) m(x, τ ) = − χ(0,+∞) (τ ) (−∞,0) 2 |τ |1/2 2 |τ |1/2 First we treat Term I for all s and all admissible pairs q, r. Pairing Term I with f (x, t) such that kf k
− 2s+1 4
L1x Ht
≤ 1, we are left to show that
Z
(sgn t0 )e−it0 ∂x2 w(x, t0 ) dt0
0
t
Hxs
≤ ckwkLq0 W s,r0 t
x
and
Z
θT (t)e−it∂x2 f (x, t) dt
Hx−s
t
≤ ckf k
− 2s+1 4
L1x Ht
The first of these follows from the proof of (a), while the second is obtained by duality and Lemma 10(b). We address Term II separately for r0 = 2, q 0 = 1, and r0 = 1, q 0 = 43 ; the intermediate cases follow by interpolation. For the case 12
r0 = 2, q 0 = 1, we use the first representation of Term II with Lemma 1, the change of variable η = −ξ 2 , and L2 -boundedness of the Hilbert transform on A2 -weighted spaces, to obtain Z kθT (t)(Term II)k
2s+1 4
s
≤c
2
1/2
|ξ| |w(ξ, ˆ τ )| dξ
Ht
ξ
Z
s
≤c
2 !1/2 |wˆ (ξ, t)| dt dξ
Z
x
|ξ| ξ
t
where w ˆ x (ξ, t) denotes the Fourier transform in the x-variable alone. Complete the bound by applying Minkowskii’s integral inequality and the Placherel theorem. The validity of this step is restricted to − 23 < s < 12 . We shall only prove the r0 = 1, q 0 =
4 3
case for s = 0. Note that by the
second representation for Term II, k(Term II)kL∞ H 1/4 is x
Z Z
−1/2
|τ | τ
t
Z
m(x − y, τ )wˆ (y, τ ) dy
y
t
|τ |−1/2 m(x − z, τ )wˆ t (z, τ ) dz dτ
z
which is equivalent to Z K(y, s, z, t)w(y, s)w(z, t) dy ds dz dt y,s,z,t
where Z K(y, s, z, t) =
|τ |1/2 e−i(s−t)τ m(x − y, τ )m(x − z, τ ) dτ
τ
From the definition of m, we see that |K(y, s, z, t)| ≤ c|s − t|−1/2 . We conclude by applying the theorem on fractional integration (see Theorem 1 of Chapter V in [17]).
13
6
Estimates for the Duhamel boundary forcing operator
For f ∈ C0∞ (R+ ), define the boundary forcing operator Z t 0 2 i 14 π Lf (x, t) = 2e ei(t−t )∂x δ0 (x)I−1/2 f (t0 ) dt0 Z t0 1 ix2 0 −1/2 =√ (t − t ) exp I−1/2 f (t0 ) dt0 4(t − t0 ) π 0
(5) (6)
The equivalence of the two definitions is evident from the formula −i π sgn t 2 b e 4 ix 1 2 √ exp (ξ) = e−itξ 1/2 4t 2 π |t| From these two definitions, we see that 3 (i∂t + ∂x2 )Lf (x, t) = 2ei 4 π δ0 (x)I−1/2 f (t) for (x, t) ∈ R × R Lf (x, 0) = 0 for x ∈ R Lf (0, t) = f (t) for t ∈ R We now establish some continuity properties of Lf (x, t) when f is suitably nice. ∞ Lemma 12. Let f ∈ C0,c (R+ ).
(a) For fixed t, Lf (x, t) is continuous in x for all x ∈ R and ∂x Lf (x, t) is continuous in x for x 6= 0 with 1
lim ∂x Lf (x, t) = e− 4 πi I−1/2 f (t) x↑0
1
lim ∂x Lf (x, t) = −e− 4 πi I−1/2 f (t) x↓0
(7) (b) ∀ k = 0, 1, 2, . . . and for fixed x, ∂tk Lf (x, t) is continuous in t for all t ∈ R.
14
We also have the pointwise estimates, for k = 0, 1, 2, . . ., on [0, T ], |∂tk Lf (x, t)| + |∂x Lf (x, t)| ≤ chxi−N where c = c(f, N, k, T ). Proof. Let us denote “integration by parts” by IBP. It is clear from (6) and dominated convergence that, for fixed t, Lf (x, t) is continuous in x, and for 1
fixed x, Lf (x, t) is continuous in t. Let h = 2ei 4 π I−1/2 f ∈ C0∞ (R+ ) (by Lemma Rt 0 7) and φ(ξ, t) = 0 e−i(t−t )ξ h(t0 ) dt0 . By IBP in t0 , |∂ξk φ(ξ, t)| ≤ chξi−k−1 , where c = c(h, k, T ), and thus |∂ξk φ(ξ 2 , t)| ≤ chξi−k−2
(8)
We have Z Lf (x, t) =
eixξ φ(ξ 2 , t) dξ
(9)
ξ 0
2
and by IBP in ξ and (8), we have |Lf (x, t)| ≤ chxi−N . By ∂t [ei(t−t )∂x δ0 (x)] = 0
2
−∂t0 [ei(t−t )∂x δ0 (x)] and IBP in t0 in (5), ∂t Lf = L∂t f , and thus, for fixed x, 0
2
∂tk Lf (x, t) is continuous in t and |∂tk Lf (x, t)| ≤ chxi−N . By ∂x2 [ei(t−t )∂x δ0 (x)] = 0
3
2
i∂t0 [ei(t−t )∂x δ0 (x)] and IBP in t0 in (5), ∂x2 Lf (x, t) = 2ei 4 π δ0 (x)I−1/2 f (t) − iL(∂t f )(x, t). Hence i 43 π
∂x Lf (x, t) = e
Z
x
(sgn x)I−1/2 f (t) − i
L(∂t f )(x0 , t) dx0 + c(t)
x0 =0
Since all terms except c(t) are odd in x, we must have c(t) = 0. From this we obtain (7), and the bound |∂x Lf (x, t)| ≤ c. From (9), IBP in ξ and (8), we obtain that |∂x Lf (x, t)| ≤ c|x|−N . Combining the two previous bounds, we have |∂x Lf (x, t)| ≤ chxi−N . Now we provide an alternate representation of Lf (x, t). 15
∞ (R+ ). Then Lemma 13. Suppose f ∈ C0,c Z 1 1/2 Lf (x, t) = eitτ e−|x|(τ −i0) fˆ(τ ) dτ 2π τ
(10)
where 1
(τ − i0) 2 = χ(0,+∞) (τ )|τ |1/2 − iχ(−∞,0) (τ )|τ |1/2 Proof. It suffices to verify that (a) On [0, T ], |Lf (x, t)| + |∂t Lf (x, t)| ≤ chxi−N , with c = c(f, N, T ). (b) Lf (x, 0) = 0 3
(c) (i∂t + ∂x2 )Lf (x, t) = 2δ0 (x)e 4 πi I−1/2 f (t) (a) is integration by parts in τ in (10) using −2(τ − i0)1/2 |x|−1 ∂τ [e−|x|(τ −i0) −|x|(τ −i0)1/2
e
1/2
]=
∞ . To show (b), note that since f ∈ C0,c (R+ ), fˆ(τ ) extends to an
analytic function on Im τ < 0 satisfying |fˆ(τ )| ≤ chτ i−k with c = c(f, k), and thus 1 lim Lf (x, 0) = 2π γ↑0 Since |e−|x|τ
1/2
Z
e−|x|τ
1/2
fˆ(τ ) dτ
(11)
Im τ =γ
| ≤ 1 for Im τ < 0, by Cauchy’s theorem, (11) = 0. (c) is a direct
computation from (10). Denote the operator defined by (10) as L2 f (x, t) and the one given by (5)-(6) as L1 f (x, t). Setting w = L1 f − L2 f , we have w(x, 0) = 0 and (i∂t + ∂x2 )w = 0. R Compute ∂t x |w|2 dx = 0, which yields w = 0, to complete the proof. Lemma 14. Suppose q, r ≥ 2 and
1 q
+
1 2r
= 14 .
(a) Space traces: If − 12 < s < 32 , then kθT (t)Lf (x, t)kC(Rt ;Hxs ) ≤ chT i1/4 kf k
16
2s+1 4 (R+ )
H0
.
(b) Time traces: If s ∈ R, then kLf k
2s+1 4 (R+ )) t
≤ ckf k
2s+1 4 (R+ )
.
H0
C(Rx ;H0
(c) Mixed norm: If 0 ≤ s ≤ 1, r 6= ∞, we have kLf kLqt Wxs,r ≤ ckf k
2s+1 4 (R+ )
.
H0
∞ (R+ ). Proof. By density, it suffices to establish these facts for f ∈ C0,c
By pairing (a) with φ(x) such that kφkH −s ≤ 1, we see that it suffices to show Z
t
t0 =0
0 2 f (t0 )θT (t)ei(t−t )∂x φ x=0 dt0 ≤ chT i1/4 kf k
H
2s+1 4
But LHS ≤ kχ(−∞,t) f (t0 )k
0
2
kθT (t)ei(t−t )∂x φ(x)k
2s+1 Ht0 4
−2s−1 4
≤ RHS
Ht0
by Lemmas 10(b) and 2. To establish the continuity statement, write θT (t2 )Lf (x, t2 )− Rt θT (t1 )Lf (x, t1 ) = t12 ∂t [θ(t)Lf (x, t)] dt. By ∂t L = L∂t and the bound just derived, we have kθT (t2 )Lf (x, t2 ) − θT (t1 )Lf (x, t1 )k ≤ c|t2 − t1 |kf k
2s+5 4
.
H0
(b) is immediate from Lemma 13, except that we should confirm that (under ∞ (R+ ), that ∂tk Lf (x, 0) = 0 for all k = 0, 1, 2, . . .. This, the assumption f ∈ C0,c
however, follows from ∂t L = L∂t . The continuity statement follows by using Rx Lf (x2 , t) − Lf (x1 , t) = x12 ∂x Lf (x, t) dx. From Lemma 13, we have Z 1 1/2 − 41 πi ∂x Lf (x, t) = e (sgn x) eitτ e−|x|(τ −i0) [I−1/2 f ]b(τ ) dτ 2π τ and thus kLf (x2 , t) − Lf (x1 , t)k
2s+1 4 (R+ )
H0
≤ c|x2 − x1 |kf k
2s+3 4
H0
To prove (c), it suffices to establish ≤ ckf kH˙ 1/4 kLf (x, t)kL4t L∞ x
(12)
k∂x Lf (x, t)kL4t L∞ ≤ ckf kH˙ 3/4 x
(13)
and
17
Indeed, the proof of (a) in the case s = 1 shows 2 ≤ ckf k ˙ 1/4 kLf (x, t)kL∞ H t Lx
2 ≤ ckf k ˙ 3/4 k∂x Lf (x, t)kL∞ H t Lx
Interpolate (12) with the first inequality and (13) with the second inequality to obtain kLf (x, t)kLqt Lrx ≤ ckf kH˙ 1/4
k∂x Lf (x, t)kLqt Lrx ≤ ckf kH˙ 3/4
for admissible q, r. This implies kLf (x, t)kLqt Wxs,r ≤ ckf k
H
2s+1 4
,
r 6= ∞
for s = 0 and s = 1. Now interpolate over s between these two endpoints to obtain the result as stated. 4/3
By pairing LHS of (12) against w(x, t) ∈ Lt L1x , we see that it suffices to show
Z Z
1/2 itτ −|x|(τ −i0)
e e w(x, t) dx dt
˙ −1/4 x
Writing out the Z
L2τ
x,t,y,s
t
H
norm, we see that it suffices to show
K(x, t, y, s) w(x, t)w(y, s) dx dt dy ds ≤ ckwkL4/3 L1 t
x
where Z K(x, t, y, s) =
|τ |−1/2 ei(t−s)τ e−|x|(τ −i0)
1/2
e−|y|(τ +i0)
1/2
dτ
τ
By a change of contour calculation, it follows that |K(x, y, t, s)| ≤ c|t − s|−1/2 , and hence (12) follows by the theorem on fractional integration. For (13), the kernel is instead Z K(x, t, y, s) = (sgn x)(sgn y)
|τ |−1/2 ei(t−s)τ e−|x|(τ −i0)
τ
and hence the estimation of |K| is identical. 18
1/2
e−|y|(τ +i0)
1/2
dτ
7
Existence: Proof of Theorem 1
First we prove the subcritical assertion (a) in the case 0 ≤ s < 12 . Select an ˜ H s ≤ 2kφkH s (R+ ) . Set r = extension φ˜ ∈ H s of φ such that kφk q=
4(α+1) . (α−1)(1−2s)
α+1 1+(α−1)s
and
2 This is an admissible pair with r ≥ 2 and q ≥ 2( 1−2s + 1). Set 2s+1 4
Z = C(Rt ; Hxs ) ∩ C(Rx ; Ht
) ∩ Lqt Wxs,r
Take w ∈ Z. By the chain rule (Lemma 5), for α ≥ 1 (see below for details) kDs (|w|α−1 w)kLq0
r0 4T Lx
for some σ > 0. 2 f (t) − θ2T (t)eit∂x φ˜
≤ cT σ kwkαLq
(14)
r,s 4T Wx
Note that by Lemmas 10(b), 11(b), 2, if w ∈ Z, then 2s+1 4
x=0
∈ H0
2s+1 4
α−1 (R+ )(0, t) ∈ H0 t ) and θ2T (t)D(w|w|
(R+ t ),
and the evaluation at x = 0 in these statements is understood in the sense of 2s+1 4
C(Rx ; Ht
). Let
2 2 Λw(t) = θT (t)eit∂x φ˜ + θT (t)L(f − θ2T ei·∂x φ˜ x=0 )(t)
(15)
− λθT (t)D(w|w|α−1 )(t) + λθT (t)L(θ2T D(w|w|α−1 ) x=0 )(t) so that, on [0, T ], (i∂t + ∂x2 )Λw = −λw|w|α−1 for x 6= 0 in the sense of distributions. By Lemmas 10, 11, 14 and (14), kΛwkZ ≤ ckφkH s (R+ ) + ckf k
H
2s+1 4 (R+ )
+ cT σ kwkαZ
(16)
In the sense of C(Rt ; Hxs ), we have Λw x, 0) = φ(x) on R, and in the sense of 2s+1 4
C(Rx ; Ht
), we have Λw(0, t) = f (t) on [0, T ]. We therefore look to solve
Λw = w for some selection of T . By the chain rule and product rule (see below for details), for α ≥ 2, α−1 kΛw1 − Λw2 kZ ≤ cT σ (kw1 kα−1 ZT + kw2 kZ )kw1 − w2 kZ
19
(17)
Now choose T small in terms of kφkH s (R+ ) and kf k
H
2s+1 4 (R+ )
, so that, by (16)
and (17), Λ is a contraction, which yields a unique fixed point u, which on [0, T ] solves the integral equation 2 2 u(t) = eit∂x φ˜ + L(f − ei·∂x φ˜ x=0 )
(18)
− λD(u|u|α−1 ) + λL(D(u|u|α−1 ) x=0 ) Let S be the set of all times T > 0 for which (1) ∃ u ∈ Z such that u solves (18) on [0, T ] and (2) for each pair u1 , u2 ∈ Z, such that u1 solves (18) on [0, T1 ] with T1 ≤ T and u2 solves (18) on [0, T2 ] with T2 ≤ T , we have u1 = u2 on [0, min(T1 , T2 )]. We claim that T as given in the above contraction argument is in S. We need only show condition (2). But the integral equation (18) has a unique solution by the contraction argument in the space LqTm Wxs,r , where Tm = min(T1 , T2 ), by Lemmas 10(c), 11(c), 14(c) and the fact that χ[0,Tm ] Lg = χ[0,Tm ] L(θTm g), χ[0,Tm ] Dw = χ[0,Tm ] Dχ[0,Tm ] w. Let T ∗ = sup S. Define u∗ on [0, T ∗ ) by setting, for t < T ∗ , u∗ (t) = u(t) for some u ∈ Z whose existence is given by condition (1); this is well-defined by condition (2). Suppose T ∗ < ∞ and limt↑T ∗ ku(·, t)kH s (R+ ) 6= ∞. Then ∃ a and a sequence tn → T ∗ such that ku∗ (tn )kH s (R+ ) ≤ a. By the above existence argument applied at time tn for n sufficiently large, we obtain a contradiction, as follows. We shall select T = tn for n sufficiently large in a moment. We have, by assumption, u1 ∈ Z solving the integral equation 2 2 u1 (t) = eit∂x φ˜ + L(f − ei·∂x φ˜ x=0 )
(19)
− λD(u1 |u1 |α−1 ) + λL(D(u1 |u1 |α−1 ) x=0 ) on [0, T ]. Apply the above existence argument to obtain u2 ∈ Z solving, on 20
[T, T + δ], the integral equation 2 2 u2 (t) = ei(t−T )∂x u(T ) + LT (f − ei(·−T )∂x u(T ) x=0 ) − λDT (u2 |u2 |α−1 ) + λLT (DT (u2 |u2 |α−1 ) x=0 )
(20)
where LT g(t) = (g(· + T ))(t − T ) Since δ = δ(a, kf k
H
2s+1 4 (R+ )
DT v(t) = D(v(· + T ))(t − T )
), we can select n sufficiently large so that T + δ =
tn + δ > T ∗ . Now we show that we can concatenate these two integral equations. Define u(t) = u1 (t) for −∞ < t ≤ T and u(t) = u2 (t) for T ≤ t < +∞. Then clearly u ∈ Lqt Wxr,s ∩ C(Rt ; Hxs ). Evaluate (19) at t = T , substitute into (20), and apply the two identities 2 2 Lg(t) = ei(t−T )∂x Lg(T ) − LT (g − ei(·−T )∂x Lg(T ) x=0 )(t) for t ≥ T i(t−T )∂x2
Dv(t) = e
T
Dv(T ) + D v(t)
(21)
for all t
2 with v(t) = −λu|u|α−1 (t) and g(t) = f (t)−eit∂x φ˜ x=0 −Dv(0, t) on 0 ≤ t ≤ T +δ. This establishes that u solves, on [0, T + δ], the integral equation (18). Next, we show that u ∈ C(Rx ; H ψ(t) = 1 for
T 2
2s+1 4
). Let ψ ∈ C ∞ such that ψ(t) = 0 for t ≤ 0,
≤ t ≤ T + 2δ , ψ(t) = 0 for t > T + δ. It is clear from the definition 2s+1 4
of u that (1 − ψ)u ∈ C(Rx ; Ht
). Since by (18)
2 2 ψ(t)u(t) = ψ(t)eit∂x φ˜ + ψ(t)L(f − θ2(T +δ) ei·∂x φ˜ x=0 )(t)
− λψ(t)D(u|u|α−1 )(t) + λψ(t)L(θ2(T +δ) D(u|u|α−1 ) x=0 ) 2s+1 4
by Lemmas 10(b), 11(b), and 14(b) we have ψu ∈ C(Rx ; Ht
).
Next, we need to verify condition (2) in the definition of f . Now suppose u is a solution on [0, Tu ] with Tu ≤ T + δ, and v is a solution on [0, Tv ] with 21
Tv ≤ T + δ, and suppose min(Tu , Tv ) ≥ T ∗ . Then u(t) = v(t) for all t ≤ T (since T ∈ S). Then, again by (21), u solves (20) with u2 replaced by u, and v solves (20) with u2 replaced by v (u(T ) = v(T )). By uniqueness of the fixed point to (20) in Lq[T,T +δ] Wxs,r , we get that u(t) = v(t) on [T, T + δ]. We have thus established that sup S ≥ T + δ > T ∗ , which is a contradiction, so in fact limt↑T ∗ ku(·, t)kH s (R+ ) = ∞ if T ∗ < ∞. Now we move on the continuity claim. Suppose (φ, f ) gives a solution u of (18) on [0, T ∗ ), and consider (φ1 , f1 ) with kφ−φ1 kH s (R+ ) +kf −f1 k
H
2s+1 4 (R+ )
< δ.
Fix T < T ∗ . Let u1 be the solution corresponding to (φ1 , f1 ) on [0, T1 ], where T1 is the first time t such that ku1 kLq[0,t] Wxs,r = 2kukLqT Wxs,r . We claim that T1 > T provided we take δ sufficiently small. Indeed, taking the difference of the two integral equations, we find, for t ≤ min(T1 , T ) ku − u1 kLq[0,t] Wxs,r ≤ cδ + c(kukLqT Wxs,r + ku1 kLqT
1
Wxs,r )ku
− u1 kLq1
[0,t]
Wxs,r
where q1 < q, and c depends only upon operator norms. This gives, by Lemma 4, ku − u1 kLq[0,t] Wxs,r ≤ cδ
(22)
where now c depends on f , φ, and T . Now if T1 < T , then take t = T1 in (22) and δ sufficiently small to obtain a contradiction. The inequality (22) plus estimates on the difference of the integral equations for u and u1 also shows ku − u1 kC([0,T ]; Hxs ) + ku − u1 k
2s+1 4 (0,T ))
C(Rx ; H
≤ cδ
Now we remark on the proof in the subcritical case (a) for 2s+1 4
Z = C(Rt ; Hxs ) ∩ C(Rx ; Ht
22
)
1 2
< s < 32 . Let
Set r = 2, q = ∞ in the remainder of the argument above. Do note, how 2s+1 2 ever, that to show f (t) − θ2T (t)eit∂x φ˜ x=0 ∈ H 4 (R+ t ), we need to appeal to the compatibility condition f (0) = φ(0) and Lemma 3. Also, by Lemma 3, 2s+1 4
θ2T (t)D(w|w|α−1 )(0, t) ∈ H0
(R+ t )
Now we discuss the critical case (b). Let Z = Lqt Wxs,r with r = q=
4(α+1) . (α−1)(1−2s)
α+1 1+(α−1)s
and
The integral equation is
2 2 Λw(t) = θT (t)eit∂x φ˜ + θT (t)L(f − θ2T ei·∂x φ˜ x=0 )(t)
(23)
− λθT (t)D(w|w|α−1 )(t) + λθT (t)L(θ2T D(w|w|α−1 ) x=0 )(t) ˜ Lq W s,r → 0 as T ↓ 0 and kθT (t)L(f − Now, because q 6= ∞, kθT (t)eit∂x φk x t 2 θ2T ei·∂x φ˜ x=0 )(t)kLqt Wxr,s → 0 as T ↓ 0. Therefore, ∃ T > 0 such that 2
2 ˜ Lq W s,r + kθT (t)L(f − θ2T ei·∂x2 φ˜ )(t)kLq W s,r < δ kθT (t)eit∂x φk x x x=0 t t which gives kΛwkZ ≤ δ + ckwkαZ
(24)
For δ sufficiently small, there will be a fixed point in the space { w ∈ Z | kwkZ < 2δ }. From Λu = u, (23) and Lemmas 10(a), 11(a), 14(a), we can recover the bounds in C(Rt ; Hxs ), and by Lemmas 10(b), 11(b), 14(b), we can recover the 2s+1 4
bounds in C(Rx ; Ht
). Let T ∗ be the supremum of all existence times with a
uniqueness stipulation, as before. We are not able to show the blowup statement in this case. Moreover, we also can only establish the continuity assertion for some T < T ∗ .
23
7.1
Notes on applying the chain and product rule
We shall apply the chain rule (Lemma 5) with w : R → C and F : C → C given by F (w) = |w|α−1 w, for α ≥ 1. Then α−3 2 α−1 α−3 (α − 1)|w| (Re w) + |w| (α − 1)|w| (Re w)(Im w) F 0 (w) = α−3 α−3 2 α−1 (α − 1)|w| (Re w)(Im w) (α − 1)|w| (Im w) + |w| and consequently each component of F 0 (w) is bounded by |w|α−1 . Thus kDs |w|α−1 wkLrx0 ≤ cαk|w|α−1 kLrx00 kDs wkLrx where
1 r00
=
1 r0
− 1r = 1 − 2r and
1 so that (α−1)r 00 =
1 r
− s and
1 q 00
=
1 (α−1)q 00
1 q0
− 1q = 1 − 2q . Since r, q have been selected
> 1q , we have
kDs |w|α−1 wkLrx0 ≤ ckDs wkαLrx To handle differences, for w0 , w1 : R → C, set wθ = θw1 + (1 − θ)w0 . Then Z 1 α−1 α−1 |w1 | w1 − |w0 | w0 = (α − 1)|wθ |α−3 wθ (wθ ◦ (w1 − w0 ))+ |wθ |α−1 (w1 − w0 ) θ=0
where z1 ◦ z2 = (Re z1 )(Re z2 ) + (Im z1 )(Im z2 ). To this, apply Ds , and invoke the product rule (Lemma 6) and the chain rule (Lemma 5).
8
Uniqueness: Proof of Prop. 1
We shall begin by establishing uniqueness of a distributional solution with weak traces for the linear problem for s ≥ 0. Given two solutions u1 , u2 , consider the difference v = u1 − u2 . We are thus assuming v ∈ C([0, T ∗ ); L2 (R+ )) with v(x, 0) = 0 24
(25)
and lim kv(x, ·)kL2(0,T ) = 0
x→0+
(26)
Take T < T ∗ . Let θ(t) be a nonnegative smooth function supported on [−2, −1] R with θ = 1. Let θδ (t) = δ −1 θ(δ −1 t). For δ, > 0, let ZZ vδ, (x, t) = v(y, s)θδ (x − y)θ (t − s) dy ds (27) which defines, in the sense of distributions, vδ, (x, t) a smooth function on −δ < x < +∞, − < t < T − 2. Owing to the assumption (25) we can write Z Z v(y, s)θδ (x − y) dy ds vδ, (x, t) = θ (t − s) y
s
where the integrals are defined in the usual sense. From this it follows that kvδ, (·, t)kL2 (R+x ) ≤
sup t+≤s≤t+2
kv(·, s)kL2 (R+x )
Owing to the assumption (26), ∃ L > 0 such that sup0<x≤2L kv(x, ·)kL2(0,T ) ≤ 1. It follows that, for x + 2δ < 2L, (27) can be written Z Z v(y, s)θ (t − s) ds dy vδ, (x, t) = θδ (x − y) s
y
where the integrals are understood in the usual sense, and we also have kvδ, (x, ·)kL2 (0,T ) ≤
sup
kv(y, ·)kL2 (,T +2)
(28)
x+δ
