An Inverse Function Theorem in Sobolev Spaces and Applications to Quasi-Linear Schrodinger¨ Equations

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View metadata, citation and similar papers at core.ac.uk brought to you by CORE Journal of Mathematical Analysis and Applications 258, 146–170 (2001) provided by Elsevier - Publisher Connector doi:10.1006/jmaa.2000.7366, available online at http://www.idealibrary.com on

Markus Poppenberg

Fachbereich Mathematik, Universitat¨ Dortmund, D-44221 Dortmund, Germany E-mail: [email protected]

Submitted by John Horvath´

Received August 29, 2000

A Nash–Moser type inverse function theorem in Banach spaces with loss of derivatives is proved, and applications are given to singular quasi-linear Schrodinger¨ equations like the superﬂuid ﬁlm equation in plasma physics. Based on an implicit function theorem in Sobolev spaces, a linearization method is introduced for the local and global well-posedness of the Cauchy problem for nonlinear evolution equations. The technique compensates for a loss of derivatives in the linearized problem. This is illustrated by an application to strongly singular quasi-linear Schrodinger¨ equations where the nonlinearities include derivatives of second order. The local well-posedness of the Cauchy problem is proved in Sobolev spaces for arbitrary space dimension without assuming smallness assumptions on the initial value. Here the linearized problem is solved using hyperbolic semigroup theory, including evolution systems. © 2001 Academic Press Key Words: inverse function theorem; implicit function theorem; Nash–Moser; loss of derivatives; nonlinear evolution equation; quasi-linear Schrodinger¨ equation; semigroup theory; Cauchy problem; local well posedness; global well posedness.

1. INTRODUCTION

The purpose of this paper is to prove an inverse function theorem of the Nash–Moser type in Banach spaces and to give applications to nonlinear evolution equations. As an example, well-posedness results in Sobolev spaces are derived for strongly singular quasi-linear Schrodinger¨ equations. The problem of loss of derivatives was overcome in inverse function theorems of the Nash–Moser type in Fr´echet spaces [7, 16, 17, 19, 20, 26, 28, 29]. Hormander¨ proves in [8] an inverse function theorem with loss of derivatives for Holder¨ spaces. This result does not apply to Sobolev spaces k n a a 1−λa +λa H , because the interpolation property C 0 C 1 λ ∞ ⊂ C 0 1

146 0022-247X/01 $35.00 Copyright © 2001 by Academic Press All rights of reproduction in any form reserved. an inverse function theorem 147 of Holder¨ spaces (cf. [8], Theorem A.11, or [33], 2.7.2) used in [8] fails for spaces Hkn, where the interpolation space is a Besov space (cf. [33],

2.4). The related result [9] requires compact imbeddings Eb → Ea and thus does not apply to Hkn as well. The inverse function Theorem 2.8 proved in this paper applies to Sobolev spaces Hkn; the proof uses methods of [8]. Based on the implicit function Theorem 2.9, a linearization method is introduced for local and global well-posedness in Sobolev spaces of the Cauchy problem for nonlinear evolution equations. The well-posedness of the nonlinear problem is reduced to the solution of a linear problem including higher-order Sobolev norm estimates. Different from usual Banach space techniques the linearized problem may produce a loss of derivatives. As an application, we consider the quasi-linear Schrodinger¨ equation

2 2 2 iut =−u + V xu − f u u − g u g u u (1) where V , f , and g are smooth functions; note that the nonlinearity contains derivatives of second order. Such equations have been derived as models for several phenomena in plasma physics and fluid mechanics [15, 18, 24] and in the theory of Heisenberg ferromagnet [1, 11, 25, 30]. The case gs=s models the time evolution of the wave function in superfluid films [12, 13], and the case gs=1 + s1/2 describes the self–channeling of a high–power ultra short laser in matter [2, 3, 4, 5, 6, 27]. In the mathematical literature, few results are known on Eq. (1). The case gs=1 + s1/2 is considered in [3] for the space dimensions n = 1 2 3, where smallness assumptions are needed even for local well-posedness if n = 2 3 The case gs=s n = 1 is studied in [14]. The general equation (1) is treated in [4] and [22] for n = 1 and in [23] for n ≥ 1. Semilinear Schrodinger¨ equations are well studied in the literature by energy methods. Because of a loss of derivatives these standard techniques do not directly apply to quasi-linear equations (1). It seems that no result has been known concerning the local well-posedness of (1) in higher space dimensions without requiring smallness assumptions on the initial value. As a first result, the local well-posedness of (1) in the Fréchet space H∞n is proved in [23]. Using the implicit function Theorem 2.9, we obtain well- posedness results in usual Sobolev spaces Hkn, improving [23]. This paper is organized as follows. Section 2 gives the main results, the inverse and implicit function Theorems 2.8 and 2.9. In view of applications, all appearing constants are explicitly calculated; suitable scalings of the norm systems are used to get a small loss of derivatives, also simplify- ing some methods of [8]. Section 3 introduces a linearization method for the local, global, and almost global well-posedness of the Cauchy problem 148 markus poppenberg for nonlinear evolution equations of general type. Section 4 contains applications to singular quasi-linear Schrodinger¨ equations where the linearized problem is solved using hyperbolic semigroup theory, including evolution systems.

2. AN INVERSE FUNCTION THEOREM

In this section we use notations and techniques of Hormander¨ [8, 9]. Let

Eaa≥ 0 be a decreasing family of Banach spaces with injections Eb → Ea of norm ≤1 when b ≥ a Let E =∩a≥0Ea Assume that there is given a family of linear operators Sθ E0 → E for θ ≥ 1 such that b−a Sθu b ≤ cbθ u aa≤ b (2) a−b u − Sθu a ≤ cbθ u ba≤ b (3) d b−a−1 Sθu ≤ cba θ u aab≥ 0 (4) dθ b with constants cbcb a ≥ 1 (cf. [8], [9]). For c ≤ a ≤ b, we then have

λ 1−λ u λa+1−λb ≤ cb u a u b 0 <λ<1 (5)

u a u b ≤ cb u a−c u b+c (6) with constants cb ≥ 1 (cf. [8], [9]). Assume that there are another family Faa ≥ 0 and Sθ F0 → F =∩a≥0Fa also satisfying (2)–(6). Fix β d d0d1d2 ≥ 0 with β d ≥ d0 and η > 0 Let u0 ∈ Ed+ and put

Ua =u ∈ Ea u − u0 a ≤ η (7) Let ! U ∩ E →F be a nonlinear C2 map for any a ≥ 0 so that d0 a+d0 a ! U ∩ E ×E → F and ! U ∩ E ×E2 → F exist with d0 a+d0 a a d0 a+d0 a a ! uv v ≤ c1 u v v + v v + v v (8) 1 2 a a a+d0 1 0 2 0 1 a 2 0 1 0 2 a for all u ∈ U ∩ E and v v ∈ E a ≥ 0, where c1 ≥ 1 are constants. d0 a+d0 1 2 a a Assume that there is a map # Ud ∩ E×F → Eβ such that

! u#ug = g u ∈ Ud ∩ E g ∈ F (9) #ug ≤ c2 u g + g (10) a a a+d1 d2 a+d2 2 for all u ∈ Ud ∩ E g ∈ F 0 ≤ a ≤ β, where ca ≥ 1 are constants. Sup- 1 2 pose that a → caa → ca, and a → ca are nondecreasing. If (8) and (10) hold only for integers a ≥ 0, then we will use (8) and (10) only for those values of a. an inverse function theorem 149

Our purpose is to solve the equation !u=!u0+f for u near u0 and f near 0. For that, we deﬁne an iteration as follows. Choose numbers

−1 1/ κ ≥ 1θ0 ≥ max2 η cd+ u0 d+ (11) Notice that S u ∈ U For k = 0 1 2, put θ0 0 d κ 1/κ θk =θ0 + k k = θk+1 − θk (12) We take advantage of the elementary estimates

−1 1−κ −κ −1 1−κ κ κ θk 1 − θk ≤k ≤ κ θk θk+1 ≤ 2θkk ≤ 2 k+1 (13) ∞ 1 1 θ−1− ≤ θκ − 1−/κ ≤ >0 (14) j j 0 j=0 For u ∈ E f ∈ F , deﬁne e by (19) and (20) if u + w ∈ U and 0 d+ 0 0 0 0 0 d0 E = 0g= −1S f v = S u w= #v g (15) 0 0 0 θ0 0 θ0 0 0 0 0

If Ej, gj, uj, vj, wj, and ej are already deﬁned for 0 ≤ j ≤ k − 1, then we put k−1 Ek = jej (16) j=0 g = −1S − S f − E − S e (17) k k θk θk−1 k−1 k−1 θk k−1 u = u + w v= S u w= #v g (18) k k−1 k−1 k−1 k θk k k k k

ek = ek + ekek =! uk−! vkwk (19) −1 ek = k !uk + kwk−!uk−! ukkwk (20) provided that v ∈ U and u u + w ∈ U Because v ∈ E and k d k k k k d0 k gk ∈ F, it is enough that # is deﬁned for elements in the spaces E and F (smooth functions in applications). In the case where u u ∈ U v ∈ k k+1 d0 k Ud, we obtain

!uk+1−!uk=kek + gk (21) k g + S E = S f (22) j j θk k θk j=0

Lemma 2.1. Let 0 <α<β>0δ>0, and u0 ∈ Eβ. Let wk be deﬁned and

−1− wj α ≤ δθj (23) β−α−1 wj β ≤ δθj (24) 150 markus poppenberg for 0 ≤ j ≤ k With the notation t+ = maxt 0, we then get the inequalities

1 a−α+ vj a ≤ µaθj a≥ 0 (25) 2 a−α uj − vj a ≤ µaθj 0 ≤ a ≤ β (26) 1 2 for 0 ≤ j ≤ k + 1, where the constants µa and µa are deﬁned by 1 −1 1 2 µa = ca u0 α + δ µa = µa + µa (27) 2 −1 −1 µa = cβ u0 maxaα + δ maxcα 2cββ − α (28) Proof. Using (2) and (3), we obtain for all a ≥ 0 and j ≥ 0 the estimates

+ S u ≤ c θa−α u (29) θj 0 a a j 0 α u − S u ≤ c θa−α u (30) 0 θj 0 a maxaα j 0 maxa α Let U = k w and V = S U We get U ≤ δ from (14) and j=0 j j θk+1 α + U ≤ 2δ θβ−α by an integration. Hence V ≤ c δ θa−α for a ≥ 0 β β−α k+1 a a k+1 and U − V ≤ c δ θa−α for 0 ≤ a ≤ α For α ≤ a ≤ β, we obtain a α k+1

β−a/β−α a−α/β−α U − V a ≤ cβ U − V α U − V β c 2c ≤ c max α β δθa−α β β − α k+1 using (5). This proves the result.

Lemma 2.2. Let 0 <α<βα≥ d0 > 0 0 <δ≤ 1, and u0 ∈ Eβ Assume that

−1− wj α ≤ δθj (31) a−α−1 wj a ≤ δθj 0 ≤ a ≤ β (32) for 0 ≤ j ≤ k, where v ∈ U and u ∈ U are still deﬁned. Then we obtain k d k+1 d0 e ≤ c1δµ + 1µ + 3θa−2α−1 0 ≤ a ≤ β − d (33) j a a a+d0 0 j 0 for 0 ≤ j ≤ k, where µa are given by (27) and (28).

Proof. Writing zj = uj − vj, we see for 0 ≤ a ≤ β − d0 and 0 ≤ j ≤ k that 1 ej a = ! vj + tuj − vjwjuj − vj dt 0 a ≤ c1 v + z w z + w z + w z a j a+d0 j a+d0 j 0 j 0 j a j 0 j 0 j a ≤ c1δθa−2α−1µ + 1µ2 + µ2 a j a+d0 0 a an inverse function theorem 151 using Lemma 2.1 and (32) and observing α ≥ d0 In a similar way we get

1 −1 ej a = j 1 − t! uj + tjwjjwjjwj dt 0 a ≤ c1 u + w w 2 + 2 w w j a j a+d0 j j a+d0 j 0 j a j 0 ≤ c1δ2θa−2α−2µ + δ + 2 a j a+d0 observing that α ≥ d0 and j ≤ 1 Because δ ≤ 1, this proves the assertion.

Lemma 2.3. Let β>τα+ d0 and 1 <τ≤ 2 We then have

a−τα−1 κ+a a−γ−1 gk+1 a ≤ Aaδθk + 2 ca γθk f γ (34) for all a γ ≥ 0 in the situation of Lemma 2.2, where Aa is given by (45). Proof. We put S = −1 S − S Using (4) and (13), we get k k+1 θk+1 θk

κ+b b−a−1 Sku b ≤ 2 cb a θk u aab≥ 0 (35) Lemma 2.2 applied to e gives for a ≥ β − d the estimates k β−d0 0 −1 S e ≤ 2κ+ac c1 δµ + 1µ + 3θa−2α−1 (36) k k+1 θk+1 k a a β−d0 β 0 k and Lemma 2.2 applied to ek a yields for 0 ≤ a ≤ β − d0 the inequalities −1 S e ≤ 2κc c1δµ + 1µ + 3θa−2α−1 (37) k k+1 θk+1 k a a a a+d0 0 k

Because β>τα+ d0, we get, by means of straightforward calculation,

β−τα−d E ≤ 2β − τα − d −1c1 δµ + 1µ + 3θ 0 (38) k β−d0 0 β−d0 β 0 k Combining (35) and (38), we obtain

1 ca β−d c κ+a+1 0 β−d0 a−τα−1 SkEk a ≤ 2 δµβ + 1µ0 + 3θk (39) β − τα − d0 Applying (35) to the term Skf a yields the result.

Now we ﬁx u0 ∈ Eβ and numbers α β γ > 0 and 1 <τ≤ 2 such that

−1 −1 α ≥ maxd + d1 + d2 + 2τ d2 + 2τ − 1 (40)

β>τα+ d0 (41)

γ ≥ maxα d1+d2 + 2 (42) 152 markus poppenberg and deﬁne the constants (where B and B depend only on u c1 c2 ) 4 0 β β−d0 β 1 −1 1 2 µa = ca u0 α + µa = µa + µa (43) 2 −1 −1 µa = cβ u0 maxaα + maxcα 2cββ − α (44) A = 2κ+ac1 c + 2c β − τα − d −1µ + 1µ + 3 (45) a β−d0 a a β−d0 0 β 0 2 β+d 1 B = 2c 2 1 µ A + A B= maxB B (46) 1 β β+d1 d2 β+d2 2 3 −1 −1 −1 −1 2 1/ θ0 = maxB1 2 η cd+ u0 d+ 2η 2η µd (47) β+d +κ+1 2 d 1 B = 2 2 c 2 1 µ c + c (48) 2 β β+d1 d2γ β+d2γ B = −1c2 c c u + c θα+1+ (49) 3 0 β β+d1 d2 0 d1 β+d2 0 B = max8c2c1 c c 2 + u 2 + u 2η−1 (50) 4 0 d2 d0+d2 d1 0 d1 0 d0+d2 −1 Lemma 2.4. Let u0 ∈ Eβ and let f ∈ Fγf = 0 satisfy f γ ≤ B Then uk and vk ∈ Ud for all k ≥ 0, and with δ = B f γ we have the estimates a−α−1− wk a ≤ δθk 0 ≤ a ≤ β (51)

Proof. We obtain (51) for k = 0 from the choice of B3 because w ≤ −1c2θac c u + c f (52) 0 a 0 a 0 a+d1 d2 0 d1 a+d2 d2

We assume that ukvk ∈ Ud and that (51) is proved for some k ≥ 0 Then k η u − u ≤ δ θd−α−1− ≤ θ−−1 ≤ (53) k+1 0 d j j 0 2 j=0 η η v − u ≤ + µ2 θd−α ≤ + µ2 θ− ≤ η (54) k+1 0 d 2 d k+1 2 d 0 by Lemma 2.1 and thus uk+1vk+1 ∈ Ud Hence the assumptions of Lemmas 2.2, 2.3 hold and we obtain using (40) and (42) the estimates

2 wk+1 a ≤ ca vk+1 a+d gk+1 d + gk+1 a+d 1 2 2 + 2 1 a+d1−α d −τα−1 κ+d d −γ−1 ≤ c µ θ A δθ 2 + 2 2 c θ 2 f a a+d1 k+1 d2 k d2γ k γ a+d −τα−1 κ+a+d a+d −γ−1 + A δθ 2 + 2 2 c θ 2 f a+d2 k a+d2γ k γ 2 a+d 1 ≤ c 2 1 µ A + A δ a a+d1 d2 a+d2 a+d +κ d 1 a−α−1−2 + 2 2 2 1 µ c + c f θ a+d1 d2γ a+d2γ γ k which implies (51) for k + 1 by deﬁnition of B1 and B2 This gives the result. an inverse function theorem 153

Proposition 2.5 (Existence). Let u0 ∈ Eβ and let f ∈ Fγ satisfy f γ ≤ −1 B . Then uf =limk→∞ uk exists in Eα and !uk→!u0+f in Ea when a

−1 uf −u0 α ≤ B f γ (55)

Proof. By Lemma 2.4, the sequence uk is a Cauchy sequence in Eα and uf =limk→∞ uk exists in Eα; thus (55) follows because uk+1 − u0 α ≤ δ−1 by means of (51). From (21) and (22), we obtain

!u −!u =S f + e + E − S E (56) k+1 0 θk k k k θk k Note that S f → f in F if a<γBy Lemma 2.2, we have e → 0 θk a k k a+d −β in F if a<2α Because E − S E ≤ Cθ 0 E , inequality (38) a k θk k a k k β−d0 shows that E − S E → 0inF if a < τα By continuity, we get !uf = k θk k a !u0+f in Fa for a

Lemma 2.6 (Uniqueness). Let µ>maxd d0 + d2µ≥ d1 and

#u! uw = w u ∈ Ud ∩ E w ∈ E (57)

−1 Then ! is injective in W =u ∈ Eµ u − u0 µ ≤ B4 Proof. Writing w = v − u and y = !u + w−!u−! uw, we have

y ≤ c1 u + v w + 2 w w = A w (58) d2 d2 d0+d2 d0+d2 0 d2 0 0 for all u v ∈ W ∩ E Because w = #u!v−!u−y, we obtain

w ≤ c21 + u !v−!u + c22 + u A w (59) 0 0 d1 d2 0 0 d1 0 The choice of B implies that c22 + u A ≤ 1/2 and thus u − v ≤ 4 0 0 d1 0 2c21 + u !v−!u Using (2) and (3) and replacing u and v by 0 d1 d2 Sθu and Sθv, we get a similar inequality for all u v ∈ W . This proves the assertion.

We put α0 = α1 = α β0 = β1 = β γ0 = γ1 = γ, and τ0 = τ1 = τ and choose numbers 1 <τn ≤ 2 and increasing sequences αn, βn, and γn such that

+ αn+1 ≤ minτnαnτn−1αn−1 −d1 − αn −d2 (60)

βn+1 >τnαn + d0 + (61) + γn ≥ max3γn−1 − 2γ αn+1 + d2 + +d1 − αn (62)

Note that τn = τ and αn+1 = αn + satisfy (60) and αn →+∞ 154 markus poppenberg

Lemma 2.7 (Higher Regularity). Let u0 ∈ Eβ and let f ∈ Fγ satisfy f γ ≤ B−1 If u ∈ E and f ∈ F , then uf ∈E and 0 βn γn αn uf −u ≤ C f + 1 (63) 0 αn n γn a−α −1− w ≤ C f + 1θ n 0 ≤ a ≤ β (64) k a n γn k n for all k ≥ 0, where the constant C > 0 depends only on u c1 , and n 0 βn βn−d0 c2 and can be chosen uniformly for all f ∈ F with f ≤ B−1 βn γn γ Proof. The cases n = 0 1 hold by Proposition 2.5 and Lemma 2.4; the case k = 0 follows from (52). Assume that (63) and (64) are proved for n ≥ 1 Applying Lemma 2.1 to δ = C f + 1α= α β= β ,weget n γn n n+1 a−α + v ≤ c f + 1θ n a≥ 0j≥ 0 (65) j a n γn j a−α u − v ≤ c f + 1θ n 0 ≤ a ≤ β j≥ 0 (66) j j a n γn j n+1 We have f 3 ≤ c f by (5) and (62). Thus the proof of Lemma 2.2 γn γn+1 yields

a−2α −1− e ≤ C f + 1θ n 0 ≤ a ≤ β − d (67) j a γn+1 j n+1 0 As in the proof of Lemma 2.3, we get with constants C the estimates

a−2α −1− −1 S e ≤ C f + 1θ n (68) k k+1 θk+1 k a γn+1 k β −τ α −d − E ≤ C f + 1θ n+1 n n 0 (69) k βn+1−d0 γn+1 k a−τ α −1− S E ≤ C f + 1θ n n (70) k k a γn+1 k a−τ α −1− a−γ −1 g ≤ C f + 1θ n n + θ n+1 (71) k+1 a γn+1 k k

For a = d2, we use (71) with n − 1 in place of n and get the estimates a−α −1− g ≤ C f + 1θ n+1 (72) k+1 a+d2 γn+1 k d −τ α −1− d −γ −1 g ≤ C f + 1θ 2 n−1 n−1 + θ 2 n (73) k+1 d2 γn k k a+d −α + v ≤ C f + 1θ 1 n (74) k+1 a+d1 γn k Hence we get (64) for n + 1 and k + 1 This gives (63) and thus the result.

Deﬁne B and B4 for u0 ∈ Eβ by (43)–(50) and consider the sets

−1 U =u ∈ Eα u − u0 α ≤ B4 (75) −1 −1 V =g ∈ !u0+Fγ g − !u0 γ ≤ B B4 (76) an inverse function theorem 155

For unique solvability, assume the following conditions: d + d d α>max d d + d 1 2 2 α≥ d 1 <τ≤ 2 (77) 0 2 τ τ − 1 1

β>τα+ d0 γ>α+ d2 (78)

Theorem 2.8 (On Inverse Functions). Let u0 ∈ Eβ and assume that ! U ∩ E → F is a C2 map for a ≥ 0 such that (8)–(10) hold. d0 a+d0 a (i) If (40)–(42) hold, then there is 2 V → U so that !2g = g, g ∈ V (ii) If (57), (77), and (78) are supposed, then there is a unique solution map 2 V → U as in (i). The map 2 V ⊂ !u0+Fγ→Eα is continuous.

(iii) Let u0 ∈ E = 1 If (40)–(42), (57), and (77) hold, then there is a unique C2-solution map 2 V ∩ F→E between Fr´echet spaces. Thus ! U ∩ E → F is a C2 diffeomorphism near u If ! U ∩ E→F is Cn 0 d0 then 2 is Cn as well, 2 ≤ n ≤∞ Proof. Proposition 2.5 proves (i). If (57), (77), and (78) hold, then we choose >0 such that (40)–(42) are true. Lemma 2.6 yields uniqueness in (ii), and 2f −2g ≤ C f − g for f g ∈ V By Lemma 2.4, we have 0 d2 2f α+/2 ≤ C for f ∈ V Hence 2 V ⊂ !u0+Fγ→Eα is continuous by (5). By Lemma 2.7, also 2 V ∩ F→E is continuous in (iii). The remaining parts of (iii) follow as in [20], 3.11. Part (iii) of Theorem 2.8 contains classical theorems of Nash–Moser type as e.g. [7, 16, 26]. Next, we apply Theorem 2.8 to derive a theorem on implicit functions. Let EaFaGaa ≥ 0 be Banach spaces as before such that (2)–(6) hold. Fix w0 =u0v0∈Eβ × Fβη> 0, and d ≥ d0 and put

Wa =w ∈ Ea × Fa w − w0 a ≤ η (79) Let f W ∩ E × F →G a≥ 0, be a C2 map, f w =0, and d0 a+d0 a+d0 a 0 f wz ≤ d0 w z + z (80) a a a+d0 0 a f wz z ≤ d1 w z z + z z + z z (81) 1 2 a a a+d0 1 0 2 0 1 a 2 0 1 0 2 a for w ∈ W ∩ E × F zz z ∈ E × F a ≥ 0, where d0d1 ≥ 1 d0 a+d0 a+d0 1 2 a a a a are constants, f = f u v.Letg Wd ∩ E × F×G → Fβ satisfy

fvwgwy = y w ∈ Wd ∩ E × F y ∈ G (82)

gwfvwv = v w ∈ Wd ∩ E × F v ∈ F (83) gwy ≤ d2 w y + y (84) a a a+d1 d2 a+d2 156 markus poppenberg

2 for all w ∈ Wd ∩ E × F y ∈ G, and 0 ≤ a ≤ β, where da ≥ 1 and d1d2 ≥ 0 are constants. For δ1δ2 > 0, consider the sets

∗ U =u ∈ u0 + Eγ u − u0 γ ≤ δ1 (85) ∗ V =v ∈ Fα v − v0 α ≤ δ2 (86)

Theorem 2.9 (On Implicit Functions). Let f W ∩ E × F → d0 a+d0 a+d0 2 Ga be a C map for a ≥ 0fw0=0w0 ∈ Eβ × Fβ Assume (80)–(82), (84), and

α ≥ d0 + d2d1 ≥ d0 + d2 (87)

∗ ∗ (i) If (40)–(42) hold, then there are δ1δ2 > 0 and h U → V ∗ such that f u hu = 0 for u ∈ U Here δ1 and δ2 depend on w d2 d1 d0 0 β β β−d0 β+d2

(ii) If (77), (78), and (83) hold, then there exist δ1δ2 > 0 depending ∗ on the same data as in (i) and a continuous h U ⊂ u0 + Eγ→Fα such that

u v∈U∗ × V ∗ f u v=0=u hu u ∈ U∗ (88)

(iii) Let u0 ∈ E v0 ∈ F, and = 1 and assume (40)–(42), (77), and (83). Then there is a C2 map h U∗ ∩ E→F between Fr´echet spaces n n satisfying (88).Iff Wd ∩ E → F is C , then h is C as well, 2 ≤ n ≤∞

Proof. Consider ! Wd ∩ Ea+d × Fa+d → Ea × Ga!u v= 0 0 0 u f u v Then ! wu v=u fuwu + fvwv and ! =0f 1 1 Hence (8) follows for ! from (81) with ca = da The map #wu y= u −gwfuwu + gwy satisﬁes (9). By Lemma 2.4, we may assume w ≤ C and thus w ≤ C when proving (10). We get α d0+d2

gwf wu ≤ d2Cd0 + d0 w u + u (89) u a a d2 a+d2 a+d1 d2 a+d2

2 2 0 0 which proves (10) with ca = daCdd + da+d , where C depends on w0 α Hence Theorem 2.8 (i) gives 2 V 2→ U with2 !2w = w w ∈ V Here the neighborhoods U ⊂ Eα × Fα of w0 and V ⊂u0 + Eγ×Gγ of u0 0 are defined as in (75) and (76). Then define U∗ and V ∗ by (85) and (86), −1 −1 −1 ∗ putting δ1 = B B4 , and δ2 = B4 .For2 =2122 define h U → ∗ V by hu=22u 0 and obtain (i). From (83), we get #w! wz = z for w ∈ Wd ∩ E × F, and z ∈ E × F Thus (ii) and (iii) follow from Theorem 2.8. an inverse function theorem 157

3. NONLINEAR EVOLUTION EQUATIONS

Fix integers m ≥ 0n ≥ 1, and M ≥ 1 and a real q> n Equip the 2 a a n M Sobolev space H = H with its usual norm denoted by a = ∞ k+q Ha a≥ 0 Note that uk ≤ Cuk+q for u ∈ H , where ∞ α n n uk = sup ∂ ux α ∈ 0 α ≤k x ∈ (90) 0 a For a ≥ 0, put Ca0T=C0TH and ua = supt∈0T uta 1 1 a a+m Equip Ca m0T=C 0TH ∩C0TH with the norm

C1 1 u a m = max∂t ua ua+mu∈ Ca m0T (91) 1 1 0 0 n Write Ca m = Ca m0 1Ca = Ca0 1, Im=α ∈ 0 α ≤m, and ∞ a M Im H =∩a≥0H Let A ⊂ be open and convex, 0 ∈ A and a a a α n U = U A=u ∈ H ∂ ux α ≤m ∈ A a.e. in a≥ m (92) Let F 0 ∞ × n × A → M F = Ft x u be continuous such that i α β the derivatives ∂t ∂x∂u F exist and are continuous for i = 0 1 2 and every α n and β and are bounded on 0 ∞ × × A0 for any bounded A0 ⊂ A and i α n n ∂t ∂xFt x 0=0i= 0 1 2α∈ 0 t ≥ 0x∈ (93) For instance, any F ∈ C∞A with F0=0 that is independent of t and x satisﬁes these conditions. For a bounded subset A0 ⊂ A, write A0 = max y y ∈ A0 Consider the nonlinear differential operator α t ux=Ft x ∂ ux α ≤m (94) Lemma 3.10. Formula (94) deﬁnes for every integer a ≥ 0 a C2 map a+m+q a+m+q a 0 ∞×U ⊂ H →H For any bounded set A0 ⊂ A there is a constant C = Ca A0 > 0 depending only on a A0 such that ∞ t ua ≤ Cum+aum +um+a t uv ≤ Cu∞ v +v u a m+a m m+a (95) ∞ uut uv1v2a ≤ Cum+av1mv2m +v1m+av2m +v1mv2m+a a+m+q a+m for all u ∈ U A0 and v v1v2 ∈ H and all t ≥ 0 Proof. This follows from the proof of [21], 2.2, 2.3, and 2.4 (cf. [10]), where α α ut uv = F∂αut · ∂ u· α ≤m∂ v (96) α ≤m using the assumptions on F and the chain rule for higher derivatives. 158 markus poppenberg

Consider the Cauchy problem for the nonlinear evolution equation u = t ut∈t t + b t 0 0 (97) ut0=φ a For t0 ≥ 0 and an initial value φ ∈ H 0 a0 ≥ 0 we are looking for a solu- 1 a a tion u ∈ C t0t0 + bH 1 ∩Ct0t0 + bH 2 for suitable a1a2 ≥ 0 and b>0 Performing the transformation t = t0 + bs, we get the equiva- lent problem u = b t + bt ut∈0 1 t 0 (98) u0=φ We want to solve the equation !u b φ=0 0 for the nonlinear map 2 2 !u b φ=ut − b t0 + b t utu0−φ (99) using an implicit function theorem. We already have !φ 0φ=0 0 1 a 0 a For a ≥ 0, we put Ea = Ca m × × H and Fa = Ca × H The norms in E and F are denoted by = and = respectively. Let a a a Ea a Fa a a 1 a+m U = U A=u ∈ Ca m ut∈U At ∈0 1 (100) a a a a and put W = W A=U A×−1 1×H ⊂ Ea a+q 2 Lemma 3.11. The map ! W ⊂ Ea+q→Fa given by (99) is C for any integer a ≥ 0 For any bounded set A0 ⊂ A and any D>0, there is a constant C = CaD A0 > 0 depending only on a D A0 such that

!v a ≤ C v a+q

! vw a ≤ C v a+q w 0 + w a (101)

! vw1w2 a ≤ C v a+q w1 0 w2 0 + w1 a w2 0 + w1 0 w2 a a+q for all v =u b φ∈W A0 with um ≤ D and all w w1w2 ∈ Ea. Proof. As in the proof of [21], Lemma 4.3, we obtain 2 2 !uu b φz =zt − b ut0 + b t utztz0 2 2 !uuu b φz1z2=−b uut0 + b t utz1tz2t 0 and !φvψ =0 −ψ similarly for !b!ub, and !bb Lemma 3.10 implies

!u b φ a ≤ C u a+q u 0 + u a +φa

! vz d ψ a ≤ C u a+q z 0 + z a + d u a+q u 0 + u a+ψa

! vw1w2 a ≤ C u a+q z1 0 z2 0 + z1 a z2 0 + z1 0 z2 a

+ d2 u a+q z1 0 + z1 a+ d1 u a+q z2 0 + z2 a

+ d1d2 u a+q u 0 + u a an inverse function theorem 159 for v =u b φw1 =z1d1ψ1, and w2 =z2d2ψ2 and thus the assertion. m+q m+q m+q For φ ∈ U , we choose 0 <η<1 such that Uη φ⊂U , where a a Uηφ=ψ ∈ H φ − ψa ≤ η (102) Lemma 3.12. Let φ ∈ Um+q and 0 <η<1 be chosen such that m+q m+q Uη φ⊂U Let v0 =φ 0φ Then for any integer a ≥ 0 there is a constant C = Caφm+q > 0 depending only on a and on φm+q such that estimates (101) hold for all v ∈ Ea+q with v − v0 q ≤ η q Proof. For v − v0 q ≤ η, we have v ∈ W A0 for some bounded A0 ⊂ A, where A0 depends only on φm+q Lemma 3.11 gives the result. The assumptions (80) and (81) of Theorem 2.9 thus hold for ! with 0 1 d0 = q, where the constants da and da depend only on a and on φm+q Put a 1 Uηφ=u ∈ Ca m u − φa+m +ut a ≤ η (103) and consider for d ≥ q> n d d β ≥ 0 0 <η<1 the following 2 1 2 condition. Definition 3.13 (Linear Solvability Criterion). The initial value φ ∈ m+q d q η β m+q m+q H satisfies condition Ad d if Uη φ⊂U and if problem 1 2 z t=b t + bt utzt+f tt∈0 1 t u 0 (104) z0=g admits for any integer 0 ≤ a ≤ β, any t0 ≥ 0 0 0 depending only on a and φ. If the foregoing holds for all integers a ≥ 0, then we say that φ satisfies Ad q η or φ ∈ d1d2 Ad q η. d1d2 Estimate (105) must be checked in applications; it means explicitly that

zt C01Ha +zC01Ha+m

≤ Caφ f C01Ha+d2 +gHa+d2

+ut C0 1Ha+d1 +uC0 1Ha+m+d1

×f C0 1Hd2 +gHd2 160 markus poppenberg

Theorem 3.14 (Local Solutions). Let the integers α β γ d1d2 ≥ 0 and d ≥ d = q> n and real numbers 0 <η<1 > 0, and 1 <τ≤ 2 satisfy 0 2 (40)–(42) and (87). Assume that φ ∈ Hβ+m satisﬁes condition Ad q η β d1d2 (i) There exist numbers B>0 and δ>0 depending only on a bound for φβ+m + Cβφ (cf. (105)) such that the nonlinear Cauchy problem u = t u t (106) ut0=ψ

∗ γ admits for any ψ ∈ V =ψ ∈ φ + H φ − ψγ ≤ δ and any t0 ≥ 0 1 a solution u ∈ Cα mt0t0 + B If (77) and (78) hold, then the solution is ∗ 1 unique and the solution map V → Cα mt0t0 + Bψ→ u is continuous. (ii) Let = 1 and suppose (77). Let φ ∈ H∞ satisfy Ad q η and d1d2 1 choose B δ as in (i). Then (106) admits a unique solution u ∈ C t0, ∞ ∗ ∞ 1 ∞ t0 + BH . The solution map V ∩ H → C t0t0 + BH ψ → u is a C1 map.

a+q Proof. Apply Theorem 2.9 to ! W ⊂ Ea+q→Fa given by (99). The spaces Ea and Fa admit smoothing operators satisfying (2)–(4) (cf. the proofs in [26] and [8], A.10). We have !v0=0 for v0 =φ 0φ 0 Assumptions (80) and (81) of Theorem 2.9 hold with d0 = q, where da and d1 depend only on a φ Because φ satisﬁes Ad q η β, the lin- a m+q d1d2 earization !uu b ψ admits a linear right inverse #u b ψ such that # d 1 ∞ ∞ 1 ∞ Uηφ∩C 0 1H ×−η η×H ×F → C 0 1H satis- 2 ﬁes (82), (83), and (84), as required by Theorem 2.9, where da = Caφ the case b = 0 is trivial. Applying Theorem 2.9 as in [21], Theorem 4.4, we get the assertion where B and δ depend only on v0 β =φβ+m and 2 on dβ = Cβφ from (105); uniqueness follows from local uniqueness by a standard argument as in [21], Theorem 4.4. We investigate the existence of global solutions to the Cauchy problem u = t u t (107) u0=φ

Theorem 3.15 (Global Smooth Solutions). Let = 1 0 <η<1 Let ∞ d q η the initial value φ ∈ H satisfy φ ∈Ad d Suppose (40)–(42), (77), and (87). Assume that for any T>0 there is M1 2= MT > 0 so that any solution u ∈ C10TH∞ of (107) satisﬁes ut∈Ad q η for 0 ≤ t

utβ+m + Cβut ≤ M 0 ≤ t

Proof. Assume that T ∗φ < ∞, where

T ∗φ=supT>0 (107) has a solution u ∈ C10TH∞ (109)

∗ Let tj be an increasing sequence tending to T φ By Theorem 3.14 there 1 ∞ is B>0 such that (107) admits a solution uj ∈ C 0tj + BH for ∗ every j. Choosing j with tj + B>T φ we get a contradiction. Global solutions in Sobolev spaces require a regularity assumption. Let σ ≥ β + m Problem (107) is called α β σ regular if any solution u ∈ 1 σ β+m Cα m0T of (107) with u0∈H satisfies ut∈H for 0 ≤ t0 there is M = MT > 0 d1d2 1 so that any solution u ∈ Cαm0T of (107) satisfies (108) and ut∈ Ad q η β for 0 ≤ t0 such that (107) admits for every j a 1 ∗ solution u ∈ Cα m0tj + B Choosing j with tj + B>Tα φ, we get a contradiction. d q η β We write φ ∈Ad d T if 3.13 holds with 0T replacing 0 1 and b = 1. 1 2 Theorem 3.17 (Almost-Global Solutions). Let F ∈ C∞AF0=0 Suppose (40)–(42). Then for any (large) T>0 there is = T > 0 so that for any φ ∈ Hβ+m with φ ∈Ad q η β and φ ≤ , there is a solution d1d2T γ 1 u ∈ Cα m0T of problem (107). a+q 1 0 a Proof. Define ! U A⊂Ca+q m0T→Ca0T×H by !u=ut − uu0 Theorem 2.8 gives the result (cf. [21], 6.11).

4. QUASI-LINEAR SCHRODINGER¨ EQUATIONS

In this section we give applications to quasi-linear Schrodinger¨ equations essentially following [23]. We consider the Cauchy problem for the nonlinear equation iu =−u + V xu + u·u x ∈ n, t ∈0B t (110) u0x=φxx∈ n, 162 markus poppenberg where φ ∈ Hq+2nq>n/2 is a given initial value, V ∈ ∞n ; that ∞ 2 is, V ∈ C and all partial derivatives of V are bounded, = ∂j , and n n 2 n 2 n u=Gu1u2 ∂iu1i=1 ∂iu2i=1 ∂i u1i=1 ∂i u2i=1 (111) is deﬁned by G ∈ C∞A G = Gu Du D2u, where A ⊂ 4n+2 is open and convex with 0 ∈ A here u = u1 + iu2 with real-valued u1 and u2. Put a a n 2 n n U =u ∈ H ux ∂iuxi=1 ∂i uxi=1∈A a.e. in (112) for a ≥ 2. Let φ ∈ Uq+2. For all u ∈ Uq+2 and all i j, and k, we assume that

∂ 2 G = ∂ 2 G ∂ G0=∂ 2 G0=0 (113) ∂i uk ∂j uk ∂iuk ∂i uk 2 2 ∂ 2 Gu Du D u·u = ∂ 2 Gu Du D u·u (114) ∂1u1 2 ∂1u2 1 2 2 1 − φ ∂ 2 Gφ Dφ D φ−φ ∂ 2 Gφ Dφ D φ > 0 (115) 1 ∂1u1 2 ∂1u2 u ∂ G + u ∂ G u1∂∂ u G + u2∂∂ u G 1 ∂iu1 2 ∂iu2 j 1 j 2 ∂j = ∂i (116) 1 − u ∂ 2 G − u ∂ 2 G 1 − u ∂ 2 G − u ∂ 2 G 1 ∂1u1 2 ∂1u2 1 ∂1u1 2 ∂1u2 where the derivatives of G are evaluated at u Du D2u in (116); we often omit u Du D2u in the notation. Condition (115) implies that there are µ>0 and a neighborhood U of φ in Hq+2 so that for u ∈ U, we have 2 2 1 − u ∂ 2 Gu Du D u−u ∂ 2 Gu Du D u≥µ>0 (117) 1 ∂1u1 2 ∂1u2

Conditions (114), (117), and (116) mean that traceMu=0 and detMu≥ −1 µ>0 (the equation has Schrodinger¨ type) and that traceMu Gu admits a potential (the crucial integrability condition) with Mu, and Gu as below. Note that (113) implies that G = Gu Du D2u depends only on u ∇u, and u The linearization (104) of problem (110) can be written in the real form   f2t zt = bMuz + Gu ·∇z + Huz+ t∈0 1 −f1t g (118)  z0= 1 g 2 z where z = 1 ∼ z + iz and M G , and H (depending on t x) are z = 1 2 u u u given by 2 u2∂∂2u G u2∂∂2u G − 1 Mu = 1 1 1 2 (119) 1 − u1∂∂2u G −u1∂∂2u G 1 1 1 2 u2∂∇u G u2∂∇u G Gu = 1 2 (120) −u1∂∇u G −u1∂∇u G 1 2 u2∂u G u2∂u G + V + G H = 1 2 (121) u −V − G − u ∂ G −u ∂ G 1 u1 1 u2 an inverse function theorem 163 Here ∂ G =∂ Gn and ∂ G·∇z = n ∂ G∂ z Put m = 2 ∇uk ∂iuk i=1 ∇uk j i=1 ∂iuk i j and ﬁx an initial value φ ∈ Ud+2, where the integer d satisﬁes

d = max3n+q q>n/2 (122)

d d Deﬁne Uηφ and W = Uηφ by (102) and (103), where 0 <η<1is q+2 q+2 q+2 chosen such that Uη φ⊂U and (117) holds for u ∈ Uη φ We then have

utd+2 +ut td ≤ C u ∈ W t ∈0 1 (123) with C =φd+2 + 1 and det Mut ≥ µ>0 for u ∈ W and t ∈0 1. 1 n i −1 i Writing Gu =GuGu,weputsu =−1/2 traceMu Gu The vec- tor field s =s1sn satisfies s =2 det M −1u ∂ G + u ∂ G u u u u u 1 ∇u1 2 ∇u2 1 Assumption (116) implies that pux= 0 surx·xdr defines a potential for su with ∇pu = su. Thus nux=exppux solves ∇nu = sunu n 0 1 0 si 0 The matrices N = u with N 0= and Si =u satisfy the u 0 1 u 0 1 u 0 0 i 1 n equations ∂iNu = SuNu for all i Writing Su =SuSu, we see that ∇N = S N and traceSi + 1 M−1Gi =0 for all i Putting z˜ = N−1z u u u u 2 u u u and thus z = Nuz˜ we obtain from (118) for z˜ the transformed equation −1 f2t z˜t = bBu + Cu˜z + Luz˜ + Nu (124) −f1t where

−1 −1 Bu = Nu MuNu Cu = Nu MuNuTu ·∇+Du (125)

1 T = N−12S + M−1G N D= div T (126) u u u u u u u 2 u

−1 −1 Lu = bNu MuNu + Gu∇Nu + HuNu − MuNuDu+∂t Nu Nu (127)

1 n n j Here Tu =Tu Tu and div Tu = j=1 ∂jTu. The matrix Wu = Wut x is introduced by putting W = I · N−1 · M−1 · N , where I = 0 1 Then u 2 u u u 2 −1 0 Wu is symmetric since traceMu = 0 For u ∈ W t ∈0 1,weput 2 n f gt = Wut xf x·gxdx f g ∈ L (128)

The matrices I T j are symmetric because traceSj + 1 M−1Gj =0 Thus 2 u u 2 u u I2Du is symmetric as well. 164 markus poppenberg

Lemma 4.18. There exists a constant C>0 depending only on φd+2 such that for all u ∈ W ∩ C10 1H∞ and all t ∈0 1x ∈ n, we have

Mut x + Gut x + Hut x + ∂t Mut x + ∂t Gut x ≤ C (129) −1 −1 C ≤ det Mut x≤C C ≤ det Nut x≤C (130) 1 1 ∂ ···∂ S 1 +∂ ∂ ···∂ S 1 ≤ C (131) 2 n u Lx t 2 n u Lx

Nut x + ∂t Nut x + ∂t Wut x + Lut x ≤ C (132) −1 C ≤ Wut x≤C (133)

Proof. This follows from [23], Lemma 4.2, using (122) and (123).

Proposition 4.19. Problem (118) has for any u ∈ W ∩ C10 1H∞ 0 0 depending only on φd+2 so that

zC01L2 ≤ Cf C01L2 +gL2 (134)

Proof. This is proved in [23], 4.9; we include a short sketch of the proof. 2 By [23], 4.5, the operators But and Cut are skew adjoint in L with respect to the scalar product (128). By a real version of Stone’s theorem, the operators Aut=bBut+Cut + Lut generate strongly continuous groups in L2 that are stable in the sense of [31], 4.3, or [32], 7.2, with stability constants depending only on C in Lemma 4.18 (cf. [23], 4.7). Thus there is a strongly continuous evolution operator Ut sL2 → L2

0 ≤ s

t zt=Ut 0g + Ut sf sds (135) 0

(cf. [23], 4.8). This gives the assertion.

Next we investigate higher-order estimates as required in (105) by the k k linear solvability criterion 3.13. Applying to Eq. (118) and writing Gu = 1k n k Gu Gu =Gu + 2k∇Mu, we get (cf. [23], 5) the equation k k k k k k f2 zt = bMu z+Gu ·∇ z+Ru z+ (136) −f1 an inverse function theorem 165

k 2k α α where Ru z is with Pkξ= ξ and Pk ξ=∂ Pkξ given by

∂αM n ∂αGj Rkz = u Pα∂z + u Pα∂∂ z u α! k α! k j 2≤ α ≤2k 1≤ α ≤2k j=1 ∂αH + u Pα∂z (137) α! k α ≤2k

Put sk =s1 ksn k, where si k = trace− 1 M−1Gi k We can u u u u 2 u u k k k k k k choose pu with ∇pu = su pu 0=0 Then nu x=exppu x k k k k k k−1 −1 k solves ∇nu = su nu nu 0=1 Put Wu = I2Nu Mu Nu and k k i k i k k 1 k n k N =nu 0 S =su 0 , and S =S S We u 0 1 u 0 0 u u u get ∂ Nk = Si kNk and traceSi k + 1 M−1Gi k=0 for all i. i u u u u 2 u u

Lemma 4.20. For any k there is C>0 depending only on k φd+2 such that for all u ∈ W ∩ C10 1H∞t ∈0 1x∈ n, we have

1 k 1 k ∂ ···∂ Su 1 +∂ ∂ ···∂ Su 1 ≤ C (138) 2 n Lx t 2 n Lx k k k k Gu t x + ∂t Gu t x + ∂t Nu t x + ∂t Wu t x ≤ C (139) −1 k −1 k C ≤ det Nu t x≤C C ≤ Wu t x≤C (140)

Proof. This follows from [23], Lemma 5.1, using (122) and (123).

Lemma 4.21. For any i and k there is C>0 depending only on i and 1 ∞ 1 ∞ k φd+2 such that for all u ∈ W ∩ C 0 1H and z ∈ C 0 1H , we have

−1 ∞ k ∞ MuMu GuHui +Nu i ≤ C u i+q2 + 1

k ∞ k ∞ Gu i +Su i ≤ C u i+q+12 + 1

k ∞ ∂t Nu i ≤ C u i+q+22 + 1 2k−1 k Ru z0 ≤ C u j+q+22 + 1z2k−j j=0

∞ ∞ where the norms i of the space C0 1H are deﬁned using sup– ∞ C1 1 norms (90) in H and u a2 = u a 2 are the norms of the space Ca 2

Proof. This follows from Lemmas 4.18 and 4.20 using (122) and (123). 166 markus poppenberg

Lemma 4.22. For any k there is C>0 depending only on k φd+2 such that for any u ∈ W ∩ C10 1H∞f ∈ C0 1H∞g ∈ H∞ and 0 < b ≤ 1t ∈0 1, any solution z ∈ C10 1H∞ of (118) satisﬁes

2k zt2k ≤ C u 2k+q+3−i 2 + 1 f g i (141) i=0 Proof. Using the method of [23], Lemma 5.4, the case k = 0 follows from Proposition 4.19. Assume that (141) is true for k − 1 ≥ 0 Then zk = k k k−1 k z solves Eq. (136) and z˜ = Nu z satisﬁes k k k k k k−1 k k f2 ˜z t = b Bu +Cu +Lu z˜ +Nu bRu z+ (142) −f1 where

k k−1 k k k−1 k k k Bu = Nu MuNu Cu = Nu MuNu Tu ·∇+Du 1 T k = Nk−1 2Sk + M−1Gk NkDk = div T k u u u u u u u 2 u Lk = bNk−1 M Nk + Gk∇Nk − M NkDk + Nk−1 Nk u u u u u u u u u u t u

k ∞ From Lemma 4.21, we get Lu 0 ≤ C uniformly for u ∈ W As before, the k k operators Bu t+Cu t are skew adjoint with respect to the scalar product k k k deﬁned using Wu instead of Wu in (128). Thus the operators Bu t+Cu t generate an evolution operator Ukt s in L2 (cf. [23], Lemma 5.4) satisfy- k 1 ∞ ing U t sL2→L2 ≤ M uniformly for all u ∈ W ∩ C 0 1H 0 < b ≤ 1 and 0 ≤ s

k Using the interpolation estimate z2k ≤ C˜z 0 +z0,weget 2k−2 t k k ˜z t0 ≤ C u 2k+q+2−i2 + 1zi + f g 2k + ˜z s0ds i=0 0 Applying Gronwall’s lemma, we derive the inequality k−1 k ˜z t0 ≤ C u 2k+q+3−2i 2 + 1z2i + f g 2k (144) i=0 The hypothesis of the induction and (6) give the result. an inverse function theorem 167

Theorem 4.23. Problem (118) admits for any u ∈ W ∩ C10 1H∞ f ∈ C0 1H∞g∈ H∞ 0

For any integer k, there is C>0 depending only on k φd+2 such that z k 2 ≤ C u k+q+6 2 + 1 f g 3 + f g k+3 (145)

In particular, the linear solvability condition Ad q η holds for φ, where d1d2 n d = q + 6d= 3d= max3n+q d = q> (146) 1 2 0 2 Proof. By [23], Theorem 5.12, (118) admits a unique solution z ∈ C10 1H∞ Using the differential equations (118) and (141), we get

2k+2 zt2k+2 +zt t2k ≤ C u 2k+q+5−i 2 + 1 f g i (147) i=0 for t ∈0 1 By means of interpolation (6), we obtain z 2k 2 ≤ C u 2k+q+5 2 + 1 f g 2 + f g 2k+2 (148)

This gives the desired estimate (145) and thus the assertion.

Now conditions (40), (41), (42), (77), (78), and (87) hold if we put τ1 = 10 τ = τ = 11 , and τ = τ = 4 and deﬁne α γ β=α γ β for 7 2 3 8 4 5 3 n n n n ≤ 5by   8 12 13n= 1 α γ β = 9 13 15n= 2 3 (149) n n n  10 14 17n= 4 5 and if we put τ = 1 + 3/d for n ≥ 6 and (with x=largest integer ≤ x)  α = n +n/2+2n≥ 6  n γ = n +n/2+6n≥ 6 n (150)  βn = 2n + 7n≥ 6, n even βn = 2n + 6n≥ 6nodd

Theorem 4.24. Let αnγnβn be deﬁned by (149) and (150). Let φ ∈ β +2 H n satisfy (115) and assume (113), (114), and (116). Then there exist numbers B>0 and δ>0 such that the nonlinear Cauchy problem (110) ∗ γ admits for any initial value ψ ∈ V =ψ ∈ H n φ − ψ ≤ δ a unique γn 1 α α +2 solution u ∈ C 0BH n ∩C0BH n Moreover, the solution map V ∗ → C1 0Bψ→ u is continuous. αn 2 Proof. This follows from Theorem 3.14. 168 markus poppenberg

Theorem 4.25. In the situation of Theorem 4.24 for any (large) T>0 β +2 there is = T > 0 so that for any φ ∈ H n satisfying (115) and φ ≤ γn , there is a solution u ∈ C1 0T of problem (110). αn 2 Proof. This follows from Theorem 3.17. As a particular case, we consider the more specific example iu =−u + V xu − f u 2u − g u 2g u 2u t (151) u0x=φxx∈ n where f g ∈ C∞0 ∞ and V ∈ ∞n Note that conditions (113)–(116) are always satisfied for equations of type (151) (cf. [23]). Hence d+2 d q η any given initial value φ ∈ H satisfies the solvability condition Ad d Equations of the form (151) have been studied in many papers in physics.1 2 The case gs=s corresponds to the superfluid film equation in plasma physics (cf. [12] and [13] and the references in [14] and [23]); the case gs=1 + s1/2 models an ultra-short laser in matter (cf. the literature cited in [2–6, and 27]. Further references can be found in [1, 11, 15, 18, 24, 25, and 30]. In the mathematical literature well-posedness results for the Cauchy problem (151) are contained in [3, 14, 22, and 23]. Different from [3], here we can consider arbitrary space dimensions and do not need any smallness assumptions when proving local well-posedness.

β +2 Corollary 4.26. Let φ ∈ H n be an arbitrary initial value. Then there exists B>0 such that the nonlinear Cauchy problem (151) admits a unique 1 α α +2 solution u ∈ C 0BH n ∩C0BH n

β +2 Corollary 4.27. For any T>0 there is >0 so that for any φ ∈ H n with φ ≤ , problem (151) has a unique solution u ∈ C1 0T γn αn 2 The existence of global solutions requires more work. Here we give only some sufﬁcient conditions obtained from Theorems 3.15 and 3.16. Corollary 4.28. Let φ ∈ H∞ and assume that for any T>0 there is M>0 such that any solution u ∈ C10TH∞ of problem (151) satisﬁes

ut ≤ M 0 ≤ t

σ Corollary 4.29. Let φ ∈ H n σn ≥ βn + 2 Assume that for any T>0 1 α α +2 there is M>0 so that any solution u ∈ C 0TH n ∩C0TH n β +2 of (151) satisﬁes ut∈H n 0 ≤ t

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