A nonlinear inverse problem of the Korteweg-de Vries equation

Transcription

1 Bull. Math. Sci. A nonlinear inverse problem of the Korteweg-e Vries equation Shengqi Lu Miaochao Chen 2 Qilin Liu 3 Receive: 0 March 207 / Revise: 30 April 208 / Accepte: 5 May 208 The Author(s) 208 Abstract In this paper we prove the existence an uniqueness of solutions of an inverse problem of the simultaneous recovery of the evolution of two coefficients in the Korteweg-e Vries equation. Keywors Korteweg-e Vries equation Inverse problem Integral overetermination Two coefficients Mathematics Subject Classification 35K55 49K20 82D55 Introuction We consier the following nonlinear Korteweg-e Vries (KV) equation for an unknown scalar function u = u(x, t), x R, t R: Communicate by Ari Laptev. B Miaochao Chen chenmiaochao@chu.eu.cn Shengqi Lu 00336@sju.eu.cn Qilin Liu liuqlseu@26.com Department of Mathematics an Physics, Sanjiang University, Nanjing 2002, People s Republic of China 2 School of Applie Mathematics, Chaohu University, Hefei , People s Republic of China 3 School of Mathematics, Southeast University, Nanjing 289, People s Republic of China

2 S. Lu et al. u t + uu x + u xxx + α(t)u = f (t)g(x), x, t (0, T ), (.) u(x +, t) = u(x, t), x R, t R, (.2) u(x, 0) = u 0 (x), x = (0, ), (.3) with the conition of integral overeterminations: u(x, t)w (x)x = φ (t), u(x, t)w 2 (x)x = φ 2 (t), t [0, T ]. (.4) Here T is a given positive constant, u 0, g,w,w 2,φ an φ 2 are known functions, α an f are two coefficients to be etermine by (.4). Here it shoul be note that all functions are real. We will use the Hilbert space H k := { f H k ; f, f,..., f (k) L 2 }, an Hper k () := {u H k ; u(x) = u(x + ) for all x R}. When f (t) = α(t) = 0, u 0 Hper 2 (), the existence an uniqueness of solutions for the problem (.) (.3) was prove by Temam [] an later by Bona an Smith [2]. In this paper, we want to stuy the nonlinear inverse problem consists of fining a set of the functions {u,α, f } satisfying (.) (.4). This kin of inverse problem of incompressible Naiver Stokes equations an Ginzburg Lanau equations in superconuctivity has been stuie in [3 5]. Throughout this paper, we will assume (H) u 0 Hper 2 (), (H2) α, f C[0, T ], (H3) g Hper 2 (), (H4) w,w 2 Hper 3 (), (H5) φ,φ 2 C [0, T ], u 0 w (x)x = φ (0), u 0 w 2 (x)x = φ 2 (0), φ 2 g 0 φ g 20 = 0 for all t [0, T ], (H6) g(x)w (x)x = g 0 = 0, g(x)w 2 (x)x = g 20 = 0. Remark. (H5) implies that w an w 2 be linearly inepenent. Let us explain what a solution to the irect problem (.) (.3) is. Definition. Assume that (H), (H2) an (H3) hol true. A function u in L (0, T ; Hper 2 ()) C([0, T ]; L2 ()) is calle a solution to (.) (.3) if(.2) an (.3) hol true everywhere an (.) hols true in the following sense, T T (uφ t + u xx φ x )xt + (uu x + αu fg)φxt 0 0 = (u 0 φ(x, 0) u(x, T )φ(x, T ))x (.5) for all φ L (0, T ; H per ()).

3 A nonlinear inverse problem of the Korteweg-e Vries equation By a similar proof as that in [6], we can prove an existence an uniqueness result of the irect problem (.) (.3) an we here omit the etails. Theorem. Let (H) (H3) be satisfie. Then there exists a unique solution u satisfying u L (0, T ; H 2 per ()) C([0, T ], L2 ()), u t L (0, T ; H ()), T > 0. Base on Theorem., we can efine the nonlinear operator A : C([0, T ]) C([0, T ]) C([0, T ]) C([0, T ]) acting on every vector χ ={α(t), f (t)} as follows: [A(χ)](t) ={[A (α, f )](t), [A 2 (α, f )](t)}, t [0, T ] where [ A (α, f ) ] ( ) (t) := {φ φ 2 g 0 φ g g 20 φ 2 g 0 + g u2 w 2 + uw 2 x ( ) } g 20 2 u2 w + uw x, (.6) [ A2 (α, f ) ] { ( ) (t) := φ φ 2 g 0 φ g φ 2 φ φ 2 + φ 20 2 u2 w 2 + uw 2 x ( ) } φ 2 2 u2 w + uw x, (.7) an u has been alreay foun as the unique solution of the system (.) (.3). We procee to stuy the operator equation of the secon kin over the space C([0, T ]) C([0, T ]): χ = Aχ (.8) Here the space C([0, T ]) C([0, T ]) is equippe with the norm u C([0,T ]) C([0,T ]) = u C([0,T ]) + u 2 C([0,T ]), where u an u 2 are the components of the vector u an u i C = sup exp{ γ t}u i (t), t [0,T ] with γ = a positive constant to be etermine. Let us explain what a solution to the inverse problem (.) (.4) is. Definition.2 Let (H) an (H3) (H6) hol true. A vector function (u,α, f ) L (0, T ; H 2 per ()) C([0, T ]; L2 ()) C([0, T ]) C([0, T ])

4 S. Lu et al. is calle a solution to the inverse problem (.) (.4) if(.2), (.3) an (.4) hol true everywhere an (.) hols true in the sense of (.5). An interrelation between the inverse problem (.) (.4) an the nonlinear equation (.8) from the view point of their solvability is reveale in the following assertion. Theorem.2 Let (H) (H6) be satisfie. Then the following assertions are vali: (a) if the inverse problem (.) (.4) is solvable, then so is the Eq. (.8). (b) if Eq. (.8) has a solution, then there exists a solution of the inverse problem (.) (.4). Proof The proof is the same as that of [3, pp ], however, for the reaer s convenience, we present the proof. (a) Let the inverse problem (.) (.4) possess a solution, say {u,α, f }. Now testing (.) byw i (i =, 2) an using (.4), we see that φ i (t) 2 u 2 w i x uw i x + αφ i (t) = f (t)g i0. Solving the above system an using (H6) an (H5) we conclue that α an f solve the Eq. (.8). (b) We suppose that Eq. (.8) has a solution, say {α, f }. By Theorem. on the unique solvability of the irect problem we are able to recover u as the solution of (.) (.3) associate with {α, f }, so that it remains to be shown that the function u satisfies the over-etermination conition (.4), which follow from (.6), (.7), an (.8) immeiately. This provies support for the view that {u,α, f } is just a solution of the inverse problem (.) (.4). This completes the proof. Now we are in a position to state our main result. Theorem.3 Let (H) (H6) be satisfie an T be small enough. Then there exists a unique solution {u,α, f } to the inverse problem (.) (.5) with (α, f ) D where D := { (α, f ) C([0, T ]) C([0, T ]) (α, f ) C } NT e γ T an N is a large integer to be etermine in the following calculations. We will use := L 2 (). In the next Sect. 2, we give some preliminaries to the proof of Theorem.3, which is given in the final Sect Preliminaries Let u be the unique solution of the problem (.) (.3) when (α, f ) D is given. We will prove some a priori estimates for u.

5 A nonlinear inverse problem of the Korteweg-e Vries equation Lemma 2. Let N 2. Then u(, t) 2 u 0 + g =:a, t [0, T ]. (2.) Here Te γ T α C 2 an Teγ T f C 2. Proof Multiplying (.) byu an integrating by parts lea to 2 t u 2 + α u 2 = f (t) gu x f g u, an thus Diviing both sies by u, we get Integrating this inequality gives u t u α u 2 + f g u. u α u + f g. t u u 0 +Te γ T α C u L (0,T ;L 2 ()) + g Te γ T f C u u L (0,t;L 2 ()) + 2 g which gives (2.). Lemma 2.2 Let N 8. Then u x (, t) (2 u 0 + g ) 0/3 + 4 (2 u 0 + g ) g g L (2 u 0 + g ) u u 0 3 L 3 =: a2 2. (2.2) Here Te γ T α C 8 an Teγ T f C 8. Proof Let I (u) := ( u 2 x 3 u3) x, then from [6, pp ] we have t I (u) = α(t) (u 3 2u 2 x )x + f (t) (2g u 2 gu 2 )x 2 α(t) u 2 x x + α(t) u 3 x + f (t) u 2 x x + f (t) ( g 2 + g L u 2 ). (2.3)

6 S. Lu et al. Also from [6, pp. 262] we estimate u 3 x u x u 0/3 + u 3. (2.4) From (2.3) an (2.4) we euce that ( ) 3 t I (u) (3 α(t) + f(t) ) u 2 x x + α(t) 4 u 0/3 + u 3 + f (t) ( g 2 + g L u 2 ). Integrating this inequality gives u 2 x x (3 α C + f C )Te γ T u x 2 L (0,t;L 2 ()) + Te γ T α C [ 3 4 (2 u 0 + g ) 0/3 + (2 u 0 + g ) 3 ] [ + Te γ T f C g 2 + g L (2 u 0 + g ) 2] + 3 u 3 x + I (u 0 ) 2 u x 2 L (0,t;L 2 ()) + [ ] (2 u 0 + g ) 0/3 + (2 u 0 + g ) [ g 2 + g L (2 u 0 + g ) 2 ]+ 3 u 3 x + I (u 0 ) which gives (2.2) by(2.4) an (2.). Lemma 2.3 Let N 8. Then there exists a positive constant C epening only on u 0 H 2 an g H 2 such that u xx (, t) 2 C. (2.5) Here Te γ T α C 8 an Teγ T f C 8. We use the embeing inequality u L C 0 ( u + u x ) C 0 (a + a 2 ) =: a 3, an u 3 x a a a 3 =: a 4, an C := 2I 2 (u 0 ) a 3a2 2 + [ g g L a g 2 L a2 + 5 ] 9 g L a 4.

7 A nonlinear inverse problem of the Korteweg-e Vries equation Proof Let I 2 (u) := Testing (.) by ( u 2 xx 5 3 uu2 x + 5 ) 36 u4 x, we use the iea in [6, pp. 275]. L 2 (u) := 2u xxxx u2 x uu xx u3. an We obtain t I 2(u) = u t L 2 (u)x, α(t) ( 2u 2xx + 5uu2x 59 ) u4 f (t) x = (2g u xx + 53 gu2x + 03 guu xx + 59 gu3 ) x = αul 2 (u)x, fgl 2 (u)x, (uu x + u xxx )L 2 (u)x = (uu x + u xxx ) (2u xxxx + 53 u2x + 03 uu xx + 59 ) u3 x [ = ( u 2 x uu xx)2u xxx uu3 x + 0 ( 5 3 u2 u x u xx + 3 u2 x uu xx + 5 ) ] 9 u3 u xxx x [( = u2 x uu xx + 5 ) 9 u3 u xxx + 53 uu3x + 53 ] u2 ((u x ) 2 ) x x [( 2 = 3 u xu 2 xx + 2 ) 3 u(u2 xx ) x 5 3 u2 u x u xx 5 ] 3 uu3 x x = 5 (u 2 u x u xx + uu 3 x 3 )x = 5 ( uu 3 x 3 + uu3 x )x = 0. Thus we have t I 2(u) = α(t) ( 2u 2xx + 5uu2x 59 ) u4 x + f (t) (2g u xx + 53 gu2x + 03 guu xx + 59 ) gu3 x 2( α(t) + f (t) ) u 2 xx x + 5 α(t) u L u x α(t) u 4 x [ + f (t) g g L u x g 2 L u ] 9 g L u 3 L 3. (2.6)

8 S. Lu et al. Now estimating as Lemma 2.2 gives (2.5). Lemma 2.4 Let u i (i =, 2) be the corresponing solutions of the problem (.) (.3) with ata α i, f i (i =, 2), then there exists a positive constant C 2 epening only on u 0 H 2 an g H 2 such that 4 sup (u u 2 )(,τ) e γ t [ f f 2 C g +(2 u 0 + g ) α α 2 C ]. [0,t] γ 2C 2 (2.7) Here we take N 8 an γ > 2C 2 an C 2 u x + u 2x L ( (0,T )). Weuse u xx = u 2xx = 0, an the embeing inequality u x + u 2x L C 0 u xx + u 2xx 2C 0 C =: C 2. Proof From (.) we obtain (u u 2 ) t + 2 (u2 u2 2 ) x + (u u 2 ) xxx + α (u u 2 ) + (α α 2 )u 2 = ( f f 2 )g. (2.8) Multiplying (2.8) byu u 2 an integrating by parts, an noting that (u 2 2 u2 2 ) x(u u 2 )x = + u 2 ) x (u u 2 ) 2 (u 2 x + 2 = + u 2 ) x (u u 2 ) 2 (u 2 x 4 = (u + u 2 ) x (u u 2 ) 2 x, 4 we obtain (u + u 2 )(u u 2 ) x (u u 2 )x (u + u 2 ) x (u u 2 ) 2 x u 2 ) 2 t (u 2 x = (u + u 2 ) x (u u 2 ) 2 x α (t) (u u 2 ) 2 x 4 + ( f (t) f 2 (t)) g(u u 2 ) x (α (t) α 2 (t)) u 2 (u u 2 ) x 4 C 2 (u u 2 ) 2 x + α (t) (u u 2 ) 2 x + g f (t) f 2 (t) u u 2 + u 2 α (t) α 2 (t) u u 2,

9 A nonlinear inverse problem of the Korteweg-e Vries equation which gives t u u 2 C 2 u u 2 + α (t) u u 2 + g f (t) f 2 (t) + u 2 α (t) α 2 (t). Integrating the above inequality, we erive (u u 2 )(, t) C 2 t 0 + g t (u u 2 )(,τ) τ + t 0 f (τ) f 2 (τ) τ t 0 α (τ) τ sup (u u 2 )(,τ) [0,t] + u 2 L (0,t;L 2 ()) α (τ) α 2 (τ) τ 0 t C 2 (u u 2 )(,τ) τ sup (u u 2 )(,τ) [0,t] + γ eγ t ( f f 2 C g + u 2 L (0,t;L 2 ()) α α 2 C ), which leas to t sup (u u 2 )(,τ) 2C 2 sup (u u 2 )(,ξ) τ [0,t] 0 [0,τ] + 2 γ eγ t [ f f 2 C g +(2 u 0 + g ) α α 2 C ]. An application of Gronwall s inequality gives (2.7). 3 Proof of Theorem.3 Lemma 3. Let (H) (H6) be satisfie. If T is small enough, then A maps D onto itself. Proof Using Lemma 2., wehave [ Aχ C φ φ 2 g 0 φ g 20 C φ 2 C + φ 2 C φ C + ( φ 2 C + g 20 ) C ( ) 2 u 2 w L + u w ( ) ] + ( φ C + g 0 ) 2 u 2 w 2 L + u w 2 + φ C g 20 + φ 2 C g 0 φ 2 g 0 φ g 20 C { φ C φ 2 C + φ 2 C φ C

10 [ ] + ( φ 2 C + g 20 ) 2 (2 u 0 + g ) 2 w L + (2 u 0 + g ) w [ ] + ( φ C + g 0 ) 2 (2 u 0 + g ) 2 w 2 L + (2 u 0 + g ) w 2 + φ C g 20 + φ 2 C g 0 } NT e γ T S. Lu et al. if T is taken small enough. Lemma 3.2 Let (H) (H6) be satisfie. Then there exists a positive constant C 3 such that if γ = 3C 2 + C 3 an T is small enough, then the operator A is a contraction mapping in the ball D. Proof Using Lemmas 2. an 2.4, wehave e γ t Aχ Aχ 2 C φ 2 g 0 φ g 20 C [( ) 2 φ 2 C + g 20 u + u 2 L 2 w L + ( φ 2 C + g 20 ) w ( ) ] + 2 φ C + g 0 u +u 2 L 2 w 2 L + ( φ C + g 0 ) w 2 u u 2 e γ t {( ) φ 2 g 0 φ g 20 C 2 φ 2 C + g 20 (4 u 0 +2 g ) w L ( ) +( φ 2 C + g 20 ) w + 2 φ C + g 0 (4 u 0 +2 g ) w 2 L +( φ C + g 0 ) w 2 } u u 2 C 3 C 3 χ χ 2 χ χ 2 C. γ 2C 2 C 2 + C 3 This proves Lemma 3.2. Proof of Theorem.3 The proof of Theorem.3 follows easily from Lemma 3. an Lemma 3.2 by employing the contraction mapping principle. Acknowlegements The authors are inebte to the referee for careful reaing of the paper an many nice suggestions. This paper is supporte by the key project of university natural science of Anhui province (No. KJ207A453), the University Teaching Research Founation of Anhui province (No. 206 jyxm0693). Open Access This article is istribute uner the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricte use, istribution, an reprouction in any meium, provie you give appropriate creit to the original author(s) an the source, provie a link to the Creative Commons license, an inicate if changes were mae. References. Temam, R.: Sur un problème non linéaire. J. Math. Pures Appl. 48, (969) 2. Bona, J.L., Smith, R.: The initial value problem for the Korteweg-e Vries equation. Philos. Trans. Roy. Soc. Lon. Ser. A 278, (975)

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