ROTHE TIME-DISCRETIZATION METHOD APPLIED TO A QUASILINEAR WAVE EQUATION SUBJECT TO INTEGRAL pdf

ROTHE TIME-DISCRETIZATION METHOD APPLIED TO A QUASILINEAR WAVE EQUATION SUBJECT TO INTEGRAL CONDITIONS ABDELFATAH BOUZIANI AND NABIL MERAZGA Received 27 January 2004 and in revised form 12 February 2004 This paper presents a well-posedness result for an initial-boundary value problem with only integ ral conditions over the spatial domain for a one-dimensional quasilinear wave equation. The solution and some of its properties are obtained by means of a suitable application of the Rothe time-discretization method. 1. Introduction Recently, the study of initial-boundary value problems for hyperbolic equations with boundary integral conditions has received considerable attention. This kind of condi- tions has many important applications. For instance, they appear in the case where a direct measurement quantity is impossible; however, their mean values are known. In this paper, we deal with a class of quasilinear hyperbolic equations (T is a positive constant): ∂ 2 v ∂t 2 − ∂ 2 v ∂x 2 = f  x, t,v, ∂v ∂t  ,(x, t) ∈ (0,1) × [0,T], (1.1) subject to the initial conditions v(x,0)= v 0 (x), ∂v ∂t (x,0)= v 1 (x), 0  x  1, (1.2) and the boundar y integral conditions  1 0 v(x,t)dx = E(t), 0  t  T,  1 0 xv(x,t)dx = G(t), 0  t  T, (1.3) where f, v 0 , v 1 , E,andG are sufficiently regular given functions. Problems of this type were first introduced in [3], in which the first author proved the well-posedness of certain linear hyperbolic equations with integral condition(s). Later, Copyright © 2004 Hindawi Publishing Corporation Advances in Difference Equations 2004:3 (2004) 211–235 2000 Mathematics Subject Classification: 35L05, 35D05, 35B45, 35B30 URL: http://dx.doi.org/10.1155/S1687183904401071 212 On a quasilinear wave equation with integral conditions similar problems have been studied in [1, 4, 5, 6, 7, 8, 16, 24, 25] by using the energetic method, the Schauder fixed point theorem, Galerkin method, and the theory of charac- teristics. We refer the reader to [2, 9, 10, 11, 12, 13, 14, 15, 17, 21, 22, 23, 26] for other types of equations with integral conditions. Differently to these works, in the present paper, we employ the Rothe time-discre- tization method to construct the solution. This method is a convenient tool for both the theoretical and numerical analyses of the stated problem. Indeed, in addition to giving the first step towards a fully discrete approximation scheme, it provides a constructive proof of the existence of a unique solution. We remark that the application of Rothe method to this nonlocal problem is made possible thanks to the use of the so-called Bouziani space, first introduced by the first author, see, for instance, [4, 6, 20]. Introducing a new unknown function u(x,t) = v(x,t) − r(x,t), where r( x,t) = 6  2G(t) − E(t)  x − 2  3G(t) − 2E(t)  , (1.4) problem (1.1)–(1.3) with inhomogeneous integral conditions (1.3) can be equivalently reduced to the problem of finding a function u satisfying ∂ 2 u ∂t 2 − ∂ 2 u ∂x 2 = f  x, t,u, ∂u ∂t  ,(x, t) ∈ (0,1) × I, (1.5) u(x,0) = U 0 (x), ∂u ∂t (x,0)= U 1 (x), 0  x  1, (1.6)  1 0 u(x, t)dx = 0, t ∈ I, (1.7)  1 0 xu(x,t)dx = 0, t ∈ I, (1.8) where I := [0, T], f  x, t,u, ∂u ∂t  := f  x, t,u+ r, ∂u ∂t + ∂r ∂t  − ∂ 2 r ∂t 2 , U 0 (x):= v 0 (x) − r(x,0), U 1 (x) = v 1 (x) − ∂r ∂t (x,0). (1.9) Hence, instead of looking for v, we simply look for u. The solution of problem (1.1)–(1.3) will be directly obtained by the relation v = u + r. The paper is divided as follows. In Section 2, we present notations, definitions, as- sumptions, and some auxiliary results. Moreover, the concept of the required solution is stated, as well as the main result of the paper. Section 3 is devoted to the construction of approximate solutions of problem (1.5)–(1.8) by solving the corresponding linearized time-discretized problems, while in Section 4, some a priori estimates for the approxima- tions are derived. We end the paper by Section 5 whereweprovetheconvergenceofthe method and the well-posedness of the investigated problem. A. Bouziani and N. Mer azga 213 2. Preliminaries, notation, and main result Let H 2 (0,1) be the (real) second-order Sobolev space on (0,1) with norm · H 2 (0,1) and let (·,·)and·be the usual inner product and the corresponding norm, respectively, in L 2 (0,1). The nature of the boundary conditions (1.7)and(1.8) suggests introducing the following space: V :=  φ ∈ L 2 (0,1);  1 0 φ(x)dx =  1 0 xφ(x)dx = 0  , (2.1) which is clearly a Hilbert space for (·,·). Our analysis requires the use of the so-called Bouziani space B 1 2 (0,1) (see, e.g., [4, 5]) defined as the completion of the space C 0 (0,1) of real continuous functions with compact support in (0,1), for the inner product (u,v) B 1 2 =  1 0  x u · x vdx (2.2) and the associated norm v B 1 2 =  (v,v) B 1 2 , (2.3) where  x v :=  x 0 v(ξ)dξ for every fixed x ∈ (0,1). We recall that, for every v ∈ L 2 (0,1), the inequality v 2 B 1 2  1 2 v 2 (2.4) holds, implying the continuity of the embedding L 2 (0,1)  B 1 2 (0,1). Moreover, we will work in the standard functional spaces of the ty pes C(I,X), C 0,1 (I,X), L 2 (I,X), and L ∞ (I,X), where X is a Banach space, the main properties of which can be found in [19]. For a given function w(x,t), the notation w(t) is automatically used for the same func- tion considered as an abst ract function of the variable t ∈ I into some functional space on (0,1). Strong or weak convergence is denoted by → or , respectively. The Gronwall lemma in the following continuous and discrete forms will be ver y use- ful to us thereafter. Lemma 2.1. (i) Let x(t)  0,andleth(t), y(t) be real integrable functions on the interval [a,b].If y(t)  h(t)+  t a x(τ)y(τ)dτ, ∀t ∈ [a,b], (2.5) then y(t)  h(t)+  t a h(τ)x(τ)exp   t τ x(s)ds  dτ, ∀t ∈ [a,b]. (2.6) 214 On a quasilinear wave equation with integral conditions In particular , if x(τ) ≡ C is a constant and h(τ) is nondecreasing, then y(t)  h(t)e C(t−a) , ∀t ∈ [a,b]. (2.7) (ii) Let {a i } be a sequence of real nonnegative numbers satisfying a i  a+ bh i  k=1 a k , ∀i = 1, , (2.8) where a, b,andh are positive constants with h<1/b. Then a i  a 1 − bh exp  b(i − 1)h 1 − bh  , ∀i = 1,2, (2.9) Proof. Theproofisthesameasthatof[18, Lemma 1.3.19].  Throughout the paper, we will make the following assumptions: (H 1 ) f (t, w, p) ∈ L 2 (0,1) for each (t,w, p) ∈ I ×V × V and the following Lipschitz condition:   f (t,w, p) − f (t  ,w  , p  )   B 1 2  l  |t − t  | + w − w   B 1 2 + p − p   B 1 2  (2.10) is satisfied for all t,t  ∈ I and all w,w  , p, p  ∈ V, for some positive constant l; (H 2 ) U 0 ,U 1 ∈ H 2 (0,1); (H 3 ) the compatibility condition U 0 ,U 1 ∈ V, that is, concretely,  1 0 U 0 (x) dx =  1 0 xU 0 (x) dx = 0, (2.11)  1 0 U 1 (x) dx =  1 0 xU 1 (x) dx = 0. (2.12) We look for a weak solution in the following sense. Definit ion 2.2. A weak solution of problem (1.5)–(1.8) means a function u : I → L 2 (0,1) such that (i) u ∈ C 0,1 (I,V); (ii) u has (a.e. in I) strong derivatives du/dt ∈ L ∞ (I,V) ∩ C 0,1 (I,B 1 2 (0,1)) and d 2 u/dt 2 ∈ L ∞ (I,B 1 2 (0,1)); (iii) u(0) = U 0 in V and (du/dt)(0) = U 1 in B 1 2 (0,1); (iv) the identity  d 2 u dt 2 (t),φ  B 1 2 +  u(t),φ  =  f  t,u(t), du dt (t)  ,φ  B 1 2 (2.13) holds for all φ ∈ V and a.e. t ∈ I. A. Bouziani and N. Mer azga 215 Note that since u ∈ C 0,1 (I,V)anddu/dt ∈ C 0,1 (I,B 1 2 (0,1)), condition (iii) makes sense, whereas assumption (H 1 ), together with (i) and the fact that du/dt ∈ L ∞ (I,V)and d 2 u/dt 2 ∈ L ∞ (I,B 1 2 (0,1)), implies that (2.13) is well defined. On the other hand, the ful- fillment of the integral conditions (1.7)and(1.8)isincludedinthefactthatu(t) ∈ V,for all t ∈ I. The main result of the present paper reads as follows. Theorem 2.3. Under assumptions (H 1 ), (H 2 ), and (H 3 ), problem (1.5)–(1.8)admitsa unique weak solution u, in the sense of Definition 2.2, that depends continuously upon the data f , U 0 ,andU 1 . Moreover, the following convergence statements hold: u n −→ u in C(I,V), with convergence order O  1 n 1/2  , δu n −→ du dt in C  I,B 1 2 (0,1)  , d dt δu n  d 2 u dt 2 in L 2  I,B 1 2 (0,1)  , (2.14) as n→∞, where the sequences {u n } n and {δu n } n are defined in (3.18)and(3.19), respectively. 3. Construction of an approximate solution Let n be an arbitrary positive integer, and let {t j } n j =1 be the uniform partition of I, t j = jh n with h n = T/n. Successively , for j = 1, , n, we solve the linear stationary boundary value problem u j − 2u j−1 + u j−2 h 2 n − d 2 u j dx 2 = f j , x ∈ (0,1), (3.1)  1 0 u j (x) dx = 0, (3.2)  1 0 xu j (x) dx = 0, (3.3) where f j := f  t j ,u j−1 , u j−1 − u j−2 h n  , (3.4) starting from u −1 (x) = U 0 (x) − h n U 1 (x), u 0 (x) = U 0 (x), x ∈ (0,1). (3.5) Lemma 3.1. For each n ∈ N ∗ and each j = 1, ,n,problem(3.1) j –(3.3) j admits a unique solution u j ∈ H 2 (0,1). Proof. We use induction on j. For this, suppose that u j−1 and u j−2 are already known and that they belong to H 2 (0,1), then f j ∈ L 2 (0,1). From the classical theory of linear ordinary differential equations with constant coefficients, the general solution of (3.1) j 216 On a quasilinear wave equation with integral conditions which can be written in the form d 2 u j dx 2 − 1 h 2 n u j = −2u j−1 + u j−2 h 2 n − f j (3.6) is given by u j (x) = k 1 (x)cosh x h n + k 2 (x) sinh x h n , x ∈ (0,1), (3.7) where k 1 and k 2 are two functions of x satisfying the linear algebraic system dk 1 dx (x)cosh x h n + dk 2 dx (x) sinh x h n = 0, dk 1 dx (x) sinh x h n + dk 2 dx (x)cosh x h n = h n F j (x), (3.8) with F j := −2u j−1 + u j−2 h 2 n − f j . (3.9) Since the determinant of (3.8)is ∆ = cosh 2 x h n − sinh 2 x h n = 1, (3.10) then dk 1 dx (x) =         0 sinh x h n h n F j (x)cosh x h n         =− h n F j (x) sinh x h n , dk 2 dx (x) =         cosh x h n 0 sinh x h n h n F j (x)         = h n F j (x)cosh x h n , (3.11) that is, k 1 (x) =−h n  x 0 F j (ξ)sinh ξ h n dξ +λ 1 , k 2 (x) = h n  x 0 F j (ξ)cosh ξ h n dξ +λ 2 , (3.12) with λ 1 and λ 2 two arbitrary real constants. Inser ting (3.12)into(3.7), we get u j (x) = h n  x 0 F j (ξ)sinh x − ξ h n dξ +λ 1 cosh x h n + λ 2 sinh x h n . (3.13) A. Bouziani and N. Mer azga 217 Obviously, the function u j will be a solution to problem (3.1) j –(3.3) j ifandonlyifthe pair (λ 1 ,λ 2 ) is selected in such a manner that conditions (3.2) j and (3.3) j hold, that is, λ 1  1 0 cosh x h n dx + λ 2  1 0 sinh x h n dx =−h n  1 0  x 0 F j (ξ)sinh x − ξ h n dξ dx, λ 1  1 0 xcosh x h n dx + λ 2  1 0 xsinh x h n dx =−h n  1 0  x 0 xF j (ξ)sinh x − ξ h n dξ dx. (3.14) An easy computation shows that (λ 1 ,λ 2 ) is the solution of the linear algebraic system λ 1 sinh 1 h n + λ 2  cosh 1 h n − 1  =−  1 0  x 0 F j (ξ)sinh x − ξ h n dξ dx, λ 1  sinh 1 h n − h n cosh 1 h n + h n  + λ 2  cosh 1 h n − h n sinh 1 h n  =−  1 0  x 0 xF j (ξ)sinh x − ξ h n dξ dx, (3.15) whose determinant is D  h n  = 2h n − 2h n cosh 1 h n + sinh 1 h n = 2sinh 1 2h n  cosh 1 2h n − 2h n sinh 1 2h n  . (3.16) Note that D(h n ) does not vanish for any h n > 0, indeed equation D(h n ) = 0isequivalent to the equation cosh(1/2h n ) − 2h n sinh(1/2h n ) = 0, that is, to the equation tanh(1/2h n ) = 1/2h n which clearly has no solution. Therefore, for all h n > 0, system (3.15)admitsa unique solution (λ 1 ,λ 2 ) ∈ R 2 , which means that problem (3.1) j –(3.3) j is uniquely solv- able, and it is obvious that u j ∈ H 2 (0,1) since F j ∈ L 2 (0,1).  Now, we introduce the notations δu j := u j − u j−1 h n , j = 0, ,n, δ 2 u j := δu j − δu j−1 h n = u j − 2u j−1 + u j−2 h 2 n , j = 1, , n, (3.17) and construct the Rothe function u n : I → H 2 (0,1) ∩ V by setting u n (t) = u j−1 + δu j  t − t j−1  , t ∈  t j−1 ,t j  , j = 1, ,n, (3.18) 218 On a quasilinear wave equation with integral conditions and the following auxiliary functions: δu n (t) = δu j−1 + δ 2 u j  t − t j−1  , t ∈  t j−1 ,t j  , j = 1, ,n, (3.19) u n (t) =    u j for t ∈  t j−1 ,t j  , j = 1, ,n, U 0 for t ∈  − h n ,0  , (3.20) δu n (t) =    δu j for t ∈  t j−1 ,t j  , j = 1, ,n, U 1 for t ∈  − h n ,0  . (3.21) We expect that the limit u := lim n→∞ u n exists in a suitable sense, and that is the desired weak solution to our problem (1.5)–(1.8). The demonstration of this fact requires some a priori estimates whose derivation is the subject of the following section. 4. A pr i ori estimates for the approximations In what follows, c denote generic positive constants which are not necessarily the same at any two places. Lemma 4.1. There exist c>0 and n 0 ∈ N ∗ such that   u j    c, (4.1)   δu j    c, (4.2)   δ 2 u j   B 1 2  c, (4.3) for all j = 1, ,n and all n  n 0 . Proof. To derive these estimates, we need to write problem (3.1) j –(3.3) j in a weak for- mulation. Let φ be an arbit rary function from the space V defined in (2.1). One can easily find that  x 0 (x − ξ)φ(ξ)dξ = 2 x φ, ∀x ∈ (0, 1), (4.4) where  2 x φ := x   ξ φ  =  x 0 dξ  ξ 0 φ(η)dη. (4.5) This implies that  2 1 φ =  1 0 (1 − ξ)φ(ξ)dξ =  1 0 φ(ξ)dξ −  1 0 ξφ(ξ)dξ = 0. (4.6) Next, we multiply, for all j = 1, ,n,(3.1) j by  2 x φ and integrate over (0,1) to get  1 0 δ 2 u j (x) 2 x φdx−  1 0 d 2 u j dx 2 (x) 2 x φdx=  1 0 f j (x)  2 x φdx. (4.7) A. Bouziani and N. Mer azga 219 Here, we used the notations (3.17). Performing some standard integrations by par t s for each term in (4.7) and invoking (4.6), we obtain  1 0 δ 2 u j (x) 2 x φdx=  1 0 d dx   x  δ 2 u j   2 x φdx = x  δ 2 u j   2 x φ     x=1 x=0 −  1 0  x  δ 2 u j   x φdx =−  δ 2 u j ,φ  B 1 2 ,  1 0 d 2 u j dx 2 (x) 2 x φdx= du j dx (x) 2 x φ     x=1 x=0 −  1 0 du j dx (x) x φdx =−  1 0 du j dx (x) x φdx =−u j (x) x φ     x=1 x=0 +  1 0 u j (x) φ(x)dx =  u j ,φ  ,  1 0 f j (x) 2 x φdx=  1 0 d dx   x f j   2 x φdx = x f j  2 x φ     x=1 x=0 −  1 0  x f j  x φdx =−  f j ,φ  B 1 2 , (4.8) so that (4.7) becomes finally  δ 2 u j ,φ  B 1 2 +  u j ,φ  =  f j ,φ  B 1 2 , ∀φ ∈ V, ∀ j = 1, ,n. (4.9) Now, for i = 2, , j,wetakethedifference of the relations (4.9) i − (4.9) i−1 ,testedwith φ = δ 2 u i = (δu i − δu i−1 )/h n which belongs to V in view of (3.2) i − (3.3) i ,(3.2) i−1 − (3.3) i−1 ,and(H 3 ). We have  δ 2 u i − δ 2 u i−1 ,δ 2 u i  B 1 2 +  δu i ,δu i − δu i−1  =  f i − f i−1 ,δ 2 u i  B 1 2 , (4.10) then, using the identity 2  v,v − w  = v 2 −w 2 + v − w 2 (4.11) and its analog for (·,·) B 1 2 , it follows that   δ 2 u i   2 B 1 2 −   δ 2 u i−1   2 B 1 2 +   δ 2 u i − δ 2 u i−1   2 B 1 2 +   δu i   2 −   δu i−1   2 +   δu i − δu i−1   2 = 2  f i − f i−1 ,δ 2 u i  B 1 2 , (4.12) hence, omitting the third and last terms in the left-hand side, we get   δ 2 u i   2 B 1 2 +   δu i   2    δ 2 u i−1   2 B 1 2 +   δu i−1   2 +2   f i − f i−1   B 1 2   δ 2 u i   B 1 2 . (4.13) 220 On a quasilinear wave equation with integral conditions We sum up these inequalities and obtain   δ 2 u j   2 B 1 2 +   δu j   2    δ 2 u 1   2 B 1 2 +   δu 1   2 +2 j  i=2   f i − f i−1   B 1 2   δ 2 u i   B 1 2 , (4.14) hence, thanks to the Cauchy inequality 2ab  1 ε a 2 + εb 2 , ∀a,b ∈ R, ∀ε ∈ R ∗ + , (4.15) we can write, for ε = h n ,   δ 2 u j   2 B 1 2 +   δu j   2    δ 2 u 1   2 B 1 2 +   δu 1   2 + 1 h n j  i=2   f i − f i−1   2 B 1 2 + h n j  i=2   δ 2 u i   2 B 1 2 . (4.16) To majorize  j i=2  f i − f i−1  2 B 1 2 ,weremarkthat   f i − f i−1   2 B 1 2 =   f  t i ,u i−1 ,δu i−1  − f  t i−1 ,u i−2 ,δu i−2    2 B 1 2  l 2  h n +   u i−1 − u i−2   B 1 2 +   δu i−1 − δu i−2   B 1 2  2 = l 2 h 2 n  1+   δu i−1   B 1 2 +   δ 2 u i−1   B 1 2  2  3l 2 h 2 n  1+   δu i−1   2 B 1 2 +   δ 2 u i−1   2 B 1 2  , i = 2, , j. (4.17) Summing up for i = 2, , j, we may arrive at j  i=2   f i − f i−1   2 B 1 2  3l 2 ( j − 1)h 2 n +3l 2 h 2 n j  i=2    δu i−1   2 B 1 2 +   δ 2 u i−1   2 B 1 2  (4.18) or j  i=2   f i − f i−1   2 B 1 2  3l 2 ( j − 1)h 2 n +3l 2 h 2 n j−1  i=1    δ 2 u i   2 B 1 2 +   δu i   2 B 1 2  . (4.19) To e s timate δ 2 u 1  2 B 1 2 + δu 1  2 ,wetesttherelation(4.9) 1 with φ = δ 2 u 1 =(δu 1 − δu 0 )/h n = (δu 1 − U 1 )/h n which is an element of V owing to (3.2) 1 –(3.3) 1 and assumption (H 3 ). We have   δ 2 u 1   2 B 1 2 +  u 1 h n ,δu 1 − U 1  =  f 1 ,δ 2 u 1  B 1 2 (4.20) or   δ 2 u 1   2 B 1 2 +  δu 1 ,δu 1 − U 1  =  f 1 ,δ 2 u 1  B 1 2 −  U 0 ,δ 2 u 1  . (4.21) [...]... 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ROTHE TIME-DISCRETIZATION METHOD APPLIED TO A QUASILINEAR WAVE EQUATION SUBJECT TO INTEGRAL CONDITIONS ABDELFATAH BOUZIANI AND NABIL MERAZGA Received 27 January 2004 and in revised. February 2004 This paper presents a well-posedness result for an initial-boundary value problem with only integ ral conditions over the spatial domain for a one-dimensional quasilinear wave equation. . hyperbolic equations with boundary integral conditions has received considerable attention. This kind of condi- tions has many important applications. For instance, they appear in the case where a direct