Existence local and global solution of multipoint Cauchy problem for nonlocal nonlinear equations
aa r X i v : . [ m a t h . A P ] M a r Existence local and global solution of multipoint Cauchy problemfor nonlocal nonlinear equationsVeli B. Shakhmurov
Department of Mechanical Engineering, Okan University, Akfirat, Tuzla 34959Istanbul, E-mail: [email protected];
Rishad Shahmurov [email protected] of Alabama Tuscaloosa USA, AL 35487A bstract
In this paper, the multipoint Cauchy problem for nonlocal nonlinear wavetype equatıons are studied.The equation involves a convolution integral operatorwith a general kernel function whose Fourier transform is nonnegative. Weestablish local and global existence and uniqueness of solutions assuming enoughsmoothness on the initial data together with some growth conditions on thenonlinear term
Key Word:
Boussinesq equations , Hyperbolic equations, differential oper-ators, Fourier multipliers
AMS: 35Lxx, 35Qxx, 47D1 . Introduction
The aim in this paper is to study the existence and uniqueness of solution ofthe multipoint initial value problem (IVP) for nonlocal nonlinear wave equatıon u tt − a ∆ u + b ∗ u = ∆ [ g ∗ f ( u )] , x ∈ R n , t ∈ (0 , ∞ ) , (1.1) u ( x,
0) = ϕ ( x ) + m X k =1 α k u ( x, λ k ) , for a.e. x ∈ R n , (1.2) u t ( x,
0) = ψ ( x ) + m X k =1 β k u t ( x, λ k ) , for a.e. x ∈ R n , where m is an integer, λ k ∈ (0 , ∞ ) , α k , β k are complex numbers g ( x ), b ( x ) aremeasurable functions on (0 , ∞ ); a ≥ , ∆ denotes the Laplace operator in R n ,f ( u ) is the given nonlinear function, ϕ ( x ) and ψ ( x ) are the given initial valuefunctions. Note that for α k = β k = 0 we obtain classical Cauchy problem fornonlocal equation u tt − a ∆ u + b ∗ u = ∆ [ g ∗ f ( u )] , x ∈ R n , t ∈ (0 , ∞ ) , (1.3) u (0 , x ) = ϕ ( x ) , u t (0 , x ) = ψ ( x ) for a.e. x ∈ R n , The predictions of classical (local) elasticity theory become inaccurate whenthe characteristic length of an elasticity problem is comparable to the atomic1ength scale. To solution this situation, a nonlocal theory of elasticity was intro-duced (see [1 −
3] and the references cited therein) and the main feature of thenew theory is the fact that its predictions were more down to earth than thoseof the classical theory. For other generalizations of elasticity we refer the readerto [4 − − − L p − well-posedness of the classıcal Cauchy problem (1 .
3) depends cru-cially on the presence of a suitable kernel. Then the question that naturallyarises is which of the possible forms of the kernel functions are relevant for theglobal well-posedness of the multipoint initial-value problem (IVP) (1 . − (1 . . − (1 .
2) with a general class of kernel functions and provide local, globalexistence and blow-up results for the solutions of the problem (1 . − (1 .
2) infram of L p spaces. The kernel functions most frequently used in the literatureare particular cases of this general class of kernel functions. Note that nonlocalCauchy problem for wave equations were studied e.g. in [20, 21] . The strategy is to express the equation (1 .
1) as an integral equation. Totreat the nonlinearity as a small perturbation of the linear part of the equation,the contraction mapping theorem is used. Also, a priori estimates on L p normsof solutions of the linearized version are utilized. The key step is the derivation ofthe uniform estimate of the solutions of the linearized Boussinesq equation. Themethods of harmonic analysis, operator theory, interpolation of Banach Spacesand embedding theorems in Sobolev spaces are the main tools implemented tocarry out the analysis.In order to state our results precisely, we introduce some notations and somefunction spaces. Definitions and Background
Let E be a Banach space. L p (Ω; E ) denotes the space of strongly measurable E -valued functions that are defined on the measurable subset Ω ⊂ R n with thenorm k f k p = k f k L p (Ω; E ) = Z Ω k f ( x ) k pE dx p , ≤ p < ∞ , k f k L ∞ (Ω) = ess sup x ∈ Ω k f ( x ) k E . Let C denote the set of complex numbers. For E = C the L p (Ω; E ) denotesby L p (Ω) . E and E be two Banach spaces. ( E , E ) θ,p for θ ∈ (0 , , p ∈ [1 , ∞ ]denotes the interpolation spaces defined by K -method [22, § m be a positive integer. W m,p (Ω) denotes the Sobolev space, i.e. spaceof all functions u ∈ L p (Ω) that have the generalized derivatives ∂ m u∂x mk ∈ L p (Ω) , ≤ p ≤ ∞ with the norm k u k W m,p (Ω) = k u k L p (Ω) + n X k =1 (cid:13)(cid:13)(cid:13)(cid:13) ∂ m u∂x mk (cid:13)(cid:13)(cid:13)(cid:13) L p (Ω) < ∞ . Let H s,p ( R n ), ∞ < s < ∞ denotes fractionanal Sobolev space of order s which is defined as: H s,p = H s,p ( R n ) = ( I − ∆) − s L p ( R n )with the norm k u k H s,p = (cid:13)(cid:13)(cid:13) ( I − ∆) s u (cid:13)(cid:13)(cid:13) L p ( R n ) < ∞ . It clear that H ,p ( R n ) = L p ( R n ) . It is known that H m,p ( R n ) = W m,p ( R n )for the positive integer m (see e.g. [23, § . For p = 2 , the space H s,p ( R n )will be denoted by H s ( R n ) . Let S ( R n ) denote Schwartz class, i.e., the spaceof rapidly decreasing smooth functions on R n , equipped with its usual topologygenerated by seminorms. Let S ′ ( R n ) denote the space of all continuous linearoperators L : S ( R n ) → C , equipped with the bounded convergence topology.Recall S ( R n ) is norm dense in L p ( R n ) when 1 ≤ p < ∞ . Let 1 ≤ p ≤ q < ∞ . A function Ψ ∈ L ∞ ( R n ) is called a Fourier multiplierfrom L p ( R n ) to L q ( R n ) if the map B : u → F − Ψ( ξ ) F u for u ∈ S ( R n ) is welldefined and extends to a bounded linear operator B : L p ( R n ) → L q ( R n ) . Let L ∗ q ( E ) denote the space of all E − valued function space such that k u k L ∗ q ( E ) = ∞ Z k u ( t ) k qE dtt q < ∞ , ≤ q < ∞ , k u k L ∗∞ ( E ) = sup t ∈ (0 , ∞ ) k u ( t ) k E . Here, F denote the Fourier transform. Fourier-analytic representation ofBesov spaces on R n is defined as: B sp,q ( R n ) = n u ∈ S ′ ( R n ) , k u k B sp,q ( R n ) = (cid:13)(cid:13)(cid:13) F − t κ − s (cid:16) | ξ | κ (cid:17) e − t | ξ | F u (cid:13)(cid:13)(cid:13) L ∗ q ( L p ( R n )) , | ξ | = n X k =1 ξ k , ξ = ( ξ , ξ , ..., ξ n ) , p ∈ (1 , ∞ ) , q ∈ [1 , ∞ ] , κ > s } .
3t should be note that, the norm of Besov space does not depends on κ (seee.g. ( [22, § p = q the space B sp,q ( R n ) will be denoted by B sp ( R n ) . Sometimes we use one and the same symbol C without distinction in orderto denote positive constants which may differ from each other even in a singlecontext. When we want to specify the dependence of such a constant on aparameter, say α , we write C α . Moreover, for u , υ > u . υ means that there exists a constant C > u and υ such that u ≤ Cυ.
The paper is organized as follows: In Section 1, some definitions and back-ground are given. In Section 2, we obtain the existence of unique solution and apriory estimates for solution of the linearized problem (1 . − (1 . . In Section3, we show the existence and uniqueness of local strong solution of the problem(1 . − (1 . . − (1 . . Sometimes we use one and the same symbol C without distinction in orderto denote positive constants which may differ from each other even in a singlecontext. When we want to specify the dependence of such a constant on aparameter, say h , we write C h .
2. Estimates for linearized equation
In this section, we make the necessary estimates for solutions of the Cauchyproblem u tt − a ∆ u + b ∗ u = g ( x, t ) , x ∈ R n , t ∈ (0 , T ) , T ∈ (0 , ∞ ] , (2.1) u ( x,
0) = ϕ ( x ) + m X k =1 α k u ( x, λ k ) , for a.e. x ∈ R n , (2.2) u t ( x,
0) = ψ ( x ) + m X k =1 β k u t ( x, λ k ) , for a.e. x ∈ R n , Here, X p = L p ( R n ) , 1 ≤ p ≤ ∞ , Y s,p = H s,p ( R n ) , Y s,p = H s,p ( R n ) ∩ L ( R n ) , Y s,p ∞ = H s,p ( R n ) ∩ L ∞ ( R n ) , Y s,pp = H s,p ( R n ) ∩ L p ( R n ) , k u k Y s, p = k u k H s ( R n ) + k u k L p ( R n ) < ∞ , ≤ p ≤ ∞ . Let ˆ b ( ξ ) is the Fourier transformation of b ( x ) , i.e. ˆ b ( ξ ) = F b and let η = η ( ξ ) = h a | ξ | + ˆ b ( ξ ) i . κ k = κ k ( ξ ) = λ k η ( ξ ) , k = 1 , , ..., m. Condition 2.1.
Assume a ≥ , b is an integrable function whose ˆ b ( ξ ) ≥ a + ˆ b ( ξ ) > ξ ∈ R n . Moreover, let D ( ξ ) = 1 − m X k =1 ( α k + β k ) cos κ k + m X i,j =1 α i β j cos ( κ i − κ j ) = 0for all ξ ∈ R n with ξ = 0 . First we need the following lemmas:
Lemma 2.1.
Let the Condition 2.1. holds. Then, the problem (2 . − (2 . Proof.
By using of the Fourier transform, we get from (2 . − (2 . u tt ( ξ, t ) + η ( ξ ) ˆ u ( ξ, t ) = ˆ g ( ξ, t ) , (2.3)ˆ u ( ξ,
0) = ˆ ϕ ( ξ ) + m X k =1 α k ˆ u ( ξ, λ k , ) , ˆ u t ( ξ,
0) = ˆ ψ ( ξ ) + m X k =1 β k ˆ u ( ξ, λ k ) , (2.4)where ˆ u ( ξ, t ) is a Fourier transform of u ( x, t ) with respect to x and ˆ ϕ ( ξ ) , ˆ ψ ( ξ )are Fourier transform of ϕ, ψ, respectively.By using the variation of constants it is easy to see that the general solutionof (2 .
3) is represented asˆ u ( ξ, t ) = g ( ξ ) cos ηt + g ( ξ ) cos ηt + 12 η t Z sin η ( t − τ ) ˆ g ( ξ, τ ) dτ , (2.5)where g , g are general continuous differentiable functions. By taking themultipoint condition (2 .
4) from (2 .
5) we get that (2 . − (2 .
4) has a solution for ξ ∈ R n , when g , g are solution of the following system g − m X k =1 α k γ k + g cos κ k + g sin κ k = ˆ ϕ ( ξ ) , (2.6) ηg − m X k =1 β k [ η k + η ( − g sin κ k + g cos κ k )] = ˆ ψ ( ξ ) , where γ k = γ k ( ξ ) = 12 η λ k Z sin η ( λ k − τ ) ˆ g ( ξ, τ ) dτ ,µ k = µ k ( ξ ) = − λ k Z cos η ( λ k − τ ) ˆ g ( ξ, τ ) dτ .
5y Condition 2.1 we get D ( ξ ) = η ( ξ ) D ( ξ ) = 0for all ξ = 0 . Solving the system (2 . , we obtain g = D ( ξ ) D ( ξ ) , g = D ( ξ ) D ( ξ ) , (2.7)where D ( ξ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − m X k =1 α k cos κ k − m X k =1 α k sin κ k − − η m X k =1 β k sin κ k η − m X k =1 β k cos κ k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ,D ( ξ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˆ ϕ ( ξ ) + m X k =1 α k γ k − m X k =1 α k sin κ k ˆ ψ ( ξ ) + m X k =1 β k µ k η − m X k =1 β k cos κ k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ,D ( ξ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − m X k =1 α k cos κ k ˆ ϕ ( ξ ) + m X k =1 α k γ k − η m X k =1 β k sin κ k ˆ ψ ( ξ ) + m X k =1 β k µ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ;here, γ k ( ξ ) = 12 η λ k Z sin η ( λ k − τ ) ˆ g ( ξ, τ ) dτ ,µ k ( ξ ) = − λ k Z cos η ( λ k − τ ) ˆ g ( ξ, τ ) dτ . Hense, problem (2 . − (2 .
4) has a unique solution expressed as (2 . , where g and g are defined by (2 . , i.e. problem (2 . − (2 .
2) has a unique solution u ( x, t ) = F − [ C ( ξ, t ) g ( ξ )]+ (cid:2) F − S ( ξ, t ) g ( ξ ) (cid:3) + 12 η t Z F − [sin η ( t − τ ) ˆ g ( ξ, τ )] dτ . (2.8) Theorem 2.1.
Let the Condition 2.1 holds and s > n . Then for ϕ,ψ, g ( x, t ) ∈ Y s,p the solution (2 . − (2 .
2) satisfies the following uniformly in t ∈ [0 , T ] estimate k u k X ∞ + k u t k X ∞ ≤ C h k ϕ k Y s,p + (2.9)6 ψ k Y s,p + t Z (cid:0) k g ( ., τ ) k Y s,p + k g ( ., τ ) k X (cid:1) dτ , where the positive constant C depends only on initial data. Proof.
From (2 .
7) we deduced that g ( ξ ) = η − ( ξ ) D − ( ξ ) " η − m X k =1 β k cos κ k ! ˆ ϕ ( ξ ) +ˆ ψ ( ξ ) m X k =1 α k sin κ k ! + m X k =1 β k µ k ! m X k =1 α k sin κ k ! + η m X k =1 α k γ k − m X k =1 β k cos κ k ! , (2.10) g ( ξ ) = η − ( ξ ) D − ( ξ ) " − m X k =1 α k cos κ k ! ˆ ψ ( ξ ) + m X k =1 β k µ k − m X k =1 α k cos κ k ! + ˆ ϕ ( ξ ) m X k =1 β k sin κ k + η m X k =1 β k sin κ k ! m X k =1 α k γ k ! . Then, from (2 . .
8) and (2 .
10) we obtain that the solution (2 . − (2 .
2) canbe expressed as u ( x, t ) = S ( x, t ) ϕ + S ( x, t ) ψ +Φ ( g ) ( x, t )+ 12 η t Z F − [sin η ( t − τ ) ˆ g ( ξ, τ )] dτ , (2.11)where S ( x, t ) ϕ = F − ( D − ( ξ ) " − m X k =1 β k cos κ k ! sin ( ηt ) , + " η − ( ξ ) m X k =1 β k sin κ k cos ( ηt ) ˆ ϕ ( ξ ) } ,S ( x, t ) ψ = F − (" η − ( ξ ) D − ( ξ ) m X k =1 α k sin κ k ! sin ( ηt ) + η − ( ξ ) D − ( ξ ) " − m X k =1 α k cos κ k ! cos ( ηt ) ˆ ψ ( ξ ) , x ; t ) = Φ ( g ) ( x ; t ) = X j = m X k =1 F − Φ jk ( ξ ; t ) , where Φ k ( ξ ; t ) = 12 D − ( ξ ) β k A cos ( ηt ) λ k Z cos η ( λ k − τ ) ˆ g ( ξ, τ ) , Φ k ( ξ ; t ) = 12 D − ( ξ ) α k A cos ( ηt ) λ k Z sin η ( λ k − τ ) ˆ g ( ξ, τ ) , Φ k ( ξ ; t ) = 12 D − ( ξ ) β k B sin ( ηt ) λ k Z cos η ( λ k − τ ) ˆ g ( ξ, τ ) , Φ k ( ξ ; t ) = 12 D − ( ξ ) α k B sin ( ηt ) λ k Z sin η ( λ k − τ ) ˆ g ( ξ, τ ) , here, A = m X k =1 α k sin κ k , A = 1 − m X k =1 β k cos κ k ,B = − m X k =1 α k cos κ k ! , B = m X k =1 β k cos κ k . By Condition 2.1, D − ( ξ ) η − ( ξ ) , η − m X k =1 α k sin κ k ! , η − − m X k =1 α k cos κ k ! are uniformly bounded. From (2 .
11) and (2 .
8) we obtain | g ( ξ ) | . | ˆ ϕ ( ξ ) | + (cid:12)(cid:12)(cid:12) ˆ ψ ( ξ ) (cid:12)(cid:12)(cid:12) + | Φ ( ξ ) | , (2.12) | g ( ξ ) | . | ˆ ϕ ( ξ ) | + (cid:12)(cid:12)(cid:12) ˆ ψ ( ξ ) (cid:12)(cid:12)(cid:12) + | Φ ( ξ ) | , where Φ ( ξ ) = m X k =1 λ k Z ˆ g ( ξ, τ ) dτ , Let N ∈ N andΠ N = { ξ : ξ ∈ R n , | ξ | ≤ N } , Π ′ N = { ξ : ξ ∈ R n , | ξ | ≥ N } . .
8) we dedused that k S ( x, t ) g k X ∞ + k S ( x, t ) g k X ∞ . (cid:13)(cid:13) F − C ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ (Π N ) + (cid:13)(cid:13) F − S ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ (Π N ) + (2.13) (cid:13)(cid:13) F − C ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) + (cid:13)(cid:13) F − S ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) . From (2 . − (2 .
13) due to uniform boundedness of D − ( ξ ) and C ( ξ, t ), S ( ξ, t ) we have k S ( x, t ) g k X ∞ + k S ( x, t ) g k X ∞ . (cid:13)(cid:13) F − ˆ ϕ ( ξ ) (cid:13)(cid:13) X ∞ + (cid:13)(cid:13)(cid:13) F − ˆ ψ ( ξ ) (cid:13)(cid:13)(cid:13) X ∞ + (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) F − m X k =1 Φ ( ξ ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X ∞ , k S ( x, t ) g k X ∞ + k S ( x, t ) g k X ∞ . (cid:13)(cid:13) F − ˆ ϕ ( ξ ) (cid:13)(cid:13) X ∞ + (cid:13)(cid:13)(cid:13) F − ˆ ψ ( ξ ) (cid:13)(cid:13)(cid:13) X ∞ + (cid:13)(cid:13) F − Φ ( ξ ) (cid:13)(cid:13) X ∞ , In view of (2 . , by using the Minkowski’s inequality for integrals from abovewe get (cid:13)(cid:13) F − C ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ (Π N ) + (cid:13)(cid:13)(cid:13)(cid:13) F − S ( ξ, t ) g ( ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ (Π N ) . (2.14) (cid:2) k ϕ k X + k ψ k X + k g k X (cid:3) . Moreover, by (2 .
11) and (2 .
12) we have (cid:13)(cid:13) F − C ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) + (cid:13)(cid:13) F − S ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) . (cid:13)(cid:13) F − C ( ξ, t ) ˆ ϕ ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) + (cid:13)(cid:13)(cid:13) F − S ( ξ, t ) ˆ ψ ( ξ ) (cid:13)(cid:13)(cid:13) L ∞ ( Π ′ N ) + (cid:13)(cid:13) F − S ( ξ, t ) Φ ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) . = (cid:13)(cid:13)(cid:13)(cid:13) F − (cid:16) | ξ | (cid:17) − s C ( ξ, t ) (cid:16) | ξ | (cid:17) s ˆ ϕ ( ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( Π ′ N ) + (2.15) (cid:13)(cid:13)(cid:13)(cid:13) F − (cid:16) | ξ | (cid:17) − s S ( ξ, t ) (1 + | ξ | ) s ˆ ψ ( ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( Π ′ N ) + (cid:13)(cid:13)(cid:13)(cid:13) F − (cid:16) | ξ | (cid:17) − s S ( ξ, t ) (1 + | ξ | ) s Φ ( ξ ) (cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( Π ′ N ) +9y using (2 . .
7) and (2 .
12) we getsup ξ ∈ R n ,t ∈ [0 ,T ] | ξ | (cid:12)(cid:12)(cid:12)(cid:12) | α | + np D α (cid:20)(cid:16) | ξ | (cid:17) − s C ( ξ, t ) (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C , sup ξ ∈ R n ,t ∈ [0 ,T ] | ξ | (cid:12)(cid:12)(cid:12)(cid:12) | α | + np D α (cid:20)(cid:16) | ξ | (cid:17) − s S ( ξ, t ) (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (2.16)for s > np , α = ( α , α , ..., α n ), α k ∈ { , } , ξ ∈ R n and uniformly in t ∈ [0 , T ] . By multiplier theorems (see e.g. [24]) from (2 .
16) we get that thefunctions (cid:16) | ξ | (cid:17) − s C ( ξ, t ) , (cid:16) | ξ | (cid:17) − s S ( ξ, t ) are L p ( R n ) → L ∞ ( R n )Fourier multipliers. Then by Minkowski’s inequality for integrals, from (2 . . − (2 .
16) we obtain (cid:13)(cid:13) F − C ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) + (cid:13)(cid:13) F − S ( ξ, t ) g ( ξ ) (cid:13)(cid:13) L ∞ ( Π ′ N ) . (2.17)[ k ϕ k Y s,p + k ψ k Y s,p + k g k Y s,p ] . By reasoning as the above we have (cid:13)(cid:13) F − Φ ( ξ ) (cid:13)(cid:13) X ∞ ≤ C t Z (cid:0) k g ( ., τ ) k Y s + k g ( ., τ ) k X (cid:1) dτ . (2.18)Thus, from (2 .
8) and (2 .
15) we obtain k u k X ∞ ≤ C (cid:2) k ϕ k Y s,p + k ϕ k X + (2.19) k ψ k Y s,p + k ψ k X + t Z (cid:0) k g ( ., τ ) k Y s,p + k g ( ., τ ) k X (cid:1) dτ . By using (2 . , (2 .
7) and in view of (2 .
17) in similar way, we deduced theestimate of type (2 .
19) for u t , i.e. we obtain the assertion. Theorem 2.2.
Let the Condition 2.1 holds and s > n . Then for ϕ,ψ, g ( x, t ) ∈ Y s,p the solution of (2 . − (2 .
2) satisfies the following uniformestimate( k u k Y s,p + k u t k Y s,p ) ≤ C k ϕ k Y s,p + k ψ k Y s,p + t Z k g ( ., τ ) k Y s,p dτ . (2.20) Proof.
From (2 .
7) and (2 .
12) we have the following uniform estimate (cid:13)(cid:13)(cid:13)(cid:13) F − (cid:16) | ξ | (cid:17) s ˆ u (cid:13)(cid:13)(cid:13)(cid:13) X p + (cid:13)(cid:13)(cid:13)(cid:13) F − (cid:16) | ξ | (cid:17) s ˆ u t (cid:13)(cid:13)(cid:13)(cid:13) X p ! ≤ (2.21)10 (cid:26)(cid:13)(cid:13)(cid:13) F − (1 + | ξ | ) s C ( ξ, t ) ˆ ϕ (cid:13)(cid:13)(cid:13) X p + (cid:13)(cid:13)(cid:13) F − (1 + | ξ | ) s S ( ξ, t ) ˆ ψ (cid:13)(cid:13)(cid:13) X p + t Z (cid:13)(cid:13)(cid:13) (1 + | ξ | ) s ˆ g ( ., τ ) (cid:13)(cid:13)(cid:13) X p dτ . By Condition 2.1 and by virtue of Fourier multiplier theorems (see [24, § C ( ξ, t ), S ( ξ, t ) and Φ ( ξ ) are Fourier multipliers in L p ( R n ) uni-formly with respect to t ∈ [0 , T ] . So, the estimate (2 .
21) by using the Minkowski’sinequality for integrals implies (2 . .
3. Local well posedness of IVP for nonlinear nonlocal equation
In this section, we will show the local existence and uniqueness of solutionfor the Cauchy problem (1 . − (1 . . For the study of the nonlinear problem(1 . − (1 .
2) we need the following lemmas
Lemma 3.1 (Nirenberg’s inequality) [25]. Assume that u ∈ L p (Ω), D m u ∈ L q (Ω), p, q ∈ (1 , ∞ ). Then for i with 0 ≤ i ≤ m, m > nq we have (cid:13)(cid:13) D i u (cid:13)(cid:13) r ≤ C k u k − µp n X k =1 k D mk u k µq , (3.1)where 1 r = im + µ (cid:18) q − mn (cid:19) + (1 − µ ) 1 p , im ≤ µ ≤ . Lemma 3.2 [26] . Assume that u ∈ W m,p (Ω) ∩ L ∞ (Ω) and f ( u ) possessescontinuous derivatives up to order m ≥
1. Then f ( u ) − f (0) ∈ W m,p (Ω) and k f ( u ) − f (0) k p ≤ (cid:13)(cid:13)(cid:13) f (1) ( u ) (cid:13)(cid:13)(cid:13) ∞ k u k p , (cid:13)(cid:13) D k f ( u ) (cid:13)(cid:13) p ≤ C k X j =1 (cid:13)(cid:13)(cid:13) f ( j ) ( u ) (cid:13)(cid:13)(cid:13) ∞ k u k j − ∞ (cid:13)(cid:13) D k u (cid:13)(cid:13) p , 1 ≤ k ≤ m, (3.2)where C ≥ X p = L p ( R n ) , k u k p = k u k X p , Y = W ,p ( R n ) , E = ( X p , Y ) p ,p = B − p p ( R n ) . Remark 3.1.
By using J.Lions-I. Petree result (see e.g [21, § u → u ( t ), t ∈ [0 , T ] is continuous and surjective from W ,p (0 , T )onto E and there is a constant C such that k u ( t ) k E ≤ C k u k W ,p (0 ,T ) , ≤ p ≤ ∞ .11irst all of, we define the space Y ( T ) = C (cid:0) [0 , T ] ; Y ,p ∞ (cid:1) equipped with thenorm defined by k u k Y ( T ) = max t ∈ [0 ,T ] k u k Y ,p + max t ∈ [0 ,T ] k u k X ∞ , u ∈ Y ( T ) . It is easy to see that Y ( T ) is a Banach space. For ϕ , ψ ∈ Y ,p , let M = k ϕ k Y ,p + k ϕ k X ∞ + k ψ k Y ,p + k ψ k X ∞ . Definition 3.1.
For any
T > υ, ψ ∈ Y ,p ∞ and u ∈ C (cid:0) [0 , T ] ; Y ,p ∞ (cid:1) satisfies the equation (1 . − (1 .
2) then u ( x, t ) is called the continuous solutionor the strong solution of the problem (1 . − (1 . . If T < ∞ , then u ( x, t ) iscalled the local strong solution of the problem (1 . − (1 . . If T = ∞ , then u ( x, t ) is called the global strong solution of the problem (1 . − (1 . Condition 3.1.
Assume:(1) Assume that the kernel g is an integrable function whose Fourier trans-form satisfies 0 ≤ ˆ g ( ξ ) . (cid:16) | ξ | (cid:17) − for all ξ ∈ R n ;(2) The Condition 2.1 holds, ϕ, ψ ∈ Y ,p ∞ for 1 < p < ∞ and np < u → f ( x, t, u ): R n × [0 , T ] × E → E is a measurable in( x, t ) ∈ R n × [0 , T ] for u ∈ E ; f ( x, t, u ). Moreover, F ( x, t, u ) is continuousin u ∈ E and f ( x, t, u ) ∈ C (3) ( E ; E ) uniformly with respect to x ∈ R n ,t ∈ [0 , T ] . Main aim of this section is to prove the following result:
Theorem 3.1.
Let the Condition 3.1. holds. Then problem (1 . − (1 .
2) hasa unique local strange solution u ∈ C (2) (cid:0) [0 , T ) ; Y ,p ∞ (cid:1) , where T is a maximaltime interval that is appropriately small relative to M . Moreover, ifsup t ∈ [0 , T ) (cid:16) k u k Y ,p ∞ + k u t k Y ,p ∞ (cid:17) < ∞ (3.3)then T = ∞ . Proof.
First, we are going to prove the existence and the uniqueness of thelocal continuous solution of the problem (1 . − (1 .
2) by contraction mappingprinciple. Consider a map G on Y ( T ) such that G ( u ) is the solution of theCauchy problem G tt ( u ) − a ∆ G ( u ) = ∆ [ g ∗ f ( G ( u ))] , x ∈ R n , t ∈ (0 , T ) , (3.4) G ( u ) ( x,
0) = ϕ ( x ) + m X k =1 α k G ( u ) ( x, λ k ) , for a.e. x ∈ R n ,G t ( u ) (0 , x ) = ψ ( x ) + m X k =1 β k G t ( u ) ( x, λ k ) , for a.e. x ∈ R n . F ( u ) ∈ L p (cid:0) , T ; Y ,p ∞ (cid:1) for any T >
0. Thus,by Lemma 2.1, problem (3 .
4) has a solution which can be written as G ( u ) ( x, t ) = [ S ( x, t ) ϕ + S ( x, t ) ψ + Φ ( g ∗ f ( G ( u )))] + (3.5)12 η t Z F − h sin η ( t − τ ) | ξ | ˆ g ( ξ ) ˆ f ( G ( u ) ( ξ )) i dτ , where S ( x, t ), S ( x, t ) , Φ are operator functions defined by (2 .
10) and (2 . g replaced by g ∗ f ( G ( u )) . From Lemma 3.2 it is easy to see that themap G is well defined for f ∈ C (2) ( X ; C ). We put Q ( M ; T ) = n u | u ∈ Y ( T ) , k u k Y ( T ) ≤ M + 1 o . First, by reasoning as in [9] let us prove that the map G has a unique fixedpoint in Q ( M ; T ) . For this aim, it is sufficient to show that the operator G maps Q ( M ; T ) into Q ( M ; T ) and G : Q ( M ; T ) → Q ( M ; T ) is strictly contractive if T is appropriately small relative to M. Consider the function ¯ f ( ξ ): [0 , ∞ ) → [0 , ∞ ) defined by ¯ f ( ξ ) = max | x |≤ ξ n(cid:13)(cid:13)(cid:13) f (1) ( x ) (cid:13)(cid:13)(cid:13) C , (cid:13)(cid:13)(cid:13) f (2) ( x ) (cid:13)(cid:13)(cid:13) C o , ξ ≥ . It is clear to see that the function ¯ f ( ξ ) is continuous and nondecreasing on[0 , ∞ ) . From Lemma 3.2 we have k f ( u ) k Y ,p ≤ (cid:13)(cid:13)(cid:13) f (1) ( u ) (cid:13)(cid:13)(cid:13) X ∞ k u k X p + (cid:13)(cid:13)(cid:13) f (1) ( u ) (cid:13)(cid:13)(cid:13) X ∞ k Du k X p + C (cid:20)(cid:13)(cid:13)(cid:13) f (1) ( u ) (cid:13)(cid:13)(cid:13) X ∞ k u k X p + (cid:13)(cid:13)(cid:13) f (2) ( u ) (cid:13)(cid:13)(cid:13) X ∞ k u k X ∞ (cid:13)(cid:13) D u (cid:13)(cid:13) X p (cid:21) ≤ (3.6)2 C ¯ f ( M + 1) ( M + 1) k u k Y ,p . In view of the assumptıon (1) and by using Minkowski’s inequality for integrals¨owe obtain from (3 . k G ( u ) k X ∞ . k ϕ k ∞ + k ψ k ∞ + t Z k ∆ [ g ∗ f ( G ( u ))] ( x, τ ) dτ k ∞ , (3.7) k G ( u ) k Y ,p . k ϕ k Y ,p + k ψ k Y ,p + t Z k ∆ [ g ∗ f ( G ( u ))] ( x, τ ) dτ k Y ,p dτ . (3.8)Thus, from (3 . − (3 .
8) and Lemma 3.2 we get k G ( u ) k Y ( T ) ≤ M + T ( M + 1) (cid:2) C ( M + 1) ¯ f ( M + 1) (cid:3) . T satisfies T ≤ (cid:8) ( M + 1) (cid:2) C ( M + 1) ¯ f ( M + 1) (cid:3)(cid:9) − , (3.9)then k Gu k Y ( T ) ≤ M + 1 . Therefore, if (3 .
9) holds, then G maps Q ( M ; T ) into Q ( M ; T ) . Now, we aregoing to prove that the map G is strictly contractive. Assume T > u ,u ∈ Q ( M ; T ) given. We get G ( u ) − G ( u ) = t Z F − S ( t − τ , ξ ) | ξ | ˆ g ( ξ ) h ˆ f ( u ) ( ξ, τ ) − ˆ f ( u ) ( ξ, τ ) i dτ , t ∈ (0 , T ) . By using the assumption (3) and the mean value theorem, we obtainˆ f ( u ) − ˆ f ( u ) = ˆ f (1) ( u + η ( u − u )) ( u − u ) ,D ξ h ˆ f ( u ) − ˆ f ( u ) i = ˆ f (2) ( u + η ( u − u )) ( u − u ) D ξ u +ˆ f (1) ( u ) ( D ξ u − D ξ u ) ,D ξ h ˆ f ( u ) − ˆ f ( u ) i = ˆ f (3) ( u + η ( u − u )) ( u − u ) ( D ξ u ) +ˆ f (2) ( u ) ( D ξ u − D ξ u ) ( D ξ u + D ξ u ) +ˆ f (2) ( u + η ( u − u )) ( u − u ) D ξ u + ˆ f (1) ( u ) (cid:0) D ξ u − D ξ u (cid:1) , where 0 < η i < , i = 1 , , , . Thus, using H¨older’s and Nirenberg’s inequality,we have (cid:13)(cid:13)(cid:13) ˆ f ( u ) − ˆ f ( u ) (cid:13)(cid:13)(cid:13) X ∞ ≤ ¯ f ( M + 1) k u − u k X ∞ , (3.10) (cid:13)(cid:13)(cid:13) ˆ f ( u ) − ˆ f ( u ) (cid:13)(cid:13)(cid:13) X p ≤ ¯ f ( M + 1) k u − u k X p , (3.11) (cid:13)(cid:13)(cid:13) D ξ h ˆ f ( u ) − ˆ f ( u ) i(cid:13)(cid:13)(cid:13) X p ≤ ( M + 1) ¯ f ( M + 1) k u − u k X ∞ + (3.12)¯ f ( M + 1) (cid:13)(cid:13)(cid:13) ˆ f ( u ) − ˆ f ( u ) (cid:13)(cid:13)(cid:13) X p , (cid:13)(cid:13)(cid:13) D ξ h ˆ f ( u ) − ˆ f ( u ) i(cid:13)(cid:13)(cid:13) X p ≤ ( M + 1) ¯ f ( M + 1) k u − u k X ∞ (cid:13)(cid:13) D ξ u (cid:13)(cid:13) Y ,p +¯ f ( M + 1) k D ξ ( u − u ) k Y ,p k D ξ ( u + u ) k Y ,p +¯ f ( M + 1) k u − u k X ∞ (cid:13)(cid:13) D ξ u (cid:13)(cid:13) X p + ¯ f ( M + 1) k D ξ ( u − u ) k X p ≤ ¯ f ( M + 1) k u − u k X ∞ k u k X ∞ (cid:13)(cid:13) D ξ u (cid:13)(cid:13) X p + (3.13) C ¯ f ( M + 1) k u − u k X ∞ (cid:13)(cid:13) D ξ ( u − u ) (cid:13)(cid:13) X p k u + u k X ∞ (cid:13)(cid:13) D ξ ( u + u ) (cid:13)(cid:13) X p + ( M + 1) ¯ f ( M + 1) k u − u k X ∞ + ¯ f ( M + 1) (cid:13)(cid:13) D ξ ( u − u ) (cid:13)(cid:13) X p ≤ C ( M + 1) ¯ f ( M + 1) k u − u k X ∞ +2 C ( M + 1) ¯ f ( M + 1) (cid:13)(cid:13) D ξ ( u − u ) (cid:13)(cid:13) X p , where C is the constant in Lemma 3 .
1. From (3 . − (3 . X p spaces and Young’sinequality, we obtain k G ( u ) − G ( u ) k Y ( T ) ≤ t Z k u − u k X ∞ dτ + t Z k u − u k Y ,p dτ + t Z k f ( u ) − f ( u ) k X ∞ dτ + t Z k f ( u ) − f ( u ) k Y ,p dτ ≤ T h C ( M + 1) ¯ f ( M + 1) i k u − u k Y ( T ) , where C is a constant. If T satisfies (3 .
9) and the following inequality holds T ≤ h C ( M + 1) ¯ f ( M + 1) i − , (3.14)then k Gu − Gu k Y ( T ) ≤ k u − u k Y ( T ) . That is, G is a contructive map. By contraction mapping principle we knowthat G ( u ) has a fixed point u ( x, t ) ∈ Q ( M ; T ) that is a solution of (1 . − (1 . . − (2 .
11) we get that u is a solution of the following integral equation u ( x, t ) = S ( t ) g + S ( t ) g − t Z F − h S ( t − τ , ξ ) | ξ | ˆ g ( ξ ) ˆ f ( u ) ( ξ, τ ) i dτ , t ∈ (0 , T ) . Let us show that this solution is a unique in Y ( T ). Let u , u ∈ Y ( T ) aretwo solution of the problem (1 . − (1 . u − u = t Z F − h S ( t − τ , ξ ) | ξ | ˆ g ( ξ ) (cid:16) ˆ f ( u ) ( ξ, τ ) − ˆ f ( u ) ( ξ, τ ) (cid:17)i dτ . (3.15)By the definition of the space Y ( T ), we can assume that k u k X ∞ ≤ C ( T ) , k u k X ∞ ≤ C ( T ) . . k u − u k Y ,p ≤ C ( T ) t Z k u − u k Y ,p dτ . (3.16)From (3 .
16) and Gronwall’s inequality, we have k u − u k Y ,p = 0, i.e. prob-lem (1 . − (1 .
2) has a unique solution which belongs to Y ( T ) . That is, we obtainthe first part of the assertion.Now, let [0 , T ) be the maximal time interval of existence for u ∈ Y ( T ). Itremains only to show that if (3 .
3) is satisfied, then T = ∞ . Assume contrarythat, (3 .
3) holds and T < ∞ . For T ∈ [0 , T ) , we consider the following integralequation υ ( x, t ) = S ( t ) u ( x, T ) + S ( t ) u t ( x, T ) − (3.17) t Z F − h S ( t − τ , ξ ) | ξ | ˆ g ( ξ ) ˆ f ( υ ) ( ξ, τ ) i dτ , t ∈ (0 , T ) . By virtue of (3 . T ′ > T we havesup t ∈ [0 , T ) (cid:0) k u k Y ,p + k u k X ∞ + k u t k Y ,p + k u t k X ∞ (cid:1) < ∞ . By reasoning as a first part of theorem and by contraction mapping principle,there is a T ∗ ∈ (0 , T ) such that for each T ∈ [0 , T ) , the equation (3 .
17) has aunique solution υ ∈ Y ( T ∗ ) . The estimates (3 .
9) and (3 .
14) imply that T ∗ canbe selected independently of T ∈ [0 , T ) . Set T = T − T ∗ and define˜ u ( x, t ) = ( u ( x, t ) , t ∈ [0 , T ] υ ( x, t − T ) , t ∈ h T, T + T ∗ i . (3.18)By construction ˜ u ( x, t ) is a solution of the problem (1 . − (1 .
2) on h T, T + T ∗ i and in view of local uniqueness, ˜ u ( x, t ) extends u. This is against to the maxi-mality of [0 , T ), i.e we obtain T = ∞ . Consider the problem (1 . − (1 . , when ϕ, ψ ∈ H s . We first need two lemmas concerning the behaviour of the nonlinear term[8, 13, 27].
Lemma 3.3.
Let s ≥ , f ∈ C [ s ]+1 ( R ) with f (0) = 0. Then for any u ∈ H s ∩ L ∞ , we have f ( u ) ∈ H s ∩ L ∞ . Moreover there is some constant A ( M )depending on M such that for all u ∈ H s ∩ L ∞ with k u k L m ≤ M, k f ( u ) k H s ≤ A ( M ) k u ) k H s . emma 3.4. Let s ≥ , f ∈ C [ s ]+1 ( R ). Then for for any M there issome constant B ( M ) depending on M such that for all u , υ ∈ H s ∩ L ∞ with k u k L m ≤ M, k υ k L ∞ ≤ M, k u k H ≤ M, k υ k H s ≤ M, k f ( u ) − f ( υ k H s ≤ B ( M ) k u − υ k H s , k f ( u ) − f ( υ k L ∞ ≤ B ( M ) k u − υ k L ∞ . By reasoning as in [13, Lemma 3.4] we have
Corollary 3.1.
Let s > n , f ∈ C [ s ]+1 ( R ). Then for for any B there is someconstant B ( M ) depending on M such that for all u , υ ∈ H s with k u k H s ≤ M, k υ k H s ≤ M, k f ( u ) − f ( υ k H s ≤ B ( M ) k u − υ k H s . Lemma 3.5. If s >
0, then Y s, ∞ is an algebra. Moreover, for f, g ∈ Y s, ∞ , k f g k H s ≤ C [ k f k ∞ + k g k H s + k f k H s + k g k ∞ ] . Lemma 3.6 [28, Lemma X 4] . Let s ≥ , f ∈ C [ s ]+1 ( R ) and f ( u ) = O (cid:16) | u | α +1 (cid:17) for u → α ≥ u ∈ Y s, ∞ and k u k ∞ ≤ M ,then k f ( u ) k H s ≤ C ( M ) [ k u k H s k u k α ∞ ] , k f ( u ) k ≤ C ( M ) k u k k u k α − ∞ . Lemma 3.7 [13, Lemma 3.4] . Let s ≥ , f ∈ C [ s ]+1 ( R ) and f ( u ) = O (cid:16) | u | α +1 (cid:17) for u → α ≥ u, υ ∈ Y s, ∞ , k u k H s ≤ M , k υ k H s ≤ M and k u k ∞ ≤ M , k υ k ∞ ≤ M, then k f ( u ) − f ( u ) k H s ≤ C ( M ) [( k u k ∞ − k υ k ∞ ) ( k u k H s + k υ k H s )( k u k ∞ + k υ k ∞ ) α − , k f ( u ) − f ( υ k ≤ C ( M ) ( k u k ∞ + k υ k ∞ ) α − ( k u k + k υ k ) k u − υ k . By reasoning as in Theorem 3.1 and [13, Theorem 1.1] we have:
Theorem 3.2.
Let the Condition 3.1 hold. Assume f ∈ C k ( R ) , with k an integer k ≥ s > n , satisfies f ( u ) = O (cid:16) | u | α +1 (cid:17) for u → . Then there existsa constant δ > , such that for any ϕ, ψ ∈ Y s, satisfying k ϕ k Y s, + k ψ k Y s, ≤ δ, (3.19)problem (1 . − (1 .
2) has a unique local strange solution u ∈ C (2) (cid:0) [0 , ∞ ) ; Y s, (cid:1) .Moreover, sup ≤ t< ∞ ( k u k Y s, + k u t k Y s, ) ≤ Cδ, (3.20)where the constant C only depends on f and initial data.17 roof. Consider a metric space defined by W s = n u ∈ (2) (cid:0) [0 , ∞ ) ; Y s, (cid:1) , k u k W s ≤ C δ o , equipped with the norm k u k W s = sup t ≥ (cid:16) k u k Y s, ∞ + k u t k Y s, ∞ (cid:17) , where δ > .
19) and C is a constant in Theorem 2.1. It is easy toprove that W s is a complete metric space. From Sobolev imbedding theorem weknow that k u k ∞ ≤ δ is enough small. Consider the problem(3 . f ( u ) ∈ L (cid:16) , T ; Y s, (cid:17) for any T > .
4) has a unique solution which can be written as (3 . . Weshould prove that the operator G ( u ) defined by (3 .
5) is strictly contractive if δ is suitable small. In fact, by (2 .
9) in Theorem 2.1 and Lemma 3.6 we get k G ( u ) k ∞ + k G t ( u ) k ∞ ≤ C h k ϕ k Y s, + k ψ k Y s, + t Z k K ( u ) ( ., τ ) k Y s, dτ ≤ C δ + C t Z k K ( u ) ( ., τ ) k Y s, dτ ≤ C δ + C t Z h k u ( ., τ ) k k u ( ., τ ) k α − ∞ + k u ( τ ) k H s k u ( τ ) k α ∞ i dτ ≤ C δ + C k u k α +1 W s . (3.21)On the other hand, by (2 .
20) in Theorem 2.2 and Lemma 3.6 we have k G ( u ) k H s + k G t ( u ) k H s ≤ C [ k ϕ k H s + k ψ k H s + t Z k K ( u ) ( ., τ ) k H s dτ ≤ C δ + C t Z k K ( u ) ( ., τ ) k H s dτ ≤ C δ + C t Z k u ( τ ) k H s k u ( τ ) k α ∞ dτ ≤ C δ + C k u k α +1 W s . (3.22)Therefore, combining (3 .
21) with (3 .
22) yields k G ( u ) k W s ≤ C δ + C k u k α +1 W s . (3.23)18aking that δ is enough small such that C (3 C δ ) α < /
3, from (3 .
23) andfrom Theorems 2.1, 2.2 we dedused that G maps W s into W s . Then, by rea-soning as the remaining part of [13, Theorem 1.1] we obtain that G : W s → W s is strictly contractive. Using the contraction mapping principle, we know that G ( u ) has a unique fixed point u ( x, t ) ∈ C ([0 , ∞ ) , H s ) and u ( x, t ) is the solutionof the problem (1 . − (1 . u ( x, t ) of the problem (1 . − (1 .
2) is also unique in C ([0 , ∞ ) , H s ). In fact, let u and u be two solutions of the problem (1 . − (1 . u , u ∈ C ([0 , ∞ ) , H s ). Let u = u − u ; then u tt − a ∆ u + b ∗ u = ∆ [ g ∗ ( f ( u ) − f ( u ))] . This fact is derived in a similar way as in Theorem 3.2, by using Theorems2.1, 2.2 and Gronwall’s inequality.
Condition 3.2.
Let the Condition 2.1 holds. Assume f ∈ C [ s ]+1 ( R ) with f (0) = 0 for some s ≥ g is an integrable function whose Fourier trans-form satisfies 0 ≤ ˆ g ( ξ ) ≤ (cid:16) | ξ | (cid:17) − r for all ξ ∈ R n and r ≥ . Theorem 3.3.
Let the Condition 3.2 hold. Moreover, s ≥ r ≥ T > . − (1 .
2) is wellposed with solution in C ([0 , T ] ; H s ) for initial data ϕ, ψ ∈ H s . Proof.
Consider the convolution operator u → Ku = ∆ [ g ∗ f ( u )] . In viewof assumptios we have k ∆ g ∗ υ k H s . (cid:13)(cid:13)(cid:13) (1 + ξ ) s | ξ | ˆ g ( ξ ) ˆ υ ( ξ ) (cid:13)(cid:13)(cid:13) . k υ k H s , (3.24)i.e. ∆ g ∗ υ is a bounded linear operator on H s . Then by Corollary 3.1, K ( u ) is lo-cally Lipschitz on H s . Then by reasoning as in Theorem 3.2 and [13, Theorem 1.1]we obtain that G : H s → H s is strictly contractive. Using the contraction map-ping principle, we get that the operator G ( u ) defined by (3 .
5) has a uniquefixed point u ( x, t ) ∈ C ([0 , ∞ ) , H s ) and u ( x, t ) is the solution of the problem(1 . − (1 . u ( x, t ) of (1 . − (1 .
2) is alsounique in C ([0 , ∞ ) , H s ). In fact, let u and u be two solutions of the problem(1 . − (1 .
2) and u , u ∈ C ([0 , ∞ ) , H s ). Let u = u − u ; then u tt − a ∆ u + b ∗ u = ∆ [ g ∗ ( f ( u ) − f ( u ))] . This fact is derived in a similar way as in Theorem 3.2, by using Theorems2.1, 2.2 and Gronwall’s inequality.
Theorem 3.4.
Let the Condition 3.2 hold and r > n . Then thereis some T > . − (1 .
2) is well posed withsolution in C (cid:0) [0 , T ] ; Y s, ∞ (cid:1) for initial data ϕ, ψ ∈ Y s, ∞ . roof. All we need here, is to show that K ∗ f ( u ) is Lipschitz on Y s, ∞ .Indeed, by reasoning as in Theorem 3.3 we have k ∆ g ∗ υ k H s + r − . (cid:13)(cid:13)(cid:13) (1 + ξ ) s + r − | ξ | ˆ g ( ξ ) ˆ υ ( ξ ) (cid:13)(cid:13)(cid:13) . k υ k H s , Then ∆ g ∗ υ is a bounded linear map from H s into H s + r − . Since s ≥ r > n we get s + r − > n . The Sobolev embedding theorem implies that∆ g ∗ υ is a bounded linear map from Y s, ∞ into Y s, ∞ . Lemma 3.4 implies theLipschitz condition on Y s, ∞ . Then, by reasoning as in Theorem 3.3 we obtainthe assertion.The solution in theorems 3.2-3.4 can be extended to a maximal interval[0 , T max ) , where finite T max is characterized by the blow-up conditionlim sup T → T max k u k Y s, ∞ = ∞ . Lemma 3.8.
Suppose the conditions of theorems 3.4, 3.5 hold and u is thesolution of multipoint IVP (1 . − (1 . . Then there is a global solution if forany
T < ∞ we have sup t ∈ [0 , T ) (cid:0) k u k Y s,p ∞ + k u t k Y s,p ∞ (cid:1) < ∞ . (3.25) Proof.
Indeed, by reasoning as in the second part of the proof of Theorem3.1, by using a continuation of local solution of (1 . − (1 .
2) and assumingcontrary that, (3 .
25) holds and T < ∞ we obtain contradiction, i.e. we get T = T max = ∞ .
4. Conservation of energy and global existence.
In this section, we prove the existence and the uniqueness of the global strongsolution and the global classical solution for the problem (1 . − (1 . . For thispurpose, we are going to make a priori estimates of the local strong solution forthe problem (1 . − (1 . . Condition 4.1.
Let the Condition 2.1 holds. Assume that the kernel g isan integrable function whose Fourier transform satisfies0 < ˆ g ( ξ ) ≤ (cid:16) | ξ | (cid:17) − r for all ξ ∈ R n and r ≥ . Let F − denote the inverse Fourie rtransform. We consider the operator B defined by u ∈ D ( B ) = H s , Bu = F − h | ξ | − (ˆ g ( ξ )) − − ˆ u ( ξ ) i , Then it is clear to see that B − u = − ∆ g ∗ u, B − u = F − h | ξ | (ˆ g ( ξ )) − ˆ u ( ξ ) i . (4.1)20irst, we show the following Lemma 4.1.
Suppose the conditions of theorems 3.4, 3.5 hold with ˆ g ( ξ ) > . − (1 .
2) exists in C (cid:0) [0 , T ] ; Y s, ∞ (cid:1) forsome s ≥ . If Bϕ ∈ L and Bψ ∈ L , then Bu, Bu t ∈ C (cid:0) [0 , T ) ; L (cid:1) . Proof.
By Lemma 2.1, problem (1 . − (1 .
2) is equıvalent to followingintegral equation , u ( x, t ) = [ S ( x, t ) ϕ + S ( x, t ) ψ + Φ ( g ∗ f ( u ))] + (4.2)12 η t Z F − h sin η ( t − τ ) | ξ | ˆ g ( ξ ) ˆ f ( G ( u ) ( ξ )) i dτ , where S ( x, t ), S ( x, t ) , Φ are are operator functions defined by (2 .
10) and(2 . g replaced by g ∗ f ( u ) . From (4 .
2) we get that u t ( x, t ) = (cid:20) ddt S ( x, t ) ϕ + ddt S ( x, t ) ψ + ddt Φ ( g ∗ f ( u )) (cid:21) +12 t Z F − Q ( ξ, t − τ ) dτ , (4.3)where Q ( ξ, t − τ ) = cos η ( t − τ ) | ξ | ˆ g ( ξ ) ˆ f ( u ) ( ξ ) ,S ( x, t ) ϕ = F − ( D − ( ξ ) " − m X k =1 β k cos κ k ! sin ( ηt ) , + " η − ( ξ ) m X k =1 β k sin κ k cos ( ηt ) ˆ ϕ ( ξ ) } , (4.4) ddt S ( x, t ) ϕ = F − ( D − ( ξ ) η " − m X k =1 β k cos κ k ! cos ( ηt ) − " m X k =1 β k sin κ k sin ( ηt ) ˆ ϕ ( ξ ) } ,S ( x, t ) ψ = F − (" η − ( ξ ) D − ( ξ ) m X k =1 α k sin κ k ! sin ( ηt ) + η − ( ξ ) D − ( ξ ) " − m X k =1 α k cos κ k ! cos ( ηt ) ˆ ψ ( ξ ) , (4.5)21 dt S ( x, t ) ψ = F − (" D − ( ξ ) m X k =1 α k sin κ k ! cos ( ηt ) − D − ( ξ ) " − m X k =1 α k cos κ k ! sin ( ηt ) ˆ ψ ( ξ ) , Φ ( x ; t ) = Φ ( g ∗ f ( u )) = X j = m X k =1 F − Φ jk (cid:16) | ξ | ˆ g ( ξ ) ˆ f ( u ) (cid:17) ( ξ ; t ) , (4.6)Since D − ( ξ ) , − m X k =1 β k cos κ k ! sin ( ηt ) , η − ( ξ ) m X k =1 β k sin κ k cos ( ηt )are uniformly bounded for fixet t by (4 . .
4) (4 .
5) we have k BS ( x, t ) ϕ k L = (cid:13)(cid:13)(cid:13) F − h | ξ | − (ˆ g ( ξ )) − − ˆ u ( ξ ) S ( x, t ) ϕ i(cid:13)(cid:13)(cid:13) L . k ϕ k H s < ∞ , (4.7) k BS ( x, t ) ϕ k L = (cid:13)(cid:13)(cid:13) F − h | ξ | − (ˆ g ( ξ )) − − ˆ u ( ξ ) S ( x, t ) ψ i(cid:13)(cid:13)(cid:13) L . k ψ k H s < ∞ . For fixed t , we have f ( u ) ∈ H s . Since D − ( ξ ) , cos ( ηt ) , sin ( ηt ) are uni-formly bounded, from (4 . , (2 .
11) and (4 .
6) we get k B Φ k L ≤ (cid:13)(cid:13)(cid:13) F − h | ξ | − (ˆ g ( ξ )) − − | ξ | ˆ g ( ξ ) ˆ f ( u ) i(cid:13)(cid:13)(cid:13) L . k f ( u ) k H s < ∞ . (4.8)From (4 .
8) we have k B Φ k L ≤ (cid:13)(cid:13)(cid:13) F − h | ξ | − (ˆ g ( ξ )) − − Q ( ξ, t − τ ) i(cid:13)(cid:13)(cid:13) L ≤ (4.9) (cid:13)(cid:13)(cid:13) F − h | ξ | − (ˆ g ( ξ )) − − | ξ | ˆ g ( ξ ) ˆ f ( u ) i(cid:13)(cid:13)(cid:13) L . k f ( u ) k H s < ∞ . Then from (4 . , (4 . , (4 .
8) and (4 .
9) we obtain the assertion.
Remark 4.1.
Due to nonlocality of initial conditions the additional condi-tions appears in Theorem 4.1. For classical Cauchy problem this extra conditionsare not required
Lemma 4.2.
Assume the conditions of theorems 3.4, 3.5 hold with a = 0 , ˆ g ( ξ ) > b ( ξ ) = O (cid:16) | ξ | (cid:17) s − r +1 . Suppose the solution of (1 . − (1 .
2) exists in C (cid:0) [0 , T ) ; Y s, ∞ (cid:1) for some s ≥ . If Bψ ∈ L , Bu t ( x, λ k ) ∈ L , k = 1 , , ..., m, then Bu t ∈ C (cid:0) [0 , T ) ; L (cid:1) . Moreover, if Bϕ ∈ L , Bu ( x, λ k ) ∈ L then Bu ∈ C (cid:0) [0 , T ) ; L (cid:1) . roof. Integrating the equation (1 .
1) for a = 0 , twice and calculating theresulting double integral as an iterated integral, we have u ( x, t ) = ϕ ( x ) + m X k =1 α k u ( x, λ k ) + t " ψ ( x ) + m X k =1 β k u t ( x, λ k ) − t Z ( t − τ ) ( b ∗ u ) ( x, τ ) dτ + t Z ( t − τ ) ∆ ( g ∗ f ( u )) ( x, τ ) dτ , (4.10) u t ( x, t ) = ψ ( x ) + m X k =1 β k u t ( x, λ k ) − t Z ( b ∗ u ) ( x, τ ) dτ + t Z ∆ ( g ∗ f ( u )) ( x, τ ) dτ . (4.11)From (4 .
1) and (4 .
11) for fixed t and τ we get k Bu t ( x, t ) k L = k Bψ ( x ) k L + m X k =1 β k k Bu t ( x, λ k ) k L − t Z k B ( b ∗ u ) ( x, τ ) k L dτ − t Z (cid:13)(cid:13) B − f ( u ) ( x, τ ) (cid:13)(cid:13) L dτ . (4.12)By assumption on b , g and by (4 .
1) for fixed τ we have k B ( b ∗ u ) ( x, τ ) k L ≤ (cid:13)(cid:13)(cid:13) F − h | ξ | − ˆ b ( ξ ) (ˆ g ( ξ )) − − ˆ u ( ξ, τ ) i(cid:13)(cid:13)(cid:13) L . k u ( ., τ ) k H s . (4.13)Moreover, by Lemma 3.3 for all t we have f ( u ) ∈ H s . Also k B ( f ( u )) ( x, τ ) k L . k f ( u ) ( ., τ ) k H s . (4.14)Then from (4 . − (4 .
14) we obtain Bu t ∈ C (cid:0) [0 , T ) ; L (cid:1) . The secondstatement follows similarly from (4 . . From Lemma 4.2 we obtain the following result.
Result 4.1.
Assume the conditions of theorems 3.4, 3.5 hold with a = 0 , ˆ g ( ξ ) > , α k = β k = 0 andˆ b ( ξ ) = O (cid:16) | ξ | (cid:17) s − r +1 . . − (1 .
2) exists in C (cid:0) [0 , T ] ; Y s, ∞ (cid:1) for some s ≥ . If Bψ ∈ L then Bu t ∈ C (cid:0) [0 , T ) ; L (cid:1) . Moreover, if Bϕ ∈ L , then Bu ∈ C (cid:0) [0 , T ) ; L (cid:1) . Here, G ( τ ) = τ Z g ( s ) ds. Lemma 4.3.
Assume the all conditions of Lemma 4.2 are satisfied. Let Bψ ∈ L , Bu t ( x, λ k ) ∈ L , k = 1 , , ..., m and G ( ϕ ) ∈ L . Then for any t ∈ [0 , T ) the energy E ( t ) = k Bu t k L + a (cid:13)(cid:13) F − ˆ g ∗ u (cid:13)(cid:13) L + k B ( b ∗ u ) k L + 2 Z R n G ( u ) dx (4.15)is constant [0 , T ) . Proof.
By use of equation (1 . ddt E ( t ) = 2 ( Bu tt , Bu t ) + 2 a (cid:0) F − ˆ g ∗ u, (cid:0) F − ˆ g ∗ u (cid:1) t (cid:1) +2 [ B ( b ∗ u ) , B ( b ∗ u ) u t ( t )] + 2 ( f ( u ) , u t ) = 2 (cid:0) B u tt , u t (cid:1) +2 (cid:0) B ( b ∗ u ) , ( b ∗ u ) u t ( t ) (cid:1) + 2 (cid:2) B ( b ∗ u ) , ( b ∗ u ) u t ( t ) (cid:3) =2 B [( u tt − a ∆ u + b ∗ u + ∆ [ g ∗ f ( u )] , u t )] = 0 , where ( u, υ ) denotes the inner product of L space. Integrating the aboveequality with respect to t , we have (4 . Theorem 4.1.
Let the Condition 3.2 hold a = 0 , ˆ g ( ξ ) > b ( ξ ) = O (cid:16) | ξ | (cid:17) s − r +1 . Moreover, let Bψ ∈ L , Bu t ( x, λ k ) ∈ L , k = 1 , , ..., m and G ( ϕ ) ∈ L , s ≥ r > k > G ( r ) ≥ − kr for r ∈ R . Then thereis some T > . − (1 .
2) has a global solution u ∈ C (cid:0) [0 , ∞ ) ; Y s, ∞ (cid:1) . References