The first initial boundary value problem for semilinear hyperbolic equations in nonsmooth domains

In this paper we study the first initial boundary problem for semilinear hyperbolic equations in nonsmooth cylinders, where is a nonsmooth domain in Rn, n >=2. We established the existence and uniqueness of a global solution in time.

JOURNAL OF SCIENCE OF HNUE Mathematical and Physical Sci., 2013, Vol 58, No 7, pp 39-49 This paper is available online at http://stdb.hnue.edu.vn THE FIRST INITIAL-BOUNDARY VALUE PROBLEM FOR SEMILINEAR HYPERBOLIC EQUATIONS IN NONSMOOTH DOMAINS Vu Trong Luong and Nguyen Thanh Tung Faculty of Mathematics, Tay Bac University Abstract In this paper we study the first initial boundary problem for semilinear hyperbolic equations in nonsmooth cylinders Q = Ω × (0, ∞), where Ω is a nonsmooth domain in Rn , n ≥ We established the existence and uniqueness of a global solution in time Keywords: Initial boundary value problem, semilinear hyperbolic equation, global solution, non-smooth domain Introduction Let Ω ⊂ Rn be a bounded domain with non-smooth boundary ∂Ω, set ΩT = Ω × (0, T ), with < T < +∞ We use the notations H (Ω), H01 (Ω) as the usual Sobolev spaces and H −1 (Ω) as the dual space of H01 (Ω) is defined in [1] We denote L2 (Ω) as the space L2 (Ω) is defined in [2] Suppose X is a Banach space with the norm ∥ · ∥X The space Lp (0, ∞; X) consists of all measurable functions u : [0, ∞) −→ X with norm ∥u∥Lp (0,T ;X)  p1 ∞ ∫ =  ∥u(t)∥pX dt < +∞ for ≤ p < +∞ We consider the partial differential operator ( ) ∑ n n ∑ ∂u ∂u ∂ ij a (x, t) + bi (x, t) + c(x, t)u Lu = − ∂x ∂x ∂x j i i i=1 i,j=1 where (x, t) ∈ Q = Ω × (0, ∞); aij , bi , c ∈ C (Q) (1.1) (i, j = 1, · · · , n) Received March 12, 2013 Accepted June 5, 2013 Contact Nguyen Thanh Tung, e-mail address: thanhtung70tbu@gmail.com 39 Vu Trong Luong and Nguyen Thanh Tung aij (x, t) = aji (x, t) for i, j = 1, 2, · · · , n; (x, t) ∈ Q (1.2) The operator L is strongly elliptic Then there exists θ > 0, ∀ξ ∈ Rn , ∀(x, t) ∈ Q such that n ∑ aij (x, t)ξi ξj ≥ θ|ξ|2 (1.3) i,j=1 In this paper, we consider the initial-boundary value problem in the cylinder Q for semilinear PDE’s: utt + Lu + f (x, t, u, Du) = h(x, t), u(x, 0) = u0 (x), ut (x, 0) = u1 (x), u(x, t) = 0, (x, t) ∈ Q, x ∈ Ω, (x, t) ∈ ∂Ω × (0, ∞), (1.4) (1.5) (1.6) where u0 ∈ H01 (Ω), u1 ∈ L2 (Ω), h ∈ L2 (0, ∞; L2 (Ω)) and f : Q × R × Rn −→ R is continuous and satisfies the following two conditions: ( ) |f (x, t, u, Du)| ≤ C k(x, t) + |u| + |Du| , ∀(x, t) ∈ Q, k ∈ L2 (0, ∞; L2 (Ω)), (1.7) ∫ ( )( ) f (x, t, u, Du) − f (x, t, v, Dv) u − v dx ≥ 0, a.e t ∈ [0, +∞) (1.8) Ω We introduce the Sobolev space H∗1,1 (Q) which consists of all functions u defined on Q such that u ∈ L2 (0, ∞; H01 (Ω)), ut ∈ L2 (0, ∞; L2 (Ω)), and utt ∈ L2 (0, ∞; H −1 (Ω)) with the norm ∥u∥2H 1,1 (Q) = ∥u∥2L2 (0,∞;H (Ω)) + ∥ut ∥2L2 (0,∞;L2 (Ω)) + ∥utt ∥2L2 (0,∞;H −1 (Ω)) ∗ By ⟨·, ·⟩ we denote pairs of elements in H −1 (Ω) and H01 (Ω); By the notation (·, ·) we mean the inner product in L2 (Ω) Let B[u, v; t] = ∫ [∑ n Ω ij a (x, t)uxi vxj + i,j=1 n ∑ ] bi (x, t)uxi v + c(x, t)uv dx i=1 which is a bilinear form defined on H (Ω) Definition 1.1 A function u ∈ H∗1,1 (Q) is called a weak solution of the (1.4) - (1.6) if it satisfies the following conditions: ( ) ( ) - ⟨utt (t), v⟩ + B[u(t), v; t] + f (·, t, u, Du), v = h(·, t), v with each v ∈ H01 (Ω) and a.e t ∈ [0, ∞) - u(x, 0) = u0 (x), ut (x, 0) = u1 (x) with x ∈ Ω 40 The first initial-boundary value problem for semilinear hyperbolic equations Normally we write f (u, Du) instead of f (x, t, u, Du) The problem (1.4)−(1.6) in the case f −linear was considered in [5-7] in which authors proved the unique existence and regularity of a weak solution on the domain with singularity points on the boundary In [3] of A Doktor, problem (1.4) - (1.6) has been considered on smooth domains By using the results of the linear problem respectly, he proves the global solution of the problem It is noted that the method approaching is used in [3] can not be applied for the problem if the domains is nonsmooth In the present paper, we consider problem (1.4) - (1.6) with domain Ω, which is a non-smooth domain The monotonic method is used to obtain the unique existence of global solution in time The local existence and uniqueness of a weak solution In this section, we use the monotonic method to prove the existence and uniqueness of a weak solution of the problem (1.4)−(1.6) To confirm this, we first see that if {ωi (x)}∞ i=1 is a basis in H0 (Ω) ∩ L2 (Ω) and N is a positive integer chosen then existence N u (x, t) = N ∑ gi (t)ωi (x) (2.1) i=1 such that ( ) N N N (uN , ω ) + B[u , ω ; t] + f (u , Du ), ω = (h, ωi ) ≤ t ≤ T, i = 1, · · · , N (2.2) i i i tt in which gi (t) are defined on [0, ∞) such that with i = 1, · · · , N { gi (0) = (u0 , ωi ) gi′ (0) = (u1 , ωi ) (2.3) (2.4) Since (2.1) - (2.4), applying the Caratheodory theorem, the functions gi (i = 1, · · · , N ) always exist in [0, T ] Theorem 2.1 For uN (x, t) defined by (2.1) we obtain: ( ) N N 2 ∥ut ∥L2 (Ω) + ∥u ∥H (Ω) + ∥uN tt ∥L2 (0,T ;H −1 (Ω)) ) ( ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H (Ω) + ∥h∥L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) (2.5) for all t ∈ [0, T ] Proof (i) From (2.2), we multiply both sides by gi′ , sum i = 1, · · · , N , we obtain: ( ) N N N N N N (uN + (h, uN (2.6) tt , ut ) + B[u , ut ; t] = − f (u , Du ), ut t ) 41 Vu Trong Luong and Nguyen Thanh Tung We have: N (uN tt , ut ) B[uN , uN t ; t] = ∫ (∑ n d = dt ) N aij (x, t)uN xi ut,xj i,j=1 Ω ( ) N ∥u ∥ , (2.7) t L2 (Ω) ) ∫ (∑ n N N N dx dx + bi (x, t)uN xi ut + c(x, t)u ut Ω i=1 =: B1 + B2 and if setting A[uN , uN ; t] = n ∫ ∑ Ω i,j=1 N aij (x, t)uN xi uxj dx, then ( ) d N N A[u , u ; t] − C∥uN ∥2H (Ω) B1 ≥ dt ( ) |B2 | ≤ C ∥uN ∥2H (Ω) + ∥uN t ∥L2 (Ω) Using (1.7), we get: (2.8) (2.9) ( ) ∥f ∥2L2 (Ω) ≤ C ∥k∥2L2 (Ω) + ∥u∥2H (Ω) Therefore, ( ) ) 1( | f (uN , DuN ), uN ∥f ∥2L2 (Ω) + ∥uN t | ≤ t ∥L2 (Ω) ( ) ≤ C ∥k∥2L2 (Ω) + ∥uN ∥2H (Ω) + ∥uN ∥ t L2 (Ω) (2.10) Continuing, we have ) 1( ∥h∥2L2 (Ω) + ∥uN t ∥L2 (Ω) |(h, uN t )| ≤ (2.11) Combining (2.6) - (2.11) gives ) ( ) d ( N 2 ∥ ∥ut ∥ + A[uN , uN ; t] ≤ C ∥k∥2L2 (Ω) + ∥uN ∥2H (Ω) + ∥h∥2L2 (Ω) + ∥uN t L2 (Ω) dt (2.12) We have: ∫ ∫ ∑ n N N N N (2.13) θ |Du | dx ≤ aij (x, t)uN xi uxj = A[u , u ; t] Ω Ω i,j=1 by (1.3), and applying the Friedrichs theorem, we get ∥uN ∥2H (Ω) ≤ C.A[uN , uN ; t] So (2.12) becomes ) d ( N ∥ut ∥L2 (Ω) + A[uN , uN ; t] dt ) ( N N 2 ≤ C ∥k∥2L2 (Ω) + ∥uN t ∥L2 (Ω) + A[u , u ; t] + ∥h∥L2 (Ω) (2.14) 42 The first initial-boundary value problem for semilinear hyperbolic equations N N 2 Now setting η(t) = ∥uN t ∥L2 (Ω) + A[u , u ; t], ψ(t) = ∥h∥L2 (Ω) + ∥k∥L2 (Ω) , then (2.14) will be written in the form η ′ (t) ≤ C1 η(t) + C2 ψ(t) for ≤ t ≤ T, and C1 , C2 are constants Thus, applying Gronwall’s inequality, we deduce that   ∫t η(t) ≤ eC1 t η(0) + C2 ψ(s)ds , t ∈ [0, T ] (2.15) N N N It is found that η(0) = ∥uN t (0)∥ + A[u (0), u (0); 0] By (2.4) then ∥ut (0)∥L2 (Ω) ≤ C∥u1 ∥2L2 (Ω) and A[uN (0), uN (0); 0] ≤ C∥uN (x, 0)∥2H (Ω) ≤ C∥u0 ∥2H (Ω) , which is 0 obtained from (2.3) Therefore ( ) η(0) ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H (Ω) (2.16) On the other hand we see that ∫t ∫T ψ(s)ds ≤ ( ) ( ) ∥h∥2L2 (Ω) + ∥k∥2L2 (Ω) ds ≤ C ∥h∥2L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) (2.17) Using (2.15), (2.16) and (2.17), for ∀t ∈ [0, T ], we obtain: ) ( N N 2 2 ∥uN ∥ +A[u , u ; t] ≤ C ∥u ∥ + ∥u ∥ + ∥h∥ + ∥k∥ 1 t L2 (Ω) L2 (Ω) L2 (0,T ;L2 (Ω)) L2 (0,T ;L2 (Ω)) H (Ω) Applying (2.13) we get: ) ( N ∥uN ∥ + ∥u ∥ t L2 (Ω) H01 (Ω) ) ( 2 2 ≤ C ∥u1 ∥L2 (Ω) + ∥u0 ∥H (Ω) + ∥h∥L2 (0,T ;L2 (Ω)) + ∥k∥L2 (0,T ;L2 (Ω)) (2.18) for ∀t ∈ [0, T ] (ii) For each v ∈ H01 (Ω) selected, there is ∥v∥H01 (Ω) ≤ Then for each positive integer N N ⊥ there exists v1 ∈ span{ωk }N k=1 and v2 ∈ (span{ωk }k=1 ) such that v = v1 + v2 Inferred ∥v1 ∥H01 (Ω) ≤ and (v2 , ωk ) = for all k = 1, 2, · · · , N From (2.1) and (2.2), we have: N N N ⟨uN tt , v⟩ = (utt , v) = (utt , v1 ) = −(f, v1 ) + (h, v1 ) − B[u , v1 ; t] By estimates ( ) |(f, v1 )| ≤ ∥f ∥L2 (Ω) ∥v1 ∥H01 (Ω) ≤ ∥f ∥L2 (Ω) ≤ C ∥k∥L2 (Ω) + ∥uN ∥H01 (Ω) , ∫ |(h, v1 )| ≤ |hv1 |dx ≤ ∥h∥L2 (Ω) , Ω |B[u , v1 ; t]| ≤ C∥uN ∥H01 (Ω) , N 43 Vu Trong Luong and Nguyen Thanh Tung then we obtain: |⟨uN tt , v⟩| ( ) ≤ C ∥h∥L2 (Ω) + ∥u ∥ N H01 (Ω) + ∥k∥L2 (Ω) , ∀ v ∈ H01 (Ω), ∥v∥ ≤ It follows readily (2.18) that ∥uN tt ∥L2 (0,T ;H −1 (Ω)) ) ( ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H (Ω) + ∥h∥2L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) (2.19) Combining (2.18) and (2.19) we deduce (2.5) Theorem 2.2 If conditions (1.7), (1.8) are satisfied, then the problem (1.4) - (1.6) has a weak solution u ∈ H∗1,1 (ΩT ) such that ( ) ∥u∥2H 1,1 (Ω ) ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H (Ω) + ∥h∥2L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) ∗ T N ∞ N ∞ Proof (i) From (2.5) we obtain sequences {uN }∞ N =1 , {ut }N =1 , {utt }N =1 respectively bounded in L2 (0, T ; H01 (Ω)), L2 (0, T ; L2 (Ω)), L2 (0, T ; H −1 (Ω)), therefore there exists a subsequence, without the loss of generality we take the sequence {uN }∞ N =1 and u ∈ 1,1 H∗ (ΩT ) such that  N  weak in L2 (0, T ; H01 (Ω))  u ⇀ u (2.20) uN weak in L2 (0, T ; L2 (Ω)) t ⇀ ut   uN ⇀ u weak in L (0, T ; H −1 (Ω)) tt tt (ii) We prove that ⟨utt , v⟩ + B[u, v; t] + (f, v) = (h, v) for all v ∈ H01 (Ω) Indeed, for a fixed positive integer N arbitrary and we choose a function v ∈ C ([0, T ]; H (Ω)) of the N ∑ form v(x, t) = dk (t)ω(x), where {dk }N k=1 are smooth functions We select m ≥ N, k=1 N multiply (2.2) by di (t), the sum i = 1, · · · , N, then we obtain (uN tt , v) + B[u , v; t] + N (f, v) = (h, v) Setting (F (uN ), v) := (uN tt , v) + B[u , v; t] − (h, v), then ( ) (F (uN ), v) = − f (x, t, uN , DuN ), v (2.21) By ∥f (x, t, u , Du N N )∥2L2 (Ω) ≤C ( ∥k∥2L2 (Ω) + ∥uN ∥2H (Ω) ) , and also by (2.5) we have f (x, t, uN , DuN ) which is bounded in L2 (Ω) a.e ≤ t < T Therefore, there exists (ξ ∈ Ł2 (Ω) to f (x, t, )uN , DuN ) ⇀ ξ which is weak in L2 (Ω) when N −→ ∞ and hence f (x, t, uN , DuN ), v −→ (ξ, v) So when N −→ ∞ then (2.21) becomes (F (u), v) = −(ξ, v) (2.22) 44 The first initial-boundary value problem for semilinear hyperbolic equations ( ) On the other hand, from (1.8) then f (x, t, uN , DuN ) − f (x, t, ω, Dω), uN − ω ≥ for ∀ω ∈ H01 (Ω) Integrating Ω and N −→ ∞, we have: ∫ [F (u)u − ξω − f (x, t, ω, Dω)(u − ω)]dx ≥ Ω ∫ By (2.22) we also have (F (u), u) = −(ξ, u) and so [−ξu + ξω − f (x, t, ω, Dω)(u − Ω ω)]dx ≥ 0, that is ∫ [ξ − f (x, t, ω, Dω)](u − ω)dx ≤ (2.23) Ω ∫ Putting ω = u−λv for λ > 0, into (2.23), we get [ξ−f (x, t, u−λv, Du−λDv)]vdx ≤ Ω ∫ Sending λ −→ 0, yields [ξ − f (x, t, u, Du)]vdx ≤ By an argument analogous to the Ω ∫ above with ω = u + λv for λ > 0, we obtain [ξ − f (x, t, u, Du)]vdx ≥ From this we Ω deduce ( ) f (x, t, u, Du), v = (ξ, v) (2.24) Combining (2.21), (2.22) and (2.24) yields ( ) ⟨utt , v⟩ + B[u, v; t] + f (u, Du), v = (h, v) (2.25) (iii) We have to prove u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ Ω To prove this, we choose any function v ∈ C ([0, T ]; H01 (Ω)) with v(T ) = vt (T ) = With this function v, ( ) ( ) ∫T integrating by t in (2.25) on [0, T ] and by (utt , v)dt = − ut (0), v(0) + u(0), vt (0) + ∫T (vtt , u)dt so ∫T ∫T {(vtt , u) + B[u, v; t]}dt + ∫T (f, v)dt = 0 ( ) ( ) (h, v)dt + ut (0), v(0) − u(0), vt (0) (2.26) Similarly, we deduce ∫T ∫T {(vtt , u ) + B[u , v; t]}dt + N N ∫T (f, v)dt = ) ( N ) ( (h, v)dt + uN t (0), v(0) − u (0), vt (0) 45 Vu Trong Luong and Nguyen Thanh Tung When N −→ ∞ then ∫T ∫T {(vtt , u) + B[u, v; t]}dt + ∫T (f, v)dt = ( ) ( ) (h, v)dt + u1 (0), v(0) − u0 (0), vt (0) (2.27) From (2.26) and (2.27) we obtain u(x, 0) = u0 (x), ut (x, 0) = u1 (x) x ∈ Ω In order to study the uniqueness of the weak solution of problem (1.4)−(1.6), we replace condition (1.7) with the following condition |f (x, t, u, Du) − f (x, t, v, Dv)| ≤ µ(|u − v| + |Du − Dv|) (2.28) ∀u, v ∈ H01 (Ω), ∀(x, t) ∈ Q Theorem 2.3 If conditions (1.8), (2.28) are satisfied, then problem (1.4)−(1.6) has at most one weak solution in H∗1,1 (ΩT ) Proof First, suppose ω1 , ω2 are the solutions of problem (1.4)−(1.6) If setting u = ω1 − ω2 and F (ω1 , ω2 , Dω1 , Dω2 ) = f (x, t, ω1 , Dω1 ) − f (x, t, ω2 , Dω2 ), then u is a weak solution of the following problem:    utt + Lu + F (ω1 , ω2 , Dω1 , Dω2 ) = in ΩT u=0 on ∂Ω × (0, T )   u = 0, u = on Ω × {t = 0} t Next, we will prove u ≡ on ΩT (i) For each ≤ s < T that is fixed, we set v(t) = ∫s  u(τ )dτ  if ≤ t ≤ s t if s ≤ t ≤ T for each t ∈ [0, T ] then v(t) ∈ H01 (Ω) From the definition of weak solution, we have ⟨utt , v⟩ + B[u, v; t] + (F, v) = 0, integrating by t in (2.29) on [0, s] with s ∈ (0, T ), it follows further that ∫s ( 46 ∫s ) ⟨utt , v⟩ + B[u, v; t] dt + (F, v)dt = 0 (2.29) The first initial-boundary value problem for semilinear hyperbolic equations Since ut (0) = 0, v(s) = and integrating by parts twice with respect to t, we find ∫s { } − (ut , vt ) + B[u, v; t] dt + ∫s (F, v)dt = 0 Now vt = −u (0 ≤ t ≤ s), and so ∫s { ∫s } (ut , u) − B[vt , v; t] dt + (F, v)dt = 0 Thus ∫s ∫s ∫s ) { d (1 ∥u∥2L2 (Ω) − B[v, v; t] dt + (F, v)dt = − C[u, v; t] + D[v, v; t]}dt, dt 2 0 (2.30) where ∫ ( C[u, v; t] = − n ∑ i=1 Ω ∫ (∑ n D[v, v; t] = bi (x, t)uvxi − n ∑ i=1 aij t (x, t)vxi vxj i,j=1 Ω ) bi (x, t)vvt,xi dx + n ∑ ) bi,t (x, t)vxi v + ct (x, t)vv dx i=1 (2.30) is written into 1 ∥u(s)∥2L2 (Ω) + B[v(0), v(0); 0] + 2 ∫s ∫s (F, v)dt = − { C[u, v; t] + D[v, v; t]}dt (2.31) By using the Cauchy inequality and the Lipschitz condition (2.28), we obtain the following estimates: ∫s ∫s ( ) ∥v∥2H (Ω) + ∥u∥2L2 (Ω) dt, C[u, v; t] dt ≤ 0 ∫s ∫s D[v, v; t] dt ≤ ∫s ∥v∥2H (Ω) dt, 0 ∫s (F, v) dt ≤ C ∥v∥2H (Ω) dt 0 47 Vu Trong Luong and Nguyen Thanh Tung Employing the estimates above and inequality (1.3), we get from (2.31) that  s  ∫ ( ) ∥u(s)∥2L2 (Ω) + ∥v(0)∥2H (Ω) ≤ C  ∥v∥2H (Ω) + ∥u∥2L2 (Ω) dt + ∥v(0)∥2L2 (Ω)  0 (ii) Now we set w(t) := ∫t (2.32) (0 ≤ t ≤ T ), then (2.32) become u(τ )dτ ∥u(s)∥2L2 (Ω) + ∥w(s)∥2H (Ω)  s  ∫ ( ) ≤C ∥w(s) − w(t)∥2H (Ω) + ∥u(t)∥2L2 (Ω) dt + ∥w(s)∥2L2 (Ω)  (2.33) 0 Since ∥w(s) − w(t)∥2H (Ω) ≤ 2∥w(s)∥2H (Ω) + 2∥w(t)∥2H (Ω) 0 and ∫s ∥w(s)∥2L2 (Ω) ≤C ∥u(t)∥2L2 (Ω) dt, we conclude from (2.33) that ∥u(s)∥2L2 (Ω) + (1 − 2C1 s)∥w(s)∥2H (Ω) ≤ C1 ∫s ( ∥u∥2L2 (Ω) + ∥w∥2H (Ω) ) dt We choose T1 which is so small that − 2T1 C1 ≤ Then if ≤ s ≤ T1 , we will have ∥u(s)∥2L2 (Ω) + ∥w(s)∥2H (Ω) ≤ C ∫s ( ) ∥u∥2L2 (Ω) + ∥w∥2H (Ω) dt 0 Applying Gronwall’s inequality, we obtain u ≡ on [0, T1 ] (iii) We apply the same argument for the intervals [T1 , 2T1 ], [2T1 , 3T1 ], Eventually we deduce u ≡ on ΩT With the result (2.5) and condition (2.28), we have the following result Theorem 2.4 If conditions (1.8), (2.28) are satisfied, then problem (1.4)−(1.6) has a global unique solution u ∈ H∗1.1 (Q) that satisfies the condition ) ( ∥u∥2H 1,1 (Q) ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H (Ω) + ∥h∥2L2 (0,∞;L2 (Ω)) + ∥k∥2L2 (0,∞;L2 (Ω)) ∗ 48 The first initial-boundary value problem for semilinear hyperbolic equations REFERENCES [1] Evans L.C., 1997 Partial differential equations AMS [2] Robert A.Adam, 1975 Sobolev spaces Academic press New York San Francisco, London [3] Alexander Doktor, 1973 Mixed problem for semilinear hyperbolic equation of second order with Dirichlet boundary condition Czechoslovak Mathematical Journal, Vol.23, No.1, pp 95-122 [4] E.A.codington and N.Levison, 1955 Theory of odinary differential equations McGraw-Hill [5] N.M Hung, 1999 Boundary problem for nonstationary systems in domains with a non-smooth boundry Doctor dissertation, Mech Math.Department MSU, Moscow [6] Hung, N M, Luong, V T, 2008 Unique solvability of initial boundary-value problems problems for hyperbolic systems in cylinders whose base is a cups domain, Electron J Diff Eqns., Vol 2008, No 138, pp 1-10 [7] Hung, N M, Luong, V T, 2010 The Lp −Unique solvability of the first initial boundary-value problem for hyperbolic systems Taiwanese Journal of Mathematics, Vol 14, No 6, pp 2365-2381 49 ... x ∈ Ω 40 The first initial- boundary value problem for semilinear hyperbolic equations Normally we write f (u, Du) instead of f (x, t, u, Du) The problem (1.4)−(1.6) in the case f −linear was... The first initial- boundary value problem for semilinear hyperbolic equations ( ) On the other hand, from (1.8) then f (x, t, uN , DuN ) − f (x, t, ω, Dω), uN − ω ≥ for ∀ω ∈ H01 (Ω) Integrating... smooth domains By using the results of the linear problem respectly, he proves the global solution of the problem It is noted that the method approaching is used in [3] can not be applied for the problem