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

JOURNAL OF SCIENCE OF HNUE Mathematical and Physical Sci., 2013, Vol. 58, No. 7, pp. 39-49 This paper is available online at 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 ≥ 2. 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. 1. Introduction Let Ω ⊂ Rn be a bounded domain with non-smooth boundary ∂Ω, set ΩT = Ω × (0, T ), with 0 < T < +∞. We use the notations H1(Ω), H10 (Ω) as the usual Sobolev spaces and H−1(Ω) as the dual space of H10 (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) =  ∞∫ 0 ∥u(t)∥pX dt  1p < +∞ for 1 ≤ p < +∞. We consider the partial differential operator Lu = − n∑ i,j=1 ∂ ∂xj ( aij(x, t) ∂u ∂xi ) + n∑ i=1 bi(x, t) ∂u ∂xi + c(x, t)u (1.1) where (x, t) ∈ Q = Ω× (0,∞); aij, bi, c ∈ C1(Q) (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∑ i,j=1 aij(x, t)ξiξj ≥ θ|ξ|2. (1.3) 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), (x, t) ∈ Q, (1.4) u(x, 0) = u0(x), ut(x, 0) = u1(x), x ∈ Ω, (1.5) u(x, t) = 0, (x, t) ∈ ∂Ω× (0,∞), (1.6) where u0 ∈ H10 (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)∫ Ω ( f(x, t, u,Du)− f(x, t, v,Dv))(u− v)dx ≥ 0, a.e. t ∈ [0,+∞). (1.8) We introduce the Sobolev space H1,1∗ (Q) which consists of all functions u defined on Q such that u ∈ L2(0,∞;H10 (Ω)), ut ∈ L2(0,∞;L2(Ω)), and utt ∈ L2(0,∞;H−1(Ω)) with the norm ∥u∥2 H1;1∗ (Q) = ∥u∥2L2(0,∞;H10 (Ω)) + ∥ut∥ 2 L2(0,∞;L2(Ω)) + ∥utt∥2L2(0,∞;H−1(Ω)). By ⟨·, ·⟩ we denote pairs of elements in H−1(Ω) and H10 (Ω); By the notation (·, ·) we mean the inner product in L2(Ω). Let B[u, v; t] = ∫ Ω [ n∑ i,j=1 aij(x, t)uxivxj + n∑ i=1 bi(x, t)uxiv + c(x, t)uv ] dx which is a bilinear form defined on H1(Ω). Definition 1.1. A function u ∈ H1,1∗ (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 ∈ H10 (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. 2. 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 H10 (Ω) ∩ L2(Ω) and N is a positive integer chosen then existence uN(x, t) = N∑ i=1 gi(t)ωi(x) (2.1) such that (uNtt , ωi) +B[u N , ωi; t] + ( f(uN , DuN), ωi ) = (h, ωi) 0 ≤ t ≤ T, i = 1, · · · , N (2.2) in which gi(t) are defined on [0,∞) such that with i = 1, · · · , N{ gi(0) = (u0, ωi) (2.3) g′i(0) = (u1, ωi) (2.4) Since (2.1) - (2.4), applying the Caratheodory theorem, the functions gi (i = 1, 2 · · · , N) always exist in [0, T ]. Theorem 2.1. For uN(x, t) defined by (2.1) we obtain:( ∥uNt ∥2L2(Ω) + ∥uN∥2H10 (Ω) ) + ∥uNtt ∥2L2(0,T ;H−1(Ω)) ≤ C ( ∥u1∥2L2(Ω) + ∥u0∥2H10 (Ω) + ∥h∥L2(0,T ;L2(Ω)) + ∥k∥ 2 L2(0,T ;L2(Ω)) ) (2.5) for all t ∈ [0, T ]. Proof. (i) From (2.2), we multiply both sides by g′i, sum i = 1, · · · , N , we obtain: (uNtt , u N t ) +B[u N , uNt ; t] = − ( f(uN , DuN), uNt ) + (h, uNt ). (2.6) 41 Vu Trong Luong and Nguyen Thanh Tung We have: (uNtt , u N t ) = d dt ( 1 2 ∥uNt ∥2L2(Ω) ) , (2.7) B[uN , uNt ; t] = ∫ Ω ( n∑ i,j=1 aij(x, t)uNxiu N t,xj ) dx+ ∫ Ω ( n∑ i=1 bi(x, t)u N xi uNt + c(x, t)u NuNt ) dx =: B1 +B2 and if setting A[uN , uN ; t] = ∫ Ω n∑ i,j=1 aij(x, t)uNxiu N xj dx, then B1 ≥ d dt ( 1 2 A[uN , uN ; t] ) − C∥uN∥2H10 (Ω) (2.8) |B2| ≤ C ( ∥uN∥2H10 (Ω) + ∥u N t ∥2L2(Ω) ) . (2.9) Using (1.7), we get: ∥f∥2L2(Ω) ≤ C (∥k∥2L2(Ω) + ∥u∥2H10 (Ω)). Therefore, |(f(uN , DuN), uNt )| ≤ 12 (∥f∥2L2(Ω) + ∥uNt ∥2L2(Ω)) ≤ C ( ∥k∥2L2(Ω) + ∥uN∥2H10 (Ω) + ∥u N t ∥L2(Ω) ) . (2.10) Continuing, we have |(h, uNt )| ≤ 1 2 (∥h∥2L2(Ω) + ∥uNt ∥2L2(Ω)) (2.11) Combining (2.6) - (2.11) gives d dt (∥uNt ∥2 + A[uN , uN ; t]) ≤ C (∥k∥2L2(Ω) + ∥uN∥2H10 (Ω) + ∥h∥2L2(Ω) + ∥uNt ∥2L2(Ω)) . (2.12) We have: θ ∫ Ω |DuN |2dx ≤ ∫ Ω n∑ i,j=1 aij(x, t)uNxiu N xj = A[uN , uN ; t] (2.13) by (1.3), and applying the Friedrichs theorem, we get ∥uN∥2H10 (Ω) ≤ C.A[u N , uN ; t]. So (2.12) becomes d dt (∥uNt ∥2L2(Ω) + A[uN , uN ; t]) ≤ C (∥k∥2L2(Ω) + ∥uNt ∥2L2(Ω) + A[uN , uN ; t] + ∥h∥2L2(Ω)) . (2.14) 42 The first initial-boundary value problem for semilinear hyperbolic equations... Now setting η(t) = ∥uNt ∥2L2(Ω) + A[uN , uN ; t], ψ(t) = ∥h∥2L2(Ω) + ∥k∥2L2(Ω), then (2.14) will be written in the form η′(t) ≤ C1η(t) + C2ψ(t) for 0 ≤ t ≤ T, and C1, C2 are constants. Thus, applying Gronwall’s inequality, we deduce that η(t) ≤ eC1t η(0) + C2 t∫ 0 ψ(s)ds  , t ∈ [0, T ]. (2.15) It is found that η(0) = ∥uNt (0)∥2 + A[uN(0), uN(0); 0]. By (2.4) then ∥uNt (0)∥2L2(Ω) ≤ C∥u1∥2L2(Ω) and A[uN(0), uN(0); 0] ≤ C∥uN(x, 0)∥2H10 (Ω) ≤ C∥u0∥ 2 H10 (Ω) , which is obtained from (2.3). Therefore η(0) ≤ C ( ∥u1∥2L2(Ω) + ∥u0∥2H10 (Ω) ) . (2.16) On the other hand we see that t∫ 0 ψ(s)ds ≤ T∫ 0 (∥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: ∥uNt ∥2L2(Ω)+A[uN , uN ; t] ≤ C ( ∥u1∥2L2(Ω) + ∥u0∥2H10 (Ω) + ∥h∥ 2 L2(0,T ;L2(Ω)) + ∥k∥2L2(0,T ;L2(Ω)) ) . Applying (2.13) we get:( ∥uNt ∥2L2(Ω) + ∥uN∥2H10 (Ω) ) ≤ C ( ∥u1∥2L2(Ω) + ∥u0∥2H10 (Ω) + ∥h∥ 2 L2(0,T ;L2(Ω)) + ∥k∥2L2(0,T ;L2(Ω)) ) (2.18) for ∀t ∈ [0, T ]. (ii) For each v ∈ H10 (Ω) selected, there is ∥v∥H10 (Ω) ≤ 1. Then for each positive integer N there exists v1 ∈ span{ωk}Nk=1 and v2 ∈ (span{ωk}Nk=1)⊥ such that v = v1 + v2. Inferred ∥v1∥H10 (Ω) ≤ 1 and (v2, ωk) = 0 for all k = 1, 2, · · · , N. From (2.1) and (2.2), we have: ⟨uNtt , v⟩ = (uNtt , v) = (uNtt , v1) = −(f, v1) + (h, v1)−B[uN , v1; t]. By estimates |(f, v1)| ≤ ∥f∥L2(Ω)∥v1∥H10 (Ω) ≤ ∥f∥L2(Ω) ≤ C ( ∥k∥L2(Ω) + ∥uN∥H10 (Ω) ) , |(h, v1)| ≤ ∫ Ω |hv1|dx ≤ ∥h∥L2(Ω), |B[uN , v1; t]| ≤ C∥uN∥H10 (Ω), 43 Vu Trong Luong and Nguyen Thanh Tung then we obtain: |⟨uNtt , v⟩| ≤ C ( ∥h∥L2(Ω) + ∥uN∥H10 (Ω) + ∥k∥L2(Ω) ) , ∀ v ∈ H10 (Ω), ∥v∥ ≤ 1. It follows readily (2.18) that ∥uNtt ∥2L2(0,T ;H−1(Ω)) ≤ C ( ∥u1∥2L2(Ω) + ∥u0∥2H10 (Ω) + ∥h∥ 2 L2(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 ∈ H1,1∗ (ΩT ) such that ∥u∥2 H1;1∗ (ΩT ) ≤ C(∥u1∥2L2(Ω) + ∥u0∥2H10 (Ω) + ∥h∥2L2(0,T ;L2(Ω)) + ∥k∥2L2(0,T ;L2(Ω))). Proof. (i) From (2.5) we obtain sequences {uN}∞N=1, {uNt }∞N=1, {uNtt }∞N=1 respectively bounded in L2(0, T ;H10 (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 ∈ H1,1∗ (ΩT ) such that  uN ⇀ u weak in L2(0, T ;H10 (Ω)) uNt ⇀ ut weak in L2(0, T ;L2(Ω)) uNtt ⇀ utt weak in L2(0, T ;H −1(Ω)) (2.20) (ii) We prove that ⟨utt, v⟩ + B[u, v; t] + (f, v) = (h, v) for all v ∈ H10 (Ω). Indeed, for a fixed positive integer N arbitrary and we choose a function v ∈ C1([0, T ];H1(Ω)) of the form v(x, t) = N∑ k=1 dk(t)ω(x), where {dk}Nk=1 are smooth functions. We select m ≥ N, multiply (2.2) by di(t), the sum i = 1, · · · , N, then we obtain (uNtt , v) + B[uN , v; t] + (f, v) = (h, v). Setting (F (uN), v) := (uNtt , v) +B[u N , v; t]− (h, v), then (F (uN), v) = −(f(x, t, uN , DuN), v). (2.21) By ∥f(x, t, uN , DuN)∥2L2(Ω) ≤ C ( ∥k∥2L2(Ω) + ∥uN∥2H10 (Ω) ) , and also by (2.5) we have f(x, t, uN , DuN) which is bounded in L2(Ω) a.e. 0 ≤ 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) ≥ 0 for ∀ω ∈ H10 (Ω). Integrating Ω and N −→∞, we have:∫ Ω [F (u)u− ξω − f(x, t, ω,Dω)(u− ω)]dx ≥ 0. By (2.22) we also have (F (u), u) = −(ξ, u) and so ∫ Ω [−ξu + ξω − f(x, t, ω,Dω)(u − ω)]dx ≥ 0, that is ∫ Ω [ξ − f(x, t, ω,Dω)](u− ω)dx ≤ 0 (2.23) Putting ω = u−λv for λ > 0, into (2.23), we get ∫ Ω [ξ−f(x, t, u−λv,Du−λDv)]vdx ≤ 0. Sending λ −→ 0, yields ∫ Ω [ξ − f(x, t, u,Du)]vdx ≤ 0. By an argument analogous to the above with ω = u+ λv for λ > 0, we obtain ∫ Ω [ξ − f(x, t, u,Du)]vdx ≥ 0. From this we 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 ∈ C2([0, T ];H10 (Ω)) with v(T ) = vt(T ) = 0.With this function v, integrating by t in (2.25) on [0, T ] and by T∫ 0 (utt, v)dt = − ( ut(0), v(0) ) + ( u(0), vt(0) ) + T∫ 0 (vtt, u)dt so T∫ 0 {(vtt, u) +B[u, v; t]}dt+ T∫ 0 (f, v)dt = T∫ 0 (h, v)dt+ ( ut(0), v(0) )− (u(0), vt(0)) (2.26) Similarly, we deduce T∫ 0 {(vtt, uN)+B[uN , v; t]}dt+ T∫ 0 (f, v)dt = T∫ 0 (h, v)dt+ ( uNt (0), v(0) )−(uN(0), vt(0)) 45 Vu Trong Luong and Nguyen Thanh Tung When N −→∞ then T∫ 0 {(vtt, u) +B[u, v; t]}dt+ T∫ 0 (f, v)dt = T∫ 0 (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 ∈ H10 (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 H1,1∗ (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) = 0 in ΩT u = 0 on ∂Ω× (0, T ) u = 0, ut = 0 on Ω× {t = 0} Next, we will prove u ≡ 0 on ΩT . (i) For each 0 ≤ s < T that is fixed, we set v(t) =  s∫ t u(τ)dτ if 0 ≤ t ≤ s 0 if s ≤ t ≤ T for each t ∈ [0, T ] then v(t) ∈ H10 (Ω). From the definition of weak solution, we have ⟨utt, v⟩+B[u, v; t] + (F, v) = 0, (2.29) integrating by t in (2.29) on [0, s] with s ∈ (0, T ), it follows further that s∫ 0 ( ⟨utt, v⟩+B[u, v; t] ) dt+ s∫ 0 (F, v)dt = 0. 46 The first initial-boundary value problem for semilinear hyperbolic equations... Since ut(0) = 0, v(s) = 0 and integrating by parts twice with respect to t, we find s∫ 0 {− (ut, vt) +B[u, v; t]}dt+ s∫ 0 (F, v)dt = 0. Now vt = −u (0 ≤ t ≤ s), and so s∫ 0 { (ut, u)−B[vt, v; t] } dt+ s∫ 0 (F, v)dt = 0. Thus s∫ 0 d dt (1 2 ∥u∥2L2(Ω) − 1 2 B[v, v; t] ) dt+ s∫ 0 (F, v)dt = − s∫ 0 { C[u, v; t] +D[v, v; t]}dt, (2.30) where C[u, v; t] = ∫ Ω ( − n∑ i=1 bi(x, t)uvxi − n∑ i=1 bi(x, t)vvt,xi ) dx D[v, v; t] = ∫ Ω ( n∑ i,j=1 aijt (x, t)vxivxj + n∑ i=1 bi,t(x, t)vxiv + ct(x, t)vv ) dx (2.30) is written into 1 2 ∥u(s)∥2L2(Ω) + 1 2 B[v(0), v(0); 0] + s∫ 0 (F, v)dt = − s∫ 0 { 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∫ 0 C[u, v; t] dt ≤ s∫ 0 (∥v∥2H10 (Ω) + ∥u∥2L2(Ω)) dt, s∫ 0 D[v, v; t] dt ≤ s∫ 0 ∥v∥2H10 (Ω) dt, s∫ 0 (F, v) dt ≤ C s∫ 0 ∥v∥2H10 (Ω) dt. 47 Vu Trong Luong and Nguyen Thanh Tung Employing the estimates above and inequality (1.3), we get from (2.31) that ∥u(s)∥2L2(Ω) + ∥v(0)∥2H10 (Ω) ≤ C  s∫ 0 ( ∥v∥2H10 (Ω) + ∥u∥ 2 L2(Ω) ) dt+ ∥v(0)∥2L2(Ω)  . (2.32) (ii) Now we set w(t) := t∫ 0 u(τ)dτ (0 ≤ t ≤ T ), then (2.32) become ∥u(s)∥2L2(Ω) + ∥w(s)∥2H10 (Ω) ≤ C  s∫ 0 ( ∥w(s)− w(t)∥2H10 (Ω) + ∥u(t)∥ 2 L2(Ω) ) dt+ ∥w(s)∥2L2(Ω)  . (2.33) Since ∥w(s)− w(t)∥2H10 (Ω) ≤ 2∥w(s)∥ 2 H10 (Ω) + 2∥w(t)∥2H10 (Ω) and ∥w(s)∥2L2(Ω) ≤ C s∫ 0 ∥u(t)∥2L2(Ω)dt, we conclude from (2.33) that ∥u(s)∥2L2(Ω) + (1− 2C1s)∥w(s)∥2H10 (Ω) ≤ C1 s∫ 0 ( ∥u∥2L2(Ω) + ∥w∥2H10 (Ω) ) dt. We choose T1 which is so small that 1− 2T1C1 ≤ 1 2 . Then if 0 ≤ s ≤ T1, we will have ∥u(s)∥2L2(Ω) + ∥w(s)∥2H10 (Ω) ≤ C s∫ 0 ( ∥u∥2L2(Ω) + ∥w∥2H10 (Ω) ) dt. Applying Gronwall’s inequality, we obtain u ≡ 0 on [0, T1]. (iii) We apply the same argument for the intervals [T1, 2T1], [2T1, 3T1], . . . . Eventually we deduce u ≡ 0 on ΩT . With the result (2.5) and condition (2.28), we have the following result. Theorem 2.4. 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