GLOBAL SOLUTIONS FOR A NONLINEAR HYPERBOLIC
EQUATION WITH BOUNDARY MEMORY SOURCE TERM
FUQIN SUN AND MINGXIN WANG
Received 21 January 2005; Accepted 17 August 2005
We study a nonlinear hyperbolic equation with boundary memory source term. By the
use of Galerkin procedure, we prove the global existence and the decay property of solu-
tion.
Copyright © 2006 F. Sun and M. Wang. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
This paper deals with a hyperbolic equation with boundary memory source terms:
ρ(x)u′′ uu=g(u), xΩ,t>0,
u=0, xΓ0,t>0,
∂u
ν+∂u
ν+f(u)=t
0K(tτ)hτ,u(τ),xΓ1,t>0,
u(0,x)=u0(x), ρu(0,x)=ρu1(x), xΩ,
(1.1)
where u=u(t,x), Ωis a bounded domain of RN(N1) with sufficiently smooth bound-
ary Ω=Γ0Γ1,¯
Γ0¯
Γ1=∅,whereΓ0and Γ1have positive measures. u=∂u/∂t,
u′′ =2u/∂t2. Equations of type (1.1) are of interest in many applications such as in the
theory of electromagnetic materials with memory which obey the Ohms law. It can also
describe the temperature evolution in a rigid conductor with a memory. We refer to [8,9]
to see the details. In many works concerned with equations of type (1.1), we cite Aassila
et al. [1], where the following wave equation was considered:
u′′ u+f0(u)=0, xΩ,t>0,
u=0, xΓ0,t>0,
∂u
ν+gu=t
0K(tτ)hu(τ),xΓ1,t>0,
u(0,x)=u0(x), u(0,x)=u1(x), xΩ.
(1.2)
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 60734, Pages 116
DOI 10.1155/JIA/2006/60734
2 Global solutions for a nonlinear hyperbolic equation
Under some conditions on nonlinear terms, they acquired the existence and uniform
decay of solutions. Recently, Park and Park [12] generalized problem (1.2)byendowing
with some discontinuous and multivalued terms. For more related works, we refer to
[3,4,7,11,13] and the references therein. For problem (1.1) without memory source
term, we point out the work [6] of Cavalcanti et al., where they investigated the following
equation with boundary damping:
ρ(x)u′′ u=0, xΩ,t>0,
u=0, xΓ0,t>0,
∂u
ν+f(u)+g(u)=0, xΓ1,t>0,
u(0,x)=u0(x), ρu(0,x)=ρu1(x), xΩ.
(1.3)
Through a partition of boundary Γand Galerkin procedures, they acquired the existence
and decay behavior of the solution to problem (1.3). In another work of theirs [5], using
similar method, they studied problem (1.3)withρ=1 and the source term g(u)=|u|pu
coupled in the first equation. Motivated by the above works, we are devoted to study
problem (1.1). By virtue of the potential well method, and through Galerkin procedures,
we acquire the global existence and decay property of perturbed energy of solutions of
problem (1.1). The organization of this paper is as follows. In Section 2,wemakeas-
sumptions and introduce a potential well, and then state the main results. In Section 3,
making use of Galerkin procedures, we study the existence of solution of problem (1.1).
And in the last section, we derive the uniform decay by the perturbed energy method.
2. Assumptions and main results
In this section, we first give the notations used throughout this paper:
(u,v)=Ωu(x)v(x)dx,(u,v)Γ1=Γ1
u(x)v(x)dΓ,
·p=·
Lp(Ω),·=·L2(Ω),
·Γ1,p=·
Lp(Γ1),·Γ1=·
L2(Γ1),
(2.1)
and rdenotes the conjugate exponent of r>1.
Define
V=uH1(Ω):u=0onΓ0.(2.2)
Since the measure of Γ0is positive, Poincar`
e inequality holds and trace embedding theo-
rem holds (see [2]), we know that ∇uis equivalent to the norm on V.Letµ1and µ2be
the optimal constants such that
u≤µ1∇u,uΓ1µ2∇u∀uV. (2.3)
Now we make the following assumptions.
F. Sun and M. Wang 3
(A1)fC(R), f(s)s0, and there exist positive constants k1and k2such that
k1|s|q1≤|f(s)|≤k2|s|q1, (2.4)
where 2 <q<if N=1,2; 2 <q2(N1)/(N2) if N3.
(A2)gC(R), g(s)s0, and there exists positive constant k3such that
g(s)
k3|s|p, (2.5)
where 1 <p<if N=1,2; 1 <pN/(N2) if N3.
(A3)K:R+R+is a continuously differentiable function verifying
K(t)≤−k4K(t)t0, K(0) >0, 1 µ2
2
0K(s)ds L>0, (2.6)
where k4>0.
(A4)h(τ,s)ismeasurablewithτand continuous with s, and it satisfies
h(τ,s)s
K(τ)
K(0) |s|∀sR,τ0.(2.7)
(A5)ρ(x)0, ρ≡ 0andρL(Ω).
(A6) Assume that the initial data
u0,u1VH3/2(Ω) (2.8)
and satisfy the compatibility conditions
u0+u1=gu0,xΩ,
u0=0, u1=0, xΓ0,
∂u0
ν+∂u1
ν+fu1=0, xΓ1.
(2.9)
Remark 2.1. (i) The assumptions (A3)and(A
4)implythath(τ,s)(1 + K(τ))s.
(ii) Given u1VH3/2(Ω), by the assumption (A2) and the theory of elliptic prob-
lems, we see that problem (2.9) admits a weak solution u0VH3/2(Ω).
Let B>0 be the optimal constant such that
vp+1 B∇v∀vV, (2.10)
where pis the number given in the assumption (A2).
4 Global solutions for a nonlinear hyperbolic equation
If we define
Bsup
vV,v=0(1/(p+1))vp+1
p+1
∇vp+1 , (2.11)
then
BBp+1
p+1,1
p+1vp+1
p+1 B∇vp+1 vV. (2.12)
Now for some function u,wedefine
J(u)=L
2∇u2k3
p+1up+1
p+1,
E(t)=1
2
ρu
2+1
2∇u2ΩG(u)dx 1
2t
0K(τ)
u(t)
2
Γ1+1
2(Ku)(t),
(2.13)
where
G(s)=s
0g(η),(Ku)(t)=t
0K(tτ)
hτ,u(τ)u(t)
2
Γ1dτ. (2.14)
Putting
dinf
uV,u=0sup
λ>0
J(λu),H(λ)L
2λ2k3Bλp+1,λ>0.(2.15)
We have the following result.
Proposition 2.2. Let the assumptions (A2)–(A4) be fulfilled. It holds that
d=max
λ>0H(λ)=Hλ=(p1)L
2(p+1)λ2
, (2.16)
where λ=(L/(p+1)k3B)1/(p1).
If ∇u
, then
J(u)0, ∇u22(p+1)
(p1)LE(t).(2.17)
Proof. From
H(λ)= (p+1)k3Bλp=L(p+1)k3Bλp1λ, (2.18)
F. Sun and M. Wang 5
we see that λ=[L/((p+1)k3B)]1/(p1) is the maximum point of H.Hence,
max
λ>0H(λ)=Hλ=(p1)L
2(p+1)λ2
.(2.19)
Note the definition of B, by the direct computation, we have
d=inf
uV,u=0sup
λ>0
J(λu)
=L
2L
k32/(p1)
k3
p+1L
k3(p+1)/(p1)inf
uV,u=0
∇up+1
up+1
p+1
2/(p1)
=(p1)L
2(p+1)L
(p+1)k3B2/(p1)
=(p1)L
2(p+1)λ2
.
(2.20)
Thus the first conclusion is valid.
If ∇u
,thenweobtain
E(t)Ju(t)L
2∇u2k3B∇up+1 >∇u2L
2k3Bλp1
=
u2L
2L
p+1=(p1)L
2(p+1)∇u2.
(2.21)
Thus the second conclusion is valid.
Remark 2.3. The number ddefined in Proposition 2.2 is the Mountain Pass level related
to the elliptic problem
Lu=k3|u|p1u,xΩ,
u=0, xΓ0,
∂u
ν=0, xΓ1,
(2.22)
see [5]or[14]. In fact, dis equal to the number
inf
αΛ
sup
t[0,1]
Jα(t), (2.23)
where
Λ=αC[0,1];V;α(0) =0, J(α(1)) <0.(2.24)