HYPERBOLIC PROBLEMS

HYPERBOLIC PROBLEMS

RODICA LUCA

Received 1 November 2005; Revised 10 March 2006; Accepted 7 April 2006

We investigate the existence and uniqueness of solutions to a class of discrete hyperbolic systems with some nonlinear extreme conditions and initial data, in a real Hilbert space.

Copyright © 2006 Rodica Luca. This is an open access article distributed under the Cre- ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

LetHbe a real Hilbert space with the scalar product·,·and the associated norm · . In this paper we will investigate the discrete hyperbolic system

du_j

dt (t) +v_j(t)−v_j−1(t)

h_j +Au_j(t)f_j(t), dvj

dt (t) +uj+1(t)−uj(t)

hj +Bvj(t)gj(t),

j=1,N,t∈[0,T], inH, (S)

with the extreme conditions

v0(t)= −αu1(t), u_N+1(t)=βv_N(t), t∈[0,T], (EC) and the initial data

u_j(0)=u_j0, v_j(0)=v_j0, j=1,N, (ID) whereN∈N,h_j>0, j=1,N, and α,β,A,B are operators inH, which satisfy some assumptions.

Hindawi Publishing Corporation Advances in Difference Equations Volume 2006, Article ID 89260, Pages1–14 DOI 10.1155/ADE/2006/89260

This problem is a discrete version with respect tox(withH=R) of the problem

∂u

∂t(t,x) +∂v

∂x(t,x) +Au(t,x) f(t,x),

∂v

∂t(t,x) +∂u

∂x(t,x) +Bv(t,x)g(t,x),

0< x <1,t >0, inR, (S)

with the boundary conditions

v(t, 0)= −αu(t, 0), u(t, 1)=βv(t, 1), t >0, (BC) and the initial data

u(0,x)=u0(x), v(0,x)=v0(x), 0< x <1. (IC) The above problem has applications in electrotechnics (the propagation phenomena in electrical networks) and mechanics (the variable flow of a fluid)—see [7,8,13]. The system (S) subject to various boundary conditions has been studied by many authors:

Barbu, Iftimie, Moros¸anu, and Luca, in [4,5,9,11,12]. Using an idea from [14] we dis- cretize the problem (S) + (BC) + (IC) in this way: letN be a given integer (N≥1) and h=1/(N+ 1). In a first stage we approximate the system (S) and the boundary condi- tions (BC) by

∂u

∂t(t,x) +v(t,x)−v(t,x−h)

h +Au(t,x)f(t,x), x∈

h, (N+ 1)h,

∂v

∂t(t,x) +u(t,x+h)−u(t,x)

h +Bv(t,x)g(t,x), x∈(0,Nh),t >0, v(t, 0)= −αu(t, 0), u(t,Nh)=βv(t,Nh).

(1.1)

We look foruandvof the formu(t,x)=_N

j=0uj(t)ϕj(x) andv(t,x)=_N

j=0vj(t)ϕj(x), where

ϕ_j(x)=χ[jh,(j+1)h)(x)=

⎧⎨

⎩

1, x∈

jh, (j+ 1)h, 0, x∈

jh, (j+ 1)h. (1.2) We write f,g,u0,v0as

f(t,x)=

N j=0

fj(t)ϕj(x), g(t,x)=

N j=0

gj(t)ϕj(x),

u0(x)=

N j=0

uj0ϕj(x), v0(x)=

N j=0

vj0ϕj(x),

(1.3)

where fj(t)= f(t,jh),gj(t)=g(t,jh),uj0=u0(jh), andvj0=v0(jh).

Then forujandvjwe obtain the system u_j+v_j−v_j−1

h +Au_jf_j, j=1,N, v_j+uj+1−uj

h +Bvj

gj, j=0,N−1,

(1.4)

with the conditionsv0= −α(u0),u_N=β(v_N),u_j(0)=u_j0,j=1,N, andv_j(0)=v_j0, j= 0,N−1.

For a unitary writing, we takej=1,N−1 in both equations of the above system and then the extreme conditions becomev0= −α(u1) andu_N=β(v_N−1) (they do not show u0andv_N). By passingN→N+ 1 and taking different stepsh_j, we obtain the system (S) inu_j,v_j,j=1,Nwithv0= −α(u1),u_N+1=β(v_N) (α=α,β=β), andH=R.

In this way the study of the partial differential system (S) reduces to the study of the ordinary differential system (S) (withH=Randhj=h, for all j). The solution uj,vj

depends onhand it seems thatu(t,x) =

u_j(t)ϕ_j(x),v(t, x)=

v_j(t)ϕ_j(x) approximate the solutionu,vof the system (S). We will not study here the convergence of the solution

u,vtou,v, but we will investigate the well-posedness of the discrete problem (S) + (EC) + (ID).

We will also study the discrete system that corresponds to (S) forx∈(0,∞) with the boundary condition

v(t, 0)= −αu(t, 0), t >0, (BC)1

and initial data

u(0,x)=u0(x), v(0,x)=v0(x), x >0. (IC)1

More precisely we will investigate the infinite discrete hyperbolic system dun

dt +vn−vn−1

h +Au_n f_n, dv_n

dt +u_n+1−u_n

h +Bvn

gn, n=1, 2,. . .,t∈[0,T], inH,

(S)

with the extreme condition

v0(t)= −αu1(t), t∈[0,T], (EC) and initial data

u_n(0)=u_n0, v_n(0)=v_n0, n=1, 2,. . . . (ID) Although the proposed problems appeared by discretization of the problem (S) + (BC) + (IC) and the corresponding one forx∈(0,∞), our problems also cover some nonlinear differential systems in Hilbert spaces.

For other classes of difference and differential equations in abstract spaces we refer the reader to [1,2,10].

InSection 2we recall some definitions and results from the theory of maximal mono- tone operators that we need to prove our results. In Sections3and4we study the prob- lems (S) + (EC) + (ID) and (S) + ( EC) + ( ID).

2. Notations and preliminaries

Let H be a real Hilbert space with the scalar product ·,· and the associated norm · . We denote by→andthe strong and weak convergence inH, respectively. For a multivalued operatorA:H→H we denote byD(A)= {x∈H;A(x)= ∅}its domain and byR(A)= ∪{A(x); x∈D(A)}its range. The operatorAis identified with its graph G(A)= {[x,y]∈H×H;x∈D(A), y∈R(A)} ⊂H×H.

We use forAthe notationA:D(A)⊂H→H. IfA,B⊂H→Handλ∈Rthen λA=

[x,λy]; y∈A(x), D(λA)=D(A), A+B=

[x,y+z]; y∈A(x),z∈A(x), D(A+B)=D(A)∩D(B). (2.1) The operatorA:D(A)⊂H→H is monotone if for allx1,x2∈D(A) and y1∈A(x1), y2∈A(x2) we havey1−y2,x1−x2 ≥0.

An operatorA:H→Hsingle-valued and everywhere defined is hemicontinuous if for allx,y∈Hwe haveA(x+t y)A(x), ast→0. The operatorA:D(A)⊂H→His demi- continuous if it is strongly weakly continuous, that is, if ([x_n,y_n])n⊂Awithx_n→x, as n→ ∞andy_ny, asn→ ∞, then [x,y]∈A.

A demicontinuous operator is also hemicontinuous.

The operatorA:D(A)_⊂H_→His maximal monotone if it is maximal in the set of all monotone operators, that is,Ais monotone and, as subset ofH×H, it is not properly contained in any other monotone subset ofH×H.

The monotone operatorA:D(A)⊂H→H is maximal monotone if and only if for anyλ >0 (equivalently for someλ >0),R(I+λA)=H.

IfA:H→H is everywhere defined, single-valued, monotone, and hemicontinuous, then it is maximal monotone.

IfA:D(A)⊂H→H is maximal monotone andB:H→H is everywhere defined, single-valued, monotone, and hemicontinuous, thenA+Bis maximal monotone.

For a maximal monotone operatorA:D(A)⊂H→H, the operators Jλ=(I+λA)⁻¹:H−→H, λ >0, Aλ=1

λ I−Jλ

:H−→H, λ >0, (2.2)

are the resolvent and the Yosida approximation ofA.

For an operator A:D(A)⊂H→H, f : (0,∞)→H and u0∈H, we consider the Cauchy problem

du

dt(t) +Au(t) f(t), t >0,u(0)₌u0. (CP)

The functionu∈C([0,T];H) is strong solution for the problem (CP) ifuis absolutely continuous on every compact of (0,T),u(t)∈D(A), for a.a.t∈(0,T),u(0)=u0, andu satisfies (CP)1for a.a.t∈(0,T).

The functionu∈C([0,T];H) is weak solution for the problem (CP) if there exist (un)n⊂W^1,∞(0,T;H) and (fn)n⊂L¹(0,T;H) such that

dun

dt (t) +Aun(t)fn(t), for a.a.t∈(0,T),n=1, 2,. . ., (2.3) un→u, asn→ ∞, inC([0,T];H),u(0)=u0, and fn→f, asn→ ∞, inL¹(0,T;H).

For other properties of the maximal monotone operators and for the main results of existence, uniqueness of the strong and weak solutions for the nonlinear evolution equations in Hilbert spaces, we refer the reader to [3,6,13].

3. The problem (S) + (EC) + (ID)

The assumptions we will use in this section are the following.

(H1) The operatorsA:D(A)⊂H→H,B:D(B)⊂H→H are maximal monotone, possibly multivalued, withD(A)= ∅,D(B)= ∅.

(H2) The operatorsα,β:H→Hare single-valued and maximal monotone.

(H3) The constantshj>0,j=1,N.

We will write our problem as a Cauchy problem in a certain Hilbert space, for we con- sider the Hilbert spaceX=H^2N= {(u1,u2,. . .,u_N,v1,v2,. . .,v_N)^T; u_j,v_j∈H, j=1,N} with the scalar product

u1,. . .,u_N,v1,. . .,v_N^T,u1,. . .,u_N,v1,. . .,v_N^T_X=

N j=1

h_ju_j,u_j+v_j,v_j (3.1)

and the corresponding norm · X.

We introduce the operatorᏭ:D(Ꮽ)⊂X→X,

Ꮽu1,u2,. . .,u_N,v1,v2,. . .,v_N^T

=v1+αu1

h1 ,v2−v1

h2

,. . .,vN−vN−1

h_N ,u2−u1

h1 ,u3−u2

h2

,. . .,βv_N−u_N h_N

T

. (3.2)

BecauseD(α)=D(β)=H, we deduce thatD(Ꮽ)=X.

We also define the operatorᏮ:D(Ꮾ)⊂X→X,D(Ꮾ)=D(A)^N×D(B)^N, Ꮾu1,u2,. . .,u_N,v1,v2,. . .,v_N^T

=

γ1,γ2,. . .,γ_N,δ1,δ2,. . .,δ_N^T;γ_i∈Au_i,δ_i∈Bv_i,i=1,N. (3.3)

Using the operatorsᏭandᏮ, our problem can be equivalently expressed as the fol- lowing Cauchy problem in the spaceX

dU

dt (t) +ᏭU(t)+ᏮU(t)F(t), U(0)=U0, (P) where U = (u1,u2,. . .,uN,v1,. . .,vN)^T, U0 = (u10,u20,. . .,uN0,v10,. . .,vN0)^T, F =(f1,

f2,. . .,fN,g1,. . .,gN)^T.

Lemma 3.1. If the assumptions (H2) and (H3) hold, then the operatorᏭis monotone and demicontinuous; so it is maximal monotone.

Proof. The operatorᏭis defined onXand it is single-valued.Ꮽis monotone, because Ꮽ(U)−Ꮽ(U),U−U_X

=

N j=1

hj

vj−vj−1−vj+vj−1

hj ,uj−uj

+

N j=1

hj

uj+1−uj−uj+1+uj

hj ,vj−vj

=

v1+αu1

−v1−αu1

,u1−u1

+

N j=2

vj−vj,uj−uj

−

vj−1−vj−1,uj−uj

+

N−1 j=1

u_j+1−u_j+1,v_j−v_j−u_j−u_j,v_j−v_j

+βv_N−u_N−βv_N+u_N,v_N−v_N

=

v1−v1,u1−u1

+αu1

−αu1

,u1−u1

−

u1−u1,v1−v1

+v_N−v_N,u_N−u_N +βvN

−βvN

,vN−vN

−

uN−uN,vN−vN

= αu1

−αu1

,u1−u1

+βvN

−βvN

,vN−vN

≥0,

(3.4)

with v0 = −α(u1), v0 = −α(u1), u_N+1=β(v_N), u_N+1=β(v_N+1), for all U =(u1,u2, . . .,uN,v1,. . .,vN)^T,U=(u1,u2,. . .,uN,v1,. . .,vN)^T∈X.

The operatorᏭis also demicontinuous, that is, ifUⁿ→U⁰andᏭ(Uⁿ)V⁰, then V⁰=Ꮽ(U⁰). Indeed, letUⁿ=(uⁿ₁,uⁿ₂,. . .,uⁿ_N,v₁ⁿ,. . .,vⁿ_N),U⁰=(u⁰₁,u⁰₂,. . .,u⁰_N,v₁⁰,. . .,v_N⁰), Uⁿ→U⁰ and Ꮽ(Uⁿ)=((v₁ⁿ+α(uⁿ₁))/h1, (v₂ⁿ−v₁ⁿ)/h2,. . ., (v_Nⁿ −v_Nⁿ₋₁)/hN, (uⁿ₂−uⁿ₁)/

h1,. . ., (β(vⁿ_N)−uⁿ_N)/hN)^T,V⁰=(x⁰1,x2⁰,. . .,x⁰_N,y1⁰,. . .,y_N⁰)^T,Ꮽ(Uⁿ)V⁰.

FromUⁿ→U⁰we deduce that

uⁿ_j−→u⁰_j, vⁿ_j −→v⁰_j, forn−→ ∞,∀j=1,N. (3.5) BecauseᏭ(Uⁿ)V⁰, we getᏭ(Uⁿ),YX→ V⁰,YX, for allY∈X,Y=(α1,α2,. . .,α_N, β1,. . .,βN)^T, that is,

vⁿ₁+αuⁿ₁,α1

−→h1

x₁⁰,α1

, vⁿ_j−vⁿ_j−1,αj

−→hj x⁰_j,αj

, j=2,N, uⁿ_j−uⁿ_j−1,βj−1

−→hj−1

y⁰_j−1,βj−1

, j=2,N, βvⁿ_N−uⁿ_N,βN

−→hN

y_N⁰,βN

, inH,

=⇒v₁ⁿ+αuⁿ₁ h1x⁰1 , (3.6)

vⁿ_j−vⁿ_j−1 h_jx⁰_j, j=2,N,

uⁿ_j−uⁿ_j−1 h_j−1y⁰_j−1, j=2,N, (3.7) βvⁿ_N−uⁿ_N hNy_N⁰, inH, asn−→ ∞. (3.8) From the relations (3.5) and (3.7) we obtainx⁰_j=(v⁰_j−v⁰_j−1)/h_j,j=2,N,y⁰_j−1=(u⁰_j− u⁰_j−1)/hj−1, j=2,N. Becauseαandβare demicontinuous, by (3.5), (3.6), and (3.8) we deduce

h1x⁰₁−v⁰₁=αu⁰₁, h_Ny_N⁰ +u⁰_N=βv⁰_N=⇒x⁰₁=v⁰₁+αu⁰₁

h1 , y_N⁰ =βv⁰_N−u⁰_N h_N .

(3.9) ThereforeV⁰=Ꮽ(U⁰). Hence the operatorᏭis demicontinuous (so it is also hemicon- tinuous) and, by [6, Proposition 2.4] we deduce that it is maximal monotone.

Lemma 3.2. If the assumptions (H1) and (H3) hold, then the operatorᏮis maximal mono- tone inX.

Proof. The operatorᏮis evidently monotone Z−Z,U−UX=

N j=1

hj

γj−γ_j,uj−uj

+δj−δj,vj−vj

≥0, (3.10)

for allU=(u1,u2,. . .,uN,v1,. . .,vN)^T,U=(u1,u2,. . .,uN,v1,. . .,vN)^T∈D(Ꮾ),Z∈Ꮾ(U), Z∈Ꮾ(U), for allγ_j∈A(u_j),γ_j∈A(u_j),δ_j∈B(v_j),δ_j∈B(v_j), j=1,N.

It is also maximal monotone inX. Indeed, by [6, Proposition 2.2] it is sufficient (and necessary) to show that forλ >0,R(I+λᏮ)=X⇔for allY ∈X,Y=(x1,x2,. . .,x_N,y1,. . ., y_N)^Tthere existsU∈X,U=(u1,u2,. . .,u_N,v1,. . .,v_N)^T such that

U+λᏮ(U)Y. (3.11)

The last relation gives us uj+λγj=xj, j=1,N, vj+λδj=yj, j=1,N, ^=⇒

uj=

I+λA⁻¹xj

=J_λ^Axj

, j=1,N, v_j=

I+λB⁻¹y_j=J_λ^By_j, j=1,N, (3.12) whereγ_j∈A(u_j),δ_j∈B(v_j) j=1,N, andJ_λ^A andJ_λ^Bare the resolvents ofAandB, re- spectively (AandBare maximal monotone). ThenU=(u1,. . .,uN,v1,. . .,vN)^T, whereuj

andvj,j=1,N, defined above, satisfy our condition (3.11).

We give now the main result for our initial problem (S) + (EC) + (ID).

Theorem 3.3. Assume that the assumptions (H1)–(H3) hold. Ifu_j0∈D(A), for all j= 1,N, vj0∈D(B), for all j=1,N, and fj,gj∈W^1,1(0,T;H), j=1,N, then there exist unique functionsujandvj∈W^1,∞(0,T;H), j=1,N,uj(t)∈D(A),vj(t)∈D(B), for all j=1,N, for allt∈[0,T], which verify the system (S) for everyt∈[0,T), the condition (EC) for everyt∈[0,T), and the initial data (ID).

Moreoveru_j andv_j,j=1,N, are everywhere differentiable from right in the topology of Hand

d⁺uj

dt ⁼

fj−Auj

−vj−vj−1

hj

0

, j=1,N, d⁺vj

dt ⁼

gj−Bvj

−uj+1−uj

hj

0

, j=1,N,∀t∈[0,T),

(3.13)

withv0(t)= −α(u1(t)),uN+1(t)=β(vN(t)),∀t∈[0,T).

Proof. Because the operatorᏮis maximal monotone inXandᏭis single-valued, with D(Ꮽ)=X, monotone, and hemicontinuous, by [3, Corollary 1.3, Chapter II] we deduce thatᏭ+Ꮾ:D(Ꮾ)⊂X→X is maximal monotone. By [3, Theorem 2.2, Corollary 2.1, Chapter III] we deduce that, forU0∈D(Ꮾ) andF∈W^1,1(0,T;X), the problem (P) has a unique solutionU=(u1,u2,. . .,uN,v1,. . .,vN)^T ∈W^1,∞(0,T;X),U(t)∈D(Ꮾ), for all t_∈[0,T). We consider (P)1in the interval [0,T+ε],ε >0, (by extending correspondingly the functions f_jandg_j,j=1,N) and we getU(T)∈D(Ꮾ).

The solutionUis everywhere differentiable from right and d⁺U

dt (t)=

F(t)−ᏭU(t)−ᏮU(t)⁰, ∀t∈[0,T), (3.14) that is, the relations from theorem are verified. In addition we have

d⁺U dt (t)

X≤F(0)−ᏭU0

−ᏮU0

0

X+ _t

0

dF ds(s)

Xds, ∀t∈[0,T).

(3.15)

IfUandV are the solutions of (P) corresponding to (U0,F), (V0,G)∈D(Ꮾ)×W^1,1(0, T;X), then

U(t)−V(t)_X≤U0−V0

X+ _t

0

F(s)−G(s)_Xds, ∀t∈[0,T]. (3.16) Remark 3.4. IfU0∈D(Ꮾ)=D(A)^N×D(B)^NandF∈L¹(0,T;X), then, by [3, Corollary 2.2, Chapter III], the problem (P)⇔(S) + (EC) + (ID) has a unique weak solutionU∈ C([0,T];X), that is, there exist (F_n)n⊂W^1,1(0,T;X),F_n→F, asn→ ∞, inL¹(0,T;X) and (Un)n⊂W^1,∞(0,T;X),Un(0)=U0,Un→U, asn→ ∞inC([0,T];X), strong solutions for the problems

dUn

dt (t) + (Ꮽ+Ꮾ)Un(t)Fn(t), for a.a.t∈(0,T),n=1, 2,. . . . (3.17) 4. The problem (S) + ( EC) + ( ID)

We present the assumptions that we will use in this section as follows.

(H1) The operators A:D(A)⊂H→H and B:D(B)⊂H→H are maximal mono- tone, 0∈A(0), 0∈B(0), and there exista1,a2>0 such that

γ ≤a1u, ∀u∈D(A),∀γ∈A(u); δ ≤a2u, ∀u∈D(B), ∀δ∈B(u).

(4.1) (H2) The operator α:H→His single-valued and maximal monotone.

(H3) The constant h >0.

We consider the spaceY=l_h²(H)×l²_h(H), wherel²_h(H)= {(un)n⊂H;^∞_n=1un²<∞}

(=l²(H)), with the scalar product un

n,vn

n

,un

n,vn

n

Y=

un

n,un

n

l²_h(H)+vn

n,vn

n

l²_h(H)

= ^∞

n=1

hu_n,u_n+

∞ n=1

hv_n,v_n. (4.2)

We define the operatorᏭ :Y→Y, Ꮽun

n,vn

n

=

v_n−v_n−1

h

n

,

u_n+1−u_n h

n

, withv0= −αu1

, (4.3)

and the operatorᏮ :D(Ꮾ) ⊂Y→Y, Ꮾu_nn,v_nn=

γ_nn,δ_nn∈Y,γ_n∈Au_n,δ_n∈Bv_n,∀n≥1, (4.4) withD(Ꮾ) = {((un)n, (vn)n)∈Y;un∈D(A), vn∈D(B), ∀n≥1}.

Lemma 4.1. If the assumptions (H2) and (H3) hold, then the operatorᏭ is monotone and demicontinuous inY.

Proof. First we observe thatᏭ is well-defined inY. If ((un)n, (vn)n)∈Y, thenᏭ((u n)n, (v_n)n)∈Y, andD(Ꮽ) =Y.

The operatorᏭ is monotone, because

Ꮽu_nn,v_nn−Ꮽu_nn,v_nn,u_nn,v_nn−

u_nn,v_nnY

=

v_n−v_n−1

h

n−

v_n−v_n−1

h

n

,un−un

n

l²_h(H)

+

un+1−un

h

n−un+1−un

h

n,v_n−v_nn

l²_h(H)

= αu1

−αu1

,u1−u1

≥0.

(4.5)

Next we prove thatᏭ is demicontinuous, that is, if un^j

n,vn^j

n

−→

u⁰_nn,v_n⁰_n, forj−→ ∞inY, (4.6) Ꮽun^j

n,vn^j

n

x_nn,y_nn, forj−→ ∞inY, (4.7) then ((xn)n, (yn)n)=Ꮽ((u ⁰_n)n, (v⁰_n)n).

From (4.6) we deduce

hun^j

n−

u⁰_nn²_l2(H)+hvn^j

n−

v⁰_nn²_l2(H)−→0, forj−→ ∞ =⇒

un^j

n−

u⁰_nnl2(H)−→0, for j−→ ∞ vn^j

n−

v⁰_nnl2(H)−→0, forj−→ ∞=⇒

∞ n=1

un^j−u⁰_n²−→0, forj−→ ∞

∞ n=1

vn^j−v_n⁰²−→0, for j−→ ∞

=⇒un^j−→u⁰_n, forj−→ ∞,∀n, vn^j−→v_n⁰, for j−→ ∞,∀n.

(4.8)

Then by (4.7) we have Ꮽun^j

n,vn^j

n

,α_nn,β_nnY

−→

xn

n,yn

n

,αn

n,βn

n

Y, asj−→ ∞,∀ αn

n,βn

n

∈Y,

=⇒

vn^j−v_n^j₋₁ h

n

,

u_n+1^j −un^j

h

n

,αn

n,βn

n

Y

,