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HIGH ORDER OF ACCURACY DIFFERENCE SCHEMES FOR HYPERBOLIC DIFFERENTIAL EQUATIONS

ALLABEREN ASHYRALYEV AND PAVEL E. SOBOLEVSKII Received 30 March 2004

We consider the abstract Cauchy problem for differential equation of the hyperbolic type v(t) +Av(t)= f(t) (0tT),v(0)=v0,v(0)=v0 in an arbitrary Hilbert spaceH with the selfadjoint positive definite operatorA. The high order of accuracy two-step dif- ference schemes generated by an exact difference scheme or by the Taylor decomposition on the three points for the numerical solutions of this problem are presented. The stabil- ity estimates for the solutions of these difference schemes are established. In applications, the stability estimates for the solutions of the high order of accuracy difference schemes of the mixed-type boundary value problems for hyperbolic equations are obtained.

1. The Cauchy problem

We consider the abstract Cauchy problem for hyperbolic equations

v(t) +Av(t)=f(t) (0tT), v(0)=v0, v(0)=v0 (1.1) in a Hilbert spaceHwith the selfadjoint positive definite operatorA.

A functionv(t) is called a solution of the problem (1.1) if the following conditions are satisfied.

(i)v(t) is twice continuously differentiable on the segment [0,T]. The derivatives at the endpoints of the segment are understood as the appropriate unilateral derivatives.

(ii) The elementv(t) belongs toD(A) for allt[0,T], and the functionAv(t) is con- tinuous on the segment [0,T].

(iii)v(t) satisfies the equations and the initial conditions (1.1).

It is known (see, e.g., [7,9]) that various initial boundary value problems for the hy- perbolic equations can be reduced to the problem (1.1). A study of discretization, over time only, of the initial value problem also permits one to include general difference schemes in applications, if the differential operator in space variables,A, is replaced by the difference operatorsAh that act in the Hilbert spaces and are uniformly positive definite and selfadjoint inhfor 0< hh0.

In a large cycle of works on high order of accuracy difference schemes for hyperbolic partial differential equations (see, e.g., [1,10] and the references given therein), stability

Copyright©2005 Hindawi Publishing Corporation

Discrete Dynamics in Nature and Society 2005:2 (2005) 183–213 DOI:10.1155/DDNS.2005.183

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was established under the assumption that the magnitudes of the grid stepsτandhwith respect to the time and space variables are connected. In abstract terms this means, in particular, that the conditionτAh0 whenτ0 is satisfied.

Of great interest is the study of absolute stable difference schemes of a high order of accuracy for hyperbolic partial differential equations, in which stability was established without any assumptions with respect to the grid stepsτandh. In [4,5,12] the corre- sponding simple difference schemes of the first and second order of accuracy for hyper- bolic partial differential equations were studied.

In the present paper the two-step difference schemes of a high order of accuracy gen- erated by an exact difference scheme or by the Taylor decomposition on the three points for the numerical solutions of the problem (1.1) are presented. The stability estimates for the solutions of these difference schemes are established. In applications, the stability esti- mates for the solutions of the high order of accuracy difference schemes of the mixed-type boundary value problems for hyperbolic equations are obtained.

If the function f(t) is not only continuous, but also continuously differentiable on [0,T],v0D(A), andv0D(A1/2), it is easy to show that the formula

v(t)=c(t)v0+s(t)v0+ t

0s(tλ)f(λ)dλ (1.2)

gives a solution of problem (1.1). Here c(t)=eitA1/2+eitA1/2

2 , s(t)=A1/2eitA1/2eitA1/2

2i . (1.3)

Actually, obviously (1.1) can be rewritten as the equivalent initial value problem for sys- tem of the first-order linear differential equations

v(t) +iBv(t)=z(t) (0tT), v(0)=v0, v(0)=v0,

z(t)iBz(t)=f(t), (1.4)

whereB=A1/2. Integrating these, we at once obtain v(t)=eiBtv0+

t

0eiB(ts)z(s)ds, z(t)=eiBtz0+

t

0eiB(tp)f(p)dp.

(1.5)

That is,

v(t)=eiBtv0+ t

0eiB(ts)eiBsz0ds+ t

0eiB(ts) s

0eiB(sp)f(p)dp ds, (1.6) so, using the conditionv(0) +iBv(0)=z(0), we obtain

v(t)=

eiBt+iB t

0eiB(ts)eiBsds

v0+ t

0eiB(ts)eiBsdsv(0) +

t

0eiB(ts) s

0eiB(sp)f(p)dp ds.

(1.7)

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By the interchange of the order of integration, we obtain v(t)=

eiBt+eiBteiBt 2

v0+B1eiBteiBt 2i dsv(0) +

t

0B1eiB(ts)eiB(ts) 2i f(s)ds.

(1.8)

Thus, from that by the definitions ofB,c(t), ands(t) we have (1.2).

Theorem 1.1. Suppose that v0D(A),v0D(A1/2), and f(t)are continuously differ- entiable on[0,T]function. Then there is a unique solution of the problem (1.1) and the stability inequalities

0maxtT

v(t)HMv0

H+A1/2v0

H+ max

0tT

A1/2f(t)H

,

0maxtT

A1/2v(t)HMA1/2v0

H+v0

H+ max

0tT

f(t)H

,

0maxtT

d2v(t) dt2

H+ max

0tT

Av(t)HMAv0

H+A1/2v0

H+f(0)H+ T

0

f(t)Hdt

(1.9) hold, whereMdoes not depend on f(t),t[0,T], andv0,v0.

Proof. Using the formula (1.2) and estimates

c(t)HH1, A1/2s(t)HH1, (1.10) we obtain

v(t)Hc(t)HHv0

H+A1/2s(t)HHA1/2v0

H

+ t

0

A1/2s(tλ)HHA1/2f(λ)H

v0

H+A1/2v0

H+Tmax

0tT

A1/2f(t)H.

(1.11)

ApplyingA1/2to the formula (1.2) and using the estimates (1.10), in a similar manner with (1.10), we obtain

A1/2v(t)Hc(t)HHA1/2v0

H+A1/2s(t)HHv0

H

+ t

0

A1/2s(tλ)HHf(λ)H

A1/2v0

H+v0

H+T max

0tT

f(t)H.

(1.12)

Now, we obtain the estimate forAv(t)H. ApplyingAto the formula (1.2) and using an integration by parts, we can write

Av(t)=c(t)Av0+A1/2s(t)A1/2v0+f(T)c(t)f(0) t

0c(tλ)f(λ)dλ. (1.13)

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Using the last formula and estimates (1.10), we obtain Av(t)Hc(t)HHAv0

H+A1/2s(t)HHA1/2v0

H

+f(T)H+c(t)HHf(0)H +

t

0

c(tλ)HHf(λ)H

Av0

H+A1/2v0

H+ 2f(0)H+ 2Tmax

0tT

f(t)H.

(1.14)

Then from the last estimate, it follows that

0maxtT

Au(t)HAv0

H+A1/2v0

H+ 2f(0)H+ 2T max

0tT

f(t)H. (1.15)

The estimate for max0tTd2v(t)/dt2Hfollows from the last estimate and the triangle

inequality.

Remark 1.2. Theorem 1.1holds in an arbitrary Banach spaceEunder the following as- sumptions (see, e.g., [3,8,11] and the references given therein):

c(t)EEM, A1/2s(t)EEM, 0tT. (1.16) Now, we consider the application of this abstractTheorem 1.1. First, we consider the mixed problem for wave equation

utt(t,x) a(x)ux

x+u(t,x)=f(t,x), 0tT, 0xL, u(0,x)=ϕ(x), ut(0,x)=ψ(x), 0xL, u(t, 0)=u(t,L), ux(t, 0)=ux(t,L), 0tT.

(1.17)

The problem (1.17) has a unique smooth solutionu(t,x) for the smootha(x)>0 (x [0,L]),ϕ(x),ψ(x) (x[0,L]), and f(t,x) (t[0,T], x[0,L]) functions. This allows us to reduce the mixed problem (1.17) to the initial value problem (1.1) in Hilbert space H=L2[0,L] with a selfadjoint positive definite operatorAdefined by (1.17). We give a number of corollaries of the abstractTheorem 1.1.

Theorem1.3. For solutions of the mixed problem (1.17), the stability inequalities

0maxtT

u(t,·)W21[0,L]

M

0maxtT

f(t,·)L2[0,L]+ϕW21[0,L]+ψL2[0,L]

,

0maxtT

u(t,·)W22[0,L]+ max

0tT

utt(t,·)L2[0,L]

M

0maxtT

ft(t,·)L2[0,L]+f(0,·)L2[0,L]+ϕW22[0,L]+ψW21[0,L]

(1.18)

hold, whereMdoes not depend on f(t,x)andϕ(x),ψ(x).

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The proof of this theorem is based on the abstractTheorem 1.1and the symmetry properties of the space operator generated by the problem (1.17).

Second, letΩRnbe the bounded open domain with smooth boundaryS,=S.

In [0,T]×Ωwe consider the mixed boundary value problem for hyperbolic equations

utt(t,x) n

r=1

ar(x)uxr

xr= f(t,x), x=

x1,...,xn

Ω, 0tT, u(0,x)=ϕ(x), ∂u(0,x)

∂t =ψ(x), xΩ, u(t,x)=0, xS, 0tT,

(1.19)

wherear(x) (xΩ),ϕ(x),ψ(x) (xΩ), andf(t,x) (t[0,T],xΩ) are given smooth functions andar(x)>0.

We introduce the Hilbert spaceL2(Ω) which is the space of all the integrable functions defined onΩ, equipped with the norm

fL2()=

···

x

f(x)2dx1···dxn

1/2

. (1.20)

The problem (1.19) has a unique smooth solutionu(t,x) for the smoothar(x)>0 and f(t,x) functions. This allows us to reduce the mixed problem (1.19) to the initial value problem (1.1) in Hilbert spaceH=L2(Ω) with a selfadjoint positive definite operatorA defined by (1.19). We give a number of corollaries of the abstractTheorem 1.1.

Theorem1.4. For solutions of the mixed problem (1.19), the stability inequalities

0maxtT

u(t,·)W21(Ω)

M

0maxtT

f(t,·)L2()+ϕW21()+ψL2()

,

0maxtT

u(t,·)W22()+ max

0tT

utt(t,·)L2()

M

0maxtT

ft(t,·)L2(Ω)+f(0,·)L2(Ω)+ϕW22()+ψW21()

(1.21)

hold, whereMdoes not depend on f(t,x)andϕ(x),ψ(x).

The proof of this theorem is based on the abstractTheorem 1.1and the symmetry properties of the space operator generated by the problem (1.19).

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2. The high order of accuracy difference schemes generated by an exact difference scheme

We consider the initial value problem (1.1). On the segment [0,T] we consider a uniform grid

[0,T]τ=

tk=,k=0, 1,...,N,=T (2.1) with stepτ. Using the formula (1.2), we can establish the following two-step difference scheme for the solution of the initial value problem (1.1)

vtk+1

2c(τ)vtk

+vtk1

= tk+1

tk

stk+1zf(z)dz+ tk

tk1

sztk1

f(z)dz, 1kN1, v(0)=v0,

τ1v(τ)v(0)=τ1c(τ)Iv(0) +τ1s(τ)v(0) +τ1 τ

0s(τz)f(z)dz

(2.2)

or 1 τ2

uk+12uk+uk1

= 2 τ2

c(τ)Iuk+ fk, fk=τ1f1,k+1+s(τ)f2,kc(τ)f1,k

, f1,k=τ1

tk

tk1

stkzf(z)dz, f2,k=τ1 tk

tk1

ctkzf(z)dz, 1kN1, u0=v0, u1=c(τ)v(0) +s(τ)v(0) +τ f1,1.

(2.3) The latter will be referred to as the exact two-step difference scheme for the initial value problem (1.1).

Suppose the operators (Ie2τiB) and (I+eτiB) have the bounded inverses (Ie2τiB)1 and (I+eτiB)1. Then this problem is uniquely solvable, and the following formula holds:

uk=

ekτiB

Ie2τiB1ekτiBekτiBu0+eτiBIe2τiB1ekτiBekτiBu1

+τ2

k1

m=1

eτiBIe2τiB1e(km

τiBe(km)τiBfm, 2kN.

(2.4) Actually, (2.3) can be rewritten as the equivalent initial value problem for system of the first-order difference equations:

τ1ukuk1

+τ1IeτiBuk1=zk, 1kN,u0,u1are given, τ1zk+1zk

+τ1IeτiBzk=fk, 1kN1. (2.5)

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From that, the system of recursion formulas follows :

uk=eτiBuk1+τzk, 1kN,

zk+1=eτiBzk+τ fk, 1kN1. (2.6) Hence

uk=ekτiBu0+

k s=1

e(ks)τiBτzs, 1kN, zk=e(k1)iτBz1+

k1 m=1

e(k1m)τiBτ fm, 2kN.

(2.7)

From that it follows that

u1=eτiBu0+τz1, uk=ekτiBu0+e(k1)τiBτz1+

k

s=2

e(ks)τiBτ

e(s1)iτBz1+

s1

m=1

e(s1m)τiBτ fm

=ekτiBu0+

e(k1)τiB+

k

s=2

e(ks)τiBe(s1)iτB

τz1

+

k s=2

e(ks)τiB

s1 m=1

e(s1m)τiBτ2fm, 2kN.

(2.8)

Since

e(k1)τiB+

k

s=2

e(ks)τiBe(s1)iτB=

Ie2τiB1e(k1)τiBe(k+1)τiB

=

Ie2τiB1eτiBekτiBekτiB,

k

s=2

e(ks)τiB

s1

m=1

e(s1m)τiBτ2fm=k

1 m=1

k

s=m+1

e(ks)τiBe(s1m)τiBτ2fm

=k

1 m=1

Ie2τiB1eτiBe(km)τiBe(km)τiBτ2fm, (2.9) we have that

uk=ekτiBu0+Ie2τiB1eτiBekτiBekτiBτz1

+

k1

m=1

Ie2τiB1eτiBe(km)τiBe(km)τiBτ2fm, 2kN. (2.10)

Using this formula and the formulaτz1=u1eτiBu0, we can obtain the formula (2.4).

We investigate the stability of the exact two-step difference scheme (2.3).

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Theorem2.1. Suppose the operators(Ie2τiB)and(I+eτiB)have the bounded inverses (Ie2τiB)1 and (I+eτiB)1. Suppose further that u0 D(B2(I+eτiB)1), u1u0 D(B2(Ie2τiB)1), and fkD(B(Ie2τiB)1), 1kN1. Then for the solution of the exact two-step difference scheme (2.3), the following stability inequalities hold:

0maxkN

ukH

M

1maxkN1

τIe2τiB1fk

H+

Ie2τiB1u1u0

H

+

I+eτiB1u0

H

,

(2.11)

1maxkN

τ1ukuk1

H+ max

1kN

τ1IeτiBuk1

H

M

1maxkN1

τBIe2τiB1fk

H+BIe2τiB1u1u0

H

+BI+eτiB1u0

H

,

(2.12)

1maxkN1

τ2uk+12uk+uk1

H+ max

1kN1

2 τ2

c(τ)Iuk

H

M

2maxkN1

BIe2τiB1fkfk1

H+τBIe2τiB1f1

H

+B2Ie2τiB1u1u0

H+B2I+eτiB1u0

H

,

(2.13)

whereMdoes not depend onτ,fk,1kN1, andu0,u1. Proof. Using the formula (2.4) and the estimates

e±kτiBHH1, 1kN, (2.14) we obtain

uk

HI+eτiBekτiB

ekτiBekτiBHH

I+eτiB1u0

H

+ekτiBekτiBHH

Ie2τiB1u1u0

H

+τ2

k1

m=1

eτiBe(km)τiBe(km)τiBHH

Ie2τiB1fm

H

M

1maxmN1

τIe2τiB1fm

H+

Ie2τiB1u1u0

H

+

I+eτiB1u0

H

(2.15)

fork2. It obviously holds also fork=0, 1. The estimate (2.11) is established. Applying τ1(IeτiB) to the formula (2.4) and using the estimates (2.14) and

τ1IeτiBB1HH1, (2.16)

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