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A NOTE ON THE DIFFERENCE SCHEMES FOR HYPERBOLIC-ELLIPTIC EQUATIONS

A. ASHYRALYEV, G. JUDAKOVA, AND P. E. SOBOLEVSKII Received 31 October 2004; Accepted 20 January 2005

The nonlocal boundary value problem for hyperbolic-elliptic equationd2u(t)/dt2+Au(t)

= f(t), (0t1),d2u(t)/dt2+Au(t)=g(t), (1t0),u(0)=ϕ,u(1)=u(1) in a Hilbert spaceHis considered. The second order of accuracy difference schemes for ap- proximate solutions of this boundary value problem are presented. The stability estimates for the solution of these difference schemes are established.

Copyright © 2006 A. Ashyralyev et al. 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

It is known (see [14,15,19,20]) that various boundary value problems for the hyperbolic- elliptic equations can be reduced to the nonlocal boundary value problem

d2u(t)

dt2 +Au(t)=f(t) (0t1),

d2u(t)

dt2 +Au(t)=g(t) (1t0), u(0)=ϕ, u(1)=u(1)

(1.1)

for differential equation in a Hilbert spaceH, with the self-adjoint positive definite oper- atorA.

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

(i)u(t) is twice continuously differentiable in the region [1, 0)(0, 1] and contin- uously differentiable on the segment [1, 1]. The derivative at the endpoints of the segment are understood as the appropriate unilateral derivatives.

(ii) The element u(t) belongs toD(A) for allt[1, 1], and the functionAu(t) is continuous on [1, 1].

(iii)u(t) satisfies the equation and boundary value conditions (1.1).

Hindawi Publishing Corporation Abstract and Applied Analysis

Volume 2006, Article ID 14816, Pages1–13 DOI10.1155/AAA/2006/14816

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Theorem 1.1 [13]. Suppose thatϕD(A), and let f(t) be continuously differentiable on [0, 1] andg(t) be continuously differentiable on [1, 0] functions. Then there is a unique solution of the problem (1.1) and the stability inequalities

max1t1

u(t)HM

ϕH+ max

1t0

A1/2g(t)H+ max

0t1

A1/2f(t)H

,

max1t1

du dt

H+ max

1t1

A1/2u(t)H

MA1/2ϕH+ 0

1

g(t)Hdt+ 1

0

f(t)Hdt

,

max1t1

d2u dt2

H

+ max

1t1

Au(t)H

MH+g(0)H+f(0)H+ 0

1

g(t)Hdt+ 1

0

f(t)Hdt

,

(1.2)

hold, whereMdoes not depend on f(t),t[0, 1],g(t),t[1, 0] andϕ.

In the paper [13] the first order of accuracy difference scheme for approximately solv- ing the boundary value problem (1.1)

uk+12uk+uk1

τ2 +Auk+1=fk, fk=f(tk+1),tk+1=(k+ 1)τ, 1kN1, =1,

uk+12uk+uk1

τ2 +Auk=gk, gk=g(tk), tk=kτ,N+ 1k≤ −1, u0=ϕ, uN=uN, u1u0=u0u1

(1.3) was investigated.

Theorem 1.2 [6]. LetϕD(A). Then for the solution of the difference scheme (1.3) obey the stability inequalities

NmaxkN

uk

HM

ϕH+ max

N+1k≤−1

A1/2gk

H+ max

1kN1

A1/2fk

H

,

N+1maxkN

ukuk1

τ

H+ max

NkN

A1/2uk

H

M

A1/2ϕH+

1

k=−N+1

τgk

H+

N1

k=1

τfk

H

,

(3)

N+1maxkN1

uk+12uk+uk1

τ2

H+ max

NkN

AukH

M

H+g1

H+f1

H+

1

k=−N+1

gkgk1

H+

N1

k=2

fkfk1

H

,

(1.4)

whereMdoes not depend onτ,ϕ, and fk, 1kN1,gk,N+ 1k≤ −1.

Methods for numerical solutions of the nonlocal boundary value problems for partial differential equations have been studied extensively by many researches (see [1,2,5,3,4, 7–9,11,12,16–18,21,22] and the references therein).

In present paper the second order of accuracy difference schemes approximately solv- ing the boundary-value problem (1.1) are presented. The stability estimates for the solu- tion of these difference schemes are established.

2. The second order of accuracy difference schemes

Applying the second order of accuracy difference schemes of paper [10] for hyperbolic equations and the second order of accuracy difference scheme for elliptic equations we will construct the following second order of accuracy difference schemes for approxi- mately solving the boundary value problem (1.1):

uk+12uk+uk1

τ2 +Auk+τ2

4A2uk+1=fk, fk=f(tk), tk=, 1kN1,=1,

uk+12uk+uk1

τ2 +Auk=gk, gk=g(tk),tk=,N+ 1k≤ −1, u0=ϕ, uN=uN, u1u0τ2

2

f0Au0

=u0u1τ2

2(g0Au0), g0=g(0), f0=f(0),

(2.1) uk+12uk+uk1

τ2 +1 2Auk+1

4

Auk+1+Auk1

=fk,

fk=f(tk), tk=, 1kN1,=1,

uk+12uk+uk1

τ2 +Auk=gk, gk=g(tk),tk=,N+ 1k≤ −1,u0=ϕ, uN=uN,

I+τ2A

4

(u1u0)τ2 2

f0Au0

=u0u1τ2 2

g0Au0

, g0=g(0), f0=f(0).

(2.2)

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Theorem 2.1. LetϕD(A). Then for the solution of the difference scheme (2.1) obey the stability inequalities

NmaxkN

uk

HMϕH+ max

N+1k0

A1/2gk

H+ max

0kN1

A1/2fk

H

,

N+1maxkN

ukuk1

τ

H+ max

NkN

A1/2ukH

M

A1/2ϕH+

0

k=−N+1

τgkH+

N1

k=0

τfkH

,

N+1maxkN1

uk+12uk+uk1

τ2

H

+ max

NkN

Auk

H

M

H+g0

H+f0

H+

0

k=−N+1

gkgk1

H+

N1

k=1

fkfk1

H

,

(2.3)

whereMdoes not depend onτ,ϕ, and fk, 0kN1,gk,N+ 1k0.

The proof ofTheorem 2.1follows the scheme of the proof ofTheorem 1.2is based on the formulas

uk=

DτA1/2DτA1/21

×

DτA1/2IDk1τA1/2+IDτA1/2Dk1τA1/2u0

+DτA1/2DτA1/21DkτA1/2DkτA1/2u0u1

+τ2

2

DτA1/2DτA1/21DkτA1/2DkτA1/2f0g0

k1

s=1

τ

2iA1/2DksτA1/2DksτA1/2fs, 1kN1,D±τA1/2=

1±iτA1/2τ2A 2

1

, uk=Rku0+IR2N1RNkRN+kRNu0uN

+IR2N1RNkRN+k

1

s=−N+1

B1RNsRN+sR1(2 +τB)1gsτ

+

1

s=−N+1

B1R(k+s)R|sk|(2 +τB)1R1gsτ,

N+ 1k≤ −1,R=(1 +τB)1,B=+A1/2τ2A+ 4

2 ,

(5)

uN=TDτA1/2DτA1/21

×

DτA1/2IDN1τA1/2+IDτA1/2DN1τA1/2u0

+DτA1/2DτA1/21DNτA1/2DNτA1/2u0

DτA1/2DτA1/21DNτA1/2DNτA1/2

×

Ru0+IR2N1RN+1RN1RNu0+IR2N1RN+1RN1

×

1

s=−N+1

B1RNsRN+sR12 +τB1gsτ

+

1

s=−N+1

B1R1sR1+s2 +τB1R1gsτ

+τ2 2

DτA1/2DτA1/21DkτA1/2DkτA1/2f0g0

N1

s=1

τ

2iA1/2DNsτA1/2DNsτA1/2fs

, T=

I

IR2N1RN+1RN1DτA1/2DτA1/21DNτA1/2DNτA1/21 (2.4)

and on the estimates

D(±τA1/2)HH1, τA1/2D(±τA1/2)HH2, (2.5) (kτB)αRkHHM(1 +δτ)k, k1, 0α1,δ >0,M >0, (2.6)

and on the following lemmas.

Lemma 2.2. The estimate holds:

DN(±τA1/2)expiA1/2A1

HHτ

2. (2.7)

Proof. We use the identity

DN±τA1/2expiA1/2= 1

0Ψ(sτA1/2)ds, (2.8)

(6)

where

Ψ(sτA1/2)=DN±sτA1/2expi(1s)A1/2. (2.9)

The derivativeΨ(sτA1/2) is given by

ΨsτA1/2=DN+1sτA1/22s2A3/2 2

expi(1s)A1/2. (2.10)

Thus,

DN±τA1/2expiA1/2

= ∓ 1

0DN+1±sτA1/2iA3/21

2τ2s2expi(1s)A1/2ds. (2.11)

Using the last identity and estimates (2.6) and

expi(1s)A1/21, (2.12)

we obtain

DN±τA1/2expiA1/2A1

HH

1 2

1

0

DN±sτA1/2

HHτsτsA1/2D±sτA1/2HH

×expi(1s)A1/2HHds

τ 1

0s ds=τ 2.

(2.13)

Lemma 2.3. The following estimate holds:

THHM, (2.14)

whereMdoes not depend onτ.

(7)

Proof. Since

T=

IR2NIR2N+RN+1RN1

×

DτA1/2DτA1/21DNτA1/2DNτA1/21, (2.15) T

Iexp2A1/2+ 2A1/2s(1) expA1/21

=TIexp2A1/2+ 2A1/2s(1) expA1/21

×

R2Nexp2A1/2+ 2A1/2s(1) expA1/2

RN+1RN1

×

DτA1/2DτA1/21DNτA1/2DNτA1/2,

(2.16)

Iexp2A1/2+ 2A1/2s(1) expA1/21HHM, (2.17)

to prove (2.14) it suffices to establish the estimate R2Nexp2A1/2+ 2A1/2s(1) expA1/2

RN+1RN1

×

DτA1/2DτA1/21DNτA1/2DNτA1/2

HHMτ.

(2.18)

Here T=

IR2N+RN+1RN1

×

DτA1/2DτA1/21DNτA1/2DNτA1/21, s(1)=A1/2eiA1/2eiA1/2

2i .

(2.19)

The estimate (2.17) was proved in [19]. Finally, using the identity R2Nexp2A1/2+ 2A1/2s(1) expA1/2

RN+1RN1

×

DτA1/2DτA1/21DNτA1/2DNτA1/2

=R2Nexp2A1/2 +2A1/2s(1)1

i

DNτA1/2DNτA1/2expA1/2

+1 i

DNτA1/2DNτA1/2expA1/2RN

+1 i

DNτA1/2DNτA1/2

× RN

RN+1RN1DτA1/2DτA1/21

(2.20)

and the estimates (2.5), (2.6), and (2.7), we obtain the estimate (2.18).

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Theorem 2.4. LetϕD(A3/2).Then for the solution of the difference scheme (2.2) obey the stability inequalities

NmaxkN

uk

HM

I±1

2iτA1/2

ϕ

H

+ max

N+1k0

A1/2gk

H+ max

0kN1

A1/2fk

H

,

N+1maxkN

ukuk1

τ

H+ max

NkN

A1/2uk

H

M A1/2

I±1

2iτA1/2

ϕ

H

+

0

k=−N+1

τgk

H+

N1

k=0

τfk

H

,

N+1maxkN1

uk+12uk+uk1

τ2

H

+ max

NkN

Auk

H

M A

I±1

2iτA1/2

ϕ

H

+g0

H+f0

H+

0

k=−N+1

gkgk1

H+

N1

k=1

fkfk1

H

, (2.21)

whereMdoes not depend onτ,ϕ, and fk, 0kN1,gk,N+ 1k0.

The proof ofTheorem 2.4follows the scheme of the proof ofTheorem 1.2is based on the formulas

uk=

DτA1/2DτA1/21

×

IDτA1/2Dk1τA1/2+DτA1/2IDk1τA1/2u0

+DτA1/2DτA1/21DkτA1/2DkτA1/2I+τ2A 4

1 u0u1

+τ2 2

DτA1/2DτA1/21DkτA1/2DkτA1/2I+τ2A 4

1 f0g0

+

k1

s=1

I+τ2A

4 1

DτA1/2DτA1/21DksτA1/2DksτA1/2fs,

1kN1,D±τA1/2=

1iτA1/2 2

I±iτA1/2 2

1

,

(9)

uk=Rku0+IR2N1RNkRN+kRNu0uN +IR2N1RNkRN+k

1

s=−N+1

B1RNsRN+sR12 +τB1gsτ

+

1

s=−N+1

B1R(k+s)R|sk|(2 +τB)1R1gsτ,

N+ 1k≤ −1,R=(1 +τB)1,B=+A1/2τ2A+ 4

2 ,

uN=TDτA1/2DτA1/21

×

IDτA1/2DN1τA1/2+DτA1/2IDN1τA1/2u0

+DτA1/2DτA1/21DNτA1/2DNτA1/2I+τ2A 4

1

u0

DτA1/2DτA1/21DNτA1/2DNτA1/2

×

Ru0+IR2N1RN+1RN1RNu0+IR2N1RN+1RN1

×

1

s=−N+1

B1RNsRN+sR1(2 +τB)1gsτ

+

1

s=−N+1

B1R1sR1+s(2 +τB)1R1gsτ

+τ2 2

DτA1/2DτA1/21DkτA1/2DkτA1/2

× I+τ2A

4 1

f0g0

N1

s=1

τ

2iA1/2DNsτA1/2DNsτA1/2fs

,

T=

I

IR2N1RN+1RN1

×

DτA1/2DτA1/21DNτA1/2DNτA1/2I+τ2A 4

11

(2.22)

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