Numerical solution of linear wave equation with strong dissipative term

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Abstract

In this study, for the periodic problem with respect to time of wave theory, three level difference scheme is presented. The difference scheme is constructed by the method of integral identities with the use of linear basis function and interpolating quadrature rules with the remainder terms in integral form. Error of difference solution is estimated. The theoretical results are controlled on a numerical example.

Introduction

We discuss the following boundary value problemLu≡2ut2+L1ut+L0u=f(x,t),(x,t)∈D=(0,ℓ)x(0,T],u(0,t)=u(ℓ,t)=0,u(x,0)=u(x,T),ut(x,0)=ut(x,T),whereL1ut=−xa(x,t)2utx+b(x,t)ut,L0u=−xc(x,t)ux+d(x,t)uand a, b, c, d, f are sufficiently smooth functions in D.

The equations of this type arise in many areas of mathemetical physics and fluid mechanics. These are used for studies about communication lines, electron plasm waves in plasmas, ion acoustics waves and other physical models [4], [5], [9].

We presented three-level difference scheme for this problem. For wave equation with strong dissipative term, difference schemes is constructed, mathematical researches are done and approximate error is presented that the convergence is O(h2+τ2), has been proved.

Existence, uniqueness and stability of exact solution of this type problems were investigated by several mathematician [6], [7], [8], [10].

And also, the numerical solutions of this type equations in simpler models are researched [1], [2], [3].

Section snippets

Establish of difference scheme

We suppose that 0<α⩽a(x,t)⩽a, 0⩽b⩽b(x,t)⩽b, 0<c⩽c(x,t)⩽c, 0⩽d⩽d(x,t)⩽d and āa(x,t)tā, b̄b(x,t)tb̄, c̄c(x,t)tc̄, d̄d(x,t)td̄ in problem , , , .

Now, let us establish the mesh ω=ωh×ωτ in domain D, such thatωh={xi=ih,i=1,2,…,N−1,h=ℓ/N},ωτ={tj=jτ,j=1,2,…,M−1,τ=T/M}and ω̄hh∪{x=0,ℓ}, ωτ+=ωτ∪{t=T}.

We establish the scheme in two stage. Firstly, if the basis functionsϕi(x)=ϕi(1)(x)≡(x−xi−1)/h,xi−1<x<xi,ϕi(2)(x)≡(xi+1−x)/h,xi<x<xi+1,0,x∉(xi−1,xi+1),i=1,2,…,N−1is

Error estimates of approximate solution

The following difference problem for the error can be written, while z=yu,ℓz=R,(x,t)∈ωh×ωτ+,z(0,t)=z(ℓ,t)=0,t∈ω̄τ,z(x,0)=z(x,T),x∈ω̄h,z(x,τ)=z(x,T+τ),x∈ω̄h.

Lemma 3.1

When the conditionλ0c+28d140232112c̄+28d̄>0,held, the following estimation for error of difference problem∥zt2+∥ẑx̄2+∥z∥2⩽Cτ∑i=1M(∥Ri(0)2+∥Ri(1)2)exp(−C1tM−i),t∈ωτ+is true, such thatλ0<42α+b2and0=max(ā,3ā),1=max(b̄,3b̄),where C and C1, which are independent of h and τ, are positive constants.

Proof

We begin to prove by

Numerical example

The theoretical results are controlled for the following example:f(x,t)=π2sinπxesin2πt(4cos22πt−4sin2πt+2πcos2πt+1),a(x,t)=1,b(x,t)=0,c(x,t)=1,d(x,t)=0,T=1,ℓ=1.The exact solution of appropriate problem is given byu(x,t)=sinπxesin2πt.The following iteration process is applied for (2.4)yt̄t,i(n+1)Ay0tx̄(n+1)x+By0t(n+1)Cyx̄(n+1)x+Dy(n+1)=f,y(n+1)(0,t)=y(n+1)(ℓ,t)=0,y(n+1)(x,0)=y(n)(x,T),y(n+1)(x,τ)=y(n)(x,T+τ),n=0,1,2,…, y(0)(x,T) and y(0)(x,T+τ) are arbitrary functions.

The computational results

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