2011
DOI: 10.1515/gmj.2011.0044
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A one-parameter family of difference schemes for the regularized long-wave equation

Abstract: We consider an initial boundary-value problem for the regularized long-wave equation. A three level one-parameter family of conservative difference schemes is studied. Two level schemes are used to find the values of the unknown functions on the first level. A numerical method for selection of artificial boundary conditions is proposed. It is proved that the finite difference scheme converges at rate O. 2 C h 2 / when an exact solution belongs to the Sobolev space W 3 2 . and the initial boundary-value problem… Show more

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Cited by 12 publications
(12 citation statements)
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“…Note that the exact solution of the problem is u(x,t)=3csech2k0[x(c+1)tx0], where c , x 0 are arbitrary constants and k0=12cμfalse(1+cfalse).We discretize the problem by the following FD scheme: UtnμboldH1Uxx¯tn+boldH2Ux^n+12+H2Φ(boldUn+12,boldUn+12)=0,n=0,,N1, Uj0=u0(xj),j=1,,J. As it is shown in Berikelashvili and Mirianashvili, with the periodic boundary conditions for a region a ≤ x ≤ b , the RLW equation has 3 conservation laws, which can be expressed in the following form I1=abu0.1emdx, I2=ab()u2+μux2…”
Section: Numerical Experimentsmentioning
confidence: 99%
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“…Note that the exact solution of the problem is u(x,t)=3csech2k0[x(c+1)tx0], where c , x 0 are arbitrary constants and k0=12cμfalse(1+cfalse).We discretize the problem by the following FD scheme: UtnμboldH1Uxx¯tn+boldH2Ux^n+12+H2Φ(boldUn+12,boldUn+12)=0,n=0,,N1, Uj0=u0(xj),j=1,,J. As it is shown in Berikelashvili and Mirianashvili, with the periodic boundary conditions for a region a ≤ x ≤ b , the RLW equation has 3 conservation laws, which can be expressed in the following form I1=abu0.1emdx, I2=ab()u2+μux2…”
Section: Numerical Experimentsmentioning
confidence: 99%
“…We consider, in this case, the RLW equation with initial condition given by the linear sum of 2 well‐separated solitary waves of various amplitudes ufalse(x,0false)=truei=123ci0.1emsech2()kifalse(xxifalse), where ci=4ki214ki2,ci and x i are constants, i = 1,2(see Berikelashvili and Mirianashvili). In this section, we choose μ = 1, k 1 = 0.4, k 2 = 0.3, x 1 = 15, x 2 = 35, h = 0.3,and k = 0.1 with interval [0,120].…”
Section: Numerical Experimentsmentioning
confidence: 99%
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