Numerical Algorithm of Solving the Forced sin-Gordon Equation

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258 PROCEEDINGS OF THE IMCSIT. VOLUME 3, 2008(5)Substituting (3) and (5) into (4) in the first order on εwe obtainIn the second order(6)(7)To apply the Riemann’s method of solving (6) and (7) [1]we need to transform these equations to the standard formwhich does not contain the second mixed derivative. Toget rid of the mixed derivatives let us make a transformationχ = χ V χandτ = t − . In the new variables equationsa(6) and (7) read(8)Fig .2. The function f ( χ ) , a = 1 .The corrections ui( 0 0 )0 ( 0 ) ( χ , τ )χ , τ , i = 1, 2 to the kink solutionu χ at a point 0 0 can be presented through theRiemann function ω( χ, τ ) [1]:(13)and(9)with trivial initial conditions over the straight line CVC : τ = − ξ(10)a(14)where A is the characteristic triangle in the plane ( χ, τ ),bounded by the straight line C given in (10) and the characteristicsA : τ ( χ − χ0 ) = τ10 (Fig. 3).aIII. THE RIEMANN FUNCTIONEquations (8) and (9) have the same left-hand side andcan be presented as(11)(12)wheref2( χ) 2 / cosh ( χ / a)= since1χacos[4arctan( e )] = 1−212cosh ( χ)aThe graph of the function f ( χ ) for the value V = 0.5is shown in Fig. 2.Fig. 3. The characteristic triangle A.The Riemann function ω( χ, τ ) satisfies the equation(15)with ω = 1/(2a) over the characteristics and b= a .
ALEXANDRE BEZEN: NUMERICAL ALGORITHM OF SOLVING THE FORCED SIN-GORDON EQUATION 259Since outside of a finite interval ( − χlim , χlim)then one can expect that the solution of (15) can be closeto the solution of the telegrapher’s equation [1]:1which is ω( χ, τ ) =0( )2aJ b γ and2 12γ = ( τ −τ 0) − ( χ − χ2 0) . The variable γ is a hyperbolicdistance in the characteristic triangle A . The dis-atance γ2 12is positive inside A, i.e. ( τ − τ0) > ( χ − χ2 0) ,a2 12and imaginary when ( τ − τ0) < ( χ − χ2 0) . It vanishesa2 12when ( τ − τ0) = ( χ − χ2 0) .aThe equation (15) in the variables ( χ, γ ) has the formLet’s look for the solution of (16) in the form(16)(17)where the first term of the sum is the solution of thetelegrapher’s equation and the second term must be smallfor a finite value of χlim .Since1J0 ( bγ ) + J0 ( bγ ) + J0( bγ) = 0bγthen substitution of (17) into (16) givesFor χ = χ0the last equation becomesAssumeA = 0 for odd n,AAnIt follows thatand1 f ( ξ ) 1=,02 22 a ϕ( ξ0) 2f ( ξ ) ( −1)a= [ + ( ( ) −2a 2 4 ...(2 k)bk202k+ 2ϕ ξ2 2 2 2 01(1 − f ( ξ0 )) ϕ( ξ0 )) A2 k] ,2(2k+ 2) ϕ( ξ )k = 1,2,3...Ak1 ( −1)= , k = 1,2,3...2a2 4 ...(2 k)2k+ 2 2 2 20(17)1ω( χ, γ ) = [ J0 ( bγ ) + (1 − J0( bγ )) ϕ( χ)](18)2aSubstitute (18) into (16) and we obtain that the solutionϕ( χ ) satisfies the following equation−a(1 − J ( bγ )) ϕ ( χ)+0( J ( bγ ) + J ( bγ ))( χ − χ ) ϕᄁ( χ)+0 2 0(1 − f ( χ )(1 − J ( bγ ))) ϕ( χ) = J ( bγ ) f ( χ)0 0and subject to the boundary conditionsIV. NUMERICAL ALGORITHM AND CALCULATIONS(19)(20)A solution ω( χ1, τ1)of the partial differential equation(15) at any particular point ( χ1, τ1)can be reduced tosolving the boundary value problem (19), (20) with fixedvalue of γ , where2 12γ = ( τ1 −τ 0) − ( χ2 1− χ0) .aA family of curves γ = const define hyperbolaeimbedded into the characteristic triangle (Fig.4). Thesolution surface (15) can be represented as a family ofcurves over the hyperbolae in the 3D space ( χ, τ , ω ) .The surface representing the Riemann function for χ0= 0is shown in Fig. 5 and was drawn using MAPLE.and, therefore,
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Numerical Algorithm of Solving the Forced sin-Gordon Equation

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