Internatonal Journal of Mathematcal Modellng & Computatons Vol. 05, No. 04, Fall 2015, 291-305 Numercal Soluton of One-Dmensonal Heat and Wave Equaton by Non-Polynomal Quntc Splne J. Rashdna a, and M. Mohsenyzadeh a a Department of Mathematcs, Karaj Branch, Islamc Azad Unversty, Karaj, Iran.. Abstract.Ths paper present a numercal algorthm for the lnear one-dmensonal heat and wave equaton. In ths method, a fnte dfference approach had been used to dscrete the tme dervatve whle quntc splne s appled as an nterpolaton functon n the space dmenson. We dscuss the accuracy of the method by expandng the equaton based on Taylor seres and mnmze the error. The proposed method has eghth-order accuracy n space and fourth-order accuracy n tme varables. From the computatonal pont of vew, the soluton obtaned by ths method s n excellent agreement wth those obtaned by prevous works and also t s effcent to use. Numercal examples are gven to show the applcablty and effcency of the method. Receved: 28 June 2014, Revsed: 2 February 2015, Accepted: 2 August 2015. Keywords: Dfferental Equaton, Quntc Splne, Heat Equaton, Wave Equaton, Taylor Approxmaton. Index to nformaton contaned n ths paper 1 Introducton 2 Formulaton of Tenson Quntc Splne Functon 3 Numercal Technque 4 Error Estmate n Splne Approxmaton 5 Stablty Analyss 6 Numercal Example 7 Concluson 1. Introducton Consder one-dmensonal heat equaton of the form u t = c2 2 u, 0 x l, t 0, 1 x2 Correspondng author. Emal: Rashdna@ust.ac.r c 2015 IAUCTB http://www.jm2c.r
292 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. wth ntal condton ux, 0 = f 1 x, 0 x l, 2 and boundary condtons u0, t = p 1 t, t 0, and one-dmensonal wave equaton of the form ul, t = q 1 t, t 0, 3 2 u t 2 = c2 2 u, 0 x l, t 0, 4 x2 wth ntal condton ux, 0 = f 2 x, and boundary condtons ux, 0 t = f 3 t, 0 x l, 5 u0, t = p 2 t, t 0, ul, t = q 2 t, t 0, 6 where c 2 and l are postve fnte real constants and f 1 x, f 2 x, f 3 x, p 1 t, p 2 t, q 1 t and q 2 t are real contnuous functons. Some phenomena, whch arse n many felds of scentfc such as sold state physcs, plasma physcs, flud dynamcs, mathematcal bology and chemcal knetcs, can be modeled by partal dfferental equatons. The heat and wave equatons are of prmary mportance n many physcal systems such as electro-thermal analogy[1], sgnal formaton[2], dranng flm [3], water transfer n sols [4], mechanc and physcs[5-8], elastcty[9] and etc. There are several numercal schemes that have been developed for the soluton of the heat and wave equaton [10-14]. Cubc splne has been used to approxmate the one dmensonal heat conducton n[15], also the collocaton method based on hermte cubc splne appled to solve one-dmensonal heat conducton problem n [16]. In ths paper, a method based on quntc splne for second-order boundary value problems Eq.1 and Eq.4 s presented. Ths approach wll employ consstency relatons at md-knot. In Secton 2 the formulaton of non-polynomal quntc splne has been developed and the consstency relaton obtaned s useful to dscretze heat equaton Eq.1 and wave equaton Eq.4. In Secton 3, we present dscretzaton of the equaton by a fnte dfference approxmaton to obtan the formulaton of proposed method. In Secton 4, Truncaton error and stablty analyss are dscussed. In ths secton we approxmate the functons based on Taylor seres to mnmze the error term and to obtan the class of methods. In Secton 5, numercal experments are conducted to demonstrate the vablty and the effcency of the proposed method computatonally.
J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. 293 2. Formulaton of Tenson Quntc Splne Functon In recent years, many scholars have used non-polynomal splne for solvng dfferental equatons [17-19]. The splne functon s a pecewse polynomal or nonpolynomal of degree n satsfyng the contnuty of the n 1th dervatve. Tenson quntc splne s a non-polynomal functon that has sx parameters to be determned hence t can satsfy the condtons of two endponts of the nterval and contnuty of frst, second, thrd and fourth dervatves. We ntroduce the set of grd ponts n the nterval [0, l] n space drecton x = h, h = l n + 1 For each segment, quntc splne p x s defne as, = 0, 1, 2,..., n + 1. p x = a + b x x + c x x 2 + d x x 3 + e e ωx x e ωx x + f e ωx x + e ωx x, = 0, 1,..., n, 7 where a, b, c, d, e and f are the unknown coeffcents to be determned also ω s free parameter. If ω 0 then p x reduces to quntc splne n the nterval [0, l]. To derve the unknown coeffcents, we defne p x = u, p x +1 = u +1, p íx = M, p íx +1 = M +1, p 4 x = S, p 4 x +1 = S +1. 8 From Eq.7 and Eq.8, we can determne the unknown coeffcents a = u S ω 4, b = u +1 h u h + S h ω 4 + c = M 2 S 2ω 2, d = 1 6h h + S +1 6ω 2 1 hω 4 h 3ω 2 M +1 M + S ω 2 S +1 ω 2 e = S +1 ω 4 e θ e θ S e θ + e θ 2ω 4 e θ e θ, f = S 2ω 4, h 3 M h 6 M +1, where θ = ωh and = 0, 1, 2,..., n. Fnally usng the contnuty of frst dervatve at the support ponts for = 2, 3,..., n 1, we have u +1 h = S 1 2 u h + u 1 h h 6 M 1 2h 3 M h 6 M +1 = hω 4 h e θ 6ω 2 + e θ 2 2ω 3 e θ e θ + eθ e θ 2ω 3, 9
294 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. and from contnuty of thrd dervatve, we have M +1 h 2 M h + M 1 h +S 2 hω 2 + 2 e = S 1 hω 2 θ + e θ 2 2ω e θ e θ + eθ e θ 2ω + S +1 hω 2 + 2 e θ + e θ ω e θ e θ 1 ω e θ e θ. 10 From Eq.9 and Eq.10, after elmnatng S we have the followng useful relaton for = 2, 3,..., n 2 where u +2 + 2 u +1 6 u + 2 u 1 + u 2 = = h2 20 α M +2 + β M +1 + γ M + β M 1 + α M 2, 11 α = θ 4 2 θ 3 e θ e θ 1 3θ e θ e θ, β = 4 θ 4 + 4 + 2 eθ + 2 e θ θ 3 e θ e θ γ = 6 θ 4 + 4 eθ + 4 e θ 2 3θ e θ e θ + eθ + e θ 4 3θ e θ e θ, 4 + 4 eθ + 4 e θ θ 3 e θ e θ. When ω 0 so θ 0, then α, β, γ 1, 26, 66, and the relaton defned by Eq.11 reduce nto ordnary quntc splne u +2 + 2 u +1 6 u + 2 u 1 + u 2 = = h2 20 M +2 + 26 M +1 + 66 M + 26 M 1 + M 2. 12 3. Numercal Technque By usng Eq.12 for j + 1th, jth and j 1th tme level we have u j+1 +2 + 2 uj+1 +1 6 uj+1 + 2 u j+1 1 + uj+1 2 = = h2 α M j+1 +2 20 + β M j+1 +1 + γ M j+1 + β M j+1 1 + α M j+1 2, 13 u j +2 + 2 uj +1 6 uj + 2 uj 1 + uj 2 = = h2 α M j +2 20 + β M j +1 + γ M j + β M j 1 + α M j 2, 14 u j 1 +2 + 2 uj 1 +1 6 uj 1 + 2 u j 1 1 + uj 1 2 =
J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. 295 = h2 α M j 1 +2 20 + β M j 1 +1 + γ M j 1 + β M j 1 1 + α M j 1 2, 15 that we wll use these equatons to dscretze heat and wave equaton. 3.1 Heat Equaton We develop an approxmaton for Eq.1 n whch frst order tme dervatve s replaced by the followng fnte dfference approxmaton ū j t = uj+1 u j 1 2k = u j t + O k 2, 16 and the space dervatve s replaced by non-polynomal tenson splne approxmaton ū j xx = p x, t j = M j. 17 By usng Eq.16 and Eq.17 we can develop a new approxmaton for the soluton of Eq.1, So that the heat equaton Eq.1 s replaced by η M j 1 + 1 2 η M j + η M j+1 = uj+1 u j 1 2c 2 k, 18 where 0 η 1 s a free constant. If Eq.14 multpled by 1 2η added to Eq.13 and Eq.15 multpled by η and elmnate M j, then we obtan the followng relaton for heat equaton Eq.1 η u + 2, j 1 + u 2, j 1 + 1 2 η u + 2, j + u 2, j +η u + 2, j + 1 + u 2, j + 1 + 2 η [u + 1, j 1 +u 1, j 1] + 2 1 2 η u + 1, j + u 1, j +2 η u + 1, j + 1 + u 1, j + 1 6 η u, j 1 6 1 2 η u, j 6 η u, j + 1 40c 2 α [u + 2, j + 1 k u + 2, j 1 + u 2, j + 1 u 2, j 1] h2 40c 2 β [u + 1, j + 1 u + 1, j 1 + u 1, j + 1 k u 1, j 1] h2 h2 40c 2 γ u, j + 1 u, j 1 = 0, k j = 1, 2, 3,... = 2,..., N 2. 19 3.2 Wave Equaton Fnte dfference approxmaton for second order tme dervatve s ū j tt = uj+1 2 u j + uj 1 k 2 = u j tt + O k 2, 20
296 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. as we consder, the space dervatve s approxmated by the non-polynomal tenson splne ū j xx = p x, t j = M j. 21 By usng Eq.16 and Eq.17 for wave equaton Eq.4 we have η M j 1 + 1 2 η M j + η M j+1 = uj+1 2 u j + uj 1 c 2 k 2, 22 where 0 η 1 s a free constant. Agan we multply Eq.14 by 1 2η and add ths to Eq.13 and Eq.15 multpled by η and elmnate M j, then we obtan the followng relaton for wave equaton Eq.4 η u + 2, j 1 + u 2, j 1 + 1 2 η u + 2, j + u 2, j +η u + 2, j + 1 + u 2, j + 1 + 2 η [u + 1, j 1 +u 1, j 1] + 2 1 2 η u + 1, j + u 1, j +2 η u + 1, j + 1 + u 1, j + 1 6 η u, j 1 6 1 2 η u, j 6 η u, j + 1 h2 40c 2 α [u + 2, j + 1 2u + 2, j + u + 2, j 1 + u 2, j + 1 k 2u 2, j + u 2, j 1] 40c 2 β [u + 1, j + 1 2u + 1, j k +u + 1, j 1 + u 1, j + 1 2u 1, j + u 1, j 1] h2 h2 40c 2 γ u, j + 1 2u, j + u, j 1 = 0, k j = 1, 2, 3,... = 2,..., N 2. 23 4. Error Estmate n Splne Approxmaton To estmate the error for heat and wave equaton we expand Eq.19 and Eq.23 n Taylor seres about ux, t j and then we fnd the optmal values for α, β and γ. 4.1 Error Estmate for Heat Equaton By expandng Eq.19 n Taylor seres and replace the dervatves nvolvng t by the relaton +j u x t j = c2j +2j u, 24 x+2j
J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. 297 for heat equaton Eq.1, we obtan the local truncaton error. The prncpal part of the local truncaton error of the proposed method for heat equaton s T j = α10 γ20 β10 + 6 h 2 D x,x U 0, 0 + α 5 β 20 + 3 h 4 D x,x,x,x U 0, 0 2 + 6 η c 2 h 2 k 2 + α 15 + α 60 β 60 + α 30 +[ α 30 β 240 + 11 h 6 D x,x,x,x,x,x U 0, 0 60 γ c 2 h 2 k 2 D x,x,x,x,x,x U 0, 0 120 c 2 h 4 k 2 β 120 + 3 2 η β 120 + 3 2 η h 4 k 2 c 2 + + 2 225 α 1 7200 β + 43 3360 D x,x,x,x,x,x U 0, 0 h 8 ] D x,x,x,x,x,x,x,x U 0, 0 +.... 25 By choosng sutable values of parameters α, β, γ and η we obtan varous classes of the proposed method. Remark1. If θ 0 n Eq.11 we have α = 1, β = 26 and γ = 66 whch s ordnary quntc splne. Remark2. If we choose α = 7 6, β = 76 3 and γ = 67 n Eq.25 we obtan a new scheme of order Oh 8 + h 4 k 4, furthermore by choosng η = 1 6 we can optmze our scheme, too. 4.2 Error Estmate for Wave Equaton For wave equaton, we expand Eq.23 n Taylor seres and replace the dervatves nvolvng t by the relaton +j u x t j = cj +j u, 26 x+j
298 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. and then we drve the local truncaton error. The prncpal part of the local truncaton error of the proposed method for wave equaton s T j = α10 γ20 β10 + 6 h 2 D x,x U 0, 0 +[ α 5 β 20 + 3 h 4 + 2 + α 120 γ 240 β +[ α 15 + α 60 + β 240 + 11 60 β 240 + 3 2 η 120 + 6η h 6 + α 3600 γ 7200 β + 2 225 α 1 7200 β + 43 3360 h 2 k 2 c] D x,x,x,x U 0, 0 h 4 k 2 c] D x,x,x,x,x,x U 0, 0 3600 + 1 2 η h 2 k 4 c 2 D x,x,x,x,x,x U 0, 0 h 8 D x,x,x,x,x,x,x,x U 0, 0 +.... 27 Remark3. If we choose α = 7 6, β = 76 3 and γ = 67 n Eq.27 we obtan a new scheme of order Oh 8 + h 4 k 4, furthermore by choosng η = 1 12 we can optmze our scheme, too. 5. Stablty Analyss In ths secton, we dscuss stablty of the proposed method for numercal soluton of heat and wave equaton. we assume that the soluton of Eq.19 and Eq.23 at grd pont lh, jk s u j l = ξj e lθ, 28 where = 1, θ s a real number and ξ s a complex number. By substtutng Eq.28 n Eq.19 and Eq.23, we obtan a quadratc equaton as follow For heat equaton we have Qξ 2 + ϕξ + ψ = 0. 29 Q = cos 2 θ 2 η + h2 α + cos θ 4 η + h2 β 6 η + h2 γ 20ck 20ck 20ck, ϕ = cos 2 θ 1 2 η + cos θ 2 4 η 6 + 12 η,
J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. 299 ψ = cos 2 θ 2 η h2 α + cos θ 4 η h2 β 6 η h2 γ 20ck 20ck 20ck. By usng RouthHurwtz crtera and usng transformaton ξ = 1+z 1 z have n Eq.29, we Q ϕ + ψξ 2 + 2Q ψξ + Q + ϕ + ψ = 0. 30 If ξ < 1, then the dfference scheme Eq.19 s stable. It s suffcent to show that Q ϕ + ψ > 0, 2Q ψ > 0 and Q + ϕ + ψ > 0. From the above relaton we have Q ϕ + ψ = cos 2 θ 12 η 2 + cos θ 12 η 2 30 η + 7, Q ψ = h2 10ck αcos2θ + βcosθ + γ, Q + ϕ + ψ = cos 2 θ 4 η + 2 + cos θ 4 η + 2 2 η 7. If η > 1 30 6 and η > 132 then Q ϕ + ψ > 0. For α = 7 6, β = 76 3 and γ = 67 we have Q ψ > 0 and f η > 1 30 2 and η > 12 then Q + ϕ + ψ > 0. Thus our method s stable for heat equaton. For wave equaton we have Q = cos 2 θ 2 η h2 α 10ck 2 + cos θ 4 η h2 β 10ck 2 6 η h2 γ 20ck 2, ϕ = cos 2 θ 2 4 η + h2 α 5ck 2 + cos θ 4 8 η + h2 β 5ck 2 6 + 12 η + h2 γ 10ck 2, ψ = cos 2 θ 2 η h2 α 10ck 2 + cos θ 4 η h2 β 10ck 2 6 η h2 γ 20ck 2, thus we have Q ϕ + ψξ 2 + Q + ϕ + ψ = 0. In order to ξ < 1, we must have ϕ < 0 and Q + ψ > 0. Obvously we have Q + ψ > 0 for each η, f η > 1 2 + 7h2 120ck then ϕ < 0, therefore 2 our scheme wll be stable for wave equaton. 6. Numercal Example We appled the presented method to the followng heat and wave equatons. For ths purpose, we consder three examples for heat equatons and two examples for wave equaton. We appled proposed method wth α, β, γ = 1, 26, 66 method I whch s the ordnary quntc splne of wth order Oh 6 +k 4 and f we select α, β, γ = 7 6, 76 3, 67 we obtan a new method whch s of order Oh 8 + h 4 k 4 method II. Example1: We consder Eq.1 wth c = 1 π, f 1x = snπx and p 1 t = q 1 t = 0. The exact soluton for ths problem s u x, t = e t sn π x.
300 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. Table 1. Absolute errorforexample.1 x t j Method I Method II Method n [13] 0.125 0.01 4.1 10 7 3 10 7 1 10 4 0.125 0.05 8 10 7 7 10 7 2.5 10 4 0.25 0.01 5.3 10 7 4.9 10 7 4 10 4 0.25 0.05 1 10 6 9.5 10 7 4.2 10 4 0.5 0.01 5 10 7 4.9 10 7 3 10 4 0.5 0.05 1 10 6 9.6 10 7 2 10 4 Fgure 1. 1. Space-Tme graph of the soluton up to t=1 s, wth c = 1, t = 0.01 and h = 0.001 for example π Ths problem s solved by dfferent values of the step sze n the x-drecton h and tme step sze t = 0.01. The computed solutons by proposed method are compared wth the exact soluton at the grd ponts and the maxmum absolute errors are tabulated n Table 1. Also the results are compared wth the solutons obtaned n [13]. The space tme graph of the estmated soluton s gven n Fgs. 1. The maxmum absolute error of ths example by method II s 3.6 10 9 and by method I s 3 10 6. Example2: We consder Eq.1 wth c = 1, f 1 x = cos π 2 x and p 1t = e π2 4 t and q 1 t = 0. The exact soluton for ths problem s π u x, t = e π2 t 4 cos 2 x.
J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. 301 Table 2. Absolute errorforexample.2 x t j MethodI MethodII 0.1 0.0003 6.5 10 9 7 10 10 0.1 0.0005 2.5 10 9 4 10 10 0.15 0.0003 2.3 10 9 3 10 10 0.15 0.0005 2.5 10 9 1.6 10 10 0.2 0.0003 2.2 10 9 1.4 10 10 0.2 0.0005 2.1 10 9 1 10 10 Fgure 2. 2. Space-Tme graph of the soluton up to t=1 s, wth c = 1, t = 0.0001 and h = 0.01 for example Ths problem s solved by dfferent values of the step sze n the x-drecton h and tme step sze t = 0.0001. The computed solutons by proposed method are compared wth the exact soluton at the grd ponts and the maxmum absolute errors are tabulated n Table 2. The space tme graph of the estmated soluton s gven n Fgs. 2. The maxmum absolute error of ths example by method II s 1.32 10 8 and by method I s 3 10 9. Example3: We consder Eq.1 wth c = 1, f 1 x = snx, p 1 t = 0 and q 1 t = e t sn1. The exact soluton for ths problem s u x, t = e t sn x.
302 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. Table 3. Absolute errorforexample.3 x t j MethodI MethodII 0.1 0.0003 1 10 10 1 10 10 0.1 0.0005 1 10 10 1 10 10 0.15 0.0003 2 10 11 2 10 11 0.15 0.0005 4 10 11 4 10 11 0.2 0.0003 1 10 10 1 10 10 0.2 0.0005 2 10 10 1 10 10 Fgure 3. 3. Space-Tme graph of the soluton up to t=1 s, wth c = 1, t = 0.0001 and h = 0.01 for example Ths problem s solved by dfferent values of the step sze n the x-drecton h and tme step sze t = 0.0001. The computed solutons by proposed method are compared wth the exact soluton at the grd ponts and the maxmum absolute errors are tabulated n Table 3. The space tme graph of the estmated soluton s gven n Fgs. 3. The maxmum absolute error of ths example by method II s 1 10 9 and by method I s 2.1 10 9. In the followng examples we apply proposed method to the wave equaton. Example4: We consder Eq.4 wth c = 1, f 2 x = 0, f 3 x = πcosπx, p 2 t = snπt and
J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. 303 Table 4. Absolute errorforexample.4 x t j MethodI MethodII 0.05 0.03 9 10 12 3 10 12 0.05 0.05 1.6 10 11 7 10 12 0.1 0.03 1.2 10 11 1.7 10 11 0.1 0.05 1.7 10 11 1 10 12 0.2 0.03 7 10 12 1 10 12 0.2 0.05 1 10 11 1.1 10 11 Fgure 4. Space-Tme graph of the soluton up to t=1 s, wth c = 1, t = 0.01 and h = 0.01 for example 4. q 2 t = snπt. The exact soluton for ths problem s u x, t = π cos π x sn π t. Ths problem s solved by dfferent values of the step sze n the x-drecton h and tme step sze t = 0.01. The computed solutons by proposed method are compared wth the exact soluton at the grd ponts and the maxmum absolute errors are tabulated n Table 4. The space tme graph of the estmated soluton s gven n Fgs. 4. The maxmum absolute error of ths example by method II s 8.7 10 10 and by method I s 1.2 10 9. Example5: We consder Eq.4 wth c = 1, f 2 x = cosπx, f 3 x = 0, p 2 t = cosπt and
304 J. Rashdna & M. Mohsenyzadeh/ IJM 2 C, 05-04 2015 291-305. Table 5. Absolute errorforexample.5 x t j MethodI MethodII 0.05 0.0003 5 10 10 1 10 9 0.05 0.0005 1 10 10 1 10 9 0.1 0.0003 2.1 10 9 7 10 10 0.1 0.0005 9.7 10 9 4 10 10 0.2 0.0003 6.2 10 9 1 10 10 0.2 0.0005 1.3 10 8 1 10 10 Fgure 5. 5. Space-Tme graph of the soluton up to t=1 s, wth c = 1, t = 0.0001 and h = 0.01 for example q 2 t = cosπt. The exact soluton for ths problem s u x, t = 1 2 cos π x + t + 1 2 cos π x t. Ths problem s solved by dfferent values of the step sze n the x-drecton h and tme step sze t = 0.0001. The computed solutons by proposed method are compared wth the exact soluton at the grd ponts and the maxmum absolute errors are tabulated n Table 5. The space tme graph of the estmated soluton s gven n Fgs. 5. The maxmum absolute error of ths example by method II s 6.5 10 9 and by method I s 3.84 10 8.
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