The Matrix PadÃƒÂ© Approximation in Systems of Differential ... - WSEAS

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WSEAS TRANSACTIONS on MATHEMATICSC. Pestano-Gabino, C. Gonz_Lez-Concepcion, M.C. Gil-Farinaw j (x)= 2 sen πjx. In this case, the equation (10) inmatrix form is:V'(t)=A(t) V(t) (11)where V(t) = (v 1 (t) v 2 (t) ... v n (t)) t and A(t) is an nxnmatrix which elements are:3 'a ij (t)= w2 j, wi.2(4 + t )Considering the initial conditions, we have:n∑ vj(0)w j(x)= sin 2πx. Due to the fact thatj=1{w 1 ,w 2 ...w n } is an ortonormal system, then:v j (0)= sin 2π x, wj(j=1...n) (12)The expression (12) provides initial conditions forsystem (11).To illustrate the procedure, suppose that n=2 in (8).Obviously, with greater n we obtain betterapproximation.1 ⎛0 −1⎞If n=2, then A(t)=2 ⎜ ⎟t 1 0. Considering1+⎝ ⎠4A(t)=∞∑j= 0jA jtj⎛ ⎛ 1 ⎞⎜ 0 ⎜ ⎟ ( −1)4A 2j =⎜ ⎝ ⎠⎜ j⎛ −1⎞⎜⎜⎟ 0⎝⎝4 ⎠, we have that A 2j+1 =0,j+1A fundamental matrix F(t)=such that F 0 =I, verifies thatk1F k+1 = ∑ AjFk − jk≥0k + 1j=0⎞⎟⎟⎟for j=0,1,2...⎟⎠Fig. 7: Table 1 for F(t)0 1 2 3 4 50 0 0 0 0 0 01 2 2 2 2 2 22 4 4 2 2 2 23 6 6 4 2 2 24 8 8 6 4 2 25 10 10 8 6 4 2∞j∑ Fjtof the system (11)j=0It indicates that F(t)=[1/1] F . Solving the systemLS(1,1), and calculating the numerator, F(t) can berepresented as follows:⎡⎛1 0⎞ ⎛ 0 1/ 2⎞ ⎤ ⎡⎛1 0⎞ ⎛ 0 −1/ 2⎞⎤F(t) = ⎢⎜ ⎟ + ⎜ ⎟ t⎥ ⎢⎜ ⎟ + ⎜ ⎟ t⎥⎣⎝0 1⎠ ⎝ −1/ 2 0 ⎠ ⎦ ⎣⎝0 1⎠ ⎝1/ 2 0 ⎠ ⎦2⎛ t ⎞1 t1⎜ − − ⎟or equivalently F(t)= ⎜ 4 ⎟ .2 2t ⎜ t ⎟1+ t 1−4⎜⎟⎝ 4 ⎠Given initial conditions, we calculate the particular−1solution V(t)=F(t)K, where K∈ R 2 . Note that V(0)=K.⎛ 0 ⎞Taking into account (12), K=⎜1/ 2 ⎟. Therefore,⎝ ⎠⎛ −t⎞1V(t)= ⎜ 2 ⎟2 t andt 1−2(1 + ) ⎜ ⎟⎝ 4 ⎠42t−t sin π x + (1 − )sin 2πxu(x,t)=4 .2t1+4In the case that we are interested only in a particularsolution that verifies certain initial conditions, it is notnecessary to calculate a fundamental matrix. To know∞⎛ 0 ⎞jV(t), we suppose that V(t)= ∑ Vjt, with V 0 =⎜j=01/ 2 ⎟.⎝ ⎠k1Then se V k+1 = ∑ AjVk − j, k≥0.k + 1j=0Fig. 8: Table 1 for V(t)0 1 2 3 4 50 0 0 0 0 0 01 1 1 1 1 1 12 2 2 2 2 2 23 3 3 3 2 2 24 4 4 4 3 2 25 5 5 5 4 3 2This table indicates that V can be represented as leftand right approximants of the set {[2/2] V }.Considering right approximant, Table 2 is notnecessary because in Table 1 we can see taht (2,2) is apair of right m.d. (taking into account Properties 1 and2 and T1(2,2)=2). We solve the system RS(2,2) to⎛ −t / 2 ⎞⎜ ⎟ 1obtain the known solution V(t)= 2⎜ 1 t ⎟ .2⎜ − ⎟t⎝ 2 4 2 ⎠(1 + )4Out of curiosity we have calculated the Table 2 for leftapproximant in fig. 9.Fig. 9: Table 2 for left approximant0 1 20 1 1 11 1 0 02 0 0 0It indicates that (0,2) and (1,1) are two pairs of leftm.d.Let us consider the following example. In this case weneed to solve a system of first order ordinarydifferential equations, V''(t)=A(t) V(t) where A(t) is adiagonal matrix, then each equation can be solveindividually. Therefore, instead of MPA we will usescalar Padé approximation.ISSN: 1109-2769 349 Issue 6, Volume 7, June 2008
WSEAS TRANSACTIONS on MATHEMATICSC. Pestano-Gabino, C. Gonz_Lez-Concepcion, M.C. Gil-FarinaExample 4Let it be the following elliptic equation:2u tt = − u2 2 xx , t>0, t≠1, 0≤x≤1π (1 − t)u(0,t)=u(1,t)=0; u(x,0)=u t (x,0)=sinπxWe propose a solution of the form u(x,t)= 2 vj(t)w j(x)considering the same ortonormal system of Example 3.Substituting in the partial differential equation anddividing by 2 ), we obtain:'' 22v1(t)sin π x+ v2 2 1(t)( −π sin π x) +π (1−t)'' 22v2(t)sin2π x + v2 2 2(t)( −4π sin2π x) = 0π (1 −t)'' 2'' 8Then, v1(t)= v2 1(t), v2(t) = v2 2(t)(1 − t)(1 − t)We have two independent equations. Note that A(t) isa diagonal matrix.The initial conditions can be deduced by:u(x,0)=v 1 (0)w 1 (x)+v 2 (0)w 2 (x)=sin πx, therefore,v 2 (0)=0 and v 1 (0)= 1 2 ,∑j=1u t (x,0)= v ' (0)w (x) + v ' (0)w (x) = sin π x , therefore,''2 11 1 2 2v (0) = 0 and v (0) =Considering the first equation122v (t) =(1 − t)v (t)''1 2 1andits conditions v 1 (0)= 1 2 and ' 1v1(0)= , we denote2the general solution as s(t)=∞i∑ siti=0. Taking into account∞2jthat = 2 (j 1)t2 ∑ + -in the circle of convergence(1 − t) j=0of the series-. Substituting in the differential equationwe obtain:∞ ∞ ∞i i i∑ + +i+2= ∑ + ∑ ii= 0 i= 0 i=0(i 2)(i 1)s t 2( (i 1)t )( s t )Therefore, to calculate the coefficients of the series s(t)we use the following recurrence relation:is i+2 = 2( ∑ (j+ 1)si−j) /((i + 2)(i + 1)) i≥0.j=0Fig. 10: Table 1 for v 1 (t)0 1 2 3 4 50 0 0 0 0 0 01 1 1 1 1 1 12 2 1 1 1 1 13 3 2 1 1 1 14 4 3 2 1 1 15 5 4 3 2 1 1It shows that v 1 (t)=[0/1] s . Solving Padé equations for[0/1] s , v 1 (t)= 1/ 2 . In a similar way, solving1−t'' 8v2(t) = v2 2(t) , v 2 (0)= v ' 2(0) = 0 , we obtain(1 − t)v 2 (t)=0. Therefore, the solution of the partialdifferential equation is: u(x,t)= 1 sin πx.1−t4.2 Method of Finite Differences and MPAWe give the theory for the following hyperbolic partialdifferential equation (generalized wave equation):u tt =α 2 (x,t) u xx 0≤x≤1, t>0u(0,t)=f 1 (t), u(1,t)=f 2 (t); u(x,0)=g 1 (x), u t (x,0)=g 2 (x)In this method we choose n∈N, calculate h=1/n andconsider the points (x i ,t) where x i =ih for i=0,1,2...n.Note that we do not discretise variable t.Substituting these points, for i=1...n-1, we obtainu tt (x i ,t) = α 2 (x i ,t) u xx (x i ,t). If we replace second partialderivative u xx with second differences we obtain:u tt (x i ,t)=α 2 u(xi+ 1, t) − 2u(xi, t) + u(xi−1, t)(x i ,t) u2xx (x i ,t),hfor i=1...n-1Denoting2⎛ α (x1, t)f1(t)⎞⎜2 ⎟⎛ u(x1, t) ⎞ ⎜ h ⎟⎜ ⎟ ⎜ 0 ⎟u(x2, t)Y(t)= ⎜ ⎟ ⎜⎟, b(t)=⎜ ⋮ ⎟ ⎜⋮⎟and⎜ ⎟ ⎜ 0 ⎟⎝ u(xn−1, t) ⎠ ⎜⎟2α (xn−1, t)f2(t)⎜2 ⎟⎝ h ⎠2 2⎛−2 α (x,t)1α (x,t)10 ⋯ 0 ⎞⎜ 2 2 2⎟(x2,t) 2 (x2,t) (x2,t) 01⎜ α − α α ⋯⎟2 2A(t)= ⎜20 α (x3,t) −2 α(x 3,t) ⋯ 0 ⎟,h ⎜⎟⎜ ⋯ ⋯ ⋯ ⋯ ⋯ ⎟⎜2 20 0 α (xn−1 ,t) −2 α (xn−1,t) ⎟⎝⋯⎠we obtain the system of linear second-order ordinarydifferential equations:Y''(t)=A(t) Y(t)+b(t) (13)Note that the solution vector Y(t) is of dimension n-1.The initial conditions for this system are:⎛ g1(x 1)⎞ ⎛ g2(x1)⎞⎜ ⎟ ⎜ ⎟g1(x 2)Y(0)= ⎜ ⎟ , Y'(0)= ⎜g2(x2)⎟⎜ ⋮ ⎟ ⎜ ⋮ ⎟⎜ ⎟ ⎜ ⎟⎝g 1(x n−1)⎠ ⎝g 2(xn−1)⎠Let us suppose Y(t)=∞∞∑i=0iY (t − t ) , A(t)=i 0∞∑i=0iA (t − t )i 0iand b(t)= ∑ bi(t − t0) where the coefficients of thesei=0series have suitable dimensions. Substituting in (13),the expression to calculate the coefficients of aISSN: 1109-2769 350 Issue 6, Volume 7, June 2008
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