PartII Multi-step Iterative Method for Computing the

PartII Multi-step Iterative Method for Computing the Numerical Solution of System of Nonlinear Equations Associated with ODEs Fayyaz Ahmad & Malik Zaka Ullah

1 1.1

Some ODEs Bratu-problem y 00 + λey(x) = 0, 00

(1)

L(y) = y ,

(2)

L(y) + f (y) = 0.

(3)

Boundary conditions are y(0) = 0, y(1) = 0.

1.2

Frank-Kamenetzkii-problem k 0 y + λey(x) = 0, x k L(y) = y 00 + y 0 , x L(y) + f (y) = 0. y 00 +

(4) (5) (6)

Boundary conditions are y 0 (0) = 0, y(1) = 0.

2

System of nonlinear equations

Consider a nonlinear ordinary differential equation L(y) + f (y) = g(x),

(7)

where L is a linear derivative operator. (7) can be converted into system of nonlinear algebraic equations for instance by using finite differences method, Chebyshev pseudo spectral collocation method. In other words we approximate the continuous differential operator by a linear transformation. Let Matrix A be a discrete approximation of L and y = (y1 , y2 , · · · , yn )T and x = (x1 , x2 , · · · , xn )T . Finally (7) can be written as Ay + f (y) = g(x),

(8)

F (y) = Ay + f (y) − g(x) = 0,

(9)

where F (y) = (f1 (y), f2 (y), f3 (y), · · · , fn (y))T and f (y) = (f (y1 ), f (y2 ), f (y3 ), · · · , f (yn ))T and g(x) = (g(x1 ), g(x2 ), g(x3 ), · · · , g(xn ))T .

2

3

Calculation of higher-order Frechet derivatives

The Taylor’s series expansion of F (y + h) around y is written as F (y + h) = F (y) + F 0 (y)h +

1 1 00 F (y)h2 + F 000 (y)h3 + · · · 2! 3!

To make simple calculations, we just consider case n = 3. y = [y1 , y2 , y3 ]T h = [h1 , h2 , h3 ]T F (y) = [f1 (y), f2 (y), f3 (y)]T        a11 a12 a13 y1 f (y1 ) g(x1 ) F (y) = a21 a22 a23  y2  + f (y2 ) − g(x2 ) f (y3 ) g(x3 ) a31 a32 a33 y3 The first order Frechet derivative of F (y) is Jacobian of F     0 a11 a12 a13 f (y1 ) 0 0 f 0 (y2 ) 0 , F 0 (y) = a21 a22 a23  +  0 a31 a32 a33 0 0 f 0 (y3 )      a11 a12 a13 h1 h1 f 0 (y1 ) F 0 (y)h = a21 a22 a23  h2  + h2 f 0 (y2 ) a31 a32 a33 h3 h3 f 0 (y3 ) Now we calculate the 2nd-order Frechet derivative as follows ∂(F 0 (y)h) h F 00 (y)h2 = (F 0 (y)h)0 h = ∂(y1 , y2 , y2 )     h1 f 00 (y1 ) 0 0 h1  h2  0 h2 f 00 (y2 ) 0 = 0 0 h3 f 00 (y3 ) h3   2 00 h1 f (y1 ) F 00 (y)h2 = h22 f 00 (y2 ) h23 f 00 (y3 )  00   2 f (y1 ) 0 0 h1 f 00 (y2 ) 0  h22  F 00 (y)h2 =  0 0 0 f 00 (y3 ) h23  000   3 f (y1 ) 0 0 h1 f 000 (y2 ) 0  h32  F 000 (y)h3 =  0 0 0 f 000 (y3 ) h33

(10)

3

Similarly we can calculate the higher order Frechet derivatives. Now we consider F (y) = [f1 (y), f2 (y), f3 (y)]T  ∂f   ∂f1 ∂f1 1 f11 f12 ∂y1 ∂y2 ∂y3  ∂f2 ∂f2 ∂f2   f21 f22 F 0 (y) =  ∂y = ∂y2 ∂y3  1 ∂f3 ∂f3 ∂f3 f31 f32 ∂y ∂y2 ∂y3  1  h1 f11 + h2 f12 + h3 f13 F 0 (y)h = h1 f21 + h2 f22 + h3 f23  h1 f31 + h2 f32 + h3 f33

 f13 f23  f33

∂(F 0 (y)h) h ∂(y1 , y2 , y3 )    h1 f111 + h2 f121 + h3 f131 h1 f112 + h2 f122 + h3 f132 h1 f113 + h2 f123 + h3 f133 h1 F 00 (y)h2 = h1 f211 + h2 f221 + h3 f231 h1 f212 + h2 f222 + h3 f232 h1 f213 + h2 f223 + h3 f233  h2  h1 f311 + h2 f321 + h3 f331 h1 f312 + h2 f322 + h3 f332 h1 f313 + h2 f323 + h3 f333 h3  2  2 h1 f111 + h1 h2 f121 + h1 h3 f131 + h1 h2 f112 + h2 f122 + h2 h3 f132 + h1 h3 f113 + h2 h3 f123 + h23 f133 F 00 (y)h2 = h21 f211 + h1 h2 f221 + h1 h3 f231 + h1 h2 f212 + h22 f222 + h2 h3 f232 + h1 h3 f213 + h2 h3 f223 + h23 f233  h21 f311 + h1 h2 f321 + h1 h3 f331 + h1 h2 f312 + h22 f322 + h2 h3 f332 + h1 h3 f313 + h2 h3 f323 + h23 f333  2  h1 f111 + h22 f122 + h23 f133 + h1 h2 (f121 + f112 ) + h1 h3 (f131 + f113 ) + h2 h3 (f132 + f123 ) F 00 (y)h2 = h21 f211 + h22 f222 + h23 f233 + h1 h2 (f221 + f212 ) + h1 h3 (f231 + f213 ) + h2 h3 (f232 + f223 ) h21 f311 + h22 f322 + h23 f333 + h1 h2 (f321 + f312 ) + h1 h3 (f331 + f313 ) + h2 h3 (f332 + f323 )   h21 f111 + h22 f122 + h23 f133 + 2h1 h2 f121 + 2h1 h3 f113 + 2h2 h3 f123 = h21 f211 + h22 f222 + h23 f233 + 2h1 h2 f212 + 2h1 h3 f213 + 2h2 h3 (f232 + f223 ) h21 f311 + h22 f322 + h23 f333 + 2h1 h2 f312 + 2h1 h3 f313 + 2h2 h3 (f332 + f323 ) F 00 (y)h2 =

Finally we get the expression  f111 F 00 (y)h2 = f211 f311

for F 00 (y)h2   2  f122 f133 h1 f121 f222 f233  h22  + 2 f212 f322 f333 h23 f312

f113 f213 f313

  f123 h1 h2 f223 ) h1 h3  f323 ) h2 h3

Suppose f1 (y) = f (y1 ), f2 (y) = f (y2 ), f3 (y) = f (y3 ) then all cross partial derivatives are zeros and we get  f111 F 00 (y)h2 =  0 0

0 f222 0

  2   00   2 0 h1 f (y1 ) 0 0 h1 0  h22  =  0 f 00 (y2 ) 0  h22  f333 h23 0 0 f 00 (y3 ) h23

4

The Frechet derivatives of (9) w.r.t y are  0 f (y1 ) 0 ···  0 f 0 (y2 ) · · ·  F 0 (y) =  .  ..

0 0

0 0

0 0 · · · 0 f 0 (yn )  00 f (y1 ) 0 ··· 0 0 00  0 f (y ) · · · 0 0 2  F 00 (y) =  .  ..

    + A, 

(11)



  ,  00 0 0 · · · 0 f (yn )  000  f (y1 ) 0 ··· 0 0  0 f 000 (y2 ) · · · 0 0    F 000 (y) =  . .  ..  0 0 · · · 0 f 000 (yn )

(12)

(13)

It is clearly shown that the higher order Frechet derivatives w.r.t to y are diagonal matrices. The general solvers for system of nonlinear equations avoid higher order Frechet derivatives and only depend on first order Frechet derivative which is Jacobian of the system of equations. The reason is very clear for general solvers that the computational cost of higher order Frechet derivatives is much higher because in general higher order Frechet derivatives are not diagonal matrices for general system of nonlinear equations. In the case of Newton-Raphson method the calculation of Jacobian is not much expensive for (9) because It just requires the computation of a diagonal matrix but we have to invert Jacobian for each iteration which is computational expensive. So the idea is we have to design multi-step Jacobian frozen iterative methods which require only one LU decomposition of Jacobian for one complete iteration of iterative scheme. In [1], authors developed an iterative method which is computationally very efficient and has the ability of extension to construct multi-step iterative method.  sn = qn − 32 (F 0 (qn ))−1 F (qn )     t = (F 0 (q ))−1 F 0 (s ) n n (14) 23  zn = qn − ( 8 − 3t + 89 t2 )(F 0 (qn ))−1 F (qn )    qn+1 = zn − ( 25 − 32 t)(F 0 (qn ))−1 F (zn ) The computational order of convergence of (14) is six and it requires only one LU decomposition of Jacobian. The multi-step extension of (14) is   sn = qn − 32 (F 0 (qn ))−1 F (qn )    0 −1 0    t = (F (qn )) F (sn ) 9 2 0 −1 (15) zn = qn − ( 23 F (qn ) 8 − 3t + 8 t )(F (qn ))   5 3 0 −1  wn = zn − ( − t)(F (qn )) F (zn )  2 2    q 5 3 0 −1 F (wn ) n+1 = wn − ( 2 − 2 t)(F (qn )) The convergence-order of (15) is eight. Actually one can extend the proposed iterative scheme to get an even-order of convergence method. Each addition of step Ψ − ( 25 −

5 3 0 −1 F (Ψ) 2 t)(F (qn ))

increase the convergence-order by two. The iterative scheme (14) is very efficient for general system of nonlinear equations but it does not take the extra benefit from the diagonalization of higher-order Frechet derivatives of (9) and that is why avoid the consideration of higher-order Frechet derivatives. The second candidate is Chebyshev-Halley third-order iterative method ! 1 (16) qn+1 = qn − I + [I − αL]−1 L (F 0 (qn ))−1 F (qn ), 2 where L(qn ) = (F 0 )−1 F 00 (F 0 )−1 F . (16) uses the higher order Frechet derivatives but we have to invert two matrices [I − αL] and F 0 (qn ) when α 6= 0 for each iteration which is computationally not efficient. We proposed multi-step iterative method which removes the drawbacks of both methods (14) and (16) for a very particular case (9):           

sn = qn − (F 0 (qn ))−1 F (qn ) − 12 F 0 (qn ))−1 F 00 (qn )[F 0 (qn ))−1 F (qn )]2 , T = (F 0 (qnh))−1 F 0 (sn ), i zn = sn − 3 − 3T + T 2 (F 0 (qn ))−1 F (sn ) h i qn+1 = zn − 3 − 3T + T 2 (F 0 (qn ))−1 F (zn )

(17)

The convergence-order of (17) is nine. Note that T is matrix and F 00 is diagonal matrix.  0 −1 F (qn )]2 + s"n = qn − (F 0 (qn ))−1 F (qn ) − 21 F 0 (qn ))−1 F 00 (q#  n )[F (qn ))      1 0 −1 000  F (qn ) − 12 ((F 0 (qn ))−1 F 00 (qn ))2 [F 0 (qn ))−1 F (qn )]3 ,  6 (F (qn ))   T = (F 0 (qnh))−1 F 0 (sn ), i      zn = sn − 2 − 2T 2 + T 3 (F 0 (qn ))−1 F (sn )   h i    q 2 3 (F 0 (qn ))−1 F (zn ) n+1 = zn − 2 − 2T + T

(18)

h i The convergence-order of (18) is thirteen. For each inclusion of Φ− 2−2T 2 +T 3 (F 0 (qn ))−1 F (Φ) , we get an increase of three in convergence-order. In other words after step two we get an increase of three in convergence-order for a single evaluation of F .

References [1] H. Montazeri, F. Soleymani, S. Shateyi, S. S. Motsa , On a New Method for Computing the Numerical Solution of Systems of Nonlinear Equations,, Volume 2012, Article ID 751975, 15 pages.