Method of Lines for Third Order Partial Differential Equations

Journal of Applied Mathematics and Physics, 2014, 2, 33-36 Published Online January 2014 (http://www.scirp.org/journal/jamp) http://dx.doi.org/10.4236/jamp.2014.22005

Method of Lines for Third Order Partial Differential Equations Mustafa Kudu, Ilhame Amirali 1

Department of Mathematics, Faculty of Art and Science, Erzincan University, Erzincan, Turkey 2 Department of Mathematics, Faculty of Art and Science, Sinop University, Sinop, Turkey Email: [email protected], [email protected] Received December 5, 2013; revised January 5, 2014; accepted January 10, 2014

Copyright © 2014 Mustafa Kudu, Ilhame Amirali. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In accordance of the Creative Commons Attribution License all Copyrights © 2014 are reserved for SCIRP and the owner of the intellectual property Mustafa Kudu, Ilhame Amirali. All Copyright © 2014 are guarded by law and by SCIRP as a guardian.

ABSTRACT The method of lines is applied to the boundary-value problem for third order partial differential equation. Explicit expression and order of convergence for the approximate solution are obtained.

KEYWORDS Method of Lines; Partial Differential Equation; Convergence; Error Estimates

1. Introduction We consider the boundary value problem for the third order differential equation in the domain Ω {0 < x < p, 0 < y < q} : ∂ 3u ∂ 3u + = f ( x, y ) , ∂x3 ∂x∂y 2

(1)

= u ( x, 0 ) ψ= 1 ( x ) , u ( x, q ) ψ 2 ( x ) ,

(2)

= u ( 0, y ) g= g2 ( y ) , 1 ( y ) , u ( p, y )

(3)

∂u ( 0, y ) = g3 ( y ) , ∂x

(4)

where ψ 1 ( x ) ,ψ 2 ( x ) , g1 ( y ) , g 2 ( y ) , g3 ( y ) are sufficiently smooth functions. The problems of type (1)-(4) arise in many mathematical and scientific applications [1-3]. In this study, we construct first order accurate differential difference scheme for this problem and give error estimate for its solutions. The approach to the construction of the discrete problem and the error analysis for the approximate solution are similar to those in [4]. ∂7u in the domain Ω {0 ≤ x ≤ p, 0 ≤ y ≤ q} . Let the solution of the problem (1)-(4) have a bounded derivative ∂x∂y 6

2. Differential-Difference Algorithm and Convergence q   We divide the domain Ω into n + 1 stripe by lines = y y= kh = k 1, 2, , n; = h k  . On this lines the n +1  problem (1)-(4) we approximate by the following differential difference problem: OPEN ACCESS

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M. KUDU, I. AMIRALI

34

U k′′′+1 + 10U k′′′ + U k′′′−1 +

12 (U k′ +1 − 2U k′ + U k′ −1=) f k +1 ( x ) + 10 f k ( x ) + f k −1 ( x ) , h2

(5)

= U 0 ( x ) ψ= 1 ( x ) , U n +1 ( x ) ψ 2 ( x ) ,

(6)

= U k ( 0 ) g= g 2 ( yk ) , 1 ( yk ) , U k ( p )

(7)

U k ( 0 ) = g 3 ( yk ) .

(8)

Let we rewrite the problem (5)-(8) in the form

AU ′′′ −

12 MU ′ = F ( x), h2

(9)

( ) = U ( 0 ) g= g 2( ) , U ′ ( 0 ) = g3( ) , 1 , U ( p) 0

p

0

where

U = (U1 ( x ) ,U 2 ( x ) , ,U n ( x ) ) ,  2 −1 0   −1 2 −1  0 −1 2 M =     0 0 0  0 0 0

 0 0  0 0  0 0     2 −1  −1 2

 10 1 0  0 0       1 10 1  0 0    0 1 10  0 0   , A ≡ 12 I − M =  ,           0 0 0  10 1       0 0 0  1 10 

I-unit matrix,

F ( x= )

( F1 ( x ) , F2 ( x ) ,, Fn ( x ) ) ,

F1 ( x = ) f ( x, y2 ) + 10 f ( x, y1 ) + f ( x, 0 ) −ψ 1′′′( x ) −

Fk ( x ) = f ( x, yk +1 ) + 10 f ( x, yk ) + f ( x, yk −1 ) , k = 2,3,..., n − 1, Fn ( x ) = f ( x, q ) + 10 f ( x, yn ) + f ( x, yn −1 ) −ψ 2′′′( x ) − g1(

12 ψ 1′ ( x ) , h2

12 ψ 2′ ( x ) , h2

g ( y ) , g ( y ) ,  , g ( y ) ) , g ( ) ( g= ( y ) , g ( y ) ,, g ( y ) ) , g ( ) ( g ( y ) , g ( y ) ,, g ( y ) ) . (=

0)

p

1

1

1

2

1

n

2

2

1

2

2

2

n

0 3

3

1

3

2

3

n

The matrix M can be diagonalized as [5,6]

M = B −1diagonal ( λ1 , λ2 , , λn ) B, with B= B −1=

 n k +s ( bks )k , s =1=  ( −1) 

 πs  4 2 πks  cos 2  = = sin , λs  , s 1, 2, , n.  2 h n +1 n + 1 k , s =1  2 ( n + 1)  n

Multiplying Equation (9) on the left by B we have

(12 − λs ) ϕs′′′−

12 λsϕ s′ = ls ( x ) , h2

(10)

= ϕ s ( 0 ) ϕ= ϕ sp , s 0 , ϕs ( p )

(11)

′ ′ ϕ= ϕ= 1, 2, , n, s (0) s0 , s

(12)

where = ϕs ( x )

n

n

n

n

= bskU k ( x ) , ls ( x ) ∑ bsk Fk ( x ) , ϕ s 0 ∑ bsk g1 ( yk ) , ϕ sp ∑ bsk g 2 ( yk ) , = = ∑

= k 1= k 1= k 1= k 1

OPEN ACCESS

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M. KUDU, I. AMIRALI

= ϕ s′0

n

= bsk g3 ( yk ) , s ∑ k =1

35

1, 2, , n.

The solution of (10)-(12) containing the third order ordinary differential equation with constant coefficients can be explicitly found p

ϕs ( x ) = − ∫ Gs ( x, ξ )ls (ξ ) dξ +

sinh σ s p − sinh σ s ( p − x ) − sinh σ s x

0

σ s ( cosh σ s p − 1)

⋅

p   cosh σ s x − 1 cosh σ s p − cosh σ s x 1 ′ ϕ − sinh σ s ( p − ξ ) ls (ξ ) dξ  + + ϕs 0  s0 ∫ sinh σ s p 0 cosh σ s p − 1   cosh σ s p − 1

where  sinh σ s x ⋅ sinh σ s ( p − ξ ) , x ≤ξ  σ s sinh σ s p  Gs ( x, ξ ) =   sinh σ s ( p − x ) ⋅ sinh σ sξ , x ≥ ξ .  σ s sinh σ s p 

Therefore the solution of (5)-(8) can be expressed as U k ( x )=

n

∑ ( −1) s =1

k +s

p sinh σ s p − sinh σ s ( p − x ) − sinh σ s x 2 πks  sin ⋅ − ∫ Gs ( x, ξ ) ls (ξ ) dξ + n +1 n + 1  0 σ s ( cosh σ s p − 1)

p   cosh σ s p − cosh σ s x cosh σ s x − 1  1 sinh σ s ( p − ξ ) ls (ξ ) dξ  + ϕs 0 + ϕ sp  , ⋅ ϕ s′0 − ∫ sinh σ s p 0 cosh σ s p − 1 cosh σ s p − 1   

where ls ( x) =

1 x l s ( t ) dt . 12 − λs ∫0

zk ( x ) u ( x, yk ) − U k ( x ) we have Now we investigate the error of the approximate solution. For the error = the following boundary value problem: 12 ⋅ ( zk′ +1 − 2 zk′ + zk′ −1 ) = Rk ( x ) , h2 z0 ( x ) ≡ 0, zk +1 ( x ) ≡ 0, zk ( 0 ) ≡ 0, zk ( p ) ≡ 0, zk′ ( 0 ) = 0, k = 1, 2, , n, zk′′′+1 + 10 zk′′′ + zk′′′−1 +

where h4 ∂ 7u Rk ( x ) = − ⋅ const , k = 1, 2,  , n, yk −1 < yk < yk +1 . ( x, yk ) , Rk ( x ) ≤ C1h4 , C1 = 20 ∂x∂y 6

(12 − λs ) ϕs′′′−

12 ′ ls ( x ) , ϕ s ( 0= λsϕ= ) 0, ϕs ( p=) 0, ϕs′ ( 0=) 0 s h2

or

ϕ s′′ − σ s2ϕ s = l s ( x ) + K s , ϕ s ( 0 ) = 0, ϕ s ( p ) = 0, ϕ s′ ( 0 ) = 0, ϕ s ( x ) =

  ∑ bsk γ k ( x ) , l s ( x ) = ∫  ∑ bsk Rk ( t )  dt. n

x

n



= k 1= 0 k 1



Next for p

∫ sinh σ s ( p − ξ ) l s (ξ ) dξ

Ks = −0

p

∫ sinh σ s ( p − ξ ) dξ

, ( sinh σ s ( p − x ) > 0 ) .

0

By the mean value theorem we have OPEN ACCESS

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M. KUDU, I. AMIRALI

36

−l s (η ) , 0 < η < p. Ks = Then

l s ( x ) + K= l s ( x ) − l s (η = ) l s′ (ξ )( x − η =) ls (ξ )( x − η ) . s n

Since ls ( x ) = ∑ bsk Rk ( x ) , then it follows that k =1

l s ( x ) + K s ≤ ls (ξ ) p ≤ max ls ( x ) p ≤ C1 p 2qh3.5 . 0≤ x ≤ p

Further, we note that

ϕs ( x ) =

p

1

0

s

∫ G ( x, ξ )dξ ≤ σ 2

p

∫ Gs ( x, ξ ) l s (ξ ) + K s  dξ

≤

0

and

C1 p 2qh3.5 C1 p qh5.5 max l s ( x ) + K s ≤ = , s 1, 2,  , n. = 2 σ σ s (12 − λs ) 24 2 cos 2 πs 2 ( n + 1) 1

2 0≤ x ≤ p s

Hence n

2 C1 p q 5.5 n 1 h ∑ ⋅ πs n + 1 24 2 2 1= s 1 cos 2 ( n + 1)

zk ( x ) ≤ ∑ bks ϕ s ( x ) ≤

=s

Using here the inequality sin x >

n 1 ∞ 1 π2 2  π x  0 < x <  , and taking into account ∑ 2