Ocak 2010 Cilt:18 No:1 Kastamonu Eğitim Dergisi 89-98 Arzu AYKUT Atatürk Üniversitesi, Fen Fakültesi, Matematik Bölümü, Erzurum. In this paper, we applied two approximate methods for the solution of a boundary value problem for a differential equation with retarded argument: x "(t ) + a (t ) x(t − τ (t )) = f (t ) x(t ) = ϕ (t ) (λ0 ≤ t ≤ 0) x(T ) = xτ , where a(t ), f (t ),τ (t ) ≥ 0 (0 ≤ t ≤ T ) and ϕ (t ) (λ0 ≤ t ≤ 0) are known as continuous functions. Key world: ordinary differential equations, boundry value problem, succesive approximations method. Bu çalımada x "(t ) + a(t ) x(t − τ (t )) = f (t ) x(t ) = ϕ (t ) (λ0 ≤ t ≤ 0) x(T ) = xτ , Yava deikenli denklem için konulmu sınır deer probleminin çözümü için iki yaklaık a(t ), f (t ),τ (t ) ≥ 0 (0 ≤ t ≤ T ) ve metot verilmitir. Burada ϕ (t ) (λ0 ≤ t ≤ 0) önceden verilmi sürekli fonksiyonlardır. Key world: ordinary differential equations, boundry value problem, succesive approximations method January 2010 Vol:18 No:1 Kastamonu Education Journal Arzu AYKUT ... 90 A common method used for the analytical solution of the boundary value problems is the integral equation method [1,2]. With this method, we obtain an integral equation that is equivalent to the boundary value problem and the solution of the integral equation is defined as the solution of the boundary value problem. The equivalent integral equation is usually a Fredholm equation in the classical theory. In this study we obtain a Fredholm-Volterra integral equation different from classical theory for the problem x "(t ) + a(t ) x(t − τ (t )) = f (t ) (1) x(t ) = ϕ (t ) (λ0 ≤ t ≤ 0), x(T ) = xτ , and a(t ), f (t ),τ (t ) ≥ 0 (0 ≤ t ≤ T ) and 0≤t ≤T ϕ (t ) (λ0 ≤ t ≤ 0) are known as continuous functions. The Fredholm operator where included in the equivalent integral equation is an operator with a degenerated kernel. We applied the modified successive method and consecutive substitution method for problem (1). One of these methods suggested by Ja. D. Mamedov [3] has been problem (1) by Aykut and Yıldız [4]. In this study these methods were applied to the boundary value problem with retarted argument. We investigated the solution for arbitrary continuous function τ (t ) . λ (t ) = t − τ (t ) then t 0 ∈ [0, t ] is a point located at the left side of T such that conditions λ (t 0 ) = 0 and λ (t ) ≤ 0 ( 0 ≤ t ≤ t 0 ) are satisfied, where, λ 0 = min 0 ≤t ≤t λ (t ) . We assume that λ (t ) is a nondecreasing function in the interval [t 0 , t ] and the equation λ (t ) = σ has differential continuous ∗ solution t = γ (σ ) for arbitrary σ ∈ [ 0, λ (t )] . It can be seen that if x (t ) is a ∗ solution of the boundary value for problem (1) then x (t ) is also the solution of the In problem (1), if we take 0 equation t x(t ) = hˆ(t ) + T T 0 t (T − s)a( s ) x(s − τ ( s))ds − (t − s )a(s ) x( s − τ (s ))ds. (2) 0 Here, t t hˆ(t ) = ϕ (0) − ( xT − ϕ (0)) − T T T 0 Ocak 2010 Cilt:18 No:1 Kastamonu Eğitim Dergisi t (T − s ) f ( s )ds + (t − s ) f ( s )ds. 0 The Two Method on Equatıon Wıth Retarded Argument ... Let 91 σ = s − τ (s). Therefore Eq. (2) can be written as follows: x (t ) = h(t ) + − λ (t ) 0 t T λ (T ) 0 (T − γ (σ )a(γ (σ )) x(σ )γ ′(σ )dσ (3) (t − γ (σ )a(γ (σ )) x(σ )γ ′(σ )dσ , where t h(t ) = hˆ(t ) + T 0 λ (T − γ (σ )a(γ (σ ))ϕ (σ )γ ′(σ )dσ 0 0 (4) − (t − γ (σ )a (γ (σ ))ϕ (σ )γ ′(σ )dσ , λ0 K 1 (σ ) = [T − γ (σ )]a(γ (σ ))γ ′(σ ) K (t , σ ) = [t − γ (σ )]a (γ (σ ))γ ′(σ ). Therefore we write Let x (t ) = h (t ) + t T λ (T ) 0 and K1 (σ ) x(σ ) d σ − λ (t ) 0 K (t , σ ) x(σ ) dσ (5) or x (t ) + h (t ) + t Fλ x + Vλ x T (6) where Fλ x ≡ λ (T ) 0 K 1 (σ )x(σ )dσ is the Fredholm operator, Vλ x ≡ − λ (t ) 0 K (t , σ )x(σ )dσ is the Volterra operator. Eq. (6) is a Fredholm-Volterra integral equation and it is equivalent to problem (1) Now, we will define the modified successive approximations for the integral equation (5), which are different from the ordinary successive approximations, as follows: January 2010 Vol:18 No:1 Kastamonu Education Journal Arzu AYKUT ... 92 x n (t ) = h(t ) + t Fλ x n + Vλ x n−1 T (n = 1,2,...). (7) In order to obtain a solution of problem (1) using this method, we will consider the following theorem. Theorem 1. If α =1− 1 T λ (T ) 0 K 1 (σ ) σ d σ ≠ 0 , 1 λ (T ) λ (T ) q = | K 1 (σ ) | dσ + 1 | K 1 (T , σ ) | dσ < 1, |α | 0 0 then the limit of the modified successive approximations x n (t ) = h (t ) + t t Fλ V λ x n −1 Fλ h + V λ x n −1 + Tα Tα converges to the solution of the problem (1) and the convergence speed is determined by || xn − x || ≤ q n || x − x0 || or || xn − x || ≤ qn || x1 − x0 || . 1− q Proof. In order to obtain the approximation of xn (t ) we will use the auxiliary equation y(t ) = hˆ(t ) + t Fλ y. T (8) Let Fλ y = c . Then, t y(t ) = hˆ(t ) + c. T If we use Eq. (9) in Eq. (8), then we obtain Ocak 2010 Cilt:18 No:1 Kastamonu Eğitim Dergisi (9) The Two Method on Equatıon Wıth Retarded Argument ... 93 1 ˆ 1 − Fλ t c = Fλ h T that α = 1 − (1 / T ) Fλ t ≠ 0, for c . By hypothesis we know therefore c = (1 / α ) Fλ hˆ and we can write t y = hˆ(t ) + Fλ hˆ. Tα If we consider hˆ(t ) = h(t ) + Vλ x n −1 and use Eq. (7), then we have t t Fλ h + Vλ xn−1 + Fλ Vλ xn −1 (10) Tα Tα Now, we will show that the approximations {xn (t )} , which are defined by Eq. xn ( t ) = h ( t ) + (10), converge to the solution of integral equation (5). Let us write t Fλ Vλ + Vλ x. Ωx = T α Therefore, we write the approximations of Eq. (10) as xn (t ) = h + t Fλ h + Ωxn−1 Tα (n = 1, 2,3,...). (11) Thus, Eq. (5) can be written as x =h+ t Fλ h + Ωx. Tα (12) Hence, the modified succesive approximations of Eq. (10) are similar to the ordinary successive approximations (12) which are equivalent to the integral equation (5). In order to show the convergence of approximations (10) to Eq. (5) we must prove the convergence of approximations (11) to Eq. (12). If the condition || Ωx || ≤ || x || (q < 1) , is satisfied then ordinary successive approximations (11) converge to the solution of Eq. (12) and the convergence speed is determined by || xn − x || ≤ q n || x − x0 || or || xn − x || ≤ qn || x1 − x0 || . 1− q By this hypothesis we have q < 1 and January 2010 Vol:18 No:1 Kastamonu Education Journal Arzu AYKUT ... 94 | Ωx | = 1 λ (T ) | K1 (σ )| | α | 0 ( λ (σ ) 0 ) | K (σ ,τ ) | x(τ ) | dτ dσ + λ (T ) 0 | K (T , σ ) | x(σ ) dσ 1 λ (T ) λ (T ) ≤ | K1 (σ )| dσ + 1 | K (T ,τ ) | dτ || x ||, | α | 0 0 so we write || Ωx || ≤ q || x || . If we substitue right side of equation (6) instead of equation (6) then we obtain x in the operator Vλ x to t t x(t ) = h(t ) + Vλ h + + Vλ Fλ x + Vλ2 x (13) T T 2 If we rewrite the right side of equation (6) instead of x in the operator Vλ x to equation (13) then we have t t t x(t ) = h(t ) + Vλ h + V 2 λ h + + Vλ + V 2λ Fλ x + Vλ3 x T T T (14) Whene this operation is applied of n time we have n n i =0 i =0 x = Vλi h + Vλi t Fλ x + Vλn +1 x T n n If we choose hn (t ) = Vλi h , i =0 an (t ) = Vλi i= 0 t T then it becomes x = hn (t ) + an (t ) Fλ x + Vλn +1 x (15) Now we can proof that the formula it is true | Vλn x |≤ for the operator Ocak 2010 Cilt:18 No:1 Kastamonu Eğitim Dergisi [ K λ (T )] n! n (n ∈ N ) (16) The Two Method on Equatıon Wıth Retarded Argument ... Vλ x ≡ − λ (t ) 0 95 K (t , s) x ( s ) ds (0 ≤ t ≤ T ) (17) n +1 As a result of this we neclect the operator | V λ x | in equation (15) for n ’s which are big enough. Thus the consecutive approximations are formed by taking the Volterra operator into consideration. xn (t ) = hn (t ) + an (t ) Fλ xn Theorem 2: Let problem (1) and τ (t ) ≥ 0, a(t ), f (t ) (0 ≤ t ≤ T ) be αn = 1− λ (T ) 0 (18) the functions given in K1 ( s ) hn ( s ) ds ≠ 0 . Also n [ K λ (T )] =0 n! lim An such that n →∞ An = 1 + || an || λ (T ) | K1 ( s) | d s | α n | 0 Then. the limit of the approximations x n (t ) = h n (t ) + a n (t ) αn λ (T ) 0 K 1 ( s ) hn ( s ) ds converges to the solution of problem (1) and the speed of the convergence is determined by [ Kλ (T )] | x n (t ) − x(t ) | ≤ An n! n Proof: Eq. (18) is the Fredholm integral equation with a degerenated kernel.The solution of xn (t ) = hn (t ) + λ (T ) 0 an (t ) K1 (s ) xn ( s ) ds (19) is the same as the solution of eq. (15) and problem (1). Now, let us find the solution of equation (19). So, we use auxiliary equation, y(t ) = hn (t ) + a n (t ) Fλ y where Fλ y is shown as Fλ y = cn .Thus y(t ) is like that January 2010 Vol:18 No:1 Kastamonu Education Journal Arzu AYKUT ... 96 y(t ) = hn (t ) + an (t ) cn (20) Therefore, cn = λ (T ) 0 When αn =1− λ (T ) 0 K 1 ( s ) hn ( s ) ds + c n λ (T ) 0 K 1 ( s ) a n ( s) ds K 1 ( s ) hn ( s ) ds ≠ 0 is given cn is found as follows cn = If we use equation (21) in (20), for 1 αn n = 1,2,... xn (t ) = hn (t ) + λ (T ) 0 K1 ( s ) hn ( s ) ds (21) then an (t ) αn λ (T ) 0 K1 ( s ) hn ( s ) ds (22) This operation is the approximate solution of problem(1) that is, the limit of x n (t ) converges to the solution of problem (1). Now let us determine the error of the approximate solution of eq. (22). Using (15) and (18), we reach x − x n = a n ( t ) F λ ( x − x n ) + V λn + 1 x If it is acceted that ε is ε = x − x n , then we obtain the Fredholm integral equation with a degenerated kernel ε = an (t ) Fλ ε + Vλn+1 x (23) It was proven that solution of equation (23) is being found by using the following formula: ε = Vλn+1 x + a n (t ) αn λ (T ) 0 K 1 ( s ) V λn +1 x ( s ) ds Thus, we write | ε | ≤ [1 + || a n || λ (T ) | K 1 ( s ) | ds ] || V λn +1 x ( s ) || | α n | 0 Then, by the hypothesis, An = 1 + || a n || λ (T ) | K 1 ( s ) | ds | α n | 0 Ocak 2010 Cilt:18 No:1 Kastamonu Eğitim Dergisi The Two Method on Equatıon Wıth Retarded Argument ... 97 and we have n | xn (t ) − x(t ) | ≤ An [ K λ (T )] || x || n! (24) Example 1. Let us consider the boundary value problem: 1 1 3 t ) = t 2 − t 2 0 ≤ t ≤ 1), (25) 2 2 x(t ) = 0 (−1/16 ≤ t ≤ 0), x(1) = 1 x′′(t ) + tx(t − This equation can be written as the Fredholm-Volterra integral equation x (t ) = − 409 1 4 2 72 t 1/ 2 3 + 28σ − 80σ 2 t + t − t + 3 + 4σ − 16σ 2 + x(σ ) dσ 420 12 35 16 0 1 + 16σ 1 t− 16 0 t /2 (4t − 1) + (48t − 20)σ − 80σ 2 2 (4 t 1) (16 t 12) σ 16 σ − + − − + x(σ )dσ . 1 + 16σ (26) Let h (t ) = K1 (σ ) = K (t , σ ) = 409 1 4 2 72 t+ t − t , 420 12 35 1 3 + 28σ − 80σ 2 2 3 + 4 − 16 + σ σ , 16 1 + 16σ 1 (4t − 1) + (48t − 20)σ − 80σ 2 2 (4 t − 1) + (16 t − 12) − 16 + σ σ 16 1 + 16σ and 1/ 2 Fλ x ≡ K1 (σ ) x(σ ) dσ , 0 Vλ x ≡ − t− t / 2 0 K (t , σ ) x(σ )dσ . Therefore, the integral equation (19) can be written as x(t ) = h(t ) + tFλ x + Vλ x (27) and this equation is equivalent to problem (25). Some values of the solution of this equation are obtained by using the method of modified successive approximations and January 2010 Vol:18 No:1 Kastamonu Education Journal Arzu AYKUT ... 98 the method of consecutive approximations of order two which are given in Table 1, where the first approximation is x 0 (t ) = ( 409 / 420)t. Table 1. Values at some point inthe interval [ 0,1] ti x(t i ) x 2( 2) (t i ) x 2( 3) (t i ) ε 1 (t i ) ε 2 (t i ) 0.00 0.00 0.0000000 0.0000000 0.000000 0.000000 0.25 0.25 0.2501132 0.2501078 0.000113 0.000107 0.50 0.50 0.5000794 0.5000684 0.000079 0.000068 1.00 1.00 1.0000013 0.9999046 0.000001 -0.000095 x2(2) : The modified successive method; x2(3) : The concecutive substitution method. ε1 and ε 2 are speed of the modified successive method and the concecutive substitution method, respectively. 1. Norkin, S.B., Differential equations of the second orderwith retarded argument some problems of theory of vibrations of systems with retardation , A. M. S., 1972. 2. Mamedov, Ja. D., Numerical Analysis, Atatürk Üniversitesi (in Turkish), 1994. 3. Krasnosel’skij, M.A., . Lifshits, Je. AA., Sobolev ,V., Positive Linear Systems, Heldemann, Berlin, p. 153., 1989. 4. Aykut, A., Yıldız, B., On a boundary value problem for a differential equation with variant retarded argument, Appl. Math. Comput.93, 63-71, 1998 Ocak 2010 Cilt:18 No:1 Kastamonu Eğitim Dergisi