In this paper we study the existence of a unique solution for a boundary value prob-lem of a coupled system of Volterra type integro-dierential equations under nonlocal conditions

Abbas, Ashwaq

El-Sayed, A. M. A

English

KHALSA PUBLICATIONS

ISSN 2347-1921                                                           5033 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           Existence of solution for a coupled system ofVolterra type integro - differential equations with nonlocal conditions  A. M. A. El-Sayed, A. A. Hilal. Faculty of Science,Alexandria University,Egypt Faculty of Science,Zagzig University , Egypt Abstract In this paper we study the existence of a unique solution for a boundary value problemof a coupled system of Volterra  type  integro-differential equations under nonlocal conditions. Keywords: Nonlocal boundary value problems, integro - differential equation;coupled system; Lipschitz condition; Banach fixed point theorem.                       Council for Innovative Research Peer Review Research Publishing System Journal: JOURNAL OF ADVANCES IN MATHEMATICS Vol.11, No.3 www.cirjam.com , editorjam@gmail.com ISSN 2347-1921                                                           5034 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           1 INTRODUCTION The study of value problem with nonlocal conditions is of significance, since they have application in problems in physics, engineering, economics and other areas of applied Mathematics. This feature allows the study of several types of integral equations such as:Fredholm, Volterra, Hammerstein, Urysohn, for different classes of functionals [see(6),(7)]. The main object of this paper to study the existence of solution𝑥, 𝑦 ∈ 𝐶[0,1] and𝑥, 𝑦 ∈ 𝐴𝐶[0,1] for the coupled system of Volterra integro- differential equations 𝑑𝑥𝑑𝑡=  𝚏₁  𝑡, 𝑠,𝑑𝑦𝑑𝑠 𝑑𝑠,       𝑡 ∈ (0,1)                                𝑡0  (1) 𝑑𝑦𝑑𝑡=  𝚏₂  𝑡, 𝑠,𝑑𝑥𝑑𝑠 𝑑𝑠,       𝑡 ∈ (0,1)                                𝑡0 with the nonlocal boundary conditions  𝑥 𝜏 = 𝛼𝑥 𝜉 , 𝜏 ∈  0,1 , 𝜉 ∈  0,1  , 𝛼 ≠ 1   (2)   and 𝑦 𝜏 = 𝛽𝑦 𝜉 , 𝜏 ∈  0,1 , 𝜉 ∈  0,1 , 𝛽 ≠ 1    (3)     Let     𝑑𝑥𝑑𝑡= 𝑢, and 𝑑𝑦𝑑𝑡= 𝑣 in (1),  we obtain  𝑢 𝑡 =  𝚏₁ 𝑡, 𝑠, 𝑣(𝑠) 𝑑𝑠, 𝑡 ∈ (0,1)   𝑡0 (4) 𝑣 𝑡 =  𝚏₂ 𝑡, 𝑠, 𝑢(𝑠) 𝑑𝑠, 𝑡 ∈ (0,1)   𝑡0 where 𝑥(𝑡)  =  𝑥(0)  +  𝑢(𝑠)𝑡0𝑑𝑠                                                        (5) 𝑦(𝑡)  =  𝑦(0)  +  𝑣(𝑠)𝑡0𝑑𝑠                                                        (6) Using the nonlocal boundary condition (2), we obtain  𝑥(𝜏 )  =  𝑥(0)  +  𝑢(𝑠)𝜏0𝑑𝑠, and 𝑥(𝜉)  =  𝑥(0)  +  𝑢(𝑠)𝜉0𝑑𝑠, then 𝑥(0)  =  𝛼1 − 𝛼 𝑢(𝑠)𝜉0𝑑𝑠 −11 − 𝛼 𝑢 𝑠 𝜏0𝑑𝑠 Substituting in (5),  we obtain 𝑥(𝑡)  =  𝛼1−𝛼 𝑢(𝑠)𝜉0𝑑𝑠 −11−𝛼 𝑢 𝑠 𝜏0𝑑𝑠 + 𝑢(𝑠)𝑡0𝑑𝑠. (7) And using the nonlocal boundary condition (3),  we obtain 𝑦(𝜏 )  =  𝑦(0)  +  𝑣(𝑠)𝜏0𝑑𝑠, ISSN 2347-1921                                                           5035 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           and  𝑦 (𝜉)  =  𝑦(0) + 𝑣(𝑠)𝜉0𝑑𝑠, then 𝑦(0)  =  𝛽1 − 𝛽 𝑣(𝑠)𝜉0𝑑𝑠 −11 − 𝛽 𝑣 𝑠 𝜏0𝑑𝑠. Substituting in (6), we obtain 𝑦(𝑡)  =  𝛽1−𝛽 𝑣(𝑠)𝜉0𝑑𝑠 −11−𝛽 𝑣 𝑠 𝜏0𝑑𝑠 + 𝑣(𝑠)𝑡0𝑑𝑠. (8) 3. Existence of a unique continuous solution Here, we study the existence of a unique continuous solution of the coupled system of integral equations (4) , under the following assumptions:  (1) 𝚏𝑖 ∶  [0, 1]  ×  [0, 1]  𝑅+ →  𝑅 are continuous, and satisfy the Lipschitz  condition │𝚏𝑖 𝑡, 𝑠, 𝑥  −  𝚏𝑖 𝑡, 𝑠, 𝑦 │ ≤  𝑘𝑖 𝑡, 𝑠 │𝑥 −  𝑦│ ,   𝑖 =  1, 2. where 𝑘𝑖 ∶  [0, 1]  ×  [0, 1]  → 𝑅+are integral in (𝑡, 𝑠). (2) 𝑠𝑢𝑝𝑡  𝑘𝑖   (𝑡, 𝑠)𝑑𝑠 10 ≤  𝑀𝑖 ,     𝑡 ∈   [0, 1],    𝑖 =  1, 2.  Let 𝑋 =   𝑈 =   𝑢, 𝑣 : 𝑢, 𝑣 ∈  𝐶  0, 1  , and its norm defined as  ║ 𝑢, 𝑣 ║ =  ║𝑢║ +  ║𝑣║ =   𝑠𝑢𝑝𝑡  │𝑢 𝑡 │ +   𝑠𝑢𝑝𝑡  │𝑣 𝑡 │, 𝑡 ∈  [0, 1]. Now, for the existence of a unique continuous solution for the coupled system of the integral equations (4) , we have the following theorem. Theorem 1.  Let the assumption (1)-(2) be satisfied. 𝐼𝑓𝑀𝑖 <  1, 𝑖 =  1, 2 , then the coupled system of integral equations (4) has a unique solution in 𝑋. Proof.  Define the operator 𝐹 associated with the coupled system of integral equations (4) by 𝐹(𝑢, 𝑣)  =  (𝐹₁ 𝑣, 𝐹₂ 𝑢) Where                              𝐹₁ 𝑣 =  𝚏₁  𝑡, 𝑠, 𝑣 𝑠  𝑑𝑠                               𝑡0 𝐹₂ 𝑢 =  𝚏2 𝑡, 𝑠, 𝑢 𝑠  𝑑𝑠,   𝑡0 Firstly prove that𝐹 ∶  𝑋 →  𝑋. Let  𝑢, 𝑣 ∈  𝐶 0, 1 , 𝑡₁, 𝑡₂ ∈ [0,1], 𝑡₁ <  𝑡₂, and │𝑡₂_ 𝑡₁│ ≤  𝛿,  now to prove  𝐹₁ 𝑣 𝑡 : 𝐶  0, 1  →  𝐶  0, 1 ,   then │𝐹₁𝑣(𝑡₂)  −  𝐹₁𝑣(𝑡₁)│ = │  f₁ (t₂, s, v(s)) ds 𝑡20-  f₁ t₁, s, v s  ds │𝑡10  =│  f₁ (t₂, s, v(s)) ds 𝑡10+ f₁ (t₂, s, v(s)) ds 𝑡2𝑡1  - f₁ t₁, s, v s  ds │𝑡10  ≤│  f₁ (t₂, s, v(s)) ds 𝑡10−  f₁ t₁, s, v s  ds │𝑡10 ISSN 2347-1921                                                           5036 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                            +│ f₁ (t₂, s, v(s)) ds│𝑡2𝑡1  ≤  │𝚏₁  𝑡₂, 𝑠, 𝑣 𝑠  𝑡10− 𝚏₁  𝑡₁, 𝑠, 𝑣 𝑠  │𝑑𝑠  + │𝑡2𝑡1 f₁ (t₂, s, v(s))│𝑑𝑠. This prove that𝐹₁𝑣(𝑡) ∶  𝐶 [0, 1] →  𝐶 [0, 1], ∀𝑣(𝑡)  ∈  𝐶 [0, 1]. As done before, we obtain 𝐹₂𝑢(𝑡) ∶  𝐶 [0, 1]  →  𝐶 [0, 1], ∀𝑢(𝑡)  ∈  𝐶 [0, 1]. Now since 𝐹(𝑢, 𝑣)  =  (𝐹₁𝑣, 𝐹₂𝑢) 𝐹(𝑢, 𝑣)(𝑡₂) −  𝐹(𝑢, 𝑣)(𝑡₁)  =  𝐹(𝑢(𝑡₂), 𝑣(𝑡₂)) −  𝐹(𝑢(𝑡₁), 𝑣(𝑡₁)) =  (𝐹₁𝑣(𝑡₂)  −  𝐹₁𝑣(𝑡₁), 𝐹₂𝑢(𝑡₂) −  𝐹₂𝑢(𝑡₁)). Then ║𝐹(𝑢, 𝑣)(𝑡₂)  −  𝐹(𝑢, 𝑣)(𝑡₁)║ =  ║(𝐹₁𝑣(𝑡₂) −  𝐹₁𝑣(𝑡₁)║ +  ║𝐹₂𝑢(𝑡₂)  𝐹₂𝑢(𝑡₁))║. Hence  𝐹 ∶  𝑋  →   𝑋.  Secondly to prove that 𝐹 is a contraction, we have following. Let  𝑍 =  (𝑢, 𝑣)  ∈  𝑋and  𝑍₁ =  (𝑢₁, 𝑣₁)  ∈  𝑋,  we have 𝐹(𝑢, 𝑣)  =  (𝐹₁𝑣(𝑡), 𝐹₂𝑢(𝑡)) and 𝐹(𝑢₁, 𝑣₁)  =  (𝐹₁𝑣₁(𝑡), 𝐹₂𝑢₁(𝑡)), then │ 𝐹₁ 𝑣(𝑡)  −  𝐹₁ 𝑣₁(𝑡) │ = │ 𝚏₁ (𝑡, 𝑠, 𝑣(𝑠)) 𝑑𝑠 𝑡0 −   𝚏₁ (𝑡, 𝑠, 𝑣₁(𝑠)) 𝑑𝑠𝑡0 │ ≤  │f₁  t, s, 𝑣 s  −  f₁  t, s, 𝑣₁ s  │ ds𝑡0   ≤   k₁(t, s) │ 𝑣(s)  − 𝑣₁(s) │ ds𝑡0 ≤  k₁(t, s) sup𝑡  │𝑣(s)  − 𝑣₁(s) │ds𝑡0 ≤ ║ 𝑣(𝑠)  −  𝑣₁(𝑠) ║ k₁(t, s) ds𝑡0. Then ║ 𝐹₁ 𝑣(𝑡)  −  𝐹₁ 𝑣₁(𝑡) ║ ≤  𝑀₁ ║𝑣 −  𝑣₁║ . Since  𝑀₁ <  1, then 𝐹₁ is a contraction. As done before, we obtain │ 𝐹₂𝑢 (𝑡)  −  𝐹₂𝑢₁ (𝑡) │ ≤  𝑀₂ ║𝑢 −  𝑢₁║. Since  𝑀₂ <  1, then𝐹₂ is a contraction. Then ║𝐹(𝑢, 𝑣)  −  𝐹(𝑢₁, 𝑣₁) ║ =  ║ (𝐹₁𝑣, 𝐹₂𝑢)  −  (𝐹₁𝑣₁, 𝐹₂𝑢₁) ║  ISSN 2347-1921                                                           5037 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           =  ║𝐹₁𝑣 −  𝐹₁𝑣₁, 𝐹₂𝑢 −  𝐹₂𝑢₁║  =  ║𝐹₁𝑣 −  𝐹₁𝑣₁║ +  ║𝐹₂𝑢 −  𝐹₂𝑢₁║  ≤  𝑚𝑎𝑥  𝑀₁ , 𝑀₂  ║(𝑢 , 𝑣)  −  (𝑢₁ , 𝑣₁)║. 𝐿𝑒𝑡 𝑀 =  𝑚𝑎𝑥  𝑀₁ , 𝑀₂  ║𝐹(𝑢, 𝑣)  −  𝐹(𝑢₁, 𝑣₁) ║ ≤  𝑀 ║(𝑢 , 𝑣)  −  ( 𝑢₁, 𝑣₁)║. Since𝑀 <  1, then 𝐹 is a contraction, by using Banach fixed point Theorem[(5)], then there exists a unique solution in X for the coupled system of the integral equations (4). 4 Solution of the problem (1)-(3) Consider now the problem (1)-(3). Theorem 2. Let the assumption of the theorem 1 be satisfied, then there exists a unique solution 𝑥, 𝑦 ∈  𝐶[0, 1] of the problem (1)-(3) . Proof.The solution of the problem (1) and (3) is given by 𝑥 𝑡 =  𝛼1 − 𝛼 𝑢 𝑠 𝜉0𝑑𝑠 −11 − 𝛼 𝑢 𝑠 𝜏0𝑑𝑠 +   𝑢 𝑠 𝑡0𝑑𝑠      ∈ 𝐶[0, 1], and 𝑦 𝑡 =  𝛽1−𝛽 𝑣 𝑠 𝜉0𝑑𝑠 −11−𝛽 𝑣 𝑠 𝜏0𝑑𝑠 +   𝑣 𝑠 𝑡0𝑑𝑠      ∈𝐶[0, 1]. Where           𝑢(𝑡) =  𝚏₁ 𝑡, 𝑠, 𝑣 𝑠  𝑑𝑠,      ∈ C 0, 1 .            𝑡0  𝑣 𝑡 =  𝚏₂ 𝑡, 𝑠, 𝑢 𝑠  𝑑𝑠,       ∈ C 0, 1 .𝑡0 Then from Theorem 1 we can deduce that there exists a unique continuous solution of the problem (1)-(3). 5. Existence of a unique 𝑳𝟏 −solution Here, we study the existence of integrable solution of the coupled system of integral equations (4) under the following assumptions: (𝑖) 𝚏𝑖 ∶  [0, 1] ×  [0, 1] ×  𝑅+ →  𝑅 are measurable in (𝑡, 𝑠), and satisfy the Lipschitz condition │ 𝚏𝑖 𝑡, 𝑠, 𝑥 −  𝚏𝑖 𝑡, 𝑠, 𝑦 │ ≤  𝑘𝑖  │ 𝑥 −  𝑦 │,   𝑖 =  1, 2. (𝑖𝑖) 𝚏𝑖(𝑡, 𝑠 0)  ∈  𝐿₁ [0, 1], and   │𝚏𝑖 𝑡 𝑠 0 │ 𝑑𝑡 10  ≤  𝑀𝑖, 𝑡 ∈   0, 1 , 𝑖 =  1, 2. Let 𝑌 =   𝑈 =  (𝑢, 𝑣) ∶  𝑢, 𝑣 ∈ 𝐿₁ [0, 1]  ,  and its norm defined as ║(𝑢, 𝑣)║ =  ║𝑢║ +  ║𝑣║ =   │𝑢(𝑡)│𝑑𝑡 10 +   │𝑣 𝑡 │𝑑𝑡   10 Now, for the existence of integrable solution for the coupled system of the integral equations(4),  we have the following theorem. Theorem 3. Let the assumption (i)-(ii) be satisfied. 𝐼𝑓 𝑘𝑖 <  1, 𝑖 =  1, 2,  then the coupledsystem of the integral equations (4) has a unique solution in 𝑌 . ISSN 2347-1921                                                           5038 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           Proof.  Define the operator 𝐺 associated with the coupled system of integralequations (4)by 𝐺(𝑢, 𝑣)  =  (𝐺₁ 𝑣, 𝐺₂ 𝑢). Where 𝐺₁𝑣 = 𝚏₁  𝑡, 𝑠, 𝑣 𝑠  𝑑𝑠 𝑡𝑂 𝐺₂𝑢 = 𝚏₂ 𝑡, 𝑠, 𝑢 𝑠  𝑑𝑠.𝑡0 Firstly to prove that𝐺 ∶  𝑌 →  𝑌 , now to prove 𝐺₁𝑣 ∶ 𝐿₁ [0, 1]  →  𝐿₁[0, 1], then  │ 𝚏₁(𝑡, 𝑠, 𝑣) │ −  │ 𝚏₁(𝑡, 𝑠, 0) │ ≤ │ 𝚏₁(𝑡, 𝑠, 𝑣)  −  𝚏₁(𝑡, 𝑠, 0) │ ≤  𝑘₁ │ 𝑣 │ and │ 𝚏₁(𝑡, 𝑠, 𝑣) │ ≤  𝑘₁ │ 𝑣 │ +  │ 𝚏₁(𝑡, 𝑠, 0) │. Hence │𝐺₁𝑣(𝑡)│ =   │ 𝚏₁(𝑡, 𝑠, 𝑣(𝑠))𝑑𝑠│ ≤𝑡0  𝑘₁ │ 𝑣 (𝑠)│𝑑𝑠 𝑡0 +  │ 𝚏₁(𝑡, 𝑠, 0) │. Integrating both sides with respect to t, we obtain  │  f₁(t, s, 𝑣 (s))ds│dt ≤𝑡010   k₁ │ 𝑣 (s)│ds dt𝑡010+  │ f₁(t, s, 0) │10𝑑𝑡  ≤ k₁  │𝑣 s │dt +10  │ f₁(t, s, 0) │10𝑑𝑡 ≤ k₁ ║𝑣║𝐿1+  M₁. Then║𝐺₁𝑣║𝐿1 ≤  k₁ ║𝑣║𝐿1 +  𝑀₁. This proves that 𝐺₁𝑣 ∶ 𝐿₁ [0, 1]  →  𝐿₁[0, 1]. As done before, we obtain ║𝐺₂𝑢║𝐿1 ≤  k₂ ║𝑢║𝐿1 +  𝑀₂. This proves that 𝐺₂𝑢 ∶  𝐿₁ [0, 1]  →  𝐿₁[0, 1]. Hence ║𝐺 (𝑢, 𝑣) ║ =  ║ 𝐺₁𝑣, 𝐺₂𝑢 ║ =  ║ 𝐺₁𝑣 ║ +  ║ 𝐺₂𝑢 ║ =   k₁ ║𝑣║𝐿1 +  𝑀₁  + k₂ ║𝑣║𝐿1 +  𝑀₂. This proves 𝐺: 𝑌 →  𝑌. Secondly prove that 𝐺 is a contraction.  Let                    𝑍 =   𝑢, 𝑣 ∈  𝑌   and    𝑍₁ =  (𝑢₁, 𝑣₁)  ∈  𝑌 . Then                  𝐺(𝑢, 𝑣)  =  (𝐺₁𝑣, 𝐺₂𝑢)   and   𝐺(𝑢₁, 𝑣₁)  =  (𝐺₁𝑣₁, 𝐺₂𝑢₁) │ 𝐺₁𝑣 −  𝐺₁𝑣₁│  =  │ 𝚏₁ 𝑡, 𝑠, 𝑣 𝑠  𝑑𝑠 −   𝚏₁ 𝑡, 𝑠, 𝑣₁ 𝑠  𝑑𝑠│.𝑡0𝑡0 Integrating both sides with respect to t, we   │ G₁𝑣  −  G₁𝑣₁│dt10≤  │  f₁ t, s, 𝑣 s  −  f₁ t, s, 𝑣₁ s   𝑡0𝑑𝑠│𝑑𝑡10 ≤   k₁│  𝑣 s − 𝑣₁ s │ ds 𝑡010dt ISSN 2347-1921                                                           5039 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           ≤ k₁  │𝑣 s − 𝑣₁ s │dt10 ≤ 𝑘₁ ║ 𝑣 −  𝑣₁ ║𝐿₁. Then ║G₁𝑣  −  G₁𝑣₁║𝐿1≤  𝑘₁║ 𝑣 −  𝑣₁ ║𝐿₁.  Since𝑘₁ <  1, then 𝐺₁ is a contraction. As done before, we obtain ║G₂𝑢  −  G₂𝑢₁║𝐿1≤  𝑘₂║ 𝑢 −  𝑢₁ ║𝐿₁. Since 𝑘₂ <  1, then 𝐺₂ is a contraction. Hence ║𝐺(𝑢, 𝑣)  −  𝐺(𝑢₁, 𝑣₁)║ =  ║(𝐺₁𝑣, 𝐺₂𝑢)  −  (𝐺₁𝑣₁, 𝐺₂𝑢₁)║  =  ║𝐺₁𝑣 −  𝐺₁𝑣₁, 𝐺₂𝑢 −  𝐺₂𝑢₁║  ≤  𝑚𝑎𝑥 𝑘₁ , 𝑘₂ ║ 𝑢, 𝑣 − (𝑢₁, 𝑣₁)║𝐿1. Let 𝑘 =  𝑚𝑎𝑥  𝑘₁ , 𝑘₂ . Then ║𝐺 𝑢, 𝑣 −  𝐺(𝑢₁, 𝑣₁)║𝐿1≤ 𝑘║ 𝑢, 𝑣 − (𝑢₁, 𝑣₁)║𝐿1. Since𝑘 <  1, then 𝐺 𝑖s a contraction, by using Banach fixed point Theorem [(5)], then there exists of solution in 𝑌 for the coupled system of the integral equations (4). 6. Solution of the problem (1)-(3) Consider now the problem (1)-(3). Theorem 4. Let the assumption of the theorem 3 be satisfied, then there exists a uniquesolution 𝑥, 𝑦 ∈  𝐴𝐶[0, 1] of the boundary value problem (1)-(3). Proof. The solution of the problem (1) - (3) is given by 𝑥 𝑡 =  𝛼1−𝛼 𝑢 𝑠 𝜉0𝑑𝑠 −11−𝛼 𝑢 𝑠 𝜏0𝑑𝑠 +   𝑢 𝑠 𝑡0𝑑𝑠 ∈ 𝐴𝐶[0,1] ,  and 𝑦 𝑡 =  𝛽1 − 𝛽 𝑣 𝑠 𝜉0𝑑𝑠 −11 − 𝛽 𝑣 𝑠 𝜏0𝑑𝑠 +   𝑣 𝑠 𝑡0𝑑𝑠 ∈ 𝐴𝐶 0,1 . Where  𝑢(𝑡) =   𝚏₁ 𝑡, 𝑠, 𝑣 𝑠  𝑑𝑠      ∈ 𝐿₁ [0, 1]10 𝑣(t) =  𝚏₂ 𝑡, 𝑠, 𝑢 𝑠  𝑑𝑠       ∈ 𝐿₁ [0, 1] .                        10 Then from Theorem 3 we can deduce that there exists a unique solution of the problem (1) - (3). References [1] D. ORegan, M.Meehan, Existence theory for nonlinear integral and integro-differential equations, Kluwer Acad. Pulbel.  Dordrecht, 1998. [2] A. M. A. El-Sayed and E. O. Bin-Taher, An arbitraty fractional order differential equation with internal nonlocal and integral conditions, Vol.1, No.3, pp. 59-62, (2011). ISSN 2347-1921                                                           5040 | P a g e                                                   S e p t e m b e r  0 1 ,  2 0 1 5                           [3] A. M. A. El-Sayed and E. O. Bin-Taher, A nonlocal problem for a multi-term frac- tional order differential equation, Journal of Math Analysis, Vol. 5, No.29, PP.1445-1451, (2011). [4] A. M. A. El-Sayed and E. O. Bin-Taher, a multi-term fractional- order differential equation with nonlocal condition, Egy.Chin.J Comp.App.Math. , Vol. 1, No.1, PP.54-60, (2012). [5] Goebel, K. and Kirk W. A., Topics in Metric Fixed point theory, Cambridge University Press, Cambridge (1990) [6] Ibrahim Abouelfarag Ibrahim, on the existence of solution of functional integral equation of Urysohn type, Computers and Mathematics with Applications, 1609-1614, 57, (2009). [7] J. Banas, Integrable solutions of Hammerstein and Urysohn integral equation, J. Austral. Math. Soc (Series A), pp.61-68, 46, (1989). [8]R.F.Apolaya, H. R. Clark and A.J. Feitosa, on a nonlinear coupled system with internal damping, nonlinear, taxas state university, Vol., No. 64, pp.1-17,(2000). 

Existence of solution for a coupled system of Volterra type integro - dierential equations with nonlocal conditions

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Existence of solution for a coupled system of Volterra type integro - dierential equations with nonlocal conditions

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