tailieunhanh - Strong solution for a high order boundary value problem with integral condition
The present paper is devoted to a proof of the existence and uniqueness of strong solution for a high order boundary value problem with integral condition. The proof is based by a priori estimate and on the density of the range of the operator generated by the studied problem. | Turkish Journal of Mathematics Turk J Math (2013) 37: 299 – 307 ¨ ITAK ˙ c TUB Research Article doi: Strong solution for a high order boundary value problem with integral condition 1 Ahcene MERAD1, ∗, Ahmed Lakhdar MARHOUNE2 Department of Mathematics, Faculty of Sciences, Larbi Ben M’hidi University, Oum El Bouaghi,04000, Algeria 2 Department of Mathematics, Faculty of Sciences, Mentouri University, Constantine, 25000, Algeria Received: • Accepted: • Published Online: • Printed: Abstract: The present paper is devoted to a proof of the existence and uniqueness of strong solution for a high order boundary value problem with integral condition. The proof is based by a priori estimate and on the density of the range of the operator generated by the studied problem. Key words: Integral condition, energy inequality, boundary value problem 1. Introduction In the rectangular domain Q = (0, 1) × (0, T ) , with T 0, () with the constant c satisfying the region ⎧ 2 ∂a ⎪ ⎪ 1 ∂a ∂t ⎪ − ⎨ sup a ∂t − a ⎪ ⎪ ⎪ ⎩ Proof 1 ∂a a ∂t 0 2 − 1 ∂a a ∂t + 1 2 , () Denote M u = (1 − x) ∂4u ∂4u + αJ , ∂t4 ∂t4 where Ju = x u (ξ, t) dξ. 0 We consider the quadratic formula τ 1 exp (−ct) £uM udxdt, Re 0 () 0 with the constant c satisfying condition () ; obtained by multiplying equation () by exp (−ct) £uM u; and integrating over Qτ , where Qτ = (0, 1) × (0, τ ) , with 0 ≤ τ ≤ T, and by taking the real part. Integrating by parts α times in formula () with the use of boundary conditions in equations () , () , and () , we obtain τ Re 1 exp (−ct) £uM udxdt = 0 () 0 301 MERAD and MARHOUNE/Turk J Math 4 2 ∂ u exp (−ct) (1 − x) 4 dxdt + ∂t 0 0 1 ∂ 2 a (x, τ ) ∂a (x, τ ) (1 − x) ∂ ∂ α u (x, τ ) ∂ α u (x, τ ) 2 2Re exp (−cτ ) − 2c dx − + c a (x, τ .
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