Energy Methods For Initial –Boundary String Problem Aqeel F. Jaddoa Dept. of Mathematics/College of Eduction for Pure Science (Ibn Al- Haitham)/University of Baghdad Received in : 12 June 2013, Accepted in : 22 June 2014 Abstract We study one example of hyperbolic problems it's Initial-boundary string problem with two ends. In fact we look for the solution in weak sense in some sobolev spaces. Also we use energy technic with Galerkin's method to study some properties for our problem as existence and uniqueness. Keyword: Initial- Boundary String Problem, Weak Solution, Sobolev Spaces, Galerkin's Method, Energy Estimate. 220 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 1.Introduction Throughout this paper we consider energy estimate for string equation with initial –boundary conditions. We gave a little introduction for weak derivatives, weak solution in some sobolev space, for example 𝐻01(𝑢), 𝐻−1(𝑢) and others. In fact energy estimate has many results for example a speed of propagation [1], in our paper it leads us to straight forward existence, uniqueness [2], results and other properties like regularity. It yields also to well-posed for hyperbolic equations. In 2007, Tarama [3], introduced wave equations with coefficients satisfying Besov type conditions and obtained the energy estimate, also he gave an example of a wave equation with continuous and differentiable coefficients for which the 𝐿2 estimate holds. Jaipong Kasemsuwan [4], concerned with the energy decay of the global solution for initial- boundary value problem to a nonlinear damped equation of suspended string with uniform density to which a nonlinear outer force works. The string equation with initial-boundary conditions is 𝜕2𝑢 𝜕𝑡2 = 𝑎2 𝜕 2𝑢 𝜕𝑥2 + 𝑓(𝑥, 𝑡) in Q .. . (1.1)[5, p.30] Where 𝑄 = {(𝑥, 𝑡); 𝑥 ∈ (0, 𝑙), 𝑡 ∈ (0, 𝑇)}, Initial conditions: u(x, 0) = φ(x) u𝑡(x, 0) = ψ(x) � …. (1.2) Boundary conditions: u(0, t) = 0 u(𝑙, t) = 0 � … (1.3) This problem describes the vibration of the string in a finite interval [0, 𝑙] and the constant 𝑎 denotes the speed of the string. Rewrite the initial-boundary problem with the second order partial differential operator 𝐿 as: 𝜕2𝑢 𝜕𝑡2 = 𝑎2𝐿𝑢 + 𝑓(𝑥, 𝑡) 𝑖𝑛 𝑈 × (0, 𝑇] 𝑢(𝑥, 𝑡) = 0 𝑜𝑛 𝜕𝑈 × [0, 𝑇] 𝑢(𝑥, 𝑡) = 𝜑(𝑥), 𝜕 𝜕𝑡 𝑢(𝑥, 𝑡) = 𝜓(𝑥)𝑜𝑛 𝑈 × {𝑡 = 0} ⎭ ⎪ ⎬ ⎪ ⎫ …(1.4) 𝐿𝑒𝑡 𝑢𝑠 𝑑𝑒𝑛𝑜𝑡𝑒 𝑓𝑜𝑟 𝑈 = (0, 𝑙), 𝑈𝑇 = 𝑈 × (0, 𝑇], 𝑓: 𝑈 × (0, 𝑇] → ℛ and 𝜑, 𝜓: 𝑈 → ℛ and these 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛. The function 𝑢: 𝑈𝑇���� → ℛ is the unknown function u=u(x, t). The symbol 𝐿 denotes for each time 𝑡 a second–order partial differential operator, having the divergence form 𝐿𝑢 = − � (𝑎𝑖𝑗 𝑛 𝑖,𝑗=1 (𝑥, 𝑡) 𝑢𝑥𝑖)𝑥𝑗 + �𝑏𝑖 𝑛 𝑖=1 (𝑥, 𝑡)𝑢𝑥𝑖 + 𝑐(𝑥, 𝑡)𝑢 … (1.5) [6, 𝑝. 309] For given coefficients aij, bi, c ∈ ∁1(U� × [0, T]) (i, j = 1,2, … , n), f ∈ L2( U × [0, T]), φ ∈ H01( U), ψ ∈ L2( U ), And always assume that 𝑎𝑖𝑗 = 𝑎𝑗𝑖 (𝑖, 𝑗 = 1,2, … , 𝑛). Definition 1[6, p.378] The partial differential operator 𝜕 2 𝜕𝑡2 + 𝐿 is said to be (uniformly) hyperbolic if there exists a constant 𝜃𝜃 > 0 such that ∑ 𝑎𝑖𝑗(𝑥, 𝑡)𝜀𝑖𝜀𝑗 ≥ 𝜃𝜃|𝜀|2𝑛𝑖,𝑗=1 for all (x, t) ∈ 𝑈 × [0, 𝑇], 𝜀 ∈ 𝑅𝑛. Definition 2 [6, p. 285] Let 𝑢 ∈ 𝐿1(0, 𝑇, 𝑋). we say 𝑣 ∈ 𝐿1(0, 𝑇, 𝑋) is weak derivative of u, written: 𝑢′ = 𝑣 provided 221 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 �𝜑′ 𝑇 0 (𝑡) 𝑢(𝑡)𝑑𝑡 = −�𝜑 𝑇 0 (𝑡) 𝑣(𝑡)𝑑𝑡 For all scalar test functions 𝜑 ∈ 𝐶𝑐∞(0, 𝑇). Let us introduce the time-dependent bilinear form: 𝐵[𝑢, 𝑣; 𝑡] = �( � aij n i,j=1U uxivxj )dx + ��(b𝑖(𝑥, 𝑡) n i=1U uxv + c(x, t )u v )dx for all 𝑢, 𝑣 ∈ 𝐻01(𝑈) 𝑎𝑛𝑑 0 ≤ 𝑡 ≤ 𝑇. First we suppose u is smooth solution and defined the mapping: u: [0, 𝑇] → 𝐻01(𝑈), [𝑢(𝑡)](𝑥) = 𝑢(𝑥, 𝑡), (𝑥 ∈ 𝑈, 0 ≤ 𝑡 ≤ 𝑇), ( ′ = 𝑑 𝑑𝑡 ) f: [0, 𝑇] → 𝐿2 (𝑈), [𝑓(𝑡)](𝑥) = 𝑓(𝑥, 𝑡) Now we fix any function 𝑣 ∈ 𝐻01(𝑈), multiply by v the PDE (𝑢𝑡𝑡 + 𝐿𝑢 = 𝑓)𝑣 (𝑢", 𝑣) + 𝐵[𝑢, 𝑣, 𝑡] = (𝑓, 𝑣) The pairing ( , ) inner product in 𝐿2(𝑈). We see the form: 𝑢𝑡𝑡 = 𝜑0 + ∑ 𝜑𝑥𝑗 𝑗𝑛 𝑗=1 For 𝜑0 = 𝑓 − ∑ 𝑏𝑖𝑢𝑥𝑖 − 𝑐 𝑢,𝑛𝑖=1 𝜑𝑗 = 𝑓 − ∑ 𝑎𝑖𝑗𝑢𝑥𝑖 𝑛 𝑖=1 (j=1,…,n) This suggests that we should look for weak solution u with u" ∈ 𝐻−1(𝑈) for almost every where 0 ≤ 𝑡 ≤ 𝑇, 〈𝑢", 𝑣〉 means 𝑢" ∈ 𝐻−1(𝑈), 𝑣 ∈ 𝐻01(𝑈) Definition 3 [6, p.379] We say a function 𝑢 ∈ 𝐿2�0, 𝑇, 𝐻01(𝑈)�, with 𝑢′ ∈ 𝐿2�0, 𝑇, 𝐿2(𝑈)�, 𝑢" ∈ 𝐿2(0, 𝑇, 𝐻−1(𝑈)) is weak solution of the hyperbolic initial( boundary ) problem and i. 〈𝑢", 𝑣〉 + 𝐵[𝑢, 𝑣, 𝑡) = (𝑓, 𝑣) 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑣 ∈ 𝐻01(𝑈) ii.u(0)=𝜑, 𝑢′(0) = 𝜓. As known sobolev spaces are designed to contain less smooth functions that means the functions have some but not great, smoothness properties. So the idea is to make approximation for the function u(x, t) such that u(x, t)=0 in the boundary of U by the 𝑢𝑚(𝑡) = � 𝑑𝑚𝑘 (𝑡) 𝑤𝑘(𝑥) 𝑚 𝑘=1 [6, 𝑝. 353] In the right side hand we have separated variables, we note that U is a domain of the variable x only, and this is the reason to consider the variable t in respect to the sobolev space 𝐻01(𝑈) as a parameter. where wk = wk(x) ( k = 1,2, … ) are smooth functions such that {𝑤𝑘 }𝑘=1 ∞ is an orthogonal basis of H01( 𝑈), {𝑤𝑘 }𝑘=1 ∞ is an orthonormal basis of 𝐿2(𝑈 ). We are going to consider 𝑢 not as a function of x and t together, but rather as a mapping 𝑢 of t into the space 𝐻01( 𝑈) of functions of x. We will look for a function 𝑢𝑚:[0, T] → 𝐻01( 𝑈) of the form 𝑢𝑚(𝑡) = ∑ 𝑑𝑚𝑘𝑚𝑘=1 (𝑡) 𝑤𝑘 such that (𝑢𝑚″ , 𝑤𝑘) + 𝐵[𝑢𝑚, 𝑤𝑘; 𝑡] = (𝑓, 𝑤𝑘) , ( 𝑡 ∈ [0, 𝑇], 𝑘 = 1,2, … , 𝑚) … (1.6) 𝑑𝑚𝑘 (0) = (𝜑, 𝑤𝑘) 𝑑𝑚𝑘 ′(0) = (𝜓, 𝑤𝑘) � … (1.7) where 𝐵[𝑢𝑚, 𝑤𝑘; 𝑡] = �( � aij n i,j=1U um,xiwk,xj )dx + ��(b𝑖 n i=1U um,xiwk + cumwk ) dx 222 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 In other words, we construct solutions of certain finite-dimensional approximations to our problem and then passing to limits, it's called Galerkins method [6, p.380]. Indeed we need some estimates to send m to infinity and to show subsequence of our solutions 𝑢𝑚 of the approximate problem . 2.Main Result Theorem (Energy estimates)(2.1): There exists a constant ∁ , depending only on 𝑈,𝑇 and the coefficients on 𝐿 such that: max 0≤𝑡≤𝑇 �‖𝑢m(𝑡)‖𝐻01(𝑈) + ‖𝑢𝑚 ′ (𝑡)‖𝐿2(𝑈)� + ‖𝑢𝑚″ (𝑡)‖𝐿2�0,T;𝐻−1(U)� ≤ ∁�‖𝑓‖𝐿2(0,T;𝐿2(U)) + ‖𝜑‖𝐻01(𝑈) + ‖𝜓‖𝐿2(𝑈)� for m=1,2,… proof We take equation (1.6) and write it in details as: �( 𝑈 𝑢𝑚″ 𝑤𝑘)𝑑𝑥 + � � (𝑎𝑖𝑗 𝑛 𝑖,𝑗=1𝑈 𝑢𝑚,𝑥𝑖𝑤𝑘,𝑥𝑗)𝑑𝑥 + ��(𝑏𝑖 𝑛 𝑖=1𝑈 𝑢𝑚,𝑥𝑖𝑤𝑘 + 𝑐𝑢𝑚𝑤𝑘 ) 𝑑𝑥 = �𝑓 𝑤𝑘 𝑈 𝑑𝑥 Multiply by 𝑑𝑚𝑘 ′(𝑡) 𝑠𝑜: � 𝑈 �𝑢𝑚″ 𝑤𝑘 𝑑𝑚𝑘 ′(𝑡)�𝑑𝑥 + �( � 𝑎𝑖𝑗 𝑛 𝑖,𝑗=1𝑈 𝑢𝑚,𝑥𝑖𝑤𝑘,𝑥𝑗 𝑑𝑚𝑘 ′(𝑡) + �𝑏𝑖 𝑛 𝑖=1 𝑢𝑚,𝑥𝑖𝑤𝑘 𝑑𝑚𝑘 ′(𝑡) +𝑐𝑢𝑚𝑤𝑘 𝑑𝑚𝑘 ′(𝑡))𝑑𝑥 = �𝑓 𝑤𝑘 𝑑𝑚𝑘 ′(𝑡) 𝑈 𝑑𝑥. �( U um″ um′ )dx + �( � aij n i,j=1U um,xium′ + �𝑏𝑖 n i=1 um,xium′ + cum um′ ) dx = �𝑓𝑢𝑚′ 𝑈 𝑑𝑥 (𝑢𝑚″ , 𝑢𝑚′ ) + 𝐵[𝑢𝑚, 𝑢𝑚′ , ; 𝑡]= (𝑓, 𝑢𝑚′ ). And we have (𝑢𝑚″ , 𝑢𝑚′ ) = 𝑑 𝑑𝑡 �1 2 ‖𝑢𝑚′ ‖L2(U) 2 �. We write : � � (𝑎𝑖𝑗 𝑛 𝑖,𝑗=1𝑈 𝑢𝑚,𝑥𝑖𝑢′𝑚,𝑥𝑗))𝑑𝑥 + ��(𝑏𝑖 𝑛 𝑖=1𝑈 𝑢𝑚,𝑥𝑖𝑢′𝑚 + 𝑐𝑢𝑚𝑢′𝑚 ) 𝑑𝑥 = 𝐵1 + 𝐵2 where 𝐵1 = �( � 𝑎𝑖𝑗 𝑛 𝑖,𝑗=1𝑈 𝑢𝑚,𝑥𝑖𝑢𝑚𝑥𝑗 ′ 𝑎𝑛𝑑 𝐵2 = �(�𝑏𝑖 𝑛 𝑖=1𝑈 𝑢𝑚,𝑥𝑖𝑢𝑚′ + 𝑐𝑢𝑚 𝑢𝑚′ ) 𝑑𝑥 Now we define the symmetric bilinear form: 𝐴[𝑢, 𝑣; 𝑡] ≔ ∫ (∑ 𝑎 𝑖𝑗𝑛 𝑖,𝑗=1𝑈 𝑢𝑥𝑖𝑣𝑥𝑗) 𝑑𝑥 , 𝑢, 𝑣 ∈ 𝐻01( 𝑈) Then 𝑑 𝑑𝑡 � 1 2 𝐴[𝑢m, 𝑢m; t]� − 1 2 � ( � 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1𝑈 )𝑑𝑥 = 1 2 (� 𝑑 𝑑𝑡 [ � 𝑎𝑖𝑗𝑢𝑚,𝑥𝑖𝑢𝑚,𝑥𝑗] 𝑛 𝑖,𝑗=1 𝑑𝑥 − 𝑈 1 2 � �� 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 �𝑑𝑥 𝑈 223 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 = 1 2 � � 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 + 𝑣 1 2 � � 𝑎𝑖𝑗 𝑢𝑚′ 𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 𝑣 + 1 2 � � 𝑎𝑖𝑗 𝑢𝑚′ 𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 𝑣 − 1 2 � � 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 = 𝑈 � � 𝑎𝑖𝑗𝑢m𝑥𝑖 𝑢𝑚 ′ 𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 𝑈 = 𝐵1 (𝑠𝑖𝑛𝑐𝑒 𝑎𝑖𝑗 = 𝑎𝑗𝑖) 𝑇ℎ𝑒𝑛 𝐵1 = 𝑑 𝑑𝑡 � 1 2 𝐴[𝑢m, 𝑢m; t]� − 1 2 � � 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 𝑈 So 𝑑 𝑑𝑡 � 1 2 𝐴[𝑢m, 𝑢m; t]� = 𝐵1 + 1 2 � ( � 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗)𝑑𝑥 𝑛 𝑖,𝑗=1𝑈 𝑎𝑡 𝑖𝑗 ∈ ∁1(𝑈� × [0, 𝑇]), so there exists Max(𝑎𝑡 𝑖𝑗) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐵1 + 1 2 � � � 𝑎𝑡 𝑖𝑗𝑢m𝑥𝑖𝑢m𝑥𝑗 𝑛 𝑖,𝑗=1 �𝑑𝑥 𝑈 ≤ 𝐵1 + 𝐶 � ( � 𝑢m𝑥𝑖𝑢m𝑥𝑗) 𝑛 𝑖,𝑗=1 𝑑𝑥 𝑈 ≤ 𝐵1 + ∁ � �𝑢m𝑥𝑗 2 𝑛 𝑗=1 𝑑𝑥 𝑈 ≤ 𝐵1 + ∁ 𝐵1 + ∁‖𝑢𝑚‖𝐻01 2 Or we can write it as: 𝑑 𝑑𝑡 (𝐴[𝑢m, 𝑢m; t]) = 2𝐵1 + ∁‖𝑢𝑚‖𝐻01 2 … (2.1) We had (𝑢𝑚″ , 𝑢𝑚′ ) = 𝑑 𝑑𝑡 �1 2 ‖𝑢𝑚′ ‖L2(U) 2 � So 𝑑 𝑑𝑡 �‖𝑢𝑚′ ‖L2(U) 2 � = 2(𝑢𝑚″ , 𝑢𝑚′ ) = 2(𝑓, 𝑢𝑚′ ) − 2(𝐵1 + 𝐵2) ≤ ∁(‖𝑓‖𝐿2 2 + ∁‖𝑢𝑚′ ‖𝐿2 2 − 2(𝐵1 + 𝐵2) Now we obtain 𝑑 𝑑𝑡 �‖𝑢𝑚′ ‖L2(U) 2 � ≤ ‖𝑓‖𝐿2 2 + ∁‖𝑢𝑚′ ‖𝐿2 2 − 2(𝐵1 + 𝐵2) …(2.2) Bu𝑡 𝐵2 = �(�𝑏𝑖 𝑛 𝑖=1 𝑈 𝑢𝑚,𝑥𝑖𝑢𝑚′ + 𝑐𝑢𝑚 𝑢𝑚′ ) 𝑑𝑥 → |𝐵2| = ��(� bi n i=1 U um,xium′ + cum um′ ) dx � 𝑏𝑖 ∈ ∁1(𝑈� × [0, 𝑇]) 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝑀𝑎𝑥 (𝑏𝑖) such that ���(𝑏𝑖 𝑛 𝑖=1 𝑈 𝑢𝑚,𝑥𝑖𝑢𝑚′ + 𝑐𝑢𝑚 𝑢𝑚′ ) 𝑑𝑥� ≤ ∁ ��(�𝑢𝑚,𝑥𝑖𝑢𝑚′ � + |𝑢𝑚 𝑢𝑚′ |)𝑑𝑥 𝑛 𝑖=1 𝑈 ≤ ∁ � �(𝑢𝑚𝑥𝑖 2 𝑛 𝑖=1 𝑈 + 𝑢𝑚2 + 2𝑢′𝑚 2 )𝑑𝑥 224 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 ≤ ∁ (‖𝑢𝑚‖𝐻01 2 + ‖𝑢𝑚′ ‖L2(U) 2 ) Or |𝐵2| ≤ ∁ (‖𝑢𝑚‖𝐻01(𝑈) 2 + ‖𝑢𝑚′ ‖L2(U) 2 ) From (2.1) and (2.2) we obtain : 𝑑 𝑑𝑡 �𝐴[𝑢m, 𝑢m; t] + ‖𝑢𝑚′ ‖L2(U) 2 � ≤ 2𝐵1 + ∁‖𝑢𝑚‖𝐻01(𝑈) 2 + ∁‖𝑓‖𝐿2(𝑈) 2 + ∁‖𝑢𝑚′ ‖𝐿2(𝑈) 2 − 2(𝐵1 + 𝐵2) ≤ ∁‖𝑢𝑚‖𝐻01(𝑈) 2 + ∁‖𝑓‖𝐿2(𝑈) 2 + ∁‖𝑢𝑚′ ‖𝐿2(𝑈) 2 + ∁|𝐵2| ≤ ∁‖𝑢𝑚‖𝐻01(𝑈) 2 + ∁‖𝑢𝑚′ ‖𝐿2(𝑈) 2 + ∁‖𝑓‖𝐿2(𝑈) 2 + ∁‖𝑢𝑚‖𝐿2 2 ≤ ∁‖𝑢𝑚‖𝐻01(𝑈) 2 + ∁‖𝑢𝑚′ ‖𝐿2(𝑈) 2 + ∁‖𝑓‖𝐿2(𝑈) 2 So we have 𝑑 𝑑𝑡 �𝐴[𝑢m, 𝑢m; t] + ‖𝑢𝑚′ ‖L2(U) 2 � ≤ ∁‖𝑢𝑚‖𝐻01(𝑈) 2 + ∁‖𝑢𝑚′ ‖𝐿2(𝑈) 2 + ∁‖𝑓‖𝐿2(𝑈) 2 But 𝐿 𝑖𝑠 ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 𝑎𝑛𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 𝑤𝑒 ℎ𝑎𝑣𝑒 𝜃𝜃 � �𝑢m𝑥𝑖 𝑛 𝑖=1 𝑑𝑥 ≤ � � 𝑎𝑖𝑗𝑢𝑚𝑥𝑖𝑢𝑚𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 = 𝐴[𝑢𝑚, 𝑢𝑚; 𝑡]. 𝑈 𝑈 𝜃𝜃� �𝑢𝑚𝑥𝑖 𝑛 𝑖=1 𝑑𝑥 ≤ 𝐴[𝑢𝑚, 𝑢𝑚; 𝑡]. 𝑈 And ∁‖𝑢𝑚‖𝐻01(𝑈) 2 ≤ ∁ �𝑢𝑚2 𝑑𝑥 𝑈 + ∁� �𝑢𝑚𝑥𝑖 2 𝑑𝑥 𝑛 𝑖 𝑈 ≤ 𝐴[𝑢𝑚, 𝑢𝑚; 𝑡] 𝑑 𝑑𝑡 �𝐴[𝑢m, 𝑢m; t] + ‖𝑢𝑚′ ‖L2(U) 2 � ≤ ∁‖𝑢𝑚′ ‖𝐿2(𝑈) 2 + ∁‖𝑓‖𝐿2(𝑈) 2 + ∁ 𝐴[𝑢m, 𝑢m; t] We denote 𝜂𝜂(𝑡) = ‖𝑢𝑚′ (𝑡)‖L2(U) 2 + 𝐴[𝑢m(t), 𝑢m(t); t] ξ (𝑡) = ‖𝑓‖𝐿2(𝑈) 2 So we have 𝜂𝜂′(𝑡) ≤ ∁1𝜂𝜂(𝑡) + ∁2ξ(𝑡) Now we use Gronwall Inequality 𝜂𝜂(𝑡) ≤ 𝑒𝑐1𝑡( 𝜂𝜂(0) + 𝑐2 ∫ 𝜉𝜉(𝑠) 𝑑𝑠) 𝑡 0 …(2.3) But 𝜂𝜂(0) = ‖𝑢𝑚′ (0)‖𝐿2(𝑈) 2 + 𝐴[𝑢𝑚(0), 𝑢𝑚(0); 0] ≤ ‖𝑢𝑚′ (0)‖𝐿2(𝑈) 2 + � � 𝑎𝑖𝑗𝑢𝑚(0)𝑥𝑖𝑢𝑚(0)𝑥𝑗 𝑛 𝑖,𝑗=1 𝑑𝑥 𝑈 ≤ ‖𝑢𝑚′ (0)‖𝐿2(𝑈) 2 + � ( � 𝑎𝑖𝑗𝑢𝑚(0)𝑥𝑖𝑢2𝑚(0)𝑥𝑗) 𝑛 𝑖,𝑗=1 𝑑𝑥 = 𝑈 = ‖𝑢𝑚′ (0)‖𝐿2(𝑈) 2 + ‖𝑢𝑚(0)‖𝐻01(𝑈) 2 ≤ ∁(‖𝜓‖𝐿2(𝑈) 2 + ‖𝜑‖𝐻01(𝑈) 2 ) 𝑠𝑜 225 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 𝜂𝜂(0) ≤ ∁(‖𝜓‖𝐿2(𝑈) 2 + ‖𝜑‖𝐻01(𝑈) 2 ) 𝑓𝑟𝑜𝑚 (2.3)𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 ‖𝑢𝑚′ (𝑡)‖L2(U) 2 + 𝐴[𝑢m(t), 𝑢m(t); t] ≤ ∁�‖𝜓‖L2(U) 2 + ‖𝜑‖𝐻01(𝑈) 2 + � ‖𝑓‖2𝐿2(0,T;𝐿2(U))𝑑𝑡 𝑡 0 � 𝑠𝑜 ‖𝑢𝑚′ (𝑡)‖L2(U) 2 + 𝐴[𝑢m(t), 𝑢m(t); t] ≤ ∁�‖𝜓‖L2(U) 2 + ‖𝜑‖𝐻01(𝑈) 2 + ‖𝑓‖𝐿2(0,T;𝐿2(U))�. But again 𝐿 𝑖𝑠 ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 (2.3) 𝑠𝑜 ‖𝑢𝑚′ (𝑡)‖L2(U) 2 + ‖𝑢m(t)‖𝐻01(𝑈) ≤ ∁�‖𝜓‖L2(U) 2 + ‖𝜑‖𝐻01(𝑈) 2 + ‖𝑓‖𝐿2(0,T;𝐿2(U))� 𝑏𝑢𝑡 𝑡 𝑖𝑠 𝑎𝑛 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑠𝑜 max 0≤𝑡≤𝑇 (‖𝑢𝑚′ (𝑡)‖L2(U) 2 + ‖𝑢m(t)‖𝐻01(𝑈)) ≤ ∁�‖𝜓‖L2(U) 2 + ‖𝜑‖𝐻01(𝑈) 2 + ‖𝑓‖𝐿2(0,T;𝐿2(U))� … (2.4) 𝑁𝑜𝑤 𝐿𝑒𝑡 𝑣 ∈ 𝐻01(𝑈) 𝑎𝑛𝑑 ‖𝑣‖𝐻01(𝑈) ≤ 1. 𝑊𝑒 𝑤𝑟𝑖𝑡𝑒 𝑣 = 𝑣1 + 𝑣2 , where 𝑣1 ∈ 𝑠𝑝𝑎𝑛{𝑤𝑘 }𝑘=1 𝑚 𝑎𝑛𝑑 (𝑣2, 𝑤𝑘) = 0. 𝑊𝑒 ℎ𝑎𝑣𝑒 〈𝑢𝑚″ , 𝑣〉 = (𝑢𝑚″ , 𝑣) = �� 𝑤𝑘 𝑑𝑚𝑘 ″(𝑡), 𝑚 𝑘=1 𝑣1 + 𝑣2� = �� 𝑤𝑘 𝑑𝑚𝑘 ″(𝑡), 𝑚 𝑘=1 𝑣1� + �� 𝑤𝑘 𝑑𝑚𝑘 ″(𝑡), 𝑚 𝑘=1 𝑣2� (𝑢𝑚″ , 𝑣1) = (𝑓, 𝑣1) − 𝐵[𝑢𝑚, 𝑣, ; 𝑡] 𝑏𝑢𝑡 (𝑓, 𝑣1) ≤ ‖𝑓‖𝐿2(0,T;𝐿2(U)) ‖𝑣1‖L2(U) 2 ≤ ‖𝑓‖𝐿2(0,T;𝐿2(U)) So 𝐵[𝑢𝑚, 𝑣; 𝑡] ≤ ‖𝑢𝑚(𝑡)‖𝐻01(𝑈)‖𝑣1‖𝐻01(𝑈) ≤ ‖𝑢𝑚(𝑡)‖𝐻01(𝑈). |𝐵| ≤ ∁‖𝑢m(t)‖𝐻01(𝑈) and we have |(𝑓, 𝑣1)| ≤ ‖𝑓‖𝐿2(0,T;𝐿2(U)) So | < 𝑢𝑚″ , 𝑣 > | ≤ |(𝑓, 𝑣1)| + |𝐵| ≤ ∁(‖𝑓‖𝐿2(0,T;𝐿2(U)) + ‖𝑢m(t)‖𝐻01(𝑈)) By the definition of ‖𝑢𝑚″ 𝑢(t)‖𝐻−1(𝑈) = max‖𝑣‖≤1 | < 𝑢𝑚″ , 𝑣 > |. �𝑢𝑚″ �𝐻−1(𝑈) ≤ ∁�‖𝑓‖𝐿2(0,T;𝐿2(U)) + ‖𝑢m(t)‖𝐻01(𝑈)�. So � �𝑢𝑚″ �𝐻−1(𝑈)𝑑𝑡 ≤ 𝑇 0 𝐶 � �‖𝑓‖2𝐿2(0,𝑇;𝐿2(𝑈)) + ‖𝑢𝑚(𝑡)‖𝐻01(𝑈)�𝑑𝑡 𝑇 0 ≤ 𝐶‖𝑓‖2𝐿2(0,T;𝐿2(U)) + � ‖𝑢m(t)‖ 2 𝐻0 1(𝑈)𝑑𝑡 𝑇 0 226 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 ≤ 𝐶‖𝑓‖𝐿2(0,T;𝐿2(U)) + max‖𝑢m(t)‖2𝐻01(𝑈) � 𝑑𝑡 𝑇 0 From (2.4) ≤ 𝐶 �‖𝜑‖2𝐻01(𝑈) + ‖𝜓‖2𝐿2(0,T;𝐿2(U)) + ‖𝑓‖ 2 𝐿2(0,T;𝐿2 (U))�. So �𝑢𝑚″ �𝐿2�0,𝑇,𝐻−1� ≤ 𝐶 �‖𝜑‖ 2 𝐻0 1(𝑈) + ‖𝜓‖ 2 𝐿2 (0,T;𝐿2(U)) + ‖𝑓‖ 2 𝐿2(0,T;𝐿2 (U))� … (2.5) (2.4) + (2.5) = (2.3) Conclusion In our main result, we gave a good estimate to measure the regularity by using some Soblev spaces(integral spaces) which give the property of continuity to the function u which is the solution of our problem and the continuity of first derivative, second derivative more regular of course in integral sense. Also, u=0 on the boundary of the domain U is generalized sense, this means when we deal with an integral space we say that the trace of u is equal to zero on the boundary of the domain U. References 1.Hirosawa, F., (2007), On the Asymptotic Behavior of the Energy for the Wave Equations with Time Depending Coefficients, Math. Ann., 339, 819-839. 2.Faciu ,C. and Simion, N. (2000), Energy Estimates and Uniqueness of the Weak Solutions of Initial – Boundary Value Problems for Semilinear Hyperbolic Systems, Z. Angew. Math., 51, 792-805. 3.Shigeo, T. (2007), Energy Estimate for Wave Equations with Coefficients in Some Besov Type Class, Electronic Journal of Equations, 2007(85)1-12. 4.Jaipong, K. (2011), Exponential Decay for Nonlinear Damped Equation of Suspended StringProc. Of CSIT, 1, 309-313. 5.Michael, R. and Robert, C. (2000), An Introduction to Partial Differential Equations,Springer. 6.Lawrence, C. Evans, (1997), Partial Differential Equations", American Mathematical Society. 227 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014 ق الطاقة لمسألة الوتر ذات القیم األبتدائیة والحدودیةائطر عقیل فالح جدوع قسم الریاضیات /كلیة التربیة للعلوم الصرفة (أبن الھیثم)/جامعة بغداد 2014حزیران 22، قبل البحث في : 2013 حزیران 12استلم البحث في : الخالصة نھایتین ذات القیم االبتدائیة والحدودیة. في الحقیقة ذو النا أحد أمثلة مسائل القطوع الزائدة وھي مسألة الوتر درس ایضا (Galerkin)ق الطاقة مع طریقةائاستخدمنا طروبولف ونحن نبحث عن حل بمعنى الحل الضعیف لبعض فضاءات س لدراسة بعض الخصائص لمسألتنا مثل الوجود والوحدانیة. (Galerkin : مسأ؟لة الوتر ذات القیم االبتدائیة والحدودیة، الحل الضعیف، فضاءات سوبولف، طریقةالكلمات المفتاحیة تخمین الطاقة 228 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I2@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (2) 2014