The Finite Element Methods for a Nonlinear Problem

and Pk for K = 0,m are from C(k) class and Pk depends only on d ku dxk and a, b finite. We assume that 1. functions Pk satisfy conditions d dt [tPk(t)] ≥ Ck > 0, for all k = 0,m and there is at least one index k0 that d dt [tPk0(t)] ≥ Ck0 > 0. satisfy conditions em: oblem. pply the variational methods developed in (Peter, 2008) and the finite element methods approximation for We consider the Hilbert space H = L2(a, b) and Q = {u ∈ Cn[a, b]\u(a) = u (a) = ... = u(m−1)(a) = u(b) = u′(b) = ... = u(m−1)(b) = 0}, where n = 2m and P : Q→ L2(a, b).


Introduction
We consider the problem: and there is at least one index k 0 that d dt [tPk0(t)] ≥ C k0 > 0. satisfy conditions em: oblem.pply the variational methods developed in (Peter, 2008) and the finite element methods approximation for We consider the Hilbert space H = L 2 (a, b) and , where n = 2m and P : Q → L 2 (a, b).
Theorem 1 The operator P is potential operator with its Gateaux differential linear in h and positive definite.
Proof: There exists a functional (3) and using the formula integration by parts and the limit condition we obtain: For every u, h ∈ Q, there exists DP(u)h linear in h ] and the symmetry and the positive definiteness results respectively from: 2) has a unique solution, this is the unique solution and minimizes the functional (4), and conversely, if there is an u 0 ∈ Q that minimizes the functional (4) on Q, then u 0 is the solution of problem ( 1), ( 2) and, according to the first part of the theorem, the solution is unique.
b) The functional ( 4) is lower bounded on Q. c) Any minimized sequence for the functional (4) has limit in L 2 (a,b), (the existence of generalized solution of problem (1), (2).d) All minimized sequences have the same limit in L 2 (a, b).
Proof: The first inequality results from The second results from ( 7) and We consider H 0 the energetic space of problem (1), (2), the completion of Q by virtue of inner product ,2 (a,b) .This fact results from the inequalities ( 5) and ( 6) so Proposition 4 If condition (A) is true for all k = 0, m, then P is a strong monotone operator.
Proof: The strong monotony results from relations where Pk(0) and we have use the inequality (5).
Theorem 5 If condition (A) is true for all k = 0, m, then: a) Any minimized sequence for functional (4) has a limit in H 0 , b) All the minimized sequences for functional (4) have the same limit in H 0 , c) The limit in H 0 for any minimized sequence for functional (4) is the generalized solution of problem ( 1), (2), d) The generalized solution of problem ( 1), (2) has generalized derivatives to and including m degree.

Finite Element Approximation
We introduce on the interval [a,b] a finite element mesh consisting of nodes We introduce on the interval [a,b] a finite element mesh consisting of nodes and finite element T i = (x i , x i+1 ), i = 0, N − 1 with conditions: We consider the minimized problem: find a function u ∈ H 0 such that: and the discrete problem: find a function u h ∈ H h so that: Theorem 6 The minimization problems ( 9) and ( 10) both have one and only one solution.Their respective solution u ∈ H 0 and u h ∈ H h are also the unique solutions of the variational equations: Proof: The functional F is Gateaux differentiable and so the solution u and u h of the minimization problem ( 9) and ( 10) must satisfy relations ( 11) and ( 12) respectively.In view of the strict convexity of functional F, these relations are also sufficient for the existence of the unique solution.
Theorem 7 The discrete solutions uh are bounded independently of subspace H h .
Proof: Using the strong monotone propriety of the operator P we obtain an we use the inequality.
Taking v = θ and u h is the solution of discrete problem we obtain Theorem 8 If u h is the discrete solution of equation ( 12), the condition where σ is the greatest derivation degree that appears in definition of T, the reference finite element, then (u h ) h∈H is the minimization sequence of functional F and has an unique limit in H 0 .
Proof: The sequence (u h ) h∈H is bounded in a reflexive space W m,2 (a, b).
So, there exists a subsequence (u hi ) which weakly converges to the same elementu ∈ W m,2 (a, b).We shall prove that u in minimum point of functional (4) in W m,2 (a, b), so (u h ) h∈H is a minimized sequence.
Let be φ ∈ D(a, b), the space of testing function.By definition of the discrete problem we have, in particular ∀i ≥ 1 : Where Π hi is the operator of finite element approximation.Since the functional F is continuous and convex we have Since σ is the maximum order of derivatives occurring in interpolation than, using Theorem 2.10.3 from (Titus Petrila & Calin John Cheorghiu, 1987) and from ( 13) and ( 14) we get and summation after all T we have This last relation and the continuity of the functional (4) imply that Published by Canadian Center of Science and Education and so, using (15) it results that ∀φ ∈ D(a, b) F(u) ≤ F(φ).
The space D (a, b) being dense in W m,2 (a, b), it implies that and therefore the function u is the unique solution of minimized problem (10), and (u h ) h∈H is a minimized sequence.
Theorem 9 If ((u h ) h∈H is a minimization sequence (solution of discrete problem) and if (u h ) h∈H weakly converges to the same element u, then u ∈ H 0 and u has generalized derivatives up to the aforementioned order m inclusively.
Proof: This result is obtained from Theorem 5.
In order to have an approach similar to that of the linear case we have obtained minimization sequence (solution of discrete problem) and if lem (10), and an analogous result as Cea's Lemma.
Theorem 10 If Pk for all k = 0, m are bounded functions on [a,b], then there exists a constant C, independent of the space H h , such that: ) We subtract these equalities and we get We can obtain an error evaluation: Theorem 11 In the above conditions if u is in W m+k,2 (a,b), than there is a constant C independent of the space H h such that: = u ′ (a) = ... = u (m−1) (a) = 0, u(b) = u ′ (b) = ... = u (m−1) (b) = 0 (2) and P k for K = 0, m are from C ( k) class and P k depends only on d k u dx k and a, b finite.We assume that 1. functions P k satisfy conditions d dt [tP k (t)] ≥ C k > 0, for all k = 0, m (a,b) ≤ Ch k |u| w m+k,2 (a,b) .(17) Proof: Using the Theorem 10 we get∥ u − u h ∥ w m,2 (a,b) ≤ inf w h ∈H h ∥u − v h ∥ w m,2 (a,b) ≤ ∥ u − Π h u ∥ w m,2 (a,b) ≤ Ch k |u| w m+k,2 (a,b) .Let be w 1 , w 2 , ..., w s a basis in H h .Than we have u h = ∑ s k=0 α k w k .We solve the nonlinear system: h dx with the unknowns α 1 , α 2 , ......, α s .We shall approximate the integrals on each subinterval, so we have m