Nonconforming Mixed Finite Element Method for Nonlinear Hyperbolic Equations ()
1. Introduction
In this paper, we discuss a nonconforming mixed finite element method for the following nonlinear hyperbolic initial and boundary value problem.
(1)
In order to describe the results briefly, we suppose that Equation (1) satisfy following assumptions on the data:
1) and are smooth and there exist constants and satisfying
.
2) and are sufficiently smooth functions with bounded derivatives.
There have been many very extensive studies about this kind of hyperbolic equations. For example, [1-3] studied the linear situations and gave error estimates under semi-discrete and fully-discrete schemes by standard Galerkin methods. [4,5] considered the mixed finite element methods for linear hyperbolic equations and obtained L2 prior estimates about continuous time. In addition, [6] analyzed the mixed finite element methods for second order nonlinear hyperbolic equations. But all the above investigations are mainly about conforming situations and projections are indispensable. As we know, the nonconforming finite element methods arise because of the demands for reducing the calculation cost. [7] has pointed that the nonconforming finite element methods with degree of freedom defined on the element edges or element itself are appropriate for each degree of freedom belong to at most elements.
In the present work, we focus on the nonconforming mixed finite element approximation scheme for nonlinear hyperbolic equations. Firstly, we introduce the corresponding space and the interpolation operators. Secondly, Existence and uniqueness of the solutions to the discrete problem are proved. Finally, Priori estimates of optimal order are derived for both the displacement and the stress.
Throughout this paper, C denotes a general positive constant which is independent of, and is the diameter of the finite element K.
2. Construction of the Elements
Let be the a rectangular subdivision ofand satisfy the regular condition. For every K, let be the barycenter, the length of edges parallel to x-axis and y-axis by and. Then there exists an affine mapping where is the reference element in plane and is the edges. We define the finite element on as
,
,
.
The interpolation functions defined above are properly and can be expressed as:
,
,. And For every, the associated finite element spaces as
,
where denotes the jump of w across the boundary F, and, if. , the interpolation operators:
.
.
3. Main Results in Semi-Discrete Scheme
In this section, we will give the main results in this paper, including the existence and uniqueness of the solution to the discrete problem and priori estimates of optimal order.
Firstly, we introduce
and rewrite the Equation (1) as a system:
Secondly, for our subsequent use, we employ the classical Sobolev space with norm. When, we simply write as. Furthermore, we denote the natural inner production in by and the norm by, and let
,
.
Thus the corresponding weak formulation of Equation (1) is to find a pair of, such that
satisfying
(2)
where.
The semi-discrete mixed finite element procedure is determined:, such that
(3)
where. We define that
,
.
It can be seen that and are the norms for and, respectively.
Theorem 1. The above problem (3) has a unique solution.
Proof: Let and are bases of and, which satisfy
Then semi-discrete scheme can be rewritten as: Find and, such that for every satisfy
here
, ,
,
By the definition of the approximation spaces, we know that E is reversible and. Thus there holds that
.
Since and are Lipschitz continuous, it has a unique solution according to the theory of differential equations [8].
Lemma 1. For there hold that
Proof: Firstly, by the interpolation condition and definition, it is easy to see that and
. Secondly, for every, v is a constant, by application of Green’s formula and the interpolation definition yields that
Thus, we complete the proof of Lemma 1.
Lemma 2. [9] For, there hold that
,
Now we give the main result of this paper.
Theorem 2. Let and be the solutions of Equations (2) and (3), respectively. For , there hold that
Proof: Let,.
It is easy to see that and satisfy the following error equations
(4)
and
(5)
Using derivation about time t of Equation (4), combining Equation (5), we obtain
(6)
Choosing and in Equation (6), and integrating from 0 to t, we have
(7)
We will give the analysis result of Equation (7) in detail. Firstly, by the initial condition, it is followed that
(8)
Secondly, by Cauchy-Schwartz’s inequality and Young’s inequality, we obtain
(9)
Similarly, by the initial condition of and, we use Young’s inequality to get
(10)
Substituting the above estimates, and applying Lemma 2, we get
(11)
Then adding at both sides Equation (11), and noticing that, we obtain that
Further, using Gronwall’s inequality to yield
By the interpolation theory (see [10]), we have
.
Finally, by the triangle inequality, we complete the proof.