AN H 1 -GALERKIN MIXED FINITE ELEMENT METHOD FOR THE EXTENDED FISHER-KOLMOGOROV EQUATION

Transcription

1 INTERNATIONAL JOURNAL OF NUMERICAL ANALYSIS AND MODELING, SERIES B Volume 3, Number 4, Pages c 212 Institute for Scientific Computing and Information AN H 1 -GALERKIN MIXED FINITE ELEMENT METHOD FOR THE EXTENDED FISHER-KOLMOGOROV EQUATION L. JONES TARCIUS DOSS AND A. P. NANDINI Abstract. An H 1 -Galerkin mixed finite element method is applied to the extended Fisher- Kolmogorov equation by employing a splitting technique. The method described in this paper may also be considered as a Petrov-Galerkin method with cubic spline space as trial space and piecewise linear space as test space, since second derivative of a cubic spline is a linear spline. Optimal order error estimates are obtained without any restriction on the mesh. Fully discrete scheme is also discussed and optimal order estimates are obtained. The results are validated with numerical examples. Key words. Extended FisherKolmogorov (EFK) equation, Second-order splitting, H 1 - Galerkin method, Auxiliary projection, Semi and Fully discrete schemes, A priori bounds, Optimal order error estimates, Cubic B-splines. 1. Introduction. In this paper, we discuss an H 1 -Galerkin mixed finite element cubic spline approximation method for the following extended Fisher-Kolmogorov equation : (1.1) u t + γu xxxx u xx + f(u) =, < t < T, γ >, x I = (, 1); subject to the initial and boundary conditions (1.2) (1.3) where f(u) = u 3 u. Kolmogorov equation. u(, t) =, u(1, t) =, u xx (, t) =, u xx (1, t) = ; u(x, ) = g(x); When γ = in (1.1), it becomes the standard Fisher- The above extended Fisher-Kolmogorov equation has got significance, as many problems in physics related to phase transition and other bistable phenomena are mathematically modelled as equation (1.1). For the case γ >, it was first proposed by Dee and Van Saarloos [17 as a higher order model equation for physical systems that are bistable. Further, the extended Fisher-Kolmogorov equation (1.1) has a lot of applications in the theory of instability in nematic liquid crystals and traveling waves in reaction-diffusion systems as discussed in W. Zimmerman [39 and D.G. Aronson [9 respectively. Marginal stability for (1.1) is discussed in Van Saarloos [35, 36. Chaotic characteristics are studied for (1.1) in P. Coullet et al. [11. Steady state equation corresponding to (1.1) is analyzed by shooting methods in L. A. Peletier et al. [28, 29. Periodic solutions of the EFK equation are discussed in L. A. Peletier et al. [3 and Stepan Tersian et al. [33. Existence of a global attractor for (1.1) is proved in H k spaces for all k > in Hong Luo [19. As far as computational studies are concerned, there are some results in the literature related to the numerical approximations to (1.1) - (1.3). In Danumjaya et al. [14, a second order splitting combined with orthogonal spline collocation Received by the editors November 2, Mathematics Subject Classification. 65N3, 65N15, 65M6, 65M15, 35G2. 46

2 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 461 method for (1.1) is formulated, analyzed and the error bounds are obtained for semi discrete scheme. Further, using C 1 -conforming finite element method, optimal error estimates are established in Danumjaya et al. [15 for both the semi-discrete and fully-discrete cases for the extended Fisher-Kolmogorov equation in two space dimension. A Crank-Nicolson type finite difference scheme to approximate the extended Fisher-Kolmogorov equation is presented and the existence and uniqueness of the solution are discussed in Tlili et al. [34. Further, existence, uniqueness and convergence of CrankNicolson type finite difference solutions are discussed for the extended FisherKolmogorov (EFK) equation in two space dimension in Noomen Khiari et al. [25. In this article, an attempt has been made to discuss H 1 -Galerkin mixed finite element method with cubic B-splines for Extended Fisher-Kolmogorov equation (1.1) - (1.3). Related to fourth order evolution equations, in Pani and Chung [6, a C 1 - conforming finite element method is analyzed for the Rosenau equation. Numerical studies of one dimensional and multidimensional Cahn-Hilliard equation are discussed by Elliott et al. [12, 13, Danumjaya et al. [16 and Qiang and Nicolaides [31. High-order finite element methods for the Kuramoto-Sivashinsky equation are presented in Akrivis [3. Recently in Akrivis et al. [4, fully discrete schemes for a general class of dispersively modified KuramotoSivashinsky equations are presented in which linearly implicit schemes and spectral methods are used for the temporal and spatial discretizations respectively. Existence and numerical approximations of periodic solutions of semilinear fourth-order differential equations related to either extended Fisher-Kolmogorov equation or Swift-Hohenberg equation are discussed in Julia Chaparova [21. For biharmonic problem, mixed methods are discussed in A. Quarteroni [32. Further, a quadrature Galerkin method is analysed for biharmonic problem in R. Aitbayev [2. In general, the LBB (Ladyženskaja-Babuška-Brezzi) stability condition is required for the mixed finite element method, which restricts the choice of finite element spaces. In Pani, A. K., [5, an H 1 -Galerkin mixed finite element method is applied and error estimates are obtained for a parabolic partial differential equation. Further, the parabolic equation is split into two partial differential equations leading to a first order system after introducing intermediate function where C elements are used relaxing C 1 smoothness without requiring LBB condition on the finite element spaces. In recent years, substantial progress has been made in H 1 - Galerkin mixed finite element method. In Amiya K. Pani and Graeme Fairweather [7, 8, H 1 -Galerkin mixed finite element method is employed for evloution equation and also for parabolic partial integro-differential equations seperately. Subsequntly, this method is applied to reaction-diffusion equations Arul Veda Manickam et al. [1, hyperbolic equations Pani, A.K., et al. [27, nonlinear parabolic problems Neela Nataraj et al. [24 and regularized long wave equation Guo et al. [18. For more applications of this mixed formulation for pseudo-hyperbolic and for a class of heat transport equations, see Yang Liu et al. [37 and Zhaojie Zhou [38. Most recently, there are superconvergence results on H 1 -Galerkin mixed finite element method for parabolic problems Madhusmita et al. [22 and for second-order elliptic equations Madhusmita et al. [23. The aim of this paper is to analyze the H 1 -Galerkin mixed finite element method for the above extended Fisher-Kolmogorov equation after splitting (1.1) with an

3 462 J. DOSS AND A. NANDINI introduction of an intermediate function. The error estimates are obtained without imposing the quasi uniformity on the finite element spaces. Let us define the splitting of the above equation as follows: Set (1.4) u xx = v, x I. Then (1.1) becomes (1.5) u t + γv xx v + f(u) =, x I. In the present study, an H 1 -Galerkin method with cubic splines as trial functions is formulated, analysed for the coupled equations (1.4) and (1.5) and error bounds are obtained for both semi discrete and fully discrete schemes. 2. Preliminaries and Notations. In this paper, the error analysis will take place in the usual Sobolev space H m (I) defined on the domain I = (, 1). The Sobolev norms are given below: For an open interval I and a non negative integer m, and Given an integer N > 1, let v m = ( m v (i) 2 i= L 2(I) ) 1 2 H 1 (I) = { ϕ H 1 (I) : ϕ() = ϕ(1) = }. Π N : = x < x 1 <... < x N = 1, be an arbitrary partition of the interval [,1 with the property that h as N, where h = max 1 k N h k and h k = x k x k 1, k = 1, 2,...N. Let us now consider the following cubic spline space as trial space S h,3 = { ϕ C 2 (I) : ϕ Ik P 3 (I k ), k = 1, 2,...N }, where P 3 (I k ) is the space of polynomials of degree 3 defined over the sub interval I k. The corresponding space with zero Dirichlet boundary condition is denoted by Sh,3= {ϕ S h,3 : ϕ() = ϕ(1) = }. Throughout this paper, we use (u, v) to denote the L 2 inner product of any two functions u and v and u to denote the L 2 norm of the function u. We use the following results in our error analysis. Young s inequality: For a, b and ɛ > (2.1) ab a2 2ɛ + ɛ 2 b2. Interpolation inequality: If v Wp m (E) with p [1,, then there exists a constant C depending only on m such that for any δ satisfying < δ E 1, (2.2) v W i p (E) [δ C m i v W m p (E) + δ i v Lp(E) i m 1. where E denotes the length of E. For a detailed proof, one may refer Chapter 4 of Adams [1. Gronwall s lemma (Continuous case): Let f(t), g(t) and h(t) be piecewise

4 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 463 continuous non negative functions defined on an interval a t b, g(t) being non-decreasing and C a nonnegative constant. If for each t [a, b, then f(t) + h(t) g(t) + C t f(t) + h(t) g(t)e C(t a). a f(τ)dτ, Gronwall s lemma (Discrete case): Let {f n }, {g n } and {h n } be non negative sequences of real numbers. Further, let {g n } be non-decreasing, C a non-negative constant and t is positive. If f n + h n g n + (C t)f k, then k<n f n + h n g n ( e C t ). Approximation property: For each fixed h and for each v H m (I), there exists a constant C independent of h and v, such that, inf v χ j Ch 4 j v 4, j =, 1, 2. χ S h,3 Throughout this paper, C is a generic positive constant, whose dependence on the smoothness of the exact solutions u and v can easily be determined from the proofs. 3. Semi Discrete H 1 -Galerkin Formulation. For u, v H 2 (I) H 1 (I), let us define the following bilinear form A(λ : u, v) = (γu xx, v xx ) (u, v xx ) + λ(u, v) where λ > is chosen appropriately later so that A(λ : u, v) is coercive. This can easily be seen from the following: A(λ : u, u) = (γu xx, u xx ) (u, u xx ) + λ(u, u) = (γu xx, u xx ) + (u x, u x ) + λ(u, u) = γ u xx 2 + u x 2 + λ u 2 α u 2 2, where we have used integration by parts and the boundary conditions u() =, u(1) =. Further λ > is chosen in such a way that α = min(γ, 1, λ). It can also be shown that the bilinear form A(λ : u, v) is continuous. i.e., A(λ : u, v) K u 2 v 2, where K depends only on γ and λ. We now see the weak formulation and H 1 -Galerkin mixed finite element formulation for the split up equations (1.4) and (1.5) of the main equation (1.1). We obtain weak formulation for u by multiplying both sides of (1.4) with φ xx where φ H 2 (I) H 1 (I) and then integrating the resulting expression with respect to x over the interval [, 1. In a similar manner, multiplying both sides of (1.5) by φ xx with φ H 2 (I) H 1 (I), integrating with respect to x over the interval [, 1, then applying integration by parts twice for the time derivative part and then adding and subtracting λ(v, φ) in the resulting equation, we obtain the following weak formulation for v. Hence, the weak formulation corresponding to the split up

5 464 J. DOSS AND A. NANDINI equations (1.4) and (1.5) is given below: Weak Formulation: Find u, v : [, T H 2 (I) H 1 (I), such that (3.1) (u xx, φ xx ) = (v, φ xx ), φ H 2 (I) H 1 (I), u(x, ) = g(x); (v t, φ) + A(λ : v, φ) λ(v, φ) + (f(u), φ xx ) =, φ H 2 (I) (3.2) H 1 (I), v(x, ) = g xx (x). The Semi discrete H 1 -Galerkin mixed finite element formulation corresponding to the above weak formulation (3.1) and (3.2) is defined respectively as follows: The Semi discrete H 1 -Galerkin mixed finite element formulation: Find U, V Sh,3 such that (3.3) (U xx, φ hxx ) = (V, φ hxx ), φ h Sh,3, U(x, ) = g(x); (3.4) (V t, φ h ) + A(λ : V, φ h ) λ(v, φ h ) + (f(u), φ hxx ) =, φ h Sh,3, V (x, ) = g xx (x). The above formulation leads to a system of coupled equations. This method may be regarded as a Petrov-Galerkin method with cubic spline space as trial space and piecewise linear space as test space, since the second derivative of a cubic spline is a piecewise linear spline. The advantage of this method is that the size of the stiffness matrix is reduced nearly to half of the size of the matrix as in the method described in Danumjaya et al. [14. This is clearly explained in section 6 where linear fully discrete scheme is described. 4. Auxiliary projection. The error analysis is generally carried over with the help of a comparison function. In this analysis, the comparison function is the usual auxiliary projection v Sh,3 of the weak solution v onto the finite dimensional subspace Sh,3 through the elliptic part appearing in the weak formulation. Let v : [, T Sh,3 be the auxiliary projection of v defined by (4.1) A(λ : v v, φ h ) =, φ h Sh,3 and v(x, ) = g xx. We initially find the error involved in the auxiliary projection, i.e., the error between the weak solution v and the intermediate comparison function v. In the next section, we compute the error between the comparison function v and the mixed H 1 -Galerkin approximation V. Let v v = ρ. Error estimates for a similar auxiliary projection with complicated non linear terms are proved in Jones et al. [2 and Pani et al. [26. In the following lemma, we obtain the error estimates for the auxiliary projection. Lemma 4.1. There exists a constant C such that for sufficiently small h and i =, 1, 2 ρ i Ch 4 i v 4 ρ t i Ch 4 i v t 4

6 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 465 Proof: We have from (4.1), that (4.2) A(λ : ρ, φ h ) =, φ h Sh,3. Choosing φ h = ρ (v χ) for χ Sh,3, we have A(λ : ρ, ρ) = A(λ : ρ, v χ). Using coercivity and continuity of A(λ : u, v), we obtain α ρ 2 2 C ρ 2 v χ 2. Taking infimum over χ Sh,3 first and then applying the approximation property, we obtain Hence, we obtain (4.3) α ρ 2 C inf χ Sh,3 v χ 2 Ch 2 v 4. ρ 2 Ch 2 v 4. We now compute the error estimate of ρ in L 2 norm using the following duality argument. Let ψ H 4 (I) be the solution of the auxiliary problem with boundary conditions Lψ := γψ xxxx ψ xx + λψ = ρ, ψ xxx () = ψ xxx (1) = ψ xx () = ψ xx (1) =. The above problem has the regularity property It is easy to see that ψ 4 ρ A(λ : ρ, ψ) = γ(ρ xx, ψ xx ) (ρ, ψ xx ) + λ(ρ, ψ) = γ(ρ x, ψ xxx ) + γρ x ψ xx 1 (ρ, ψ xx ) + λ(ρ, ψ) = γ(ρ, ψ xxxx ) γψ xxx ρ 1 + γρ x ψ xx 1 (ρ, ψ xx ) + λ(ρ, ψ) = (ρ, γψ xxxx ψ xx ) + λ(ρ, ψ) γψ xxx ρ 1 + γρ x ψ xx 1 = (ρ, Lψ) where we have used integration by parts and the boundary conditions of the auxiliary problem. Using the above and (4.2), we have that (ρ, ρ) = (ρ, Lψ) = A(λ : ρ, ψ) = A(λ : ρ, ψ χ) for χ Sh,3. Hence, applying the continuity of A(λ : u, v), the approximation property and regularity of ψ, we obtain ρ 2 C ρ 2 ψ χ 2 i.e., ρ 2 C ρ 2 inf ψ χ 2 C ρ 2 h 2 ψ 4 Ch 2 ρ 2 ρ. χ Sh,3 Hence, (4.4) ρ Ch 2 ρ 2. On an application of (4.3) in the above, we have (4.5) ρ Ch 4 v 4.

7 466 J. DOSS AND A. NANDINI Using (4.3) and (4.5) in (2.2) with m = 2 and i = 1, we have ρ 1 C [ h 1 ρ + h ρ 2 (4.6) i.e., ρ 1 Ch 3 v 4. To obtain similar error estimates for the temporal derivative of ρ, we differentiate the projection equation with respect to time variable t. Hence, we obtain A(λ : ρ t, φ h ) =, φ h Sh,3. It can be easily verified from the above analysis that ρ t Ch 4 v t 4 ; ρ t 1 Ch 3 v t 4 ; ρ t 2 Ch 2 v t 4. Thus, we obtain the required error estimates for ρ and ρ t in L 2, H 1 and H 2 norms. 5. Error analysis of semi discrete scheme. In this section, we obtain a priori error estimate for the error between the comparison function v and the H 1 -Galerkin solution V. We also discuss the error analysis of the error between the weak solution u and its corresponding H 1 -Galerkin approximation U. For this, we first write the error equation related to the H 1 - Galerkin approximation. Subtracting (3.4) from (3.2), we obtain the following error equation: (5.1) (v t V t, φ h ) + A(λ : v V, φ h ) λ(v V, φ h ) + (f(u) f(u), φ hxx ) =, φ h Sh,3. In a similar manner, subtracting (3.3) from (3.1), we obtain (5.2) (u xx U xx, φ hxx ) = (v V, φ hxx ), φ h Sh,3. Let e 1 be the error between u and U and e 2 be that between v and V. Then we have e 1 = u U and e 2 = v V = v v + v V = ρ + ζ, where ζ = v V. Then, equation (5.1) can also be written as (ρ t + ζ t, φ h ) + A(λ : ρ + ζ, φ h ) λ(ρ + ζ, φ h ) + (f(u) f(u), φ hxx ) =. Using projection (4.2) in the above equation, we obtain (ρ t + ζ t, φ h ) + A(λ : ζ, φ h ) λ(ρ + ζ, φ h ) (5.3) + (f(u) f(u), φ hxx ) = for φ h Sh,3 i.e., (ζ t, φ h ) + A(λ : ζ, φ h ) = (ρ t, φ h ) + λ(ρ, φ h ) (5.4) + λ(ζ, φ h ) (f(u) f(u), φ hxx ) for φ h Sh,3. In a similar manner, equation (5.2) can also be written as (5.5) (u xx U xx, φ hxx ) = (ρ + ζ, φ hxx ) for φ h Sh,3. In the following lemma, We compute e 1 (x) where x is an arbitrary point in [, 1. This result is required for proving the main theorem in this section.

8 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 467 Lemma 5.1. Let u and v be the weak solution of the given problem defined as in (3.1) and (3.2) respectively. Further let U, V Sh,3 be the H 1 -Galerkin approximations of u and v as defined in (3.3) and (3.4) respectively. Then the error e 1 = u U satisfies e 1 (x) C [ h 2 e 1xx + e 2 where x is an arbitrary point in [, 1. Proof: For a given x [, 1, let Φ be an element of L 2 (I) C(I) satisfying the following auxiliary problem: Φ xx =, x I {x}, Φ() = Φ(1) =, Φ x (x) Φ + x (x) = 1. The above problem has a solution. For example, { (x 1)x, x x, Φ(x) = x(x 1), x x 1 satisfies the above differential equation and the boundary conditions. Let us define Ψ as follows: { Φxx, x I {x}, Ψ(x) = at x = x. Then Ψ = a.e. on I. We first multiply e 1 with Ψ and then integrate over I. On applying integration by parts twice, using the fact that e 1 () = e 1 (1) = and then using given jump condition, we now obtain = (e 1, Ψ) = x e 1 Ψ + 1 x e 1 Ψ = x e 1 Φ xx + 1 x e 1 Φ xx x 1 = [e 1 Φ x x e 1x Φ x + [e 1 Φ x 1 x e 1x Φ x = e 1 (x) [ Φ x (x) Φ + x (x) x 1 x e 1x Φ x e 1x Φ x x { x } 1 = e 1 (x) [e 1x Φ x e 1xx Φ + [e 1x Φ 1 x e 1xx Φ = e 1 (x) + (e 1xx, Φ) i.e., e 1 (x) = (e 1xx, Φ). We have used the zero boundary conditions and the continuity of Φ for evaluating [e 1x Φ x and [e 1xΦ 1 x. Let Φ h be the linear interpolant of Φ. Then we have, (5.6) We know that e 1 (x) = (e 1xx, Φ Φ h ) + (e 1xx, Φ h ) e 1 (x) (e 1xx, Φ Φ h ) + (e 1xx, Φ h ) T A + T B. (5.7) Φ h 1 Φ Φ h 1 + Φ 1 Ch Φ 2 + Φ 2 C Φ 2. We now compute the estimates for the terms T A and T B as follows: T A = (e 1xx, Φ Φ h ) e 1xx Φ Φ h Ch 2 e 1xx Φ 2. x

9 468 J. DOSS AND A. NANDINI Since Φ h is a linear interpolant of Φ vanishing at boundaries, Φ h can be considered as the second derivative of some χ h Sh,3. Using (5.7) and (5.2), we have T B = (e 1xx, Φ h ) = (e 2, Φ h ) C e 2 Φ h C e 2 Φ 2. For Φ satisfying the auxiliary problem, it is easy to verify that Φ 2 K where K is a constant not depending on h. Using T A and T B in (5.6), we have e 1 (x) C [ h 2 e 1xx + e 2. This completes the proof. Below, we shall discuss a priori error estimates for the semi discrete mixed H 1 - Galerkin procedure. Initially we obtain the error u U in the following lemma: error bound for ζ. Lemma 5.2. Let u and v be the weak solution of the given problem defined as in (3.1) and (3.2) respectively. Further let U, V Sh,3 be the H 1 -Galerkin approximations of u and v as defined in (3.3) and (3.4) respectively. Also, let v be the auxiliary projection of v as in (4.1). Then, we have u U 2 C [ h 8 u ρ 2 + ζ 2, where ρ = v v, ζ = v V. Proof: Choose φ h = (u U) (u χ) in (5.2) for some χ Sh,3. Then, it becomes (u xx U xx, u xx U xx ) = (u xx U xx, u xx χ xx ) + (v V, u xx U xx ) (v V, u xx χ xx ) u xx U xx 2 u xx U xx u xx χ xx + v V u xx U xx + v V u xx χ xx. Taking infimum over χ Sh,3 first and then applying the approximation property, we obtain u xx U xx 2 u xx U xx inf u xx χ xx + v V u xx U xx + v V inf χ Sh,3 χ Sh,3 u xx χ xx C [ u xx U xx h 2 u 4 + v V u xx U xx + v V h 2 u 4. Using Young s inequality, we have [ ɛ u xx U xx 2 C 2 u xx U xx ɛ h4 u ɛ 2 u xx U xx ɛ v V 2 + ɛ 2 v V 2 + h 4 1 2ɛ u 2 4 [ ( 1 1 (1 Cɛ) u xx U xx 2 C ɛ h4 u ɛ + ɛ ) v V 2. 2 Choosing ɛ > (for example ɛ = 1/2C) properly so that inequality is maintained, we obtain that (5.8) e 1xx 2 = u xx U xx 2 C [ h 4 u v V 2.

10 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 469 Now we compute the estimate of e 1 in L 2 norm. For that, we apply the following duality argument: Let Φ H 4 (I) be the solution of the auxiliary problem satisfying the boundary conditions Φ xxxx = u U = e 1, x I Φ xxx () = Φ xxx (1) =, Φ xx () = Φ xx (1) =. Then, using integration by parts and (5.5), we obtain (e 1, e 1 ) = (Φ xxxx, e 1 ) = (Φ xx, e 1xx ) = (e 1xx, Φ xx ) = (e 1xx, Φ xx χ xx ) + (e 1xx, χ xx ) = (e 1xx, Φ xx χ xx ) + (ρ + ζ, χ xx ) = (e 1xx, Φ xx χ xx ) + (ρ + ζ, χ xx Φ xx ) + (ρ + ζ, Φ xx ) for χ Sh,3. e 1 2 [ e 1xx Φ xx χ xx + ρ + ζ χ xx Φ xx + ρ + ζ Φ xx. Taking infimum over χ Sh,3 first and then applying the approximation property, we obtain e 1 2 C [ h 2 e 1xx Φ 4 + h 2 ρ + ζ Φ 4 + ρ + ζ Φ 4 C [ h 2 e 1xx + h 2 ρ + ζ + ρ + ζ Φ 4. Using the regularity condition Φ 4 e 1 of the auxiliary problem, we have that e 1 2 C [ h 2 e 1xx + h 2 ρ + ζ + ρ + ζ e 1. i.e., e 1 C [ h 2 e 1xx + h 2 ρ + ζ + ρ + ζ. i.e., e 1 C [ h 2 e 1xx + v V. Squaring both sides and using Hölder s inequality, we obtain e 1 2 C [ h 4 e 1xx 2 + v V 2. Using (5.8) in the above equation, we obtain (5.9) e 1 2 C [ h 4 (h 4 u v V 2 ) + v V 2. i.e., u U 2 = e 1 2 C [ h 8 u v V 2. i.e., u U 2 = e 1 2 C [ h 8 u ρ 2 + ζ 2. We have now computed the error bound for u U in terms of the estimates of ρ and ζ. Below, we shall obtain a priori bound for ζ. This will help us in computing error bounds for u U and v V in terms of h, since the estimates of ρ and ρ t have already been obtained. Theorem 5.1. Let u and v be the weak solution of the given problem defined as in (3.1) and (3.2) respectively. Further, let U and V Sh,3 be the H 1 -Galerkin approximations of u and v as defined in (3.3) and (3.4) respectively. Let u, v, v t

11 47 J. DOSS AND A. NANDINI L 2 (H 4 ). Then, for a sufficiently small h, we have [ v V 2 L (L 2) Ch 8 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) + v 2 L (H 4 [ ) v V 2 L 2(H 2 ) Ch 4 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 [ ) u xx U xx 2 L (L 2) Ch 4 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) + v 2 L (H 4 ) and u U 2 L (L 2) Ch 8 [ u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) + v 2 L (H 4 ) Proof: Set φ h = ζ in (5.4). Then, we have (ζ t, ζ) + A(λ : ζ, ζ) = (ρ t, ζ) + λ(ρ, ζ) + λ(ζ, ζ) (f(u) f(u), ζ xx ) (ρ t, ζ) + λ(ρ, ζ) + λ(ζ, ζ) + (f(u) f(u), ζ xx ) I 1 + I 2 + I 3 + I 4. Using coercivity of A(λ : u, v) and since (ζ t, ζ) = 1 d (ζ, ζ), we obtain 2 dt 1 d (5.1) 2 dt ζ 2 + α ζ 2 2 I 1 + I 2 + I 3 + I 4. We now estimate the terms on the right hand side of the above equation. Using Young s inequality for I 1 and I 2, we obtain I 1 = (ρ t, ζ) C 2ɛ ρ t 2 + ɛ 2 ζ 2 ; I 2 = λ(ρ, ζ) C 2ɛ ρ 2 + ɛ 2 ζ 2. It is easy to see that I 3 = λ(ζ, ζ) C ζ 2. For the computation of I 4 we now temporarily assume that U L (I) K. After having obtained the relevant error estimate, we can prove that this assumption is no longer a strong condition. With this temporary assumption and using Young s inequality, we now estimate I 4 as follows: I 4 = (f(u) f(u), ζ xx ) f(u) f(u) ζ xx C 2ɛ f(u) f(u) 2 + ɛ 2 ζ xx 2 C 2ɛ (u3 u) (U 3 U) 2 + ɛ 2 ζ xx 2 C 2ɛ (u U)(u2 + uu + U 2 1) 2 + ɛ 2 ζ xx 2 C(K ) 1 2ɛ (u U) 2 + C ɛ 2 ζ xx 2. (5.11) C(K ) 1 2ɛ e C ɛ 2 ζ xx 2 Using the estimates of I 1, I 2, I 3 and I 4 in (5.1), [ 1 d 1 2 dt ζ 2 + α ζ 2 ( 2 C ρt 2 + ρ 2) + ζ 2 + ɛ ζ 2 + C(K ) 1 2ɛ 2ɛ e C ɛ 2 ζ xx 2.

12 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 471 Using the inequalities ζ ζ 2 and ζ xx ζ 2, we obtain ( 1 d 2 dt ζ 2 + α 3Cɛ ) ζ 2 ( 2 C ρt 2 + ρ 2) + C(K ) 1 2 2ɛ 2ɛ e C ζ 2. Using the result of Lemma 5.2, we obtain ( 1 d 2 dt ζ 2 + α 3Cɛ 2 1 d 2 dt ζ 2 + ) ζ 2 2 C ( ρt 2 + ρ 2) 2ɛ + C(K ) 1 2ɛ (h8 u ρ 2 + ζ 2 ) + C ζ 2. ( α 3Cɛ ) [ 1 ζ 2 2 C(K ) 2 2ɛ (h8 u ρ t 2 + ρ 2 + ζ 2 ) + ζ 2. Now, ɛ > can be chosen appropriately (for example ɛ < 2α 1 ) in such a way 3C that the above inequality is maintained. Then, we have d (5.12) dt ζ 2 + ζ 2 2 C(K ) [ h 8 u ρ t 2 + ρ 2 + ζ 2 + C(K ) ζ 2. Integrate (5.12) with respect to time variable t from to τ with τ T, to obtain (5.13) ζ 2 + τ ζ 2 2dt C ζ() 2 + C(K ) + C(K ) τ ζ 2 dt. τ (h 8 u ρ t 2 + ρ 2 )dt From the definition of auxiliary projection, we have ζ() = v(x, ) V (x, ) =. Then an application of Gronwall s lemma in (5.13) and usage of estimates for ρ, ρ t from Lemma 4.1, we have that [ (5.14) ζ 2 L + (L 2) ζ 2 L 2(H 2 ) C(K )h 8 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ). Using the triangle inequality, estimate for ρ from lemma 4.1 and the above expression, we obtain v V 2 L (L 2) = e 2 2 L C (L 2) [ ρ 2 L + (L 2) ζ 2 L (L 2) [ (5.15) C(K )h 8 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) + v 2 L (H 4 ) and (5.16) [ v V 2 L 2(H 2 ) = e 2 2 L 2(H 2 ) C(K ) ρ 2 L 2(H 2 ) + ζ 2 L 2(H 2 ) C(K )h 4 [ u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) Applying (5.15) in (5.8), we obtain [ u xx U xx 2 = e 1xx 2 C(K )h 4 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) (5.17) + v 2 L (H 4 ). Applying (5.15) in (5.9), we obtain [ u U 2 = e 1 2 C(K )h 8 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) (5.18) + v 2 L (H 4 ).

13 472 J. DOSS AND A. NANDINI From (5.15), (5.16), (5.17)and (5.18), we obtain the required result. To complete our argument, we have to show that C can be chosen independent of K for sufficiently small h. For that, using Lemma 5.1, we have U L (I) U u L (I) + u L (I) e 1 L (I) + u L (I) C[h 2 e 1xx + e 2 + u L (I) C[h 2 u xx U xx + v V + u L (I) C(K )h 4 u 4 + u L (I) C(K )h 4 + C. Now, for sufficiently small h, the above expression can be written as U L (I) K. Once we obtain an error estimate for u xx U xx and v V, the boundedness of U L (I) can automatically be obtained. Therefore, the temporary assumption is not a strange or strong condition. Hence, C can be chosen independent of K. This completes the proof. Remark: Using the interpolation inequality (2.2) with m = 2 and i = 1, and the estimates of v V in L 2 norm and in H 2 norm, we can easily obtain the following result. [ v V 2 L 2(H 1 ) Ch6 u 2 L 2(H 4 ) + v 2 L 2(H 4 ) + v t 2 L 2(H 4 ) + v 2 L (H 4 ) 6. Error analysis of fully discrete Euler backward scheme. In this section, we see the fully discrete approximation for the split up equations (1.4) and (1.5). We retain the H 1 -Galerkin mixed method in the spatial direction and we replace the time derivative by backward finite difference in the time direction. For a time step k = T/M with M a positive integer, let t n = nk be the time levels for n =, 1, 2...M. For a given continuous function ψ, let d t ψ n+1 = (ψ n+1 ψ n )/k. Linearised fully discrete scheme: For the weak solutions v mentioned in (3.2) and u in (3.1), we consider a linearised fully discrete Euler backward approximations W and Z: {t, t 1,...t M } Sh,3 respectively defined as follows: (d t W n+1, φ h ) + A(λ : W n+1, φ h ) λ(w n+1, φ h ) + (f(z n ), φ hxx ) =, (6.1) φ h Sh,3, n =, 1, 2...M 1 with W = g xx and (6.2) (Z n+1 xx, φ hxx ) = (W n+1, φ hxx ), φ h Sh,3, n =, 1, 2...M 1 with Z = g. The scheme of numerical computation is discussed in Section 7 in detail. We can see that the the above linearised fully discrete scheme gives rise to a system of decoupled equations. Before discussing the error estimates we first obtain the error equations for the approximations of u and v. Evaluating (3.2) at t n+1 and then subtracting (6.1) from the resulting equation, we obtain (vt n+1 d t W n+1, φ h ) + A(λ : v n+1 W n+1, φ h ) λ(v n+1 W n+1, φ h ) + ( f(u n+1 ) f(z n ) (6.3) ), φ hxx =, φ h Sh,3.

14 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 473 In a similar manner, evaluating (3.1) at t n+1 and then subtracting (6.2) from the resulting equation, we obtain (6.4) (u n+1 xx Z n+1 xx, φ hxx ) = (v n+1 W n+1, φ hxx ), φ h Sh,3. Let us denote the error between u n and Z n by ε n 1 and that between v n and W n by ε n 2, i.e., ε n 1 = u n Z n and ε n 2 = v n W n. Let v n W n = v n v n + v n W n = ρ n + ζ n, where v n is defined earlier as in (4.1). Then (6.3) can be written as (6.5) (v n+1 t d t W n+1, φ h ) + A(λ : ρ n+1 + ζ n+1, φ h ) λ(ρ n+1 + ζ n+1, φ h ) But the term v n+1 t + ( f(u n+1 ) f(z n ), φ hxx ) =, φh Sh,3. d t W n+1 can be written as v n+1 t d t W n+1 = v n+1 t d t v n+1 + d t v n+1 d t W n+1 = σ n+1 + d t (ρ n+1 + ζ n+1 ), where σ n+1 = v n+1 t d t v n+1. Using the projection (4.2) in (6.5) and the above expression, we obtain the following error equation: (6.6) (σ n+1, φ h ) + (d t [ρ n+1 + ζ n+1, φ h ) + A(λ : ζ n+1, φ h ) λ(ρ n+1 + ζ n+1, φ h ) + ( f(u n+1 ) f(z n ), φ hxx ) =, φh Sh,3; i.e., (d t ζ n+1, φ h ) + A(λ : ζ n+1, φ h ) = (d t ρ n+1, φ h ) (σ n+1, φ h ) +λ(ρ n+1 + ζ n+1, φ h ) ( f(u n+1 ) f(z n ), φ hxx ), φh Sh,3. In a similar way, (6.4) can be written as (6.7) (u n+1 xx Z n+1 xx, φ hxx ) = (ρ n+1 + ζ n+1, φ hxx ), φ h Sh,3. In the following lemma, we compute ε n 1 (x) where x is an arbitrary point in [, 1, the proof of which is similar to the proof of Lemma 5.1. This result is required for the proof of the main theorem in this section. Lemma 6.1. Let u and v be the weak solution of the given problem defined as in (3.1) and (3.2). Further let Z n and W n S h,3 be fully discrete H 1 -Galerkin approximation of u and v as defined in (6.2) and (6.1) respectively. Then the error ε n 1 = u n Z n satisfies ε n 1 (x) C [ h 2 ε n 1xx + ε n 2 where x is an arbitrary point in [, 1. Below, we discuss the error analysis of the error involved in the fully discrete scheme. We first compute the error estimate u n Z n in the following lemma. Lemma 6.2. Let u and v be the weak solution of the given problem defined as in (3.1) and (3.2). Further let Z n and W n S h,3 be fully discrete H 1 -Galerkin approximation of u and v as defined in (6.2) and (6.1) respectively. Also, let v n be the auxiliary projection of v n as in (4.1). Then, we have u n Z n 2 C [ h 8 u n ρ n 2 + ζ n 2, where ρ n = v n v n, ζ n = v n W n, n =, 1, 2,..., M.

15 474 J. DOSS AND A. NANDINI Proof: In (6.4), choose φ h = (u n+1 Z n+1 ) (u n+1 χ) for some χ Sh,3. Then it becomes (u n+1 xx Zxx n+1, u n+1 xx Zxx n+1 ) = (u n+1 xx Zxx n+1, u n+1 xx χ xx ) +(v n+1 W n+1, u n+1 xx Zxx n+1 ) (v n+1 W n+1, u n+1 xx χ xx ). u n+1 xx Zxx n+1 2 u n+1 xx Zxx n+1 u n+1 xx χ xx + v n+1 W n+1 u n+1 xx Zxx n+1 + v n+1 W n+1 u n+1 xx χ xx. Taking infimum over χ Sh,3 first and then applying the approximation property, we obtain u n+1 xx Z n+1 xx 2 u n+1 xx Z n+1 xx inf χ Sh,3 u n+1 xx χ xx + v n+1 W n+1 u n+1 xx Zxx n+1 + v n+1 W n+1 inf u n+1 xx χ S h,3 C [ u n+1 xx Zxx n+1 h 2 u n v n+1 W n+1 u n+1 xx Zxx n+1 + v n+1 W n+1 h 2 u n+1 4. χ xx Using Young s inequality, we have [ ɛ u n+1 xx Zxx n+1 2 C 2 un+1 xx Zxx n ɛ h4 u n ɛ 2 un+1 xx Zxx n ɛ vn+1 W n ɛ 2 vn+1 W n h 4 1 2ɛ un [ ( 1 1 (1 Cɛ) u n+1 xx Zxx n+1 2 C ɛ h4 u n ɛ + ɛ ) v n+1 W n Choosing ɛ > properly (for example, ɛ = 1 2C ) so that inequality is maintained, we obtain (6.8) ε n+1 1xx 2 = u n+1 xx Z n+1 xx 2 C [ h 4 u n v n+1 W n+1 2. We now apply the following duality argument to obtain the estimate of ε n+1 1 in L 2 norm. Let Φ be the solution of the auxiliary problem with the boundary conditions Φ xxxx = u n+1 Z n+1 = ε n+1 1, x I Φ xxx () = Φ xxx (1) =, Φ xx () = Φ xx (1) =. Then, using integration by parts, the boundary conditions and (6.7), we obtain (ε n+1 1, ε n+1 1 ) = (Φ xxxx, ε n+1 1 ) = (Φ xx, ε n+1 1xx ) = (εn+1 1xx, Φ xx) = (ε n+1 1xx, Φ xx χ xx ) + (ε n+1 1xx, χ xx) = (ε n+1 1xx, Φ xx χ xx ) + (ρ n+1 + ζ n+1, χ xx ) = (ε n+1 1xx, Φ xx χ xx ) + (ρ n+1 + ζ n+1, χ xx Φ xx ) + (ρ n+1 + ζ n+1, Φ xx ), for χ Sh,3 ε n+1 1xx Φ xx χ xx + ρ n+1 + ζ n+1 χ xx Φ xx + ρ n+1 + ζ n+1 Φ xx

16 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 475 Taking infimum over χ Sh,3 first and then applying the approximation property, we obtain ε n C [ h 2 ε n+1 1xx Φ 4 + h 2 ρ n+1 + ζ n+1 Φ 4 + ρ n+1 + ζ n+1 Φ 4 C [ h 2 ε n+1 1xx + h2 ρ n+1 + ζ n+1 + ρ n+1 + ζ n+1 Φ 4. Using the regularity of the auxiliary problem Φ 4 ε n+1 1, we have ε n C [ h 2 ε n+1 1xx + h2 ρ n+1 + ζ n+1 + ρ n+1 + ζ n+1 ε n+1 1. i.e., ε n+1 1 C [ h 2 ε n+1 1xx + h2 ρ n+1 + ζ n+1 + ρ n+1 + ζ n+1. i.e., ε n+1 1 C [ h 2 ε n+1 1xx + ρn+1 + ζ n+1. Squaring both sides and using Hölder s inequality, we obtain ε n C [ h 4 ε n+1 1xx 2 + ρ n ζ n+1 2. Using (6.8) in the above equation, we obtain ε n C [ h 4 (h 4 u n v n+1 W n+1 2 ) + ρ n ζ n+1 2. i.e., u n+1 Z n+1 2 = ε n C [ h 8 u n ρ n ζ n+1 2, n =, 1, 2,..., M 1. Now, since u = Z, we have that (6.9) u n Z n 2 = ε n 1 2 C [ h 8 u n ρ n 2 + ζ n 2, n =, 1, 2,..., M. We have now computed the error bound of u n Z n in terms of the estimates of ρ n and ζ n at time level n. Below, we shall obtain a priori bound for ζ n. This will help us in computing the error bounds of (u n Z n ) and (v n W n ) in terms of h and the regularity condition, since the estimates of ρ n has already been obtained. Theorem 6.1. Let u and v be the weak solution of the given problem defined as in (3.1) and (3.2). Further let Z n and W n Sh,3 be the backward Euler approximation of u and v as defined in (6.2) and (6.1) respectively. Then, for a sufficiently small h, the error in the fully discrete approximation of u and v by the backward Euler scheme is given by u J+1 Z J+1 C [k h 4 u J+1 xx Z J+1 [ xx C v J+1 W J+1 C k h 2 and [ k h 4 for J =, 1, 2,..., M 1, where C is a generic constant depending only on u and v. Proof: Substituting φ h = ζ n+1 in (6.6) and using coercivity of A(λ : u, v), we obtain (d t ζ n+1, ζ n+1 ) + α ζ n (d t ρ n+1, ζ n+1 ) + (σ n+1, ζ n+1 ) + λ(ρ n+1, ζ n+1 ) + λ(ζ n+1, ζ n+1 ) + ( ) f(u n+1 ) f(z n ), ζxx n+1 (d t ζ n+1, ζ n+1 ) + α ζ n (d t ρ n+1, ζ n+1 ) + (σ n+1, ζ n+1 ) + λ(ρ n+1, ζ n+1 ) + λ(ζ n+1, ζ n+1 ) + ( ) f(u n+1 ) f(u n ), ζxx n+1 + ( ) f(u n ) f(z n ), ζxx n+1 (6.1) I1 n+1 + I2 n+1 + I3 n+1 + I4 n+1 + I5 n+1 + I6 n+1.

17 476 J. DOSS AND A. NANDINI For the first term on the left hand side, we have (d t ζ n+1, ζ n+1 ) = 1 k (ζn+1 ζ n, ζ n+1 ) = 1 k { ζn+1 2 (ζ n+1, ζ n )} 1 2 d t ζ n+1 2. where we have used Young s inequality with ɛ = 1 for (ζ n+1, ζ n ). Hence, (6.1) becomes ζ n+1 2 ζ n 2 + α ζ n I1 n+1 + I2 n+1 + I3 n+1 + I4 n+1 + I5 n+1 + I6 n+1. 2k Summing the above from n=, 1, 2, 3,...J after multiplying both sides by 2k, we obtain ζ J+1 2 ζ 2 + 2kα ζ n (6.11) 2k n= n= ( I n I n I n I n I n I n+1 6 ). We now estimate the terms on the right hand side of the above expression. For the term I1 n+1 = (d t ρ n+1, ζ n+1 ), we use Young s inequality to obtain k I1 n+1 C 2ɛ k d tρ n ɛ (6.12) 2 k ζn+1 2. We compute the error bound for I n+1 2 = (σ n+1, ζ n+1 ) as follows. Recall that σ n+1 = v n+1 t d t v n+1. Using Taylor series expansion for the first term, we obtain v n+1 t = v n t + k 1! vn tt(x, θ 1 ) for t n < θ 1 < t n+1. In a similar way, using Taylor series expansion, we have that Therefore, v n+1 = v n + k 1! vn t + k2 2! vn tt(x, θ 2 ) for t n < θ 2 < t n+1. d t v n+1 = vn+1 v n = vt n + k k 2! vn tt(x, θ 2 ) for t n < θ 2 < t n+1. Hence, we obtain Therefore, σ n+1 = kv n tt(x, θ 1 ) k 2 vn tt(x, θ 2 ) for t n < θ 1 < t n+1, t n < θ 2 < t n+1. σ n+1 k v n tt(x, θ 1 ) + k 2 vn tt(x, θ 2 ) for t n < θ 1 < t n+1, t n < θ 2 < t n+1. Hence, we have that (6.13) σ n+1 2 Ck 2 v n tt 2 L (L 2). For the term I2 n+1, using Young s inequality, we obtain I2 n+1 C 2ɛ σ n ɛ 2 ζn+1 2. On substituting (6.13) in the above, we have that (6.14) k I2 n+1 C 2ɛ k3 vtt n 2 L + k ɛ (L 2) 2 ζn+1 2.

18 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 477 Similarly, an application of Young s inequality gives the estimates of the term I3 n+1 = λ(ρ n+1, ζ n+1 ) as follows: (6.15) k I n+1 3 C 2ɛ k ρn k ɛ 2 ζn+1 2 Further, for I n+1 4 = λ(ζ n+1, ζ n+1 ), we obtain (6.16) k I4 n+1 = Ck ζ n+1 2. For the estimate of I5 n+1 = ( ) f(u n+1 ) f(u n ), ζxx n+1, we observe that, I5 n+1 = ( f(u n+1 ) f(u n ), ζxx n+1 ) C f(u n+1 ) f(u n ) ζ n+1. We now estimate the first expression appearing on the right hand side of the above expression. Using Taylor series expansion for u n+1 and then Young s inequality, we obtain as follows: f(u n+1 ) f(u n ) = [(u n+1 ) 3 u n+1 [(u n ) 3 u n = [(u n+1 ) 3 (u n ) 3 (u n+1 u n ) u n+1 u n [(u n+1 ) 2 + u n+1 u n + (u n ) 2 1 Ck u n t (x, θ 3 ) for t n < θ 3 < t n+1. We therefore have that f(u n+1 ) f(u n ) Ck u n t L (L 2). Hence the estimate of I5 n+1 can be written as I5 n+1 C 2ɛ k2 u n t 2 L + ɛ (L 2) 2 ζn+1 xx 2. i.e., k I5 n+1 C 2ɛ k3 u n t 2 L + ɛ (L 2) 2 k ζn+1 xx 2. We now temporarily assume that Z n L (I) K. After having obtained the relevant error estimate, we can prove that this assumption is no longer a strong condition. With this temporary assumption and using Young s inequality, we now estimate I n+1 6 as follows: (6.4) I n+1 6 = (f(u n ) f(z n ), ζxx n+1 ) f(u n ) f(z n ) ζxx n+1 C 2ɛ f(un ) f(z n ) 2 + ɛ 2 ζn+1 xx 2 C 2ɛ [(un ) 3 u [(Z n ) 3 Z n 2 + ɛ 2 ζn+1 xx 2 C 2ɛ (un Z n )[u n ) 2 + u n Z n + (Z n ) ɛ 2 ζn+1 xx 2 k I n+1 C(K ) 1 2ɛ (un Z n ) 2 + C ɛ 2 ζn+1 xx 2 xx 2 6 C(K ) 1 2ɛ k εn C ɛ 2 k ζn+1. xx

19 478 J. DOSS AND A. NANDINI Substituting (6.12) and (6.14) to (6.4) in (6.11), we obtain that [ ( ζ J+1 2 ζ 2 + 2k α ζ n C k d t ρ n+1 2 2ɛ n= n= ) + k 3 vtt n 2 L + k (L 2) ρ n k 3 u n t 2 L (L 2) n= +kc(k ) 1 2ɛ n= ε n C n= [ 3ɛ n= 2 k ζ n k n= ζ n ɛk n= n= ζxx n+1 2. Using the inequalities ζ n+1 ζ n+1 2, ζxx n+1 ζ n+1 2 and rearranging the terms, we obtain ζ J+1 2 ζ 2 + k(2α 5C 2 ɛ) ζ n (6.-1) [ ( C(K 1 ) k 2ɛ + k n= d t ρ n k 3 n= ρ n k 3 n= vtt n 2 L (L 2 ) n= ) u n t 2 L + k (L 2) ε n 1 2 k n= n= ζ n+1 2. Using the result of Lemma 6.2 and bringing kc(k ) ζ J+1 2 term to the left hand side and rearranging the terms, we obtain (1 kc(k )) ζ J+1 2 ζ 2 + k(2α 5C 2 ɛ) ζ n [ ( C(K 1 ) k 2ɛ d t ρ n k 3 n= + k 3 u n t 2 L + (L 2) kh8 n= n= n= vtt n 2 L (L 2 ) + k ρ n+1 2 n= n= ) u n k( ɛ ) ζ n 2. n= Choose ɛ > properly (for example, ɛ < 4α ) so that the inequality is maintained. 5C After having chosen such an ɛ, select k so that (1 kc(k )) >. Then, using discrete version of Gronwall s inequality and the fact that ζ =, we obtain [ ζ J k ζ n C(K ) k d t ρ n k 3 + k n= ρ n k 3 n= n= u n t 2 L + (L 2) kh8 n= n= n= u n 2 4. n= v n tt 2 L (L 2) The estimates of ρ n+1 and kd t ρ n+1 can be obtained from the results of Lemma 4.1 since kd t ρ n+1 = ρ n+1 ρ n. Thus and therefore, we have that (6.-6) ζ J+1 2 O [ k 3 + h 8 ; ζ J+1 O [ k h 4.

20 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 479 Using the estimate for ρ n+1 and triangle inequality, we obtain the required result for ε n+1 2 as follows: (6.-5) v J+1 W J+1 = ε J+1 2 O [ k h 4. Using (6.-5) in (6.8), we get [ u J+1 xx Zxx J+1 O k h 2 (6.-4). Further, using (6.-6) in (6.9), we obtain that [ u J+1 Z J+1 O k h 4 (6.-3). We obtain the required result from (6.-5) to (6.-3). To complete our argument, we have to show that C can be chosen independent of K for sufficiently small h. For that, using Lemma 6.1, we have Z n L (I) Z n u n L (I) + u n L (I) ε n 1 L (I) + u n L (I) C[h 2 ε n 1xx + ε n 2 + u n L (I) C[h 2 u n xx Z n xx + v n W n + u n L (I) C(K )h 4 u n 4 + u n L (I) C(K )h 4 + C. Now, for sufficiently small h, the above expression can be written as Z n L (I) K. Once we obtain an error estimate for u n xx Zxx n and v n W n, the boundedness of Z n L (I) can automatically be obtained. Therefore, the temporary assumption is not a strange or strong condition. Hence, C can be chosen independent of K. This completes the proof. Remark: From (6.-4), we also have that ζ n O [ k 2 + h 8. Using the estimate for ρ n and triangle inequality, we obtain following result. 7. Numerical Experiments. v n+1 W n O [ k 2 + h 4. In this section, we present a study on numerical implementation of the H 1 - Galerkin mixed finite element cubic spline approximation method which is discussed in this paper. Though we do not impose any assumption on the partition of the interval, for the purpose of implementation we consider a partition with uniform spacing. Given an integer N > 4, let Π N : = x < x 1 <... < x N = 1, be an arbitrary partition of the interval [,1 with the property that h as N, where h = max 1 j N h j and h j = x j x j 1, j = 1, 2,...N.

21 48 J. DOSS AND A. NANDINI The j th basis B j (x) of the cubic splines space S h,3 for j = 1,, 1, 2,, N, N + 1 is given below: if x x j 2 B j (x) = 1 6h 3 (x x j 2 ) 3 if x j 2 x x j 1 1 6h 3 ( h 3 + 3h 2 (x x j 1 ) + 3h(x x j 1 ) 2 3(x x j 1 ) 3) if x j 1 x x j 1 6h 3 ( h 3 + 3h 2 (x j+1 x) + 3h(x j+1 x) 2 3(x j+1 x) 3) if x j x x j+1 1 6h 3 (x j+2 x) 3 if x j+1 x x j+2 if x x j+2. For j = 1, and j = N, N +1, the basis functions are defined as in the above form, after extending the partition by introducing fictitious nodal points x 3, x 2, x 1 on the left hand side and x N+1, x N+2, x N+3 on the right hand side. The trial space and test space with these cubic B-Splines as basis functions are of dimension N + 3. In order to include the zero Dirichlet boundary condition in the trial and test spaces, we modify the above cubic B-splines space as follows. The j th basis B j (x) of this modified cubic splines Sh,3 for j =, 1, 2,, N is given by B (x) = B (x) B (x ) B 1 (x ) B 1(x) B 1 (x) = B 1 (x) B 1(x ) B 1 (x ) B 1(x) B j (x) = B j (x) if j = 2, 3,...N 2 B N 1 (x) = B N 1 (x) B N 1(x N ) B N+1 (x N ) B N+1(x) B N (x) = B N (x) B N(x N ) B N+1 (x N ) B N+1(x). Let us now describe the numerical scheme as follows: Express W n+1 = N j= αn+1 j B j and Z n+1 = N j= βn+1 j B j, where { B j } N j= is a basis of Sh,3. Step 1: Knowing Z n and W n, we compute W n+1 (i.e., α n+1 j, j =, 1,... N) using equation (6.1) as follows: N j= [ ( B j, B i ) + ka(λ, B j, B i ) kλ( B j, B i ) α n+1 j = (W n, B i ) k(f(z n ), B ixx ), i =, 1,...N. Step 2: With the recent value of W n+1, we solve for Z n+1 (i.e., β n+1 j, j =, 1,... N) using equation (6.2) as follows: N ( B jxx, B ixx )β n+1 = (W n+1, B ixx ), i =, 1,...N. j= j

22 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 481 The above two steps can equivalently be posed as solving the following systems of linear equations: We first solve the system Bα = C for α = (α n+1, α1 n+1,...α n+1 )T where B = {b ij } N i,j= with N and C = (c, c 1...c N ) T with We then solve the second system b ij = ( B j, B i ) + ka(λ, B j, B i ) kλ( B j, B i ) c i = (W n, B i ) k(f(z n ), B ixx ). Dβ = δ for β = (β n+1, β1 n+1,...β n+1 N )T, where D = {d ij } N i,j= with d ij = ( B jxx, B ixx ) and δ = (δ, δ 1...δ N ) T with δ i = (W n+1, φ ixx ). In the implementation, the size of the combined linear system is (8n + 4) in Danumjaya et al. [14 where as the size of the decoupled system in the present method is (n + 1) each (i.e. totally 2n + 2). Example 1: We consider the following problem for the extended Fisher- Kolmogorov equation (7.-6) u t + γu xxxx u xx + f(u) =, < t < T, x I = (, 1); subject to the initial and boundary conditions u(, t) =, u(1, t) =, u xx (, t) =, u xx (1, t) = ; u(x, ) = sin(πx); where f(u) = u 3 u. The profile of the approximate solution of equation (7.-6) for γ = 1 is given in the following figures Figure 1 and Figure 2. It can be concluded that the numerical approximations at different time levels are having similar pattern of respective initial conditions. Example 2: We now consider the following non homogenous extended Fisher- Kolmogorov equation u t + γu xxxx u xx + f(u) = φ(x, t), < t < T, x I = (, 1); with initial condition and boundary conditions where u(x, ) = sin(2πx) u(, t) =, u(1, t) =, u xx (, t) =, u xx (1, t) = ; φ(x, t) = e t sin(2πx) [ e 2t sin 2 (2πx) π 4 + 4π 2 and f(u) = u 3 u (Noomen et al. [25). The exact solution of the above problem for γ = 1 is u(x, t) = e t sin(2πx). It has been theoretically proved that the error of this approximation is of order (k h 4 ).

23 482 J. DOSS AND A. NANDINI W values for h=(1/1) x-values W values for h= (1/2) x-values W-values t= t=.2154 t=.439 t=.6463 t=.8618 t=.1775 t= t=.1586 t= t= t= W-values t= t=.3393 t=.6786 t=.1179 t= t= t=.2358 t= t= t= t=.3393 Figure 1. Approximate solution of v(x) = u(xx)(x) (i.e., W (x)) at different time levels taking h = 1/1 and h = 1/2 respectively. Z-values for h=(1/1) Z-values for h=(1/2) Z-values x-values t= t= t=.4389 t= t= t= t= t=.1581 t= t= t= Z-values x-values t= t=.3393 t=.6786 t=.1179 t= t=.1697 t=.2358 t= t= t=.3537 t=.3393 Figure 2. Approximate solution of u(x) (i.e., Z(x)) at different time levels taking h = 1/1 and h = 1/2 respectively. i.e., u J+1 Z J+1 C [k h 4 [ u J+1 xx Zxx J+1 C k h 2 and [ v J+1 W J+1 C k h 4 for J =, 1, 2,..., M 1,

24 H 1 -GALERKIN MIXED METHOD FOR EFK EQUATION 483 The numerical computations with number of sub intervals 1, 2,4 and 8 are considered separately. i.e., with h = 1 1, h = 1 2, h = 1 4 and h = 1 8 respectively. For each of these spacial mesh length, the corresponding time step lengths k s are taken satisfying k 3 2 = h 4. With these time ) step length choice, the error of convergence becomes O(h 4 ) instead of O (k h 4. We denote h 1 = 1 1, h 2 = 1 2, h 3 = 1 4 and h 4 = 1 8. To reach the final time value T = , the total number of required time levels are 5, 32, 22 and 128 respectively for h = 1 1, 1 2, 1 4 and 1 8. Let Zh n i and Wh n i be the approximate solution in the space Sh,3 of the exact solution u(x, t) and v(x, t) respectively at time t = t n, taking the spacial mesh length h i for i=1, 2 and 3. The order of convergence for this method for u is calculated by the formula ( ) Order = log un Z n h i Lp u n Z n h i+1 Lp /(log 2) at the n th time level t n. Similarly the order of convergence of this method for v is calculated by the formula ( ) Order = log vn W n h i Lp v n W n h i+1 Lp /(log 2) at the n th time level t n. Order of errors in L 2 and L norms are computed and tabulated as follows. Table 1: L 2 errors in Z and W at time t = N h u Z L2 order v W L2 order Table 2: L 2 errors in Z and W at time t = N h u Z L2 order v W L2 order Table 3: L errors in Z and W at time t = N h u Z L order v W L order Table 4: L errors in Z and W at time t = N h u Z L order v W L order

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