J Sci Comput (2012) 52:113–152 DOI 10.1007/s10915-011-9537-8
A Superconvergent Local Discontinuous Galerkin Method for Elliptic Problems Slimane Adjerid · Mahboub Baccouch
Received: 10 November 2010 / Revised: 24 June 2011 / Accepted: 4 September 2011 / Published online: 23 September 2011 © Springer Science+Business Media, LLC 2011
Abstract In this manuscript we investigate the convergence properties of a minimal dissipation local discontinuous Galerkin(md-LDG) method for two-dimensional diffusion problems on Cartesian meshes. Numerical computations show O(hp+1 ) L2 convergence rates for the solution and its gradient and O(hp+2 ) superconvergent solutions at Radau points on enriched p-degree polynomial spaces. More precisely, a local error analysis reveals that the leading term of the LDG error for a p-degree discontinuous finite element solution is spanned by two (p + 1)-degree right Radau polynomials in the x and y directions. Thus, LDG solutions are superconvergent at right Radau points obtained as a tensor product of the shifted roots of the (p + 1)-degree right Radau polynomial. For tensor product polynomial spaces, the first component of the solution’s gradient is O(hp+2 ) superconvergent at tensor product of the roots of left Radau polynomial in x and right Radau polynomial in y while the second component is O(hp+2 ) superconvergent at the tensor product of the roots of the right Radau polynomial in x and left Radau polynomial in y. Several numerical simulations are performed to validate the theory. Keywords Local discontinuous Galerkin method · Elliptic problems · Superconvergence
1 Introduction The discontinuous Galerkin (DG) finite element method was first used to solve the neutron equation [16]. Cockburn and Shu [14] extended the method to solve first-order hyperbolic partial differential equations of conservation laws. They also developed the Local Discontinuous Galerkin (LDG) method for convection-diffusion problems [15]. Consult [13] and S. Adjerid () Department of Mathematics, Virginia Polytechnic Institute and State University, Blacksburg, VA 24061, USA e-mail:
[email protected] M. Baccouch Department of Mathematics, University of Nebraska at Omaha, Omaha, NE 68182, USA
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the references cited therein for a detailed discussion of the history of DG method and a list of important citations on the DG method and its applications. DG methods allow discontinuous bases, which simplify both h-refinement (mesh refinement and coarsening) and p-refinement (method order variation). However, for DG methods to be used in an adaptive framework one needs a posteriori error estimates to guide adaptivity and stop the refinement process. The solution space consists of piecewise continuous polynomial functions relative to a structured or unstructured mesh. As such, it can sharply capture solution discontinuities relative to the computational mesh. It maintains local conservation on an elemental basis. The success of the DG method is due to the following properties: (i) does not require continuity across element boundaries, (ii) is locally conservative, (iii) is well suited to solve problems on locally refined meshes with hanging nodes, (iv) exhibits strong superconvergence that can be used to estimate the discretization error, (v) has a simple communication pattern between elements with a common face that makes it useful for parallel computation and (vi) it can handle problems with complex geometries to high order. The LDG finite element method for solving one-dimensional convection-diffusion partial differential equations was introduced by Cockburn and Shu in [15] and is based on the work by Bassi and Rebay [5]. Castillo [6] investigated the superconvergence behavior of the LDG method applied to a two-point elliptic boundary-value problem using the numerical flux proposed in [7]. He showed that on each element the p-degree LDG solution gradient is O(hp+1 ) superconvergent at the shifted roots of the p-degree Legendre polynomial. Adjerid and Klauser [3] studied the LDG method of Cockburn and Shu [15] for transient convectiondiffusion problems in one dimension. They showed that p-degree piecewise polynomial discontinuous finite element solutions of convection-dominated problems are O(hp+2 ) superconvergent at the shifted Radau points on each element. For diffusion-dominated problems, they further showed that the derivative of the LDG solution is O(hp+2 ) superconvergent at the roots of the derivative of Radau polynomial of degree p + 1. They used these results to construct asymptotically exact a posteriori finite element error estimates. Adjerid and Issaev [2] presented new superconvergence results for the LDG method applied to transient diffusion problems and investigated the effect of numerical fluxes on superconvergence. They showed that the gradient of the p-degree discontinuous finite element solution is superconvergent at the roots of the derivative of (p + 1)-degree Radau polynomial. Castillo et al. [7] presented the first a priori error analysis for the LDG method for a model elliptic problem. They considered arbitrary meshes with hanging nodes and elements of various shapes and studied general numerical fluxes. They showed that, for smooth solutions, the L2 errors in u and ∇u are of order p + 1/2 and p, respectively, when polynomials of total degree not exceeding p are used. Cockburn et al. [12] presented a superconvergence result for the LDG method for a model elliptic problem on Cartesian grids. They identified a special numerical flux for which the L2 -norms of the gradient and the potential are of orders p + 1/2 and p + 1, respectively, when tensor product polynomials of degree at most p are used. Zhang et al. [17] studied the superconvergence properties for the LDG method applied to the linear convection-diffusion equation −u + bu = f . They used the superconvergence properties of numerical fluxes at element nodes established in [9] to identify superconvergence points for the approximations of u and u . For the minimal dissipation md-LDG method, they proved that the leading terms of the discretization errors for u and u using pdegree polynomial approximations are proportional to the (p + 1)-degree right Radau and (p + 1)-degree left Radau polynomials, respectively. In particular, they showed that the pdegree discontinuous finite element solution is O(hp+2 ) superconvergent at the roots of the
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(p + 1)-degree right Radau polynomial and the derivative of the LDG solution is O(hp+2 ) superconvergent at the roots of the (p + 1)-degree left Radau polynomial. In this manuscript we investigate the superconvergence properties of the minimal dissipation local discontinuous Galerkin (md-LDG) method for two-dimensional diffusion problems on rectangular meshes. Optimal md-LDG methods for one-dimensional convectiondiffusion problems were investigated in [8] and was extended to two-dimensions on triangular meshes in [11]. In [11] the authors proved O(hp+1 ) and O(hp ) L2 convergence rates, respectively, for the potential and the flux using the complete p-degree polynomial space Pp . Here we investigate the two-dimensional md-LDG method on rectangular meshes which reduces to the one-dimensional md-LDG method proposed in [8] using the spaces Pp , the tensor product spaces Qp and the modified spaces Vp = Qp ∩ Pp+1 . Let us recall that two-dimensional LDG methods [11, 12] are obtained by introducing an auxiliary nonzero vector v = [α, β], where we assume α, β > 0. Numerical computations using the spaces Pp on Cartesian meshes yield O(hp+1 ) convergent md-LDG solution in L2 while the md-LDG gradient is O(hm ), m = p + 1/2, p = 1, 2 and m = p for p > 2. Furthermore, numerical computations exhibit optimal O(hp+1 ) convergent md-LDG solutions and gradient for both Vp and Qp spaces. In this manuscript we present a local error analysis on one element to show that the leading term of the discretization error for the discontinuous finite element solution in Vp and Qp is spanned by two (p + 1)-degree right Radau polynomials in the x and y directions, respectively. In particular, we show that the p-degree md-LDG solution is superconvergent at right Radau points obtained as a tensor product of the roots of (p + 1)-degree right Radau polynomial. Computational results for the tensor product spaces Qp further show that the error in ux is O(hp+2 ) superconvergent at the tensor product of the roots of (p + 1)-degree left Radau polynomial in x and (p + 1)-degree right Radau polynomial in y while the error in uy is O(hp+2 ) superconvergent at the tensor product of the roots of (p + 1)-degree right Radau polynomial in x and left Radau polynomial in y. These results suggest that the leading term of the md-LDG error in ux is spanned by (p + 1)-degree left Radau polynomial in x and right Radau polynomial in y while the leading term of the md-LDG error in uy is spanned by (p + 1)-degree right Radau polynomial in x and left Radau polynomial in y. Here we note that when the modified enriched spaces Vp , p > 1, the pointwise superconvergence of the solution gradient is lost. Next, we note that Pp ⊂ Vp ⊂ Qp and Vp ∪ span{x p+1 , y p+1 } = Pp+1 . The use of Vp leads to a very efficient procedure for computing an O(hp+2 ) correction to the md-LDG solution which consists of solving a local 2 × 2 problem on each element for the two missing terms x p+1 and y p+1 . This correction may be used either as an a posteriori error estimate for guiding adaptive refinement algorithms or as a correction to the md-LDG solution which is O(hp+2 ) convergent everywhere. This approach for computing O(hp+2 ) approximations is more efficient than using the space Pp+1 . We will present a detailed discussion of this approach in a forthcoming paper [1]. The spaces Qp lead to O(hp+2 ) superconvergent solutions and gradients and may be equipped with efficient a posteriori error estimation procedures for both the solution and its gradient. However, Qp may be less efficient than Pp and Vp for high degree p. For all the reasons mentioned above we recommend the use of Vp . This manuscript is organized as follows: In Sect. 2 we present a model problem and recall the LDG formulation. In Sect. 3 we present the local error analysis and new superconvergence results. In Sect. 4 we present numerical results for several two-dimensional linear diffusion problems. In Sect. 5, we conclude with a few remarks.
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2 The Local Discontinuous Galerkin Method The LDG method for one-dimensional elliptic problems was introduced by Cockburn and Shu in [15] where the first a priori error estimates were established. Later, Cockburn et al. [12] derived a priori error estimates of the LDG method on Cartesian grids for elliptic problems. Here, we consider the classical model elliptic problem on a bounded domain ⊆ R2 . Let ∂ and n, respectively, denote the domain boundary and the outward unit normal vector to ∂. Let us consider the following second-order elliptic problem −u = f (x, y),
(x, y) ∈ = (−1, 1)2 ,
(2.1a)
subject to the Dirichlet and Neumann boundary conditions u = uD ,
(x, y) ∈ ∂D ,
(2.1b)
∂u = uN , ∂n
(x, y) ∈ ∂N ,
(2.1c)
for some given functions uD and uN , where ∂ = ∂D ∪ ∂N . We assume the measure of ∂D is nonzero. In our analysis we select the boundary conditions and the source term f (x, y) such that the exact solution, u(x, y), is a smooth function. In order to construct the LDG formulation of Cockburn et al. [12], we introduce an auxiliary variable q and rewrite our elliptic model problem (2.1) as the following system of first-order equations −∇ · q = f (x, y),
(x, y) ∈ ,
(2.2a)
q = ∇u,
(x, y) ∈ ,
(2.2b)
subject to (2.1b) and (2.1c). In order to obtain the LDG weak formulation, we partition the domain = (−1, 1)2 into a Cartesian mesh consisting of N = n × n square elements ij = [xi−1 , xi ] × [yj −1 , yj ], i, j = 1, . . . , n, where xk = yk = −1 + kh, k = 0, . . . , n and the parameter h denotes the mesh size of the mesh given by h = n2 . Although the analysis is performed on square elements our results hold for quasi-uniform meshes. Let us multiply (2.2a) and (2.2b) by arbitrary smooth test functions v and r, respectively, integrate over an arbitrary element , and apply Green’s theorem to write q · ∇v dxdy = vq · n ds + f v dxdy, (2.3a)
q · r dxdy =
ur · n ds −
u∇ · r dxdy,
(2.3b)
where denotes the boundary of . Next, let Pp and Qp , respectively, denote the complete polynomial space of degree not exceeding p and the tensor product spaces consisting of polynomials where the degree in each variable does not exceed p. We also use the modified polynomial space Vp Vp = Pp+1 \ span{x p+1 , y p+1 } = Pp+1 ∩ Qp .
(2.4a)
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Global discontinuous finite element spaces for each of the above spaces are needed. For instance, for Vp , we define V p = {q ∈ [L2 ()]2 : q| ∈ Vp × Vp , ∀ ∈ Th },
(2.4b)
where Th is a rectangular mesh for . Let Pp denote the space of polynomials of degree not exceeding p in R. We note that Pp ⊂ Vp ⊂ Qp and Pp leads to O(hp+1 ) and O(hp ), respectively, 2 L -convergence rates for the solution and its gradient while both Vp and Qp yield O(hp+1 ) L2 -convergence rates for both the solution and its gradient with the solution exhibiting O(hp+2 ) superconvergence rates at Radau points. Since Qp can be significantly bigger than Vp for high values of p. We recommend Vp because of the following reasons: (i) the leading term of the error is simple to estimate by solving a 2 × 2 small problem on each element, (ii) the computed estimate is either used as an a posteriori error estimate that is combined with an adaptive refinement algorithm or may be added to correct the LDG solution in Vp which leads to an O(hp+2 ) solution everywhere [1]. A more expensive approach will be to use Pp+1 by adding 2 degrees of freedom x p+1 and y p+1 per element to Vp which yields O(hp+2 ) DG solution and O(hp+1 ) approximation of the gradient. The finite element space Vp is suboptimal for the solution u, i.e., it contains p + 1degree terms that do not contribute to the solution global convergence rate, however, these extra polynomial terms increase the convergence rate of the gradient from p to p + 1 and yield O(hp+2 ) superconvergence rates for the solution at Radau points which simplifies the a posteriori error estimation of eu described in [1]. Next, we approximate the exact solution q and u by a piecewise polynomials Q and U whose restriction to are in V p and Vp , respectively. Here U and Q are piecewise polynomials not necessarily continuous across inter-element boundaries. The discrete LDG method consists of finding Q ∈ V p and U ∈ Vp such that ˆ Q · ∇V dxdy = V Q · n ds + f V dxdy, ∀V ∈ Vp , (2.5a)
Uˆ R · n ds −
Q · R dxdy =
U ∇ · R dxdy,
∀R ∈ V p ,
(2.5b)
ˆ and Uˆ , respectively, are the discrete apfor all elements ∈ Th . The numerical fluxes Q proximations of the traces of q and u on element boundaries. ˆ and In order to complete the definition of the LDG method we need to select the fluxes Q ˆ U on . For the sake of completeness we recall the LDG method for elliptic problems introduced by Cockburn and Shu in [15]. Following [12], we use an auxiliary nonzero vector v, and define the mean values and jumps as follows: The mean values {·} and jumps J·K of a function U and a vector Q at (x, y) on an edge of are defined as 1 {U }(x, y) = (U + (x, y) + U − (x, y)), 2
1 {Q}(x, y) = (Q+ (x, y) + Q− (x, y)), 2
JU K(x, y) = (U + (x, y) − U − (x, y))n,
JQK(x, y) = (Q+ (x, y) − Q− (x, y)) · n, (2.5d)
(2.5c)
where U + is the limit of the solution on and U − is the limit of solution of an adjacent element sharing , i.e., for (x, y) ∈ , we have U − (x, y) = lim U ((x, y) + sn), s→0+
U + (x, y) = lim U ((x, y) + sn). s→0−
(2.5e)
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Since the trial function is discontinuous, the fluxes on are replaced by the following numerical fluxes defined in [12], where on all interior edges we define ˆ = {Q} − c11 JU K − c12 JQK, Q
(2.5f)
Uˆ = {U } + c12 · JU K.
(2.5g)
The auxiliary parameter c12 is defined on each face by 1 c12 · n = sign(v · n), 2
(2.5h)
where the sign function is defined as sign(x) = 1, for x ≥ 0 and sign(x) = −1, otherwise. We note that v is an arbitrary but fixed vector with nonzero components. Note that with this choice of the parameter c12 , (2.5g) can be written as 1 1 Uˆ (x, y) = (1 + sign(v · n))U + + (1 − sign(v · n))U − . 2 2
(2.5i)
Motivated by the minimal dissipation LDG method [15] for one-dimensional diffusion problems, we construct a multi-dimensional minimal dissipation LDG method that reduces to the md-LDG defined in [15] for one-dimensional problems. This multi-dimensional minimal dissipation LDG method, also investigated in [11] on triangular meshes, applies stabilization only on the artificial outflow boundary induced by v and defined as ∂+ = {(x, y) ∈ ∂ | v · n ≥ 0}.
(2.5j)
We further define the artificial inflow boundary as ∂− = ∂\∂+ .
(2.5k)
Thus, if h is the length of an edge e ⊂ ∂+ , the stabilization parameter c11 for our LDG method is given by 1 or ph , if e ⊂ ∂D ∩ ∂+ , c11 = (2.5l) 0, otherwise. In summary, if is an arbitrary element and the components of v are positive, then, on vertical edges Uˆ is the trace of U from the left, and on horizontal edges Uˆ is the trace of U ˆ is the trace of Q from the right on vertical edges and the trace of from below. Similarly, Q Q from above on horizontal edges. Thus, the numerical fluxes on interior edges are denoted ˆ = Q+ . Q
(2.5m)
uD , (x, y) ∈ ∂D , Uˆ = U − , (x, y) ∈ ∂N ,
(2.5n)
Uˆ = U −
and
On boundary edges e ⊂ ∂, we use
and
⎧ + ⎨Q , ˆ = Q− − c11 (U − − uD )n, Q ⎩ uN n,
(x, y) ∈ ∂D ∩ ∂− , (x, y) ∈ ∂D ∩ ∂+ , (x, y) ∈ ∂N ,
(2.5o)
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Fig. 1 A rectangular element and the reference square
where c11 is defined in (2.5l). If Lp (ξ ) denotes the p-degree Legendre polynomial on [−1, 1], then the (p + 1)-degree right Radau polynomial + (ξ ) = Lp+1 (ξ ) − Lp (ξ ), Rp+1
−1 ≤ ξ ≤ 1,
(2.6a)
+ has p + 1 real distinct roots, −1 < ξ0+ < ξ1+ < · · · < ξp−1 < ξp+ = 1. The (p + 1)-degree left Radau polynomial − (ξ ) = Lp+1 (ξ ) + Lp (ξ ), Rp+1
−1 ≤ ξ ≤ 1,
− has p + 1 real distinct roots, ξ0− = −1 < ξ1− < ξ2− < · · · < ξp−1 < ξp− < 1. 2 In this paper we use the standard L inner product on u(x, y)v(x, y) dydx, (u, v) =
(2.6b)
(2.7)
and its induced norm ||u||L2 =
√
(u, u).
3 Local Error Analysis We will present an error analysis on the element = [0, h]2 shown in Fig. 1 where 1− , 1+ , 2− , and 2+ , respectively, denote the left, right, bottom, and top edges. The boundary of the − − ∪ 0,2 and reference element 0 = [−1, 1]2 is defined by 0 = 0− ∪ 0+ , where 0− = 0,1 + + + 0 = 0,1 ∪ 0,2 . Let the local LDG errors eu and eq be defined as eu = u − U,
eq = q − Q = [eq,1 , eq,2 ]t .
Subtract (2.5a) from (2.3a) with v = V and (2.5b) from (2.3b) with r = R to obtain the LDG orthogonality conditions for the local errors eu and eq ˆ · n ds − V (q − Q) eq · ∇V dxdy = 0, ∀V ∈ Vp , (3.1a)
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(u − Uˆ )R · n ds −
eu ∇ · R dxdy −
eq · R dxdy = 0,
∀R ∈ V p .
(3.1b)
Mapping = [0, h] × [0, h] onto 0 = [−1, 1]2 by the transformation x(ξ, h) =
h (ξ + 1), 2
y(η, h) =
h (η + 1), 2
(3.2)
and letting u(ξ, η, h) = u(x(ξ, h), y(η, h)), q(ξ, η, h) = q(x(ξ, h), y(η, h)), the LDG orthogonality condition (3.1) becomes
ˆ · n dσ − V (q − Q) 0
(u − Uˆ )R · n dσ − 0
eq · ∇V dξ dη = 0,
0
eu ∇ · R dξ dη − 0
h 2
∀V ∈ Vp ,
(3.3a)
eq · R dξ dη = 0,
∀R ∈ V p . (3.3b)
0
If the exact solution u(x, y) is analytic, the local DG solution U (x, y, h) on is also analytic with respect to x, y since U is polynomial in x and y. We further note that U is analytic with respect to h by transforming the local LDG weak problem to the reference element and solving for the finite element coefficients which are analytic functions of h. Thus, we can expand the local errors eu and eq in Maclaurin series with respect to h as eu (x(ξ, h), y(η, h), h) =
∞
Uk (ξ, η)hk ,
(3.4)
Qk (ξ, η)hk ,
(3.5)
k=0
eq (x(ξ, h), y(η, h), h) =
∞ k=0
where Uk ∈ Pk and Qk ∈ [Pk ]2 are obtained by applying the chain rule as 1 d k eu (x(ξ, 0), y(η, 0), h = 0) k! dhk 1 i j i j k−i−j ξ η ∂x ∂y ∂h eu (0, 0, 0), = i+j 2 i+j ≤k,i,j ≥0
Uk (ξ, η) =
(3.6)
and 1 d k eq (x(ξ, 0), y(η, 0), h = 0) k! dhk 1 i j i j k−i−j ξ η ∂x ∂y ∂h eq (0, 0, 0). = i+j 2 i+j ≤k,i,j ≥0
Qk (ξ, η) =
(3.7)
A direct computation reveals that Up+1 can be split as d p+1 (u − U ) 1 (x(ξ, h), y(η, h), h) = φp+1 (ξ, η) + U¯ p (ξ, η), Up+1 (ξ, η) = (p + 1)! dhp+1 h=0 (3.8a)
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where U¯ p (ξ, η) ∈ Vp and φp+1 (ξ, η) =
p
∂ p+1 u 1 (0, 0) (ξ − ξi+ ) 2p+1 (p + 1)! ∂x p+1 i=0
+
p
∂ p+1 u 1 (0, 0) (η − ξi+ ) 2p+1 (p + 1)! ∂y p+1 i=0
+ + (ξ ) + bp+1 Rp+1 (η), = ap+1 Rp+1
(3.8b)
+ where ξi+ , i = 0, 1, . . . , p, are the roots of Rp+1 (ξ ) on [−1, 1]. In the next lemmas we state and prove few preliminary results needed in our analysis. − , respectively, defined by Lemma 3.1 Let φp+1 ∈ Pp+1 and Up+1 + + φp+1 (ξ, η) = αp+1 Rp+1 (ξ ) + βp+1 Rp+1 (η), + − αp+1 Rp+1 (ξ ), on 0,2 , − (ξ, η) = Up+1 + − βp+1 Rp+1 (η), on 0,1 .
Then 0−
− Up+1 R · n dσ +
(3.9) (3.10)
φp+1 R · n dσ −
0+
φp+1 ∇ · R dξ dη = 0,
∀R ∈ V p . (3.11)
0
Proof Substituting (3.9) and (3.10) in (3.11) with R(ξ, η) = [r(ξ, η), s(ξ, η)]t , the left-hand side of (3.11) becomes −αp+1
1
+ Rp+1 (ξ )s(ξ, −1)dξ − βp+1
−1
+ αp+1 −
1
−1
−1
1
1 −1
−1
+ Rp+1 (ξ )s(ξ, 1)dξ + βp+1
1
+ Rp+1 (η)r(−1, η)dη 1
−1
+ Rp+1 (η)r(1, η)dη
+ + (ξ ) + βp+1 Rp+1 (η) rξ + sη dξ dη, αp+1 Rp+1
(3.12)
where the polynomials s(ξ, ±1) and r(±1, η) are in the space Pp . Next, we write (3.12) as I1 + I2 + I3 , where I1 = −αp+1 I2 = αp+1 I3 = −
1 −1
1 −1
1 −1
+ Rp+1 (ξ )s(ξ, −1) dξ
− βp+1
+ Rp+1 (ξ )s(ξ, 1) dξ + βp+1 1
−1
1
1 −1
+ Rp+1 (η)r(−1, η) dη,
(3.13)
+ Rp+1 (η)r(1, η) dη,
(3.14)
+ + (ξ ) + βp+1 Rp+1 (η) rξ + sη dξ dη. αp+1 Rp+1
(3.15)
−1
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+ Using the definition of Rp+1 = Lp+1 − Lp and the fact that Lp+1 is orthogonal to all polynomials in Pp , equations (3.13) and (3.14) simplify to
I1 = αp+1
1 −1
Lp (ξ )s(ξ, −1) dξ + βp+1
I2 = −αp+1
1
−1
Lp (ξ )s(ξ, 1) dξ − βp+1
1 −1
Lp (η)r(−1, η) dη,
(3.16)
Lp (η)r(1, η) dη.
(3.17)
1 −1
Similarly, we simplify (3.15) as follows 1 1
I3 = − −αp+1 Lp (ξ ) − βp+1 Lp (η) rξ + sη dξ dη −1
−1
= αp+1
1
−1
−1
+ βp+1
Lp (ξ )rξ + Lp (ξ )sη dξ dη
1
1
−1
1
−1
Lp (η)rξ + Lp (η)sη dξ dη.
(3.18)
Since rξ and sη are in Vp−1 we have
1
−1
1 −1
Hence, (3.18) becomes 1 I3 = αp+1 = αp+1 = αp+1
−1
Lp (ξ )rξ dξ dη =
1 −1
1 −1
−1
1
−1
Lp (η)sη dξ dη = 0.
Lp (ξ )
1
−1
1
−1
sη dη dξ + βp+1
1
−1
Lp (η)rξ dξ dη
1 −1
Lp (η)
1 −1
rξ dξ dη
Lp (ξ ) [s(ξ, 1) − s(ξ, −1)] dξ
+ βp+1
Lp (ξ )sη dξ dη + βp+1
1 −1
1
1 −1
Lp (η) r(1, η) − r(−1, η) dη.
Adding (3.16), (3.17), and (3.19) we get I1 + I2 + I3 = 0 which completes the proof.
(3.19)
Lemma 3.2 Let w(x) ∈ C ∞ (0, h) and πw be a p-degree polynomial that interpolates w at the roots of (p + 1)-degree right Radau polynomial on [0, h]. Let x(ξ, h) = h2 (ξ + 1) be the + (ξ ) on [−1, 1], mapping of [0, h] onto [−1, 1]. If ξi+ , i = 0, 1, . . . , p, are the roots of Rp+1 then the interpolation error can be written as w(x(ξ )) − πw(x(ξ )) =
∞
Wk (ξ )hk ,
(3.20)
k=p+1
where w(p+1) (0)
+ (ξ − ξi+ ) = cp+1 Rp+1 (ξ ), p+1 2 (p + 1)! i=0 p
Wp+1 (ξ ) =
(3.21)
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and + Wk (ξ ) = Rp+1 (ξ )rk−p−1 (ξ ),
k > p + 1,
(3.22)
with rk (ξ ) being a polynomial of degree k.
Proof Cf. [4].
± Lemma 3.3 Let u(x, y) ∈ C ∞ (0, h)2 , uD |± = u± i , i = 1, 2 and πui be p-degree polynoi mials interpolating u± i at the roots of (p + 1)-degree right Radau polynomial shifted to the edges i± , i = 1, 2. If we apply linear transformation: ξ ∈ [−1, 1] → [x(ξ, h), a] = [ h2 (ξ + 1), a], a = 0, h to map 2± to [−1, 1]. Similarly we map 1± to [−1, 1] by η ∈ [−1, 1] → [a, y(η, h)] = [a, h2 (η + 1)], a = 0, h. Then the interpolation errors can be written as
± u± 1 (y) − πu1 (y) =
∞
± U1,k (η)hk ,
U1,k (η) ∈ Pk ,
(3.23a)
± U2,k (ξ )hk ,
U2,k (ξ ) ∈ Pk .
(3.23b)
k=p+1 ± u± 2 (x) − πu2 (x) =
∞ k=p+1
Furthermore, the leading terms of the interpolation error on − and + can be written as ⎧ p ± + ∂ p+1 u 1 ⎨U1,p+1 (η) = 2p+1 (p+1)! (0, 0) i=0 (η − ξi+ ) = ap+1 Rp+1 (η), on 1± , ∂y p+1 ± Up+1 (ξ, η) = p + ∂ p+1 u 1 ⎩U ± (ξ ) = (0, 0) i=0 (ξ − ξi+ ) = bp+1 Rp+1 (ξ ), on 2± . 2,p+1 2p+1 (p+1)! ∂x p+1 (3.24) Proof By the standard interpolation error there exist s1 (x, h) and s2 (x, h) such that − u− 2 (x) − πu2 (x) =
p
∂ p+1 u 1 (s (x, h), 0) (x − xi+ ), 1 (p + 1)! ∂x p+1 i=0
x ∈ [0, h],
+ u+ 2 (x) − πu2 (x) =
p
∂ p+1 u 1 (s (x, h), h) (x − xi+ ), 2 (p + 1)! ∂x p+1 i=0
x ∈ [0, h], (3.26)
(3.25)
h(1+ξ + )
+ i where xi+ = and ξi+ , i = 0, 1, . . . , p, are the roots of right Radau polynomial Rp+1 2 in [−1, 1]. Mapping the interpolation errors (3.25) and (3.26) to [−1, 1] we write
− u− 2 (x(ξ, h)) − πu2 (x(ξ, h)) =
p
∂ p+1 u hp+1 (s (x(ξ ), h), 0) (ξ − ξi+ ), 1 2p+1 (p + 1)! ∂x p+1 i=0
ξ ∈ [−1, 1],
(3.27)
∂ p+1 u hp+1 (s2 (x(ξ ), h), h) (ξ − ξi+ ), p+1 + 1)! ∂x i=0 p
+ u+ 2 (x(ξ, h)) − πu2 (x(ξ, h)) =
ξ ∈ [−1, 1].
2p+1 (p
(3.28)
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The Maclaurin series of h are
∂ p+1 u (s (x(ξ, h), h), 0) ∂x p+1 1
and
∂ p+1 u (s (x(ξ, h), h), h) ∂x p+1 2
with respect to
∞ ∂ p+1 u ∂ p+1 u − (s (x(ξ, h), h), 0) = (0, 0) + hk U2,k (ξ ), 1 ∂x p+1 ∂x p+1 k=1
(3.29a)
∞ ∂ p+1 u ∂ p+1 u + (s (x(ξ ), h), h) = (0, 0) + hk U2,k (ξ ), 2 ∂x p+1 ∂x p+1 k=1
(3.29b)
where by the chain rule we have for k > 0 ∂ p+1 u , (s (x(ξ, h), h), 0) 1 p+1 ∂x h=0 k p+1 1 d ∂ u + U2,k (ξ ) = (s2 (x(ξ ), h), h) , k! dhk ∂x p+1 h=0
− U2,k (ξ ) =
1 dk k! dhk
are polynomials of degree k. Combining (3.27), (3.28) and (3.29) completes the proof. ± The proof for u± 1 − πu1 is similar and is omitted.
(3.29c)
3.1 Mixed Boundary Conditions Here we consider the element = [0, h]2 as shown in Fig. 1. For simplicity, we consider the special problem (2.1a) such that ∂D = ∂− and ∂N = ∂+ . We compute U − on − by interpolating uD as πuD , if (0, y) ∈ 1− , − (3.30) U (x, y) = πuD , if (x, 0) ∈ 2− , and compute Q+ · n on + by interpolating uN as πuN , if (h, y) ∈ 1+ , + Q · n(x, y) = πuN , if (x, h) ∈ 2+ ,
(3.31)
where πw is the p-degree polynomial that interpolates w at the roots of (p + 1)-degree right Radau polynomial. Lemma 3.4 If Uk ∈ Pk and Qk ∈ [Pk ]2 , k = 0, . . . , p, satisfy V Qk · n dσ − Qk · ∇V dξ dη = 0, k = 0, . . . , p, ∀V ∈ Vp ,
0−
0+
0+
(3.32)
0
U0 R · n dσ −
U0 ∇ · R dξ dη = 0,
∀R ∈ V p ,
0
Uk+1 R · n dσ −
Uk+1 ∇ · R dξ dη − 0
k = 0, . . . , p − 1, ∀R ∈ V p ,
1 2
(3.33)
Qk · R dξ dη = 0, 0
(3.34)
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125
then Uk = 0,
0 ≤ k ≤ p,
(3.35)
Qk = 0,
0 ≤ k ≤ p − 1.
(3.36)
Furthermore, if for all R ∈ V p − Up+1 R · n dσ + 0−
−
1 2
0+
Up+1 R · n dσ −
Up+1 ∇ · R dξ dη 0
Qp · R dξ dη = 0,
(3.37)
0
where + + Up+1 = αp+1 Rp+1 (ξ ) + βp+1 Rp+1 (η) + U¯ p ,
and
− Up+1
=
+ αp+1 Rp+1 (ξ ),
− on 0,2 ,
+ βp+1 Rp+1 (η),
− on 0,1 .
U¯ p ∈ Vp ,
(3.38a)
(3.38b)
Then Qp = 0.
(3.39)
Proof Testing against R = Qk in (3.34) gives 1 |Qk |2 dξ dη = Uk+1 Qk · n dσ − Uk+1 ∇ · Qk dξ dη, 2 0 0+ 0 which by Green’s theorem yields 1 2 |Qk | dξ dη = Qk · ∇Uk+1 dξ dη − Uk+1 Qk · n dσ, 2 0 0 0− Testing against V = Uk+1 in (3.32) yields Qk · ∇Uk+1 dξ dη = Uk+1 Qk · n dσ, 0
0−
k = 0, . . . , p − 1, (3.40)
k = 0, . . . , p − 1. (3.41)
k = 0, . . . , p − 1.
Substituting (3.42) into (3.41) gives 1 2 |Qk | dξ dη = Uk+1 Qk · n dσ − Uk+1 Qk · n dσ = 0, 2 0 0− 0−
(3.42)
k = 0, . . . , p − 1. (3.43)
which, in turn, leads to Qk = 0,
0 ≤ k ≤ p − 1.
(3.44)
Next, we use induction to prove that Uk = 0, 0 ≤ k ≤ p. Testing against R = [1, 1] in (3.33) establishes that U0 = 0. Now we assume that Uk = 0, 0 ≤ k ≤ p − 1 and prove that Up = 0. t
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Combining (3.34) for k = p − 1 and (3.44) yields
0+
Up R · n dσ −
Up ∇ · R dξ dη = 0,
∀R ∈ V p ,
(3.45)
0
which by Green’s formula leads to
−
0−
Up R · n dσ +
R · ∇Up dξ dη = 0,
∀R ∈ V p .
(3.46)
0
Adding (3.45) to (3.46) with R = [Up , Up ]t = Up a, a = [1, 1]t , we obtain
Up2 a · n dσ 0+
−
Up2 a · n dσ 0−
=
|a · n|Up2 dσ = 0.
(3.47)
0
This leads to Up = 0 on 0 . Combining this with (3.46) for R = ∇Up yields ∇Up = 0 on 0 which leads to Up = 0 on 0 . In order to show that Qp = 0, we substitute (3.8) in (3.37) to obtain 1 2
Qp · R dξ dη =
0
− Up+1 R · n dσ +
0−
+
0+
U¯ p R · n dσ −
0+
φp+1 R · n dσ −
φp+1 ∇ · R dξ dη 0
U¯ p ∇ · R dξ dη,
∀R ∈ V p .
(3.48)
U¯ p ∇ · R dξ dη,
∀R ∈ V p .
(3.49)
0
By Lemma 3.1 and (3.24) we write (3.48) as 1 2
Qp · R dξ dη = 0
0+
U¯ p R · n dσ −
0
Applying Green’s theorem to the last term, (3.49) can be written as 1 2
Qp · R dξ dη = − 0
0−
U¯ p R · n dσ +
R · ∇ U¯ p dξ dη,
∀R ∈ V p .
(3.50)
0
Testing against R = Qp we obtain 1 2
|Qp |2 dξ dη = − 0
0−
U¯ p Qp · n dσ +
Qp · ∇ U¯ p dξ dη.
(3.51)
0
From the assumption (3.32) for V = U¯ p , the right-hand side of (3.51) is 0 which leads to 1 2
|Qp |2 dξ dη = 0.
(3.52)
0
Thus, Qp = 0 which completes the proof. Now we are ready to state the main result of this section.
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Theorem 1 Let (u, q) and (U, Q), respectively, be the solutions of (2.2) and (2.5) such that ∂D = ∂− and ∂N = ∂+ . Then, the local finite element error can be written as eu (x(ξ, h), y(η, h), h) =
∞
Uk (ξ, η)hk ,
(3.53a)
Qk (ξ, η)hk ,
(3.53b)
+ + Up+1 (ξ, η) = ap+1 Rp+1 (ξ ) + bp+1 Rp+1 (η).
(3.53c)
k=p+1
eq (x(ξ, h), y(η, h), h) =
∞ k=p+1
where
Proof Applying the mixed boundary conditions, the DGM orthogonality conditions (3.1a) and (3.1b) become for all V ∈ Vp and R ∈ V p ˆ V eq · n dσ + V (q − Q) · n dσ − eq · ∇V dξ dη = 0, (3.54a) 0−
0+
(u − Uˆ )R · n dσ +
0−
−
h 2
0+
0
eu R · n dσ −
eu ∇ · R dξ dη 0
eq · R dξ dη = 0.
(3.54b)
0
Using (3.23a) and (3.23b) we write the interpolation errors as ∞
u − Uˆ =
Uk− hk ,
on
0− ,
where
Uk−
=
− U1,k
− on 0,1 ,
− U2,k
− on 0,2 ,
k=p+1
ˆ ·n= (q − Q)
∞
k Q+ k · nh ,
(3.55)
on 0+ ,
(3.56)
k=p+1 + ∈ Pk , i = 1, 2. where Q+ k · n|0,i Substituting the series (3.56) and (3.5) in the DGM orthogonality condition (3.54a) and the series (3.55), (3.4), and (3.5) in the DGM orthogonality condition (3.54b), and collecting terms having the same powers of h we write for all V ∈ Vp and R ∈ V p p k h V Qk · n dσ − Qk · ∇V dξ dη
0−
k=0
∞
+
h
0−
0+
+
k
k=p+1
0=
0
V Qk · n dσ +
U0 R · n dσ −
p k=1
· n dσ −
Qk · ∇V dξ dη = 0, (3.57) 0
U0 ∇ · R dξ dη 0
h
V Q+ k 0+
k 0+
Uk R · n dσ −
1 Uk ∇ · R dξ dη − 2 0
Qk−1 · R dξ dη 0
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+
∞
h
k=p+1
− 0
k 0−
Uk− R · n dσ +
0+
Uk R · n dσ
1 Uk ∇ · R − Qk−1 · R dξ dη . 2
(3.58)
Thus, we have for all V ∈ Vp and R ∈ V p
V Qk · n dσ −
0−
Qk · ∇V dξ dη = 0,
U0 R · n dσ −
0+
U0 ∇ · R dξ dη = 0,
Uk+1 R · n dσ −
1 2
Uk+1 ∇ · R dξ dη − 0
k = 0, . . . , p − 1, − Up+1 R · n dσ +
0−
−
1 2
(3.59a) (3.59b)
0
0+
k = 0, . . . , p,
0
Qk · R dξ dη = 0, 0
(3.59c)
Up+1 R · n dσ −
0+
Up+1 ∇ · R dξ dη 0
Qp · R dξ dη = 0.
(3.59d)
0
By Lemma 3.4, (3.59) leads to Uk = 0, 0 ≤ k ≤ p, Qk = 0, 0 ≤ k ≤ p, and establishes (3.53a) and (3.53b). Next the O(hp+1 ) terms lead to
0−
V Qp+1 · n dσ +
∀V ∈ Vp , 0−
− Up+1 R · n dσ +
V Q+ p+1 0+
· n dσ −
Qp+1 · ∇V dξ dη = 0, 0
(3.60a)
0+
Up+1 R · n dσ −
Up+1 ∇ · R dξ dη = 0, 0
∀R ∈ V p .
(3.60b)
Substituting (3.8) in (3.60b) leads to 0−
− Up+1 R · n dσ +
+
0+
0+
U¯ p R · n dσ −
φp+1 R · n dσ −
φp+1 ∇ · R dξ dη 0
U¯ p ∇ · R dξ dη = 0,
∀R ∈ V p ,
(3.61)
0
which by Lemma 3.1 and (3.24), we obtain 0+
U¯ p R · n dσ −
0
U¯ p ∇ · R dξ dη = 0,
∀R ∈ V p .
(3.62)
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Applying Green’s theorem we write (3.62) as ¯ − Up R · n dσ + R · ∇ U¯ p dξ dη = 0, 0−
∀R ∈ V p .
(3.63)
0
Adding (3.62) to (3.63) with R = U¯ p a, a = [1, 1]t , we obtain 2 2 ¯ ¯ Up a · n dσ − Up a · n dσ = |a · n|Up2 dσ = 0. 0+
0−
(3.64)
0
This leads to U¯ p = 0 on 0 . Testing against R = ∇ U¯ p in (3.63) yields ∇ U¯ p = 0 on 0 which, in turn, leads to U¯ p = 0. Using (3.8) we establish (3.53c). In the previous theorem we have proved that the p-degree LDG solution is O(hp+2 ) superconvergent at the roots of (p + 1)-degree right Radau polynomial (ξi+ , ξj+ ), i, j = 0, . . . , p. 3.2 Dirichlet Boundary Conditions We consider (2.1a) on = [0, h]2 subject to Dirichlet boundary conditions uD and let U − on − and U + on + be the interpolant of uD as πuD , if (0, y) ∈ 1− , πuD , if (h, y) ∈ 1+ , − + (x, y) = U (x, y) = U (3.65) πuD , if (x, 0) ∈ 2− , πuD , if (x, h) ∈ 2+ , where πw is the p-degree polynomial that interpolates w at the roots of (p + 1)-degree right Radau polynomial. Next we state and prove several preliminary lemmas needed in our analysis. ± defined by Lemma 3.5 Let φp+1 ∈ Pp+1 and Up+1 + + (ξ ) + βp+1 Rp+1 (η), φp+1 (ξ, η) = αp+1 Rp+1
− (ξ, η) Up+1
Then 0−
=
+ αp+1 Rp+1 (ξ ),
− on 0,2 ,
+ βp+1 Rp+1 (η),
− on 0,1 ,
− Up+1 R · n dσ +
0+
+ Up+1 R · n dσ −
+ (ξ, η) Up+1
=
(3.66)
+ αp+1 Rp+1 (ξ ), + βp+1 Rp+1 (η),
+ on 0,2 , + on 0,1 . (3.67)
φp+1 ∇ · R dξ dη = 0,
∀R ∈ V p . (3.68)
0
Proof Substituting (3.66) and (3.67) in (3.68) with R(ξ, η) = [r(ξ, η), s(ξ, η)]t , the lefthand side of (3.68) becomes 1 1 + + Rp+1 (ξ )s(ξ, −1) dξ − βp+1 Rp+1 (η)r(−1, η) dη −αp+1 −1
+ αp+1
−1
1 −1
+ Rp+1 (ξ )s(ξ, 1) dξ + βp+1
1 −1
+ Rp+1 (η)r(1, η) dη
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−
1
−1
1
−1
+ + (ξ ) + βp+1 Rp+1 (η) rξ + sη dξ dη. αp+1 Rp+1
(3.69)
which is exactly (3.12). The rest of the proof is the same as for Lemma 3.1 and is omitted. Lemma 3.6 If Uk ∈ Pk and Qk ∈ [Pk ]2 , k = 0, . . . , p, satisfy V Qk · n dσ − V Uk dσ − Qk · ∇V dξ dη = 0, 0+
0
0
k = 0, . . . , p, ∀V ∈ Vp , − U0 ∇ · R dξ dη = 0,
(3.70) ∀R ∈ V p ,
0
Uk+1 ∇ · R dξ dη −
− 0
1 2
(3.71)
Qk · R dξ dη = 0, 0
k = 0, . . . , p − 1, ∀R ∈ V p ,
(3.72)
then Uk = 0,
0 ≤ k ≤ p,
(3.73)
Qk = 0,
0 ≤ k ≤ p − 1.
(3.74)
Furthermore, if for all R ∈ V p 0−
− Up+1 R · n dσ +
−
1 2
0+
+ Up+1 R · n dσ −
Up+1 ∇ · R dξ dη 0
Qp · R dξ dη = 0,
(3.75)
0
where + + (ξ ) + βp+1 Rp+1 (η) + U¯ p , Up+1 = αp+1 Rp+1
U¯ p ∈ Vp ,
(3.76a)
and − Up+1
=
+ − αp+1 Rp+1 (ξ ), on 0,2 , + βp+1 Rp+1 (η),
+ Up+1
− on 0,1 ,
=
+ αp+1 Rp+1 (ξ ),
+ on 0,2 ,
+ βp+1 Rp+1 (η),
+ on 0,1 .
(3.76b)
then Qp = 0.
(3.77)
Proof By induction we will prove that Uk = 0, 0 ≤ k ≤ p and Qk = 0, 0 ≤ k ≤ p − 1. Testing against R = [ξ, 0]t in (3.71) yields U0 = 0. Testing against V = ξ in (3.70) for k = 0 leads to 2q1,0
1
−1
dη + 2q2,0
1 −1
ξ dξ − q1,0
1
−1
1
−1
ξ dξ dη = 0,
(3.78)
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131
where Q0 = [q1,0 , q2,0 ]t . Since the second and third integrals are equal to zero, (3.78) yields q1,0 = 0. Similarly, testing against V = η in (3.70) for k = 0 yields q2,0 = 0. Thus Q0 = 0. Next, we note that (3.72) for k = 0 becomes U1 ∇ · R dξ dη = 0, ∀R ∈ V p . (3.79) − 0
Writing U1 = a1 + a2 ξ + a3 η and testing against R = [ξ i , 0]t , i = 1, 2, 3 we obtain ai = 0, i = 1, 2, 3, i.e., U1 = 0. Now, we assume that Uk = 0,
0 ≤ k ≤ p − 1,
Qk = 0,
and
0 ≤ k ≤ p − 2,
(3.80)
and use induction to show that Up = 0 and Qp−1 = 0. Testing against R = Qp−1 in (3.72) for k = p − 1 and applying Green’s theorem lead to 1 2
|Qp−1 | dξ dη =
Qp−1 · ∇Up dξ dη −
2
0
0
Up Qp−1 · n dσ.
(3.81)
0
Testing against V = Up in (3.70) for k = p − 1 and using (3.80) yields
Qp−1 · ∇Up dξ dη =
0
Up Qp−1 · n dσ.
(3.82)
0
Thus, (3.81) becomes 1 2
|Qp−1 |2 dξ dη = 0,
(3.83)
Qp−1 = 0.
(3.84)
0
and yields
Combining (3.72) with k = p − 1 and (3.84) we obtain − Up ∇ · R dξ dη = 0, ∀R ∈ V p .
(3.85)
0
Applying Green’s theorem we write Up R · n dσ + − 0
R · ∇Up dξ dη = 0,
∀R ∈ V p .
(3.86)
0
Testing against R = Qp , we write (3.86) as
Up Qp · n dσ − 0
Qp · ∇Up dξ dη = 0.
(3.87)
0
Testing against V = Up in (3.70) with k = p yields
Up Qp · n dσ −
0
0+
(Up )2 dσ −
Qp · ∇Up dξ dη = 0. 0
(3.88)
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Combining (3.87) and (3.88) we obtain 0+
(Up )2 dσ = 0,
(3.89)
which leads to Up = 0 on 0+ . Adding (3.86) to (3.85) with R = Up a, a = [1, 1]t we obtain (Up )2 a · n dσ = 0.
(3.90)
Combining (3.89) and (3.90) to obtain (Up )2 a · n dσ = 0,
(3.91)
0
0−
which shows that Up = 0 on 0− . Thus, Up = 0 on 0 . Testing against R = ∇Up in (3.86) yields ∇Up = 0 on 0 which establishes that Up = 0. The final step consists of proving that Qp = 0. Substituting (3.76) in (3.75) leads to 1 − + Qp · R dξ dη = Up+1 R · n dσ + Up+1 R · n dσ − φp+1 ∇ · R dξ dη 2 0 0− 0+ 0 (3.92) U¯ p ∇ · R dξ dη, ∀R ∈ V p . − 0
By Lemma 3.5, (3.92) becomes 1 U¯ p ∇ · R dξ dη, Qp · R dξ dη = − 2 0 0
∀R ∈ V p .
(3.93)
Applying Green’s theorem and testing against R = Qp , (3.93) can be written as 1 2
U¯ p Qp · n dσ +
|Qp | dξ dη = − 2
0
0
Qp · ∇ U¯ p dξ dη.
(3.94)
0
Testing against V = U¯ p in (3.70) for k = p, we have
Qp · ∇ U¯ p dξ dη = 0
U¯ p Qp · n dσ.
(3.95)
0
Combining (3.94) and (3.95) yields 1 2
|Qp |2 dξ dη = 0,
(3.96)
0
which leads to Qp = 0 and completes the proof. Now we are ready to state the main result of this section.
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Theorem 2 Let (u, q), (U, Q), respectively, be the solutions of (2.2), (2.5) on = [0, h]2 with Dirichlet boundary condition. Then (3.53a), (3.53b), and (3.53c) hold. Proof The DGM orthogonality conditions (3.1a) and (3.1b) with Dirichlet boundary conditions become ˆ · n dσ − V eq · n dσ + V (q − Q) eq · ∇V dξ dη = 0, ∀V ∈ Vp , (3.97a)
0−
0−
0+
(u − Uˆ )R · n dσ +
0
0+
(u − Uˆ )R · n dσ −
eu ∇ · R + 0
h eq · R dξ dη = 0, 2
∀R ∈ V p ,
(3.97b)
where ˆ = Q − c11 (U − Uˆ )n, Q
c11 = 1
p . h
c11 =
or
(3.97c)
Since the proofs for c11 = 1 and c11 = p/ h are similar, we only consider the case c11 = 1. Using (3.23) we write the interpolation errors as (u − Uˆ )|± =
∞
0
Uk± hk
(3.98a)
k=p+1
where Uk± =
⎧ ± k ⎨ ∞ k=p+1 U1,k h , ⎩∞
± k k=p+1 U2,k h ,
± on 0,1 ,
(3.98b)
± on 0,2 .
ˆ on + as Using (3.97c), we write q − Q ˆ = q − Q + (u − Uˆ + U − u)n = eq + (u − Uˆ − eu )n q−Q =
∞ ∞ (Qk − Uk n)hk + Uk+ nhk . k=0
(3.99)
k=p+1
Substituting the series (3.99) and (3.5) in the DGM orthogonality condition (3.97a) and the series (3.98a), (3.4), and (3.5) in the DGM orthogonality condition (3.97b), and collecting terms having the same powers of h we write for all V and R p
h
V Qk · n dσ −
k 0
k=0
∞
+
k=p+1
− 0
h
0+
V Uk dσ −
Qk · ∇V dξ dη 0
V Qk · n dσ +
k 0
V (Uk+ 0+
U0 ∇ · R dξ dη + hk −
− Uk ) dσ −
p
k=1
Uk ∇ · R dξ dη − 0
Qk · ∇V dξ dη = 0, 0
1 2
(3.100) Qk−1 · R dξ dη
0
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+
k=p+1
h
− 0
k 0−
Uk− R · n dσ +
0+
Uk+ R · n dσ
1 Uk ∇ · R + Qk−1 · R dξ dη = 0, 2
(3.101)
where we have used n · n = 1. Setting each term of the power series zero yields the orthogonality conditions V Qk · n dσ − V Uk dσ − Qk · ∇V dξ dη = 0, 0+
0
0
k = 0, . . . , p, ∀V ∈ Vp , − U0 ∇ · R dξ dη = 0,
∀R ∈ V p ,
0
Uk+1 ∇ · R dξ dη −
− 0
(3.102a)
1 2
(3.102b)
Qk · R dξ dη = 0, 0
(3.102c) k = 0, . . . , p − 1, ∀R ∈ V p , 1 − + Up+1 R · n dσ + Up+1 R · n dσ − Up+1 ∇ · R + Qp · R dξ dη = 0, − + 2 0 0 0 ∀R ∈ V p .
(3.102d)
By Lemma 3.6, (3.102) leads to Uk = 0, Qk = 0, 0 ≤ k ≤ p, and establishes (3.53a) and (3.53b). Next the O(hp+1 ) terms lead to + V Qp+1 · n dσ + V (Up+1 − Up+1 ) dσ − Qp+1 · ∇V dξ dη = 0, 0+
0
∀V ∈ Vp , 0−
− Up+1 R · n dσ +
0+
0
+ Up+1 R · n dσ −
(3.103a)
Up+1 ∇ · R dξ dη = 0, 0
∀R ∈ V p .
(3.103b)
Next, we note by (3.8) that Up+1 can be split as φp+1 + U¯ p and prove that U¯ p = 0. Substituting (3.8) in (3.103b) leads to − + Up+1 R · n dσ + Up+1 R · n dσ − φp+1 ∇ · R dξ dη 0−
0+
U¯ p ∇ · R dξ dη = 0,
−
0
∀R ∈ V p .
(3.104)
0
Combining (3.24), (3.8) and Lemma 3.5 we write (3.104) as − U¯ p ∇ · R dξ dη = 0, ∀R ∈ V p . 0
(3.105)
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135
Applying Green’s theorem to (3.105) yields −
U¯ p R · n dσ +
0
R · ∇ U¯ p dξ dη = 0,
∀R ∈ V p ,
(3.106)
0
which, after testing against R = Qp , becomes
U¯ p Qp · n dσ −
0
Qp · ∇ U¯ p dξ dη = 0.
(3.107)
0
Testing against V = U¯ p in (3.70) for k = p yields 0
U¯ p Qp · n dσ −
U¯ p2 dσ −
0+
Qp · ∇ U¯ p dξ dη = 0.
(3.108)
0
Combining (3.107) with (3.108) we obtain 0+
U¯ p2 dσ = 0,
(3.109)
which leads to U¯ p = 0 on 0+ . Adding (3.105) to (3.106) with R = U¯ p a, a = [1, 1]t , we obtain
U¯ p2 a · n dσ = 0,
(3.110)
U¯ p2 a · n dσ = 0,
(3.111)
0
which reduces to
0−
and leads to U¯ p = 0 on 0− . Thus U¯ p = 0 on 0 . Testing against R = ∇ U¯ p in (3.106) yields ∇ U¯ p = 0 on 0 which leads to U¯ p = 0 and establishes (3.53c). ˆ = Q − p (U − Uˆ )n is the same and is omitted. The proof for the numerical flux Q h In the previous theorem we have proved that the p-degree DG solution is O(hp+2 ) superconvergent at the roots of (p + 1)-degree right Radau polynomial (ξi+ , ξj+ ), i, j = 0, . . . , p. Furthermore, numerical results suggest that when the spaces Qp are used with appropriate boundary approximations the leading term in eq may be written as Qp+1 (ξ, η) =
− + c1 Rp+1 (ξ ) + c2 Rp+1 (η) + − (ξ ) + d2 Rp+1 (η) d1 Rp+1
.
(3.112)
Now let us note that a global superconvergence error analysis is yet to be performed and will be investigated in the future. We expect that a two-dimensional version of the superconvergence results of Cheng and Shu [10] will be needed.
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J Sci Comput (2012) 52:113–152
4 Computational Examples In this section, we present several numerical examples on uniform Cartesian meshes with the spaces Pp , Vp , Qp , p ≥ 1 and v = [1, 1]t . Numerical results for c11 = p/ h and c11 = 1 are similar, thus, we present results for c11 = p/ h. We shift right and left Radau points ξk+ , k = 0, . . . , p and ξk− , k = 0, . . . , p, respectively, to an arbitrary square element to obtain xk+ , k = 0, . . . , p and xk− , k = 0, . . . , p. We compute the maximum LDG error ||eu ||∗∞ at shifted Radau points on each element and take the maximum over all elements as ||eu ||∗∞ = max max |eu (xi+ , xj+ )|.
i,j =0,...,p
The maximum LDG errors ||eq,k ||∗∞ , k = 1, 2 are computed as ||eq,1 ||∗∞ = max max |eq,1 (xi− , xj+ )|,
i,j =0,...,p
||eq,2 ||∗∞ = max max |eq,2 (xi+ , xj− )|.
i,j =0,...,p
For all examples Radau points are marked by × and level curve plots and results for eq,1 and eq,2 are identical. Thus, we show results for eu and eq,1 in tables below. Example 4.1 We consider the linear diffusion problem −u = f (x, y),
(x, y) ∈ = (−1, 1)2 ,
(4.1a)
where the artificial inflow boundary ∂− is the union of the bottom and left edges of and ∂+ = ∂ \ ∂− . We assume the Dirichlet and Neumann boundary conditions u|∂− = uD , ∂u = uN , ∂n ∂+
(4.1b) (4.1c)
and select uD , uN and f (x, y) such that the true solution is u(x, y) = ex+y .
(4.1d)
We perform several numerical tests on this example to study the effect of boundary conditions and investigate the pointwise superconvergence of local discontinuous Galerkin solutions. We begin by testing the minimal dissipation LDG method using the optimal polynomial spaces Pp , p = 1, . . . , 4 on uniform Cartesian meshes having 16, 64, 144, 256 square elements and present the true errors ||eu ||L2 and ||eq,1 ||L2 = ||eq,2 ||L2 with their orders of convergence in Table 1. We observe that ||eu ||L2 = O(hp+1 ) while ||eq,i ||L2 = O(hm ) where m = p + 1 for p = 1, 2 and m = p for p = 3, 4. Now we solve (4.1) on uniform Cartesian meshes having 16, 64, 144, 256, 400 elements and compute finite element solutions in the spaces Vp , p = 1, . . . , 4. We apply the true boundary conditions uD , uN and plot the zero-level curves of the true errors eu , eq,1 , and eq,2 , respectively, in Figs. 2, 3 and 4 where Radau points are marked by ×. We observe that the zero-level curves of the true error eu pass close to the Radau points marked by ×. Moreover, the errors shown in Table 2 indicate that the LDG error eu is O(hp+2 ) superconvergent at Radau points. For p = 1, the zero-level curves of eq,i , i = 1, 2, pass close to Radau points
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137
Table 1 L2 errors and orders of convergence for Example 4.1 on uniform meshes having N = 16, 64, 144, 256 elements using Pp , p = 1 to 4 with true boundary conditions N
p=1 ||eu ||L2
p=2 Order
||eu ||L2
p=3 Order
||eu ||L2
9.4029e−3
p=4 Order
||eu ||L2
5.8628e−4
Order
16
1.0450e−1
64
2.6817e−2
1.9623
1.2122e−3
2.9555
3.8318e−5
3.9355
9.2375e−7
4.9444
144
1.1988e−2
1.9857
3.6169e−4
2.9827
7.6676e−6
3.9680
1.2292e−7
4.9743
256
6.7587e−3
1.9920
1.5302e−4
2.9902
2.4383e−6
3.9825
2.9308e−8
4.9834
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
1.1127e−2
2.8443e−5
16
1.2947e−1
64
3.3867e−2
1.9347
1.4748e−3
2.9155
1.6932e−4
1.3918e−3 3.0392
1.1697e−6
3.7395e−7 3.9498
144
1.5378e−2
1.9472
4.4760e−4
2.9407
5.0286e−5
2.9942
5.8145e−6
3.9550
256
8.7980e−3
1.9411
1.9202e−4
2.9417
2.1221e−5
2.9990
8.9851e−5
3.9639
Fig. 2 Zero-level curves of eu for Example 4.1 on uniform meshes having N = 64 elements using Vp , p = 1 to 4 (upper left to lower right) with true boundary conditions
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Fig. 3 Zero-level curves of eq,1 for Example 4.1 on uniform meshes having N = 64 elements using Vp , p = 1 to 4 (upper left to lower right) with true boundary conditions
Table 2 Maximum errors and orders of convergence of eu and eq at Radau points for Example 4.1 on uniform meshes having N = 16, 64, 144, 256, 400 elements using Vp , p = 1 to 4 with true boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3 Order
||eu ||∗∞
||eu ||∗∞
1.0336e−4
Order
16
4.4774e−2
64
7.4005e−3
2.5970
1.3875e−4
3.7764
3.8451e−6
4.7484
8.7645e−8
5.6364
144
2.4862e−3
2.6903
2.8845e−5
3.8739
5.3660e−7
4.8569
8.3664e−9
5.7935
256
1.1336e−3
2.7300
9.3619e−6
3.9115
1.3108e−7
4.8993
1.5525e−9
5.8549
400
6.1324e−4
2.7532
3.8935e−6
3.9317
4.3705e−8
4.9222
4.1725e−10
5.8883
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
1.9012e−3
p=4 Order
1.1621e−2
4.3595e−6
16
9.2124e−2
64
4.7850e−2
0.9451
1.8757e−3
2.6313
6.9488e−5
8.6168e−4 3.6323
4.1790e−6
1.0735e−4 4.6830
144
3.6054e−2
0.6981
8.3495e−4
1.9962
1.5558e−5
3.6910
5.9400e−7
4.8116
256
2.8724e−2
0.7901
4.6997e−4
1.9977
5.2422e−6
3.7815
1.4654e−7
4.8649
400
2.3822e−2
0.8385
3.0089e−4
1.9984
2.4553e−6
3.3991
4.9167e−8
4.8942
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139
Fig. 4 Zero-level curves of eq,2 for Example 4.1 on uniform meshes having N = 64 elements using Vp , p = 1 to 4 (upper left to lower right) with true boundary conditions
Table 3 L2 errors and orders of convergence for Example 4.1 on uniform meshes having N = 16, 64, 144, 256 elements using Vp , p = 1 to 4 with true boundary conditions N
p=1 ||eu ||L2
p=2 Order
||eu ||L2
p=3 Order
||eu ||L2
Order
||eu ||L2
7.4624e−2
64
1.9180e−2
1.9600
3.8837e−4
2.9832
6.0227e−6
4.0128
7.6066e−8
5.0812
144
8.5842e−3
1.9828
1.1540e−4
2.9931
1.1877e−6
4.0040
9.8976e−9
5.0295
256
4.8432e−3
1.9895
4.8742e−5
2.9958
3.7562e−7
4.0017
2.3386e−9
5.0151
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
7.7621e−2
9.7224e−5
Order
16
16
3.0710e−3
p=4
4.2007e−3
2.5751e−6
1.7137e−4
5.7816e−6
64
2.0791e−2
1.9005
5.7220e−4
2.8760
1.1726e−5
3.8693
1.9613e−7
4.8816
144
9.6249e−3
1.8995
1.7621e−4
2.9048
2.4031e−6
3.9093
2.6682e−8
4.9197
256
5.6006e−3
1.8822
7.6326e−5
2.9083
7.7780e−7
3.9211
6.4451e−9
4.9384
140
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Fig. 5 Zero-level curves of eu for Example 4.1 on uniform meshes having N = 64 elements using Vp , p = 1 to 4 (upper left to lower right) with interpolated boundary conditions
on all elements except on those near the inflow boundary. The errors and their orders presented in Table 2 indicate that the LDG errors eq,i , i = 1, 2, are only O(hp ) which are not superconvergent at Radau points. Finally, the L2 LDG errors ||eu ||L2 , ||eq,1 ||L2 = ||eq,2 ||L2 shown in Table 3 suggest optimal O(hp+1 ) convergence rates for both u and q. In order to reduce the pollution in eq,i , i = 1, 2, shown in Figs. 3 and 4 near the inflow edges for p = 1 and guided by our theory we repeat the previous experiment with all parameters kept unchanged except for the boundary condition where we interpolate the true boundary conditions uD and uN at Radau points as in (3.30) and (3.31). From Figs. 5, 6 and 7, we observe that the zero-level curves for eu pass close to Radau points for all degrees p = 1 to 4 while the zero-level curves of eq,i , i = 1, 2, pass close to Radau points for p = 1 only. Moreover, maximum errors for eu of Table 4 indicate that the LDG error eu is O(hp+2 ) superconvergent at Radau points. The maximum errors for eq,i , i = 1, 2, indicate that the LDG errors eq,i , i = 1, 2, are only O(hp+1 ) convergent at Radau points for p = 2, 3, 4. However, for p = 1 the error eq,1 is close to O(h3 ) for coarse meshes but it deteriorate with finer meshes. The L2 errors ||eu ||L2 and ||eq,1 ||L2 = ||eq,2 ||L2 presented in Table 5 show optimal O(hp+1 ) convergence rates for u and q. Next we approximate the boundary conditions by LDG solutions of one-dimensional second-order elliptic problems on the two inflow edges of − . On the bottom edge of we solve uxx = z while on the left edge we solve uyy = g. Dirichlet boundary conditions and source terms z and g are selected such that uD is the true solution for both one-dimensional boundary problems. We apply the minimal dissipation LDG method for one-dimensional
J Sci Comput (2012) 52:113–152
141
Fig. 6 Zero-level curves of eq,1 for Example 4.1 on uniform meshes having N = 64 elements using Vp , p = 1 to 4 (upper left to lower right) with interpolated boundary conditions
Table 4 Maximum errors and orders of convergence at Radau points for Example 4.1 on uniform meshes having N = 16, 64, 144, 256, 400 elements using Vp , p = 1 to 4 with interpolated boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3 Order
||eu ||∗∞
||eu ||∗∞
1.0001e−4
Order
16
1.6058e−2
64
2.7393e−3
2.5514
1.0824e−4
3.7466
3.7353e−6
4.7429
8.7649e−8
5.6363
144
8.9990e−4
2.7454
2.2628e−5
3.8603
5.2188e−7
4.8540
8.3670e−9
5.7935
256
3.9965e−4
2.8215
7.3618e−6
3.9032
1.2756e−7
4.8974
1.5526e−9
5.8549
400
2.1099e−4
2.8626
3.0656e−6
3.9259
4.2543e−8
4.9207
4.1728e−10
5.8884
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
1.4529e−3
p=4 Order
9.8271e−3
4.3595e−6
16
1.7733e−2
64
2.5243e−3
2.8125
1.7832e−3
2.4623
6.8999e−5
7.9881e−4 3.5332
4.1434e−6
1.0630e−4 4.6812
144
8.1952e−4
2.7746
5.9846e−4
2.6927
1.5452e−5
3.6905
5.8918e−7
4.8106
256
3.6202e−4
2.8400
2.6872e−4
2.7833
5.2068e−6
3.7811
1.4538e−7
4.8642
400
2.4273e−4
1.7914
1.4283e−4
2.8323
2.2150e−6
3.8304
4.8783e−8
4.8936
142
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Fig. 7 Zero-level curves of eq,2 for Example 4.1 on uniform meshes having N = 64 elements using Vp , p = 1 to 4 (upper left to lower right) with interpolated boundary conditions
Table 5 L2 errors and orders of convergence for Example 4.1 on uniform meshes having N = 16, 64, 144, 256 elements using Vp , p = 1 to 4 with interpolated boundary conditions N
p=1 ||eu ||L2
p=2 Order
||eu ||L2
p=3 Order
||eu ||L2
||eu ||L2
7.9601e−2
64
1.9509e−2
2.0286
3.8931e−4
2.9921
6.0291e−6
4.0167
7.6861e−8
5.1142
144
8.6508e−3
2.0057
1.1553e−4
2.9961
1.1884e−6
4.0054
9.9462e−9
5.0431
256
4.8646e−3
2.0010
4.8777e−5
2.9974
3.7574e−7
4.0024
2.3453e−9
5.0221
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
7.4041e−2
9.7593e−5
Order
16
16
3.0975e−3
p=4 Order
4.0012e−3
2.6621e−6
1.6624e−4
5.7310e−6
64
1.8989e−2
1.9632
5.4143e−4
2.8856
1.1343e−5
3.8734
1.9284e−7
4.8933
144
8.5159e−3
1.9778
1.6467e−4
2.9355
2.3084e−6
3.9265
2.6115e−8
4.9311
256
4.8120e−3
1.9842
7.0382e−5
2.9548
7.4139e−7
3.9480
6.2876e−9
4.9496
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143
Table 6 Maximum errors and orders of convergence at Radau points for Example 4.1 on uniform meshes having N = 16, 64, 144, 256, 400 elements using Vp , p = 1 to 4 with LDG boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3 Order
||eu ||∗∞
1.4525e−3
p=4 Order
||eu ||∗∞
1.0002e−4
Order
16
1.6609e−2
64
2.8136e−3
2.5615
1.0823e−4
3.7463
3.7353e−6
4.7429
8.7649e−8
5.6363
144
9.2244e−4
2.7504
2.2626e−5
3.8602
5.2188e−7
4.8540
8.3671e−9
5.7934
256
4.0927e−4
2.8248
7.3614e−6
3.9031
1.2756e−7
4.8974
1.5527e−9
5.8548
400
2.1595e−4
2.8651
3.0655e−6
3.9259
4.2543e−8
4.9207
4.1732e−10
5.8881
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
9.8271e−3
4.3595e−6
16
1.7732e−2
64
2.5243e−3
2.8124
1.7832e−3
2.4623
6.8999e−5
7.9881e−4 3.5332
4.1434e−6
1.0630e−4 4.6812
144
8.1953e−4
2.7746
5.9846e−4
2.6927
1.5452e−5
3.6905
5.8918e−7
4.8106
256
3.6202e−4
2.8400
2.6872e−4
2.7833
5.2068e−6
3.7811
1.4538e−7
4.8642
400
1.9056e−4
2.8758
1.4283e−4
2.8323
2.2150e−6
3.8304
4.8783e−8
4.8937
Table 7 L2 errors and orders of convergence for Example 4.1 on uniform meshes having N = 16, 64, 144, 256 elements using Vp , p = 1 to 4 with LDG boundary conditions N
p=1 ||eu ||L2
p=2 Order
||eu ||L2
p=3 Order
||eu ||L2
3.0974e−3
p=4 Order
||eu ||L2
9.7586e−5
Order
16
9.5752e−3
64
2.4811e−3
1.9483
3.8931e−4
2.9921
6.0291e−6
4.0167
7.6111e−8
5.0831
144
1.1185e−3
1.9650
1.1553e−4
2.9961
1.1884e−6
4.0054
9.9006e−9
5.0303
256
6.3381e−4
1.9743
4.8777e−5
2.9974
3.7574e−7
4.0024
2.3391e−9
5.0154
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
4.0021e−3
2.5800e−6
16
8.3399e−3
64
2.3572e−3
1.8230
5.4147e−4
2.8858
1.1344e−5
1.6627e−4 3.8735
1.9248e−7
5.6839e−6 4.8841
144
1.0848e−3
1.9140
1.6468e−4
2.9356
2.3085e−6
3.9266
2.6083e−8
4.9295
256
6.2023e−4
1.9433
7.0384e−5
2.9548
7.4140e−7
3.9481
6.2760e−9
4.9518
problems [15] to obtain the LDG boundary conditions U (x, −1), −1 ≤ x ≤ 1 on the bottom edge and U (−1, y), −1 ≤ y ≤ 1 on the left edge. We approximate the Neumann boundary condition by Q(x, 1), −1 ≤ x ≤ 1 on the top edge and Q(1, y), −1 ≤ y ≤ 1 on the right edge using the interpolation of uN at the roots of (p + 1)-degree right Radau polynomial. We solve (4.1) and show the maximum errors at Radau points and their orders in Table 6. We observe that eu is still O(hp+2 ) superconvergent at Radau points and eq,i , i = 1, 2, is close to O(h3 ) for p = 1 with rates converging to 3 under h-refinement. Thus, for better superconvergence rates for eq , p = 1, we recommend the LDG boundary approximation. We observe that the errors of Table 7 still show optimal O(hp+1 ) convergence rates for u and q.
144
J Sci Comput (2012) 52:113–152
Table 8 L2 errors and orders of convergence for Example 4.1 using Qp , p = 1 to 4 with LDG boundary conditions N
p=1 ||eu ||L2
p=2 Order
||eu ||L2
p=3 Order
||eu ||L2
3.0706e−3
p=4 Order
||eu ||L2
9.4346e−5
Order
16
9.5752e−3
64
2.4811e−3
1.9483
3.8842e−4
2.9828
5.9748e−6
3.9810
7.4041e−8
4.9803
144
1.1185e−3
1.9650
1.1541e−4
2.9930
1.1835e−6
3.9931
9.7769e−9
4.9932
256
6.3381e−4
1.9743
4.8749e−5
2.9958
3.7487e−7
3.9962
2.3224e−9
4.9966
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
3.0030e−3
2.3372e−6
16
8.3399e−3
64
2.3572e−3
1.8230
3.8427e−4
2.9662
5.9274e−6
9.2831e−5 3.9691
7.3570e−8
2.3076e−6 4.9711
144
1.0848e−3
1.9140
1.1460e−4
2.9840
1.1773e−6
3.9865
9.7366e−9
4.9877
256
6.2023e−4
1.9433
4.8492e−5
2.9896
3.7339e−7
3.9916
2.3368e−9
4.9607
Table 9 Maximum errors and orders of convergence at Radau points for Example 4.1 using Qp , p = 1 to 4 with LDG boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3
Order
||eu ||∗∞
||eu ||∗∞
2.6092e−6
Order
16
1.6609e−2
64
2.8136e−3
2.5615
1.4615e−5
3.4255
1.1578e−7
4.4942
9.7235e−10
5.4448
144
9.2244e−4
2.7504
3.3050e−6
3.6664
1.7412e−8
4.6725
9.8348e−11
5.6508
256
4.0927e−4
2.8248
1.1188e−6
3.7650
4.4267e−9
4.7604
5.2012e−11
2.2144
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
1.5702e−4
p=4 Order
4.0770e−4
4.2352e−8
16
1.7732e−2
64
2.5243e−3
2.8124
2.8815e−5
3.8226
3.5172e−7
9.4352e−6 4.7456
3.3796e−9
1.8833e−7 5.8002
144
8.1953e−4
2.7746
6.1063e−6
3.8267
4.9084e−8
4.8569
4.7533e−9
−0.8412
256
3.6202e−4
2.8400
2.0014e−6
3.8775
1.1923e−8
4.9189
6.6823e−9
−1.1840
Since for V1 = Q1 we observe superconvergence at Radau points for both u and q, we repeat the previous computations using the tensor spaces Qp , p = 1, . . . , 4 with LDG boundary approximations and show the errors in Tables 8 and 9 where ||eq,1 ||∗∞ = ||eq,2 ||∗∞ . For the tensor product spaces we observe that the L2 errors for both u and q are O(hp+1 ) while the maximum errors for u and q at Radau points are close to O(hp+2 ) superconvergent. For p = 4, eq is polluted by round-off errors. The zero-level curves for the errors in Figs. 8, 9 and 10 pass close to Radau points. For the spaces Pp and Vp we recommend to use the interpolated boundary conditions which yield optimal convergence rates while for the spaces Qp one should use the LDG boundary approximation which ensures O(hp+2 ) superconvergence rates for both the solution and its gradient. Example 4.2 Let us consider the linear diffusion problem −u = f (x, y),
(x, y) ∈ = (−1, 1)2 ,
(4.2a)
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145
Fig. 8 Zero-level curves of eu for Example 4.1 on uniform meshes having N = 64 elements using Qp , p = 1 to 4 (upper left to lower right) with LDG boundary conditions
Table 10 L2 errors and orders of convergence for Example 4.2 on uniform meshes having N = 16, 64, 144, 256 elements using Pp , p = 1 to 4 with true boundary conditions N
p=1 ||eu ||L2
p=2 Order
||eu ||L2
p=3 Order
||eu ||L2
Order
||eu ||L2
9.8988e−2
64
2.6747e−2
1.8879
1.1989e−3
2.8684
3.8258e−5
3.8711
9.3106e−7
4.8981
144
1.2062e−2
1.9641
3.6111e−4
2.9595
7.7016e−6
3.9533
1.2424e−7
4.9674
256
6.8074e−3
1.9885
1.5314e−4
2.9819
2.4542e−6
3.9753
2.9693e−8
4.9753
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
1.3900e−1
5.5981e−4
Order
16
16
8.7554e−3
p=4
1.2554e−2
2.7763e−5
9.8830e−4
5.7435e−5
64
5.1729e−2
1.4261
2.2415e−3
2.4855
1.3190e−4
2.9055
4.3430e−6
3.7252
144
2.9392e−2
1.3942
8.3000e−4
2.4502
4.1958e−5
2.8249
9.5130e−7
3.7451
256
1.9649e−2
1.3998
4.1070e−4
2.4456
1.8508e−5
2.8451
3.1393e−7
3.8538
146
J Sci Comput (2012) 52:113–152
Fig. 9 Zero-level curves of eq,1 for Example 4.1 on uniform meshes having N = 64 elements using Qp , p = 1 to 4 (upper left to lower right) with LDG boundary conditions
subject to the Dirichlet boundary conditions u|∂ = uD .
(4.2b)
We select uD and f (x, y) such that the true solution is u(x, y) = ex+y .
(4.2c)
First, we solve (4.2) using Pp , p = 1, . . . , 4 on uniform Cartesian meshes having 16, 64, 144, 256 square elements and present the true errors ||eu ||L2 and ||eq,1 ||L2 = ||eq,2 ||L2 with their orders of convergence in Table 10. We observe that ||eu ||L2 = O(hp+1 ) while ||eq,i ||L2 = O(hm ) where m ≈ p + 1/2 for p = 1, 2 and m = p for p = 3, 4. Next, we solve (4.2) on uniform Cartesian meshes having 16, 64, 144, 256, 400 elements and compute LDG finite element solutions in the spaces Vp , p = 1, . . . , 4 with an LDG boundary approximation of uD . The maximum errors for eu at Radau points and their order of convergence shown in Table 11 indicate that u is O(hp+2 ) superconvergent at Radau points. The errors for the spaces Qp shown in Table 12 exhibit O(hp+1 ) convergence in L2 for both eu and eq . The maximum errors at Radau points presented in Table 13 show O(hp+2 ) superconvergence at Radau points for both eu and eq except for high p and fine meshes where results are polluted by round-off errors. The error zero-level curves plotted in Figs. 11, 12 and 13 pass close to Radau points.
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147
Fig. 10 Zero-level curves of eq,2 for Example 4.1 on uniform meshes having N = 64 elements using Qp , p = 1 to 4 (upper left to lower right) with LDG boundary conditions
Table 11 Maximum errors and orders of convergence at Radau points for Example 4.2 on uniform meshes having N = 16, 64, 144, 256, 400 elements using Vp , p = 1 to 4 with LDG boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3 Order
||eu ||∗∞
||eu ||∗∞
2.3188e−4
Order
16
2.3890e−1
64
3.3768e−2
2.8227
4.9007e−4
3.8667
7.5935e−6
4.9325
1.3148e−7
6.0479
144
1.0440e−2
2.8952
9.9645e−5
3.9287
1.0089e−6
4.9781
1.1305e−8
6.0514
256
4.5004e−3
2.9249
3.1975e−5
3.9511
2.4017e−7
4.9891
2.0622e−9
5.9143
400
2.3346e−3
2.9413
1.3206e−5
3.9628
7.8845e−8
4.9917
7.4028e−10
4.5913
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
7.1492e−3
p=4 Order
4.5042e−3
8.6988e−6
16
9.9657e−2
64
1.6392e−2
2.6040
1.3201e−3
1.7706
4.3312e−5
2.9670e−4 2.7761
2.6301e−6
4.9702e−5 4.2401
144
5.3464e−3
2.7632
5.1949e−4
2.3001
1.1428e−5
3.2860
4.4453e−7
4.3844
256
2.3686e−3
2.8299
2.5257e−4
2.5068
4.1814e−6
3.4949
1.2384e−7
4.4427
400
1.2493e−3
2.8670
1.4081e−4
2.6185
1.8696e−6
3.6072
4.9079e−8
4.1476
148
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Table 12 L2 errors and orders of convergence for Example 4.2 on uniform meshes having N = 16, 64, 144, 256 elements using Qp , p = 1 to 4 with LDG boundary conditions N
p=1 ||eu ||L2
p=2
p=3
Order
||eu ||L2
Order
||eu ||L2
2.6058e−3
p=4 Order
||eu ||L2
8.4440e−5
Order
16
9.5752e−3
64
2.4811e−3
1.9483
3.6823e−4
2.8230
5.7675e−6
3.8719
7.2132e−8
4.8982
144
1.1185e−3
1.9650
1.1243e−4
2.9259
1.1633e−6
3.9484
9.6599e−9
4.9585
256
6.3381e−4
1.9743
4.7998e−5
2.9588
3.7109e−7
3.9717
2.3103e−9
4.9729
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
||eq,1 ||L2
Order
2.7880e−3
2.1510e−6
16
8.3399e−3
64
2.3572e−3
1.8230
3.7430e−4
2.8970
5.8237e−6
8.8042e−5 3.9182
7.2716e−8
2.2162e−6 4.9297
144
1.0848e−3
1.9140
1.1311e−4
2.9514
1.1671e−6
3.9644
9.9865e−9
4.8964
256
6.2023e−4
1.9433
4.8115e−5
2.9712
3.7147e−7
3.9794
3.4286e−9
3.7162
Table 13 Maximum errors and orders of convergence at Radau points for Example 4.2 using Qp , p = 1 to 4 with LDG boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3
Order
||eu ||∗∞
||eu ||∗∞
1.4296e−4
Order
16
2.3890e−1
64
3.3768e−2
2.8227
4.2404e−4
3.8230
5.0524e−6
4.8225
5.2155e−8
5.8359
144
1.0440e−2
2.8952
8.7323e−5
3.8973
6.9312e−7
4.8991
4.3098e−9
6.1493
256
4.5004e−3
2.9249
2.8215e−5
3.9271
1.6755e−7
4.9357
7.9722e−10
5.8660
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
Order
||eq,1 ||∗∞
6.0013e−3
p=4 Order
2.8666e−3
2.9791e−6
16
9.9657e−2
64
1.6392e−2
2.6040
2.2087e−4
3.6981
2.7799e−6
7.3972e−5 4.7339
3.5969e−8
1.6116e−6
144
5.3464e−3
2.7632
4.6866e−5
3.8235
3.9624e−7
4.8047
2.4470e−8
0.9501
256
2.3686e−3
2.8299
1.5374e−5
3.8744
1.0482e−7
4.6225
3.1717e−8
−0.9018
5.4856
Example 4.3 Here we consider Poisson’s equation from Example 4.2 where f and uD are selected such that the exact solution is u(x, y) =
3 + x + y.
(4.3)
We solve this problem on uniform Cartesian meshes having 25, 100, 225, 400, 625 elements and compute LDG finite element solutions in the spaces Vp , p = 1, . . . , 4, with LDG boundary conditions. We present the maximum errors for eu at Radau points and their orders of convergence in Table 14 which, again, indicate that eu is O(hp+2 ) superconvergent at Radau points which is in agreement with our local error analysis. For p = 1, we present the maximum errors for eq,i , i = 1, 2 at Radau points and their orders of convergence in Table 15 which indicate the LDG errors eq,i , i = 1, 2 are O(hp+2 ) superconvergent at Radau points.
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149
Fig. 11 Zero-level curves of eu for Example 4.2 on uniform meshes having N = 64 elements using Qp , p = 1 to 4 (upper left to lower right) with LDG boundary conditions Table 14 Maximum errors and orders of convergence of eu at Radau points for Example 4.3 on uniform meshes having N = 25, 100, 225, 400, 625 elements using Vp , p = 1 to 4 with LDG boundary conditions N
p=1
||eu ||∗∞
p=2 Order
||eu ||∗∞
p=3 Order
||eu ||∗∞
1.6149e−5
p=4 Order
||eu ||∗∞
3.0695e−6
Order
25
3.8360e−4
100
5.8453e−5
2.7143
1.7684e−6
3.1909
1.9532e−7
3.9741
2.4920e−8
7.8920e−7 4.9850
225
1.8644e−5
2.8183
4.3284e−7
3.4712
3.4435e−8
4.2804
2.8211e−9
5.3730
400
8.1751e−6
2.8657
1.5347e−7
3.6043
9.5074e−9
4.4737
5.6343e−10
5.5993
625
4.2866e−6
2.8933
6.7465e−8
3.6832
3.4190e−9
4.5832
1.5212e−10
5.8677
5 Conclusion In this manuscript we investigated the superconvergence properties of a minimal dissipation local discontinuous Galerkin method applied to diffusion problems on rectangular meshes with smooth solutions using the spaces Pp , Qp and Vp . Computational results show that the md-LDG solutions for the potential and the flux are O(hp+1 ) convergent in L2 for both Vp and Qp spaces. The L2 errors for potential and flux in Pp are O(hp+1 ) convergent while the error in the flux is O(hp+1/2 ) for p = 1, 2 and O(hp ) for p > 2. The leading term of the LDG discretization error for the potential is proportional to two (p + 1)-degree Radau
150
J Sci Comput (2012) 52:113–152
Fig. 12 Zero-level curves of eq,1 for Example 4.2 on uniform meshes having N = 64 elements using Qp , p = 1 to 4 (upper left to lower right) with LDG boundary conditions Table 15 Maximum errors and orders of convergence of eq at Radau points for Example 4.3 on uniform meshes having N = 25, 100, 225, 400, 625 elements using V1 = Q1 with LDG boundary conditions N
||eq,1 ||∗∞
Order
||eq,2 ||∗∞
Order
25
5.3426e−4
100
1.0073e−4
2.4071
1.0073e−4
5.3426e−4 2.4071
225
3.4831e−5
2.6190
3.4831e−5
2.6190
400
1.5939e−5
2.7174
1.5939e−5
2.7174
625
8.5807e−6
2.7751
8.5807e−6
2.7751
polynomials in x and y when the md-LDG solution is in either Vp or Qp . Computational results further suggest that the leading term in the md-LDG error for the solution gradient is proportional to Radau polynomials in x and y if the md-LDG solution is in Qp . Thus, only the md-LDG solution is O(hp+2 ) superconvergent at Radau points for Vp while both the md-LDG solution and the gradient are O(hp+2 ) superconvergent at Radau points for Qp . Here, we present an analysis on one element and a more general analysis on all elements is needed and plan to carry out a global analysis using superconvergence results similar to the ones developed by Cheng and Shu [10]. In a forthcoming paper we will apply our results to compute efficient and asymptotically exact a posteriori error estimates needed to
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151
Fig. 13 Zero-level curves of eq,2 for Example 4.2 on uniform meshes having N = 64 elements using Qp , p = 1 to 4 (upper left to lower right) with LDG boundary conditions
steer adaptive algorithms. These results extend readily to quasi-uniform rectangular meshes. However, extending this error analysis to locally refined meshes with hanging nodes and to triangular meshes can be a more challenging task. Acknowledgement This research was partially supported by the National Science Foundation (Grant Number DMS 0511806, DMS 0809262).
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