In this section, we will briefly describe an alternative space-time finite element scheme, which was first introduced by Toulopoulos in [32]. For this section, we will restrict ourselves to a constant reluctivityν =const >0. Moreover, let V0h1 be the conforming finite element space of piecewise linear functions (c.f. Section 4.1). The starting point for this scheme is the following space-time finite element scheme for the model problem (1.1) – (1.3): finduh ∈V0h1, s.t.
Z
Q
∂tuhvh+ν∇xuh∇xvh dxdt= Z
Q
f vh dxdt, for all vh ∈V0h1. (4.47) However, for small values of ν, this scheme does not perform well [32]. To regain stability, we enrich the space V0h1 by so-called bubble functions, and obtain the larger space-time finite element space
Vh,b :={vh ∈H0,01,1(Q) :vh|E ∈P1(E)⊕Vb(E), for all E ∈ Th}, (4.48)
where Vb(E) = Vb|E for all E ∈ Th. Here, Vb denotes the space of bubble functions, which vanish entirely on the boundary of an element E ∈ Th and have exactly one degree of freedom for each element. For our case of piecewise linear basis functions, this space consists of cubic functions, i.e.,Vb :={vb :vb|E =cEp(E,1)p(E,2)p(E,3), for all E ∈ Th}, where we can choose cE such that maxx∈Evb(x) = 1 for each E ∈ Th. Based on (4.48), we can split every function vh ∈ Vh,b into a linear part v1h ∈ V0h1 and a bubble function vhb ∈ Vb(E), i.e., vh = vh1 +vbh. Now we can introduce a space-time finite element method, where the discrete problem reads as follows: finduh ∈Vh,b, s.t.
ˆ
ah(uh, vh) =l(vh), for all vh ∈Vh,b, (4.49) with
ˆ
ah(uh, vh) :=
Z
Q
∂tuhvh+ν∇xuh∇xvh dxdt+θh Z
Q
∂tubh∂tvbh dxdt, lh(v) =
Z
Q
f vh dxdt.
The bilinear form ˆah is bounded and coercive in some norm, see [32] for further details.
Moreover, Toulopoulos was able to derive a prior error estimates in the norm kuk2h,b :=νk∇xukL2(Q)+θhk∂tukL2(Q)+ 1
2kukL2(ΣT), (4.50) which we summarize in the next theorem.
Theorem 4.20. Let u ∈ H0,01,1(Q)∩Hs(Q), with s ≥ 2, be the exact solution and uh ∈ Vh,b be the solution of the finite element scheme (4.49). Then there holds the a priori estimate
ku−uhkh,b ≤chkukHs(Q), for θ=O(h) and h < ν. (4.51)
Numerical Results
In this chapter, we present and analyze the results of the numerical tests we performed.
We implemented the FEM with MFEM1, a C++library for finite elements. The result-ing linear systems were solved by means of the solver library HYPRE2, where we used the GMRES method, with BoomerAMG as preconditioner. This library is already fully parallelized. Hence, we were able to reach a high number of dofs. However, we observed very poor performance of the iterative solver for the 1 + 1-dimensional case.
Therefore, we decided to solve the linear system by means of a direct solver. In par-ticular, we used the MUMPS direct solver, which we included via the PETSc suite3. Any visualization of the approximate solution is done via GLVis4. The initial meshing for d = 1 and d = 2 was done via NETGEN [25], whereas the initial mesh for d = 3 was included by MFEM. The finer meshes were then obtained by subsequent uniform refinements.
To obtain a convergence result, we computed the error in theL2- and thek.kh-norm, i.e., let u be the exact solution of (4.8) and let uh be the approximate solution of (4.11), then we computedku−uhkL2(Q) and ku−uhkh, respectively.
Example 1. For this example we consider the d-dimensional unit cube as our spatial domain, and (0,1) as our time interval, i.e., Q = (0,1)d+1. We require ν ≡ 1 and choose the function
u(x, t) = sin(x1π)· · ·sin(xdπ) sin(tπ) (5.1) as our exact solution of the model problem (1.1) – (1.3), where the right hand side is computed accordingly. As the exact solution is very smooth, we expect optimal convergence rates for this example.
Example 2. For this example, we consider only d = 1 and d = 2. Our space-time cylinder is again the unit-cube, i.e., Q = (0,1)2 and Q = (0,1)3, respectively. We
1http://www.mfem.org
2https://computation.llnl.gov/projects/hypre-scalable-linear-solvers-multigrid-methods
3https://www.mcs.anl.gov/petsc/
4http://glvis.org/
39
choose the right hand side as in Example 1. However, in contrast to Example 1, we now allow discontinuities for ν. The finite element mesh is generated such that the discontinuities are only on the element boundary, i.e., there are no jumps in ν in the interior of an element E. We treated three cases of discontinuities:
(a) jumps only in the spatial directions, i.e., ν(x, t) = ν(x) for (x, t) ∈ Q, see Figure 5.2a and Figure 5.4a,
(b) jumps in space and time, see Figure 5.2b and Figure 5.4b, and
(c) jumps in space and time, resulting in a non-smooth space-time interface, see Figure 5.2c and Figure 5.4c.
For this problem, we do not know the exact solution. Hence, we have to work around this fact. An alternative way to obtain convergence rates without the knowledge of the exact solution is to perform as many uniform refinements as possible and solve the problem on this very fine mesh. Then we consider this solution u(f ms) on the finest mesh as the ”exact” solution and use it to compute error rates. However, we have to treat the obtained rates with care, as the approximate solutions tend to naturally converge to the finest mesh solution.
Another challenge that arises for this example is the loss of regularity in the exact solution, which we are not able to quantify. This means we are not able to guarantee that the a priori error estimate (4.35) holds. We therefore do not expect to obtain the optimal rates.
Example 3. Let once moreQ= (0,1)d+1. As exact solution, we choose the function u(x, t) =|t−0.5|λ sin(x1π)· · ·sin(xdπ),
with λ > 0, and we require ν ≡ 1. The choice of λ now determines the regularity of the solution. If we choose λ small enough, e.g. λ = 1.45, then u /∈ H2(Q), but u∈H1.95−(Q), with ε >0.
As for the choice of the parameterθE, we have seen in Remark 4.9 and Remark 4.14, that for p = 1, the choice of θ does not influence the stability of the method. It still influences the consistency error. For higher polynomial degree, e.g., p = 2 or p = 3, the choice θE = O(hE) is essential to ensure the stability (c.f. e.g. Table 5.3 or Table 5.20). For the time being, we do not compute the optimal constant given in Lemma 4.8, but instead use θE = c hE, which turned out to be sufficient for most of our experiments, provided that the constantcis sufficiently small. This ccan be seen as an approximation to 1/(c2I,3νE) from the inverse inequality (4.20).
5.1 1D Examples
We start with the simplest case, i.e.,d= 1, which results in a two dimensional space-time finite element method. As mentioned in the beginning of this chapter, the per-formance of the preconditioned GMRES method was really poor. Thus, all examples in this section are solved by means of the direct solver MUMPS. For this case, we did not include any comparison in solving time, as the sparsity pattern of the system matrices for different θE does not change, hence the solving time is almost the same for each case.
Example 1
For the problem with a high regularity solution, we deduce that the error rates in the k.kh-norm are indeed optimal. We also see that for p = 1, the stability of the method is not affected by choosing θE = 1 = O(1). However, the absolute value of the error is influenced by the choice of θE. As for the choice of c, for p= 1,2, c= 1 was sufficient, whereas for piecewise cubic functions, we chose c = 10−5. For higher polynomial degree, i.e.,p= 2 orp= 3, we note that the rate of convergence is slowed down or no convergence is attained, respectively (c.f. Table 5.2 and Table 5.3). This confirms the theoretical results of Chapter 4.
dofs θE= 0 θE=hE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 289 5.37×10−3 0 1.13×10−2 0 9.42×10−2 0 1089 1.34×10−3 2.00 2.95×10−3 1.93 5.12×10−2 0.88 4225 3.36×10−4 2.00 7.49×10−4 1.98 2.67×10−2 0.94 16 641 8.40×10−5 2.00 1.76×10−4 2.09 1.37×10−2 0.96 66 049 2.10×10−5 2.00 4.49×10−5 1.97 6.94×10−3 0.98 263 169 5.25×10−6 2.00 1.11×10−5 2.02 3.49×10−3 0.99 1 050 625 1.31×10−6 2.00 2.78×10−6 1.99 1.75×10−3 0.99 4 198 401 3.28×10−7 2.00 6.94×10−7 2.00 8.78×10−4 1.00 16 785 409 8.21×10−8 2.00 1.74×10−7 2.00 4.39×10−4 1.00 67 125 249 1.97×10−8 2.06 4.31×10−8 2.01 2.20×10−4 1.00
Error rates in theL2-norm.
dofs θE= 0 θE=hE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 289 1.53×10−1 0 1.56×10−1 0 3.73×10−1 0 1089 7.70×10−2 0.99 7.74×10−2 1.01 1.96×10−1 0.92 4225 3.85×10−2 1.00 3.86×10−2 1.00 9.95×10−2 0.98 16 641 1.93×10−2 1.00 1.93×10−2 1.00 5.02×10−2 0.99 66 049 9.64×10−3 1.00 9.64×10−3 1.00 2.52×10−2 0.99 263 169 4.82×10−3 1.00 4.82×10−3 1.00 1.26×10−2 1.00 1 050 625 2.41×10−3 1.00 2.41×10−3 1.00 6.33×10−3 1.00 4 198 401 1.20×10−3 1.00 1.20×10−3 1.00 3.17×10−3 1.00 16 785 409 6.02×10−4 1.00 6.02×10−4 1.00 1.58×10−3 1.00 67 125 249 3.01×10−4 1.00 3.01×10−4 1.00 7.93×10−4 1.00
Error rates in thek.kh-norm.
Table 5.1: Error rates for Example 1 with d= 1 andp= 1.
dofs θE= 0 θE=hE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 1089 5.28×10−4 0 4.60×10−4 0 3.27×10−2 0 4225 1.32×10−4 2.00 3.60×10−3 −2.97 4.40×10−2 −0.43 16 641 3.30×10−5 2.00 4.73×10−4 2.93 5.08×10−3 3.12 66 049 8.25×10−6 2.00 4.91×10−5 3.27 9.82×10−4 2.37 263 169 2.06×10−6 2.00 5.58×10−6 3.14 2.13×10−4 2.20 1 050 625 5.16×10−7 2.00 6.31×10−7 3.14 4.75×10−5 2.17 4 198 401 1.29×10−7 2.00 8.03×10−8 2.97 1.11×10−5 2.10 16 785 409 3.21×10−8 2.01 1.10×10−8 2.87 2.64×10−6 2.07 67 125 249 7.71×10−9 2.06 2.21×10−9 2.32 6.43×10−7 2.04
Error rates in theL2-norm.
dofs θE= 0 θE=hE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 1089 5.21×10−3 0 6.69×10−3 0 1.56×10−1 0 4225 1.31×10−3 1.99 3.07×10−2 −2.20 5.28×10−1 −1.76 16 641 3.28×10−4 2.00 4.60×10−3 2.74 7.26×10−2 2.86 66 049 8.20×10−5 2.00 6.24×10−4 2.88 1.73×10−2 2.06 263 169 2.05×10−5 2.00 9.54×10−5 2.71 4.71×10−3 1.88 1 050 625 5.13×10−6 2.00 1.52×10−5 2.65 1.29×10−3 1.86 4 198 401 1.28×10−6 2.00 2.60×10−6 2.54 3.73×10−4 1.79 16 785 409 3.21×10−7 2.00 4.89×10−7 2.41 1.06×10−4 1.82 67 125 249 8.02×10−8 2.00 1.03×10−7 2.25 3.12×10−5 1.76
Error rates in thek.kh-norm.
Table 5.2: Error rates for Example 1 with d= 1 andp= 2.
Figure 5.1: Decomposition of the 2D space-time cylinderQ = (0,1)2 into 256 subdomains, for parallel computation with 256 cores.
The strange behaviour of the error rates for piecewise cubic basis functions, i.e. p= 3 is most likely due to our direct solver. As the system matrixKh has a high condition number for 37 761 025 dofs, this can lead to a loss in precision.
dofs θE= 0 θE=chE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 2401 2.68×10−6 0 2.68×10−6 0 2.55×101 0 9409 1.69×10−7 3.99 1.85×10−7 3.85 3.22×103 −6.98 37 249 1.06×10−8 3.99 1.12×10−8 4.05 - -148 225 6.65×10−10 3.99 1 ×10−9 3.53 - -591 361 4.16×10−11 4.00 7.04×10−11 3.79 - -2 36-2 369 1.95×10−11 1.10 1.67×10−11 2.08 - -9 443 32-9 1.46×10−11 0.41 5.81×10−11−1.80 - -37 761 025 8.61×10−11−2.56 1.93×10−11 1.59 -
-Error rates in theL2-norm.
dofs θE= 0 θE=chE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 2401 1.45×10−4 0 1.45×10−4 0 5.40×102 0 9409 1.81×10−5 3.00 1.81×10−5 3.00 1.42×105 −8.04 37 249 2.26×10−6 3.00 2.26×10−6 3.00 - -148 225 2.83×10−7 3.00 2.83×10−7 3.00 - -591 361 3.54×10−8 3.00 3.54×10−8 3.00 - -2 36-2 369 4.42×10−9 3.00 4.42×10−9 3.00 - -9 443 32-9 5.54×10−10 3.00 5.84×10−10 2.92 - -37 761 025 3.07×10−10 0.85 1.12×10−10 2.39 -
-Error rates in thek.kh-norm.
Table 5.3: Error rates for Example 1 with d= 1 andp= 3.
Example 2
For this example, we considered a piece-wise constant, but globally discontinuous ν, with
ν(x.t) =
(ν1, for (x, t)∈ Q1∪ Q3, ν2, for (x, t)∈ Q2.
The jump in ν is ranging from a moderate height, i.e., ν1 = 10 and ν2 = 1, to big jumps, i.e., ν1 = 795 774 and ν2 = 165. The latter resembles an example where ν1 around the reluctivity of air, whereas ν2 is close to the reluctivity of iron. Moreover, we also tested for switchedν1 and ν2, where we deduced a different behaviour. Since we only considered p = 1, the choice c = 1 was sufficient. However, we consider now a coefficientν different from 1. Thus, we chose θE =hE/νE (c.f. (4.20)).
(a) (b) (c)
Figure 5.2: Plots of the space time cylinder Q used for Example 2 ford= 1.
We start with Example 2(a) and the case ν1 = 10 and ν2 = 1. Here, the reference solutionu(f ms) is calculated on a mesh with 25 176 065 dofs. We deduce that the error rates in the L2- and the k.kh-norm are very similar for all choices of θE. Moreover, for θE ∈ {0, hE}, the convergence rates in the k.k-norm are better than expected, where we almost reach an order of O(h2). This might be because we do not compute the error against the exact solution, but only a solution on a finer mesh. For constant θE = 1, the convergence rates are almost cut in half.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 7.79×10−4 0 9.41×10−4 0 1.22×10−2 0 1617 4.36×10−4 0.84 4.78×10−4 0.98 6.67×10−3 0.87 6305 2.16×10−4 1.01 2.31×10−4 1.05 3.56×10−3 0.90 24 897 7.52×10−5 1.52 7.99×10−5 1.53 1.85×10−3 0.95 98 945 2.14×10−5 1.81 2.27×10−5 1.82 9.28×10−4 0.99 394 497 5.62×10−6 1.93 5.92×10−6 1.94 4.44×10−4 1.06 1 575 425 1.37×10−6 2.04 1.44×10−6 2.04 1.93×10−4 1.20 6 296 577 2.77×10−7 2.31 2.91×10−7 2.31 6.50×10−5 1.57
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 9.32×10−3 0 1.01×10−2 0 1.25×10−1 0 1617 4.93×10−3 0.92 5.16×10−3 0.97 7.03×10−2 0.83 6305 2.26×10−3 1.13 2.36×10−3 1.13 3.92×10−2 0.84 24 897 7.66×10−4 1.56 7.97×10−4 1.57 2.11×10−2 0.90 98 945 2.30×10−4 1.74 2.37×10−4 1.75 1.08×10−2 0.96 394 497 6.35×10−5 1.85 6.51×10−5 1.86 5.26×10−3 1.04 1 575 425 1.61×10−5 1.98 1.64×10−5 1.99 2.29×10−3 1.20 6 296 577 3.37×10−6 2.25 3.43×10−6 2.26 7.65×10−4 1.58
Error rates in thek.kh-norm.
Table 5.4: Error rates for Example 2(a) with d= 1,p= 1, andν1 = 10 and ν2 = 1.
Now we increase the height of the discontinuity in ν and consider ν1 = 100, while ν2
is the same as before. The error rates in both norms are again almost identical for θE = 0 and θE =hE/νE (c.f. Table 5.5).
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 1.98×10−4 0 2.43×10−4 0 3.01×10−3 0 1617 8.97×10−5 1.14 9.85×10−5 1.30 1.71×10−3 0.82 6305 7.53×10−5 0.25 7.89×10−5 0.32 9.49×10−4 0.85 24 897 4.90×10−5 0.62 5.06×10−5 0.64 5.12×10−4 0.89 98 945 2.03×10−5 1.27 2.09×10−5 1.28 2.65×10−4 0.95 394 497 7.53×10−6 1.43 7.67×10−6 1.44 1.29×10−4 1.03 1 575 425 2.90×10−6 1.38 2.92×10−6 1.39 5.72×10−5 1.17 6 296 577 7.79×10−7 1.90 7.83×10−7 1.90 1.96×10−5 1.55
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 4.01×10−3 0 4.16×10−3 0 5.51×10−2 0 1617 1.77×10−3 1.18 1.76×10−3 1.24 3.25×10−2 0.76 6305 1.36×10−3 0.37 1.40×10−3 0.33 1.94×10−2 0.75 24 897 8.40×10−4 0.70 8.63×10−4 0.70 1.13×10−2 0.77 98 945 3.87×10−4 1.12 3.94×10−4 1.13 6.34×10−3 0.84 394 497 2.28×10−4 0.76 2.29×10−4 0.78 3.32×10−3 0.93 1 575 425 1.07×10−4 1.10 1.07×10−4 1.10 1.55×10−3 1.10 6 296 577 2.93×10−5 1.87 2.93×10−5 1.87 5.43×10−4 1.51
Error rates in thek.kh-norm.
Table 5.5: Error rates for Example 2(a) with d= 1,p= 1, andν1 = 100 andν2 = 1.
By interchangingν1 and ν2, i.e., ν1 = 1 and ν2 = 100, we notice that the error rates are now much better as before. For θE = 0 and θE = hE, we immediately reach a convergence order of 2, which is again better than expected. For θE = 1, we deduce only reduced convergence (c.f. Table 5.6).
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 2.15×10−3 0 5.54×10−3 0 8.10×10−2 0 1617 5.18×10−4 2.06 1.37×10−3 2.01 4.33×10−2 0.91 6305 1.23×10−4 2.07 3.47×10−4 1.99 2.23×10−2 0.96 24 897 2.83×10−5 2.12 8.28×10−5 2.07 1.12×10−2 0.99 98 945 6.38×10−6 2.15 2.07×10−5 2.00 5.48×10−3 1.03 394 497 1.44×10−6 2.15 5.05×10−6 2.04 2.57×10−3 1.09 1 575 425 3.23×10−7 2.15 1.20×10−6 2.07 1.10×10−3 1.22 6 296 577 6.65×10−8 2.28 2.45×10−7 2.30 3.69×10−4 1.58
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 1.30×10−2 0 2.35×10−2 0 2.97×10−1 0 1617 3.34×10−3 1.95 6.36×10−3 1.89 1.53×10−1 0.95 6305 8.45×10−4 1.99 1.56×10−3 2.03 7.69×10−2 1.00 24 897 2.12×10−4 1.99 3.74×10−4 2.06 3.81×10−2 1.01 98 945 5.33×10−5 1.99 9.37×10−5 2.00 1.85×10−2 1.04 394 497 1.34×10−5 1.99 2.31×10−5 2.02 8.65×10−3 1.10 1 575 425 3.41×10−6 1.98 5.64×10−6 2.04 3.71×10−3 1.22 6 296 577 8.25×10−7 2.05 1.22×10−6 2.20 1.24×10−3 1.58
Error rates in thek.kh-norm.
Table 5.6: Error rates for Example 2(a) with d= 1,p= 1, andν1 = 1 and ν2= 100.
At last, we consider the most difficult case, withν1 = 795 774 and ν2 = 165. Here, we notice a different behaviour. In the beginning, the convergence rates for θE = 0 and θE =hE behave very good. However, once we reach a certain number of dofs, in our case at 394 497, the error starts growing again. This behaviour might be due to the bad condition number of the matrix Kh. The height of the discontinuity has direct influence on the condition number, increasing it further by a factor of ∼ 8000. But for θE = 1, the situation is different. Although the quantity of the absolute errors is worse than for the other choices ofθE, the overall convergence rates remain stable (c.f.
Table 5.7).
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 9.87×10−7 0 9.85×10−7 0 1.42×10−5 0 1617 3.43×10−7 1.52 3.42×10−7 1.52 7.30×10−6 0.96 6305 1.19×10−7 1.53 1.19×10−7 1.53 3.73×10−6 0.97 24 897 4.14×10−8 1.53 4.12×10−8 1.53 1.92×10−6 0.96 98 945 1.49×10−8 1.47 1.49×10−8 1.47 9.96×10−7 0.94 394 497 8.51×10−9 0.81 8.50×10−9 0.81 5.22×10−7 0.93 1 575 425 1.02×10−8−0.26 1.02×10−8 −0.26 2.65×10−7 0.98 6 296 577 1.29×10−8−0.34 1.29×10−8 −0.34 1.16×10−7 1.20
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 2.44×10−4 0 2.43×10−4 0 2.95×10−3 0 1617 8.50×10−5 1.52 8.47×10−5 1.52 1.51×10−3 0.97 6305 2.96×10−5 1.52 2.95×10−5 1.52 7.75×10−4 0.96 24 897 1.03×10−5 1.52 1.03×10−5 1.52 4.05×10−4 0.94 98 945 3.73×10−6 1.47 3.71×10−6 1.47 2.16×10−4 0.91 394 497 2.11×10−6 0.82 2.11×10−6 0.82 1.17×10−4 0.88 1 575 425 2.51×10−6 −0.25 2.51×10−6 −0.25 6.27×10−5 0.90 6 296 577 3.16×10−6 −0.33 3.16×10−6 −0.33 2.80×10−5 1.17
Error rates in thek.kh-norm.
Table 5.7: Error rates for Example 2(a) withd= 1, p= 1, andν1 = 795774 andν2 = 1.
We continue with Example 2(b), where the discontinuity now depends on space and time. The finest mesh solution u(f ms) is computed on a mesh with 25 176 065 dofs.
For the simplest case, i.e. ν1 = 10 and ν2 = 1, we observe again better than expected convergence rates, but slightly worse than for Example 2(a).
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 1.05×10−3 0 1.22×10−3 0 1.39×10−2 0 1617 5.25×10−4 0.99 5.71×10−4 1.10 7.75×10−3 0.84 6305 2.12×10−4 1.31 2.27×10−4 1.33 4.20×10−3 0.88 24 897 6.51×10−5 1.70 6.96×10−5 1.71 2.20×10−3 0.93 98 945 1.83×10−5 1.83 1.95×10−5 1.84 1.11×10−3 0.99 394 497 5.29×10−6 1.79 5.55×10−6 1.81 5.32×10−4 1.06 1 575 425 1.47×10−6 1.85 1.52×10−6 1.86 2.32×10−4 1.20 6 296 577 3.28×10−7 2.17 3.37×10−7 2.18 7.81×10−5 1.57
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 1.15×10−2 0 1.25×10−2 0 1.33×10−1 0 1617 5.29×10−3 1.12 5.53×10−3 1.17 7.53×10−2 0.82 6305 1.98×10−3 1.41 2.07×10−3 1.41 4.17×10−2 0.85 24 897 6.52×10−4 1.61 6.75×10−4 1.62 2.22×10−2 0.91 98 945 2.51×10−4 1.38 2.55×10−4 1.40 1.14×10−2 0.97 394 497 9.46×10−5 1.41 9.54×10−5 1.42 5.47×10−3 1.05 1 575 425 2.97×10−5 1.67 2.99×10−5 1.67 2.38×10−3 1.20 6 296 577 7.10×10−6 2.07 7.13×10−6 2.07 7.91×10−4 1.59
Error rates in thek.kh-norm.
Table 5.8: Error rates for Example 2(b) withd= 1, p= 1, andν1 = 10 andν2= 1.
The choice ν1 = 100 and ν2 = 1 behaves similarly as in Example 2(a).
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 2.00×10−4 0 2.42×10−4 0 3.00×10−3 0 1617 8.90×10−5 1.17 9.71×10−5 1.32 1.72×10−3 0.80 6305 7.28×10−5 0.29 7.60×10−5 0.35 9.66×10−4 0.83 24 897 4.86×10−5 0.58 4.99×10−5 0.61 5.27×10−4 0.88 98 945 2.34×10−5 1.06 2.38×10−5 1.07 2.75×10−4 0.94 394 497 9.39×10−6 1.31 9.48×10−6 1.33 1.35×10−4 1.02 1 575 425 3.01×10−6 1.64 3.03×10−6 1.64 6.01×10−5 1.17 6 296 577 7.33×10−7 2.04 7.36×10−7 2.04 2.06×10−5 1.55
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 4.43×10−3 0 4.68×10−3 0 5.70×10−2 0 1617 1.90×10−3 1.22 1.92×10−3 1.29 3.43×10−2 0.73 6305 1.39×10−3 0.45 1.42×10−3 0.43 2.07×10−2 0.73 24 897 9.31×10−4 0.58 9.48×10−4 0.59 1.23×10−2 0.76 98 945 5.43×10−4 0.78 5.47×10−4 0.79 6.93×10−3 0.82 394 497 2.82×10−4 0.94 2.83×10−4 0.95 3.64×10−3 0.93 1 575 425 1.27×10−4 1.15 1.27×10−4 1.15 1.68×10−3 1.11 6 296 577 4.15×10−5 1.61 4.16×10−5 1.61 5.85×10−4 1.52
Error rates in thek.kh-norm.
Table 5.9: Error rates for Example 2(b) withd= 1, p= 1, and ν1= 100 and ν2= 1.
For the converse choice, i.e., ν1 = 1 and ν2 = 100, we note slightly worse convergence rates compared to Example 2(a), but still better than expected.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 2.07×10−3 0 4.44×10−3 0 7.29×10−2 0 1617 5.00×10−4 2.05 1.09×10−3 2.02 3.91×10−2 0.90 6305 1.18×10−4 2.08 2.91×10−4 1.91 2.02×10−2 0.95 24 897 2.86×10−5 2.05 6.84×10−5 2.09 1.02×10−2 0.99 98 945 7.12×10−6 2.01 1.75×10−5 1.97 5.00×10−3 1.03 394 497 1.81×10−6 1.98 4.31×10−6 2.02 2.35×10−3 1.09 1 575 425 4.57×10−7 1.98 1.04×10−6 2.05 1.01×10−3 1.22 6 296 577 1.10×10−7 2.06 2.17×10−7 2.26 3.37×10−4 1.58
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 1.34×10−2 0 2.20×10−2 0 2.76×10−1 0 1617 3.62×10−3 1.89 6.03×10−3 1.86 1.44×10−1 0.94 6305 9.58×10−4 1.92 1.60×10−3 1.91 7.26×10−2 0.99 24 897 2.59×10−4 1.89 3.89×10−4 2.04 3.61×10−2 1.01 98 945 7.34×10−5 1.82 1.04×10−4 1.90 1.76×10−2 1.04 394 497 2.28×10−5 1.69 2.91×10−5 1.84 8.24×10−3 1.09 1 575 425 7.93×10−6 1.53 9.03×10−6 1.69 3.54×10−3 1.22 6 296 577 2.74×10−6 1.54 2.86×10−6 1.66 1.18×10−3 1.58
Error rates in thek.kh-norm.
Table 5.10: Error rates for Example 2(b) with d= 1,p= 1 and ν1= 1 and ν2 = 100.
The rates for the last case,ν1 = 795774 andν2 = 165, we have again a certain number of dofs where the error rate stagnates, however it seems that for this example, the rates start improving again, at least in thek.kh-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 425 7.92×10−7 0 7.90×10−7 0 1.30×10−5 0 1617 2.71×10−7 1.55 2.71×10−7 1.55 6.68×10−6 0.96 6305 9.35×10−8 1.54 9.31×10−8 1.54 3.40×10−6 0.98 24 897 3.23×10−8 1.53 3.22×10−8 1.53 1.73×10−6 0.97 98 945 1.17×10−8 1.46 1.17×10−8 1.46 8.90×10−7 0.96 394 497 6.91×10−9 0.76 6.91×10−9 0.76 4.58×10−7 0.96 1 575 425 8.40×10−9−0.28 8.40×10−9 −0.28 2.28×10−7 1.00 6 296 577 1.05×10−8−0.32 1.05×10−8 −0.32 9.76×10−8 1.23
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 425 1.97×10−4 0 1.96×10−4 0 2.72×10−3 0 1617 6.79×10−5 1.53 6.77×10−5 1.53 1.39×10−3 0.97 6305 2.38×10−5 1.51 2.37×10−5 1.51 7.08×10−4 0.97 24 897 8.96×10−6 1.41 8.93×10−6 1.41 3.65×10−4 0.96 98 945 4.84×10−6 0.89 4.83×10−6 0.89 1.92×10−4 0.93 394 497 4.83×10−6 0.00 4.83×10−6 0.00 1.02×10−4 0.92 1 575 425 4.24×10−6 0.19 4.24×10−6 0.19 5.21×10−5 0.96 6 296 577 2.99×10−6 0.50 2.99×10−6 0.50 2.25×10−5 1.21
Error rates in thek.kh-norm.
Table 5.11: Error rates for Example 2(b) with d= 1, p= 1 and ν1 = 795774 andν2= 165.
Finally, we consider Example 2(c), where we have a non-smooth space-time interface for the discontinuity. Here we expect the most troubles. The reference solutionu(f ms) has 33 564 673 dofs. For the first two choices forν, i.e.,ν1 ∈ {10,100}and ν2 = 1, the observed convergence rates are similar, but slightly reduced, as in the previous two examples.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 553 1.20×10−3 0 1.56×10−3 0 1.33×10−2 0 2129 6.61×10−4 0.86 7.52×10−4 1.05 7.68×10−3 0.80 8353 3.09×10−4 1.10 3.34×10−4 1.17 4.24×10−3 0.86 33 089 1.25×10−4 1.31 1.31×10−4 1.35 2.25×10−3 0.92 131 713 4.92×10−5 1.34 5.05×10−5 1.37 1.14×10−3 0.98 525 569 1.91×10−5 1.37 1.93×10−5 1.39 5.49×10−4 1.05 2 099 713 6.97×10−6 1.45 7.02×10−6 1.46 2.40×10−4 1.20 8 393 729 2.02×10−6 1.79 2.03×10−6 1.79 8.08×10−5 1.57
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 553 2.11×10−2 0 2.35×10−2 0 1.38×10−1 0 2129 1.10×10−2 0.94 1.15×10−2 1.03 7.92×10−2 0.80 8353 5.24×10−3 1.07 5.36×10−3 1.10 4.39×10−2 0.85 33 089 2.55×10−3 1.04 2.57×10−3 1.06 2.34×10−2 0.91 131 713 1.21×10−3 1.07 1.21×10−3 1.08 1.19×10−2 0.97 525 569 5.17×10−4 1.23 5.17×10−4 1.23 5.74×10−3 1.06 2 099 713 1.93×10−4 1.42 1.93×10−4 1.42 2.49×10−3 1.21 8 393 729 5.48×10−5 1.82 5.49×10−5 1.82 8.28×10−4 1.59
Error rates in thek.kh-norm.
Table 5.12: Error rates for Example 2(c) withd= 1, p= 1, andν1 = 10 and ν2= 1.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 553 2.22×10−4 0 3.16×10−4 0 3.29×10−3 0 2129 1.06×10−4 1.07 1.25×10−4 1.34 1.90×10−3 0.79 8353 8.38×10−5 0.33 8.90×10−5 0.49 1.07×10−3 0.83 33 089 5.71×10−5 0.55 5.90×10−5 0.59 5.81×10−4 0.88 131 713 2.92×10−5 0.97 2.97×10−5 0.99 3.03×10−4 0.94 525 569 1.31×10−5 1.15 1.33×10−5 1.16 1.50×10−4 1.02 2 099 713 5.40×10−6 1.28 5.42×10−6 1.29 6.68×10−5 1.16 8 393 729 1.90×10−6 1.51 1.90×10−6 1.51 2.29×10−5 1.54
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 553 9.93×10−3 0 1.06×10−2 0 6.24×10−2 0 2129 6.82×10−3 0.54 6.89×10−3 0.62 3.78×10−2 0.72 8353 4.62×10−3 0.56 4.65×10−3 0.57 2.27×10−2 0.73 33 089 2.83×10−3 0.71 2.84×10−3 0.71 1.33×10−2 0.77 131 713 1.65×10−3 0.78 1.65×10−3 0.78 7.50×10−3 0.83 525 569 8.30×10−4 0.99 8.30×10−4 0.99 3.91×10−3 0.94 2 099 713 3.56×10−4 1.22 3.56×10−4 1.22 1.80×10−3 1.12 8 393 729 1.18×10−4 1.60 1.18×10−4 1.60 6.23×10−4 1.53
Error rates in thek.kh-norm.
Table 5.13: Error rates for Example 2(c) withd= 1, p= 1, andν1 = 100 andν2= 1.
If we interchange the values of ν1 and ν2, we reach now at most linear convergence rates in thek.kh-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 553 3.05×10−3 0 4.25×10−3 0 7.37×10−2 0 2129 9.33×10−4 1.71 1.19×10−3 1.83 3.99×10−2 0.89 8353 3.08×10−4 1.60 3.62×10−4 1.72 2.08×10−2 0.94 33 089 1.14×10−4 1.43 1.23×10−4 1.56 1.05×10−2 0.98 131 713 4.63×10−5 1.30 4.78×10−5 1.36 5.16×10−3 1.03 525 569 1.88×10−5 1.30 1.91×10−5 1.33 2.43×10−3 1.09 2 099 713 7.29×10−6 1.37 7.32×10−6 1.38 1.04×10−3 1.22 8 393 729 2.27×10−6 1.68 2.27×10−6 1.69 3.49×10−4 1.58
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 553 2.03×10−2 0 2.41×10−2 0 2.82×10−1 0 2129 8.78×10−3 1.21 9.52×10−3 1.34 1.48×10−1 0.93 8353 4.82×10−3 0.87 4.90×10−3 0.96 7.49×10−2 0.98 33 089 2.89×10−3 0.74 2.89×10−3 0.76 3.74×10−2 1.00 131 713 1.62×10−3 0.83 1.62×10−3 0.83 1.83×10−2 1.03 525 569 8.15×10−4 0.99 8.15×10−4 0.99 8.57×10−3 1.09 2 099 713 3.59×10−4 1.18 3.59×10−4 1.18 3.68×10−3 1.22 8 393 729 1.19×10−4 1.59 1.19×10−4 1.59 1.23×10−3 1.58
Error rates in thek.kh-norm.
Table 5.14: Error rates for Example 2(c) withd= 1, p= 1, andν1 = 1 andν2 = 100.
For the highest jump in the coefficient, the situation is now different than before.
Here, we do not see any stagnation in the k.kh-norm and also have almost linear convergence. However, in theL2-norm, the rates are as in the previous examples, i.e., stagnating and diverging.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 553 8.17×10−7 0 8.15×10−7 0 1.58×10−5 0 2129 2.84×10−7 1.52 2.83×10−7 1.53 8.14×10−6 0.96 8353 9.82×10−8 1.53 9.78×10−8 1.53 4.15×10−6 0.97 33 089 3.40×10−8 1.53 3.39×10−8 1.53 2.13×10−6 0.97 131 713 1.24×10−8 1.45 1.24×10−8 1.45 1.10×10−6 0.95 525 569 7.24×10−9 0.78 7.24×10−9 0.77 5.68×10−7 0.95 2 099 713 8.38×10−9−0.21 8.38×10−9 −0.21 2.85×10−7 1.00 8 393 729 1.04×10−8−0.31 1.04×10−8 −0.31 1.22×10−7 1.22
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 553 2.27×10−4 0 2.26×10−4 0 3.30×10−3 0 2129 9.91×10−5 1.19 9.89×10−5 1.19 1.69×10−3 0.96 8353 5.38×10−5 0.88 5.37×10−5 0.88 8.65×10−4 0.96 33 089 3.84×10−5 0.49 3.84×10−5 0.49 4.50×10−4 0.95 131 713 3.08×10−5 0.32 3.08×10−5 0.32 2.38×10−4 0.91 525 569 1.79×10−5 0.78 1.79×10−5 0.78 1.27×10−4 0.91 2 099 713 8.47×10−6 1.08 8.47×10−6 1.08 6.54×10−5 0.96 8 393 729 3.82×10−6 1.15 3.82×10−6 1.15 2.81×10−5 1.22
Error rates in thek.kh-norm.
Table 5.15: Error rates for Example 2(c) withd= 1,p= 1, andν1= 795774 and ν2 = 165.
Example 3
In this example, we consider again a solution with reduced regularity. However, here we know exactly to which space the exact solution belongs. We chooseλ= 1.45 from which we deduce that the analytical solutionu belongs to the space H1.95−(Q).
dofs θE= 0 θE=hE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 81 5.54×10−3 0 5.49×10−3 0 4.47×10−2 0 289 1.41×10−3 1.97 1.40×10−3 1.97 2.88×10−2 0.63 1089 3.51×10−4 2.00 3.48×10−4 2.01 1.70×10−2 0.76 4225 8.76×10−5 2.00 8.68×10−5 2.00 9.18×10−3 0.89 16 641 2.19×10−5 2.00 2.17×10−5 2.00 4.76×10−3 0.95 66 049 5.47×10−6 2.00 5.42×10−6 2.00 2.42×10−3 0.98 263 169 1.37×10−6 2.00 1.36×10−6 2.00 1.22×10−3 0.99 1 050 625 3.42×10−7 2.00 3.39×10−7 2.00 6.13×10−4 0.99 4 198 401 8.56×10−8 2.00 8.49×10−8 2.00 3.07×10−4 1.00 16 785 409 2.14×10−8 2.00 2.12×10−8 2.00 1.54×10−4 1.00
(a) Error rates in theL2-norm.
dofs θE= 0 θE=hE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 81 9.68×10−2 0 9.68×10−2 0 2.94×10−1 0 289 4.92×10−2 0.98 4.92×10−2 0.98 1.74×10−1 0.76 1089 2.47×10−2 0.99 2.47×10−2 0.99 9.93×10−2 0.81 4225 1.24×10−2 1.00 1.24×10−2 1.00 5.38×10−2 0.88 16 641 6.19×10−3 1.00 6.19×10−3 1.00 2.82×10−2 0.93 66 049 3.09×10−3 1.00 3.09×10−3 1.00 1.45×10−2 0.96 263 169 1.55×10−3 1.00 1.55×10−3 1.00 7.35×10−3 0.98 1 050 625 7.73×10−4 1.00 7.73×10−4 1.00 3.71×10−3 0.99 4 198 401 3.87×10−4 1.00 3.87×10−4 1.00 1.86×10−3 0.99 16 785 409 1.93×10−4 1.00 1.93×10−4 1.00 9.32×10−4 1.00
(b) Error rates in the k.kh-norm.
Table 5.16: Error rates for Example 3 withd= 1, λ= 1.45 and p= 1.
For piecewise linear basis functions, we deduce no change in any of the convergence rates.
dofs θE= 0 θE=hE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 289 3.93×10−4 0 3.79×10−4 0 3.34×10−2 0 1089 1.06×10−4 1.89 1.05×10−4 1.85 1.81×10−2 0.88 4225 2.81×10−5 1.91 2.76×10−5 1.93 3.95×10−3 2.20 16 641 7.36×10−6 1.93 7.11×10−6 1.96 9.24×10−4 2.10 66 049 1.91×10−6 1.94 1.86×10−6 1.93 2.35×10−4 1.97 263 169 4.94×10−7 1.95 4.80×10−7 1.96 5.74×10−5 2.04 1 050 625 1.27×10−7 1.96 1.24×10−7 1.96 1.44×10−5 2.00 4 198 401 3.26×10−8 1.96 3.17×10−8 1.96 3.57×10−6 2.01 16 785 409 8.37×10−9 1.96 8.13×10−9 1.96 8.92×10−7 2.00
(a) Error rates in theL2-norm.
dofs θE= 0 θE=hE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 289 6.25×10−3 0 6.27×10−3 0 1.66×10−1 0 1089 1.61×10−3 1.96 1.62×10−3 1.95 1.68×10−1 −0.01 4225 4.12×10−4 1.97 4.14×10−4 1.97 3.24×10−2 2.37 16 641 1.05×10−4 1.97 1.05×10−4 1.97 7.09×10−3 2.19 66 049 2.68×10−5 1.97 2.69×10−5 1.97 1.97×10−3 1.85 263 169 6.85×10−6 1.97 6.87×10−6 1.97 4.87×10−4 2.01 1 050 625 1.75×10−6 1.97 1.75×10−6 1.97 1.43×10−4 1.77 4 198 401 4.47×10−7 1.97 4.48×10−7 1.97 3.63×10−5 1.98 16 785 409 1.14×10−7 1.97 1.15×10−7 1.97 1.11×10−5 1.71
(b) Error rates in the k.kh-norm.
Table 5.17: Error rates for Example 3 withd= 1, λ= 1.45 and p= 2.
For increased polynomial degree, i.e.,p= 2, we indeed note a difference. For all three values of θE, the convergence rates in both norms are reduced, and, for θE = 0 and θE = hE, are of order O(h1.95). This directly resembles the regularity of our exact solution, which belongs to the space H1.95−(Q).
dofs θE= 0 θE=hE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 625 3.25×10−5 0 3.04×10−5 0 2.13×10−1 0 2401 6.83×10−6 2.25 3.64×10−5 −0.26 5.35×100 −4.65 9409 1.57×10−6 2.12 3.56×10−6 3.35 4.26×103 −9.64 37 249 3.77×10−7 2.06 5.68×10−7 2.65 - -148 225 9.34×10−8 2.01 1.02×10−7 2.48 - -591 361 2.36×10−8 1.98 2.43×10−8 2.07 - -2 36-2 369 6.00×10−9 1.97 6.05×10−9 2.01 - -9 443 32-9 1.55×10−9 1.96 1.56×10−9 1.96 -
-Error rates in theL2-norm.
dofs θE= 0 θE=hE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 625 3.85×10−4 0 3.94×10−4 0 2.13×10−1 0 2401 7.34×10−5 2.39 2.05×10−4 0.94 5.35×100 −4.65 9409 1.68×10−5 2.13 3.26×10−5 2.65 4.26×103 −9.64 37 249 4.20×10−6 2.00 6.46×10−6 2.34 - -148 225 1.08×10−6 1.96 1.35×10−6 2.25 - -591 361 2.79×10−7 1.95 3.14×10−7 2.11 - -2 36-2 369 7.22×10−8 1.95 7.77×10−8 2.02 - -9 443 32-9 1.87×10−8 1.95 1.98×10−8 1.97 -
-Error rates in thek.kh-norm.
Table 5.18: Error rates for Example 3 withd= 1, λ= 1.45 and p= 3.
For the highest polynomial degree p= 3, we observe the same rates as for p= 2, i.e., the convergence order in both norms is O(h1.95), provided that we choose θE = 0 or θE =hE. For θE = 1, the error diverges.
5.2 2D Examples
In this section, we analyze the numerical results for d = 2. This results in an three dimensional finite element method, where the number of dofs is higher and also the sparsity pattern of the system matrix Kh is more dense. Hence, we switch from a direct solver to an iterative solving method. As already mentioned, our bilinear form ah(. , .) is non symmetric and thus also the system matrix Kh has a non-symmetric, but positive definite, structure. This prevents us from using the powerful conjugate gradient (CG) method [24]. For our non-symmetric linear problem (4.39), we could either transform it to the normal equation
K>hKhuh =K>hfh,
and use here the CG-method. This is called the CG for the normal equation (CGNE).
Or we could try to minimize the residual
kKhuh−fhk,
over a special subspace of RNh. This is called generalized minimal residual method (GMRES). We decided to use the latter. As a preconditioner, we used the algebraic multigrid method described in [23], which is included in the HYPRE solver package.
Example 1
We begin with the highly smooth example. The convergence rates are again optimal for each polynomial degree, if we chooseθE = 0 or θE =chE. The necessity to choose θE in such a way is now more apparent than for d= 1, as there is no convergence for θE = 1 and p≥2 (compare Table 5.2 and Table 5.20). For p= 2, we chose c= 0.001, while forp= 3, the choice c= 10−5 was made.
dofs θE= 0 θE=hE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 125 1.11×10−1 0 1.86×10−1 0 2.57×10−1 0 729 3.25×10−2 1.77 6.24×10−2 1.57 1.73×10−1 0.57 4913 8.83×10−3 1.88 1.70×10−2 1.88 1.01×10−1 0.78 35 937 2.25×10−3 1.97 4.30×10−3 1.98 5.42×10−2 0.89 274 625 5.60×10−4 2.01 1.07×10−3 2.01 2.81×10−2 0.94 2 146 689 1.39×10−4 2.01 2.66×10−4 2.01 1.44×10−2 0.97 16 974 593 3.45×10−5 2.01 6.65×10−5 2.00 7.27×10−3 0.98
Error rates in theL2-norm.
dofs θE= 0 θE=hE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate
125 8.38×10−1 0 1.24 0 1.75 0
729 4.38×10−1 0.94 5.04×10−1 1.30 1.00 0.80 4913 2.24×10−1 0.97 2.33×10−1 1.12 5.42×10−1 0.89 35 937 1.12×10−1 0.99 1.13×10−1 1.04 2.81×10−1 0.95 274 625 5.62×10−2 1.00 5.63×10−2 1.01 1.43×10−1 0.97 2 146 689 2.81×10−2 1.00 2.81×10−2 1.00 7.24×10−2 0.98 16 974 593 1.40×10−2 1.00 1.40×10−2 1.00 3.64×10−2 0.99
Error rates in thek.kh-norm.
Table 5.19: Error rates for Example 1 withd= 2 andp= 1.
dofs θE= 0 θE=chE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate
729 8.84×10−3 0 8.92×10−3 0 2.89 0
4913 1.96×10−3 2.18 1.98×10−3 2.17 2.86 0.02 35 937 4.40×10−4 2.15 4.44×10−4 2.16 2.87 −0.01 274 625 1.03×10−4 2.10 1.04×10−4 2.10 2.89 −0.01 2 146 689 2.47×10−5 2.06 2.48×10−5 2.07 3.69 −0.35 16 974 593 5.98×10−6 2.04 5.97×10−6 2.06 5.28×101 −3.84
Error rates in theL2-norm.
dofs θE= 0 θE=chE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 729 1.40×10−1 0 1.40×10−1 0 4.01×101 0 4913 4.06×10−2 1.79 4.06×10−2 1.79 3.60×101 0.15 35 937 1.06×10−2 1.93 1.06×10−2 1.93 4.44×101 −0.30 274 625 2.71×10−3 1.98 2.71×10−3 1.97 7.18×101 −0.69 2 146 689 6.81×10−4 1.99 6.82×10−4 1.99 1.21×103 −4.08 16 974 593 1.71×10−4 2.00 1.71×10−4 1.99 3.49×104 −4.85
Error rates in thek.kh-norm.
Table 5.20: Error rates for Example 1 withd= 2 andp= 2.
Forp= 3 we note that forθE =chE, the error rate in the L2-norm is . However, once a certain number of dofs is reached, the absolute error is again of the same magnitude asθE = 0.
dofs θE= 0 θE=chE θ= 1
ku−uhk0 Rate ku−uhk0 Rate ku−uhk0 Rate 2197 1.70×10−3 0 1.71×10−3 0 3.53×10−1 0 15 625 1.10×10−4 3.95 1.10×10−4 3.95 3.53×10−1 −0.00 117 649 7.79×10−6 3.82 7.82×10−6 3.82 3.53×10−1 −0.00 912 673 5.50×10−7 3.82 5.63×10−7 3.80 - -7 189 05-7 3.88×10−8 3.83 5.22×10−8 3.43 -
-Error rates in theL2-norm.
dofs θE= 0 θE=chE θ= 1
ku−uhkh Rate ku−uhkh Rate ku−uhkh Rate 2197 3.13×10−2 0 3.13×10−2 0 1.74 0 15 625 3.91×10−3 3.00 3.91×10−3 3.00 1.66 0.07 117 649 4.85×10−4 3.01 4.85×10−4 3.01 1.64 0.02 912 673 6.03×10−5 3.01 6.03×10−5 3.01 - -7 189 05-7 7.52×10−6 3.00 7.53×10−6 3.00 -
-Error rates in thek.kh-norm.
Table 5.21: Error rates for Example 1 withd= 2 andp= 3.
As we are now using an iterative solver, we can compare the time needed to solve the linear system (c.f. Table 5.22). We see that for p ≤ 2, our method with performs better than θE = 0, where we need less time. However, for p = 3, θE = 0 performs slightly better than the other choices ofθE.
dofs θE= 0 θE=hE θE= 1 cores 125 0.01 0.01 0.01 128 729 0.02 0.02 0.01 128 4913 0.04 0.04 0.03 128 35 937 0.10 0.11 0.07 128 274 625 0.81 0.75 0.34 128 2 146 689 18.51 15.77 3.49 128 16 974 593 871.13 744.82 59.57 128
p= 1
dofs θE= 0 θE=chE θE= 1 cores 729 0.03 0.03 5.67 128 4913 0.06 0.06 9.10 128 35 937 0.27 0.27 4.54 128 274 625 2.28 2.13 35.67 128 2 146 689 41.78 41.01 188.04 128 16 974 593 1103.13 1032.82 1195.64 128
p= 2
dofs θE= 0 θE=chE θE= 1 cores 2197 0.07 0.07 2.49 128 15 625 0.18 0.18 3.42 128 117 649 1.02 1.01 9.32 128 912 673 12.68 12.53 51.33 128 7 189 057 189.27 194.30 - 128
p= 3 Table 5.22: Time needed to solve the linear system for Example 1.
Figure 5.3: Decomposition of the 3D space time cylinderQ = (0,1)3 into 256 subdomains for parallel computing on 256 cores.
Example 2
For this example, we considered again a piece-wise constant, but globally discontinuous ν, with
ν(x.t) =
(ν1, for (x, t)∈ Q1, ν2, for (x, t)∈ Q2,
where Q1 is the transparent part and Q2 the non-transparent part of the figures in Figure 5.4. As for the choice ofν1 andν2, we use one pair of values less than ford= 1, i.e., we consider
• ν1 = 100 and ν2 = 1,
• ν1 = 1 and ν2 = 100, and
• ν1 = 795774 and ν2 = 165.
(a) (b) (c) Figure 5.4: Plots of the space time cylinder Q used for Example 2 ford= 2.
We begin with Example 2(a) and ν1 = 100 and ν2 = 1. The reference solution u(f ms) was computed on a mesh with 21 835 009 dofs. Here we see, in contrast to d = 1, only reduced convergence rates. Moreover, the error rates in thek.kh-norm are again almost the same for both choices of θE.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 133 1.01×10−2 0 3.26×10−2 0 841 6.48×10−3 0.64 1.24×10−2 1.39 5969 2.11×10−3 1.62 3.68×10−3 1.76 44 961 6.21×10−4 1.77 1.01×10−3 1.86 348 993 1.75×10−4 1.83 2.73×10−4 1.90 2 750 081 4.45×10−5 1.97 6.34×10−5 2.10
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 133 1.34×10−1 0 3.57×10−1 0 841 1.52×10−1 −0.18 1.93×10−1 0.89 5969 8.05×10−2 0.91 8.59×10−2 1.16 44 961 4.30×10−2 0.90 4.37×10−2 0.98 348 993 2.54×10−2 0.76 2.54×10−2 0.78 2 750 081 2.21×10−2 0.20 2.20×10−2 0.21
Error rates in thek.kh-norm.
Table 5.23: Error rates for Example 2(a) with d= 2,p= 1, andν1 = 100 andν2 = 1.
If we interchange ν1 and ν2, the convergence rates in the k.kh-norm improve and are now linear, in contrast to d= 1, where we attained quadratic convergence.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 133 5.06×10−2 0 7.24×10−2 0 841 2.80×10−2 0.86 3.31×10−2 1.13 5969 1.54×10−2 0.86 1.54×10−2 1.10 44 961 8.42×10−3 0.88 8.18×10−3 0.91 348 993 3.48×10−3 1.27 3.42×10−3 1.26 2 750 081 9.42×10−4 1.88 9.32×10−4 1.87
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate
133 1.08 0 1.13 0
841 6.90×10−1 0.65 6.97×10−1 0.70 5969 3.67×10−1 0.91 3.67×10−1 0.93 44 961 1.82×10−1 1.01 1.82×10−1 1.01 348 993 8.94×10−2 1.03 8.94×10−2 1.03 2 750 081 4.44×10−2 1.01 4.44×10−2 1.01
Error rates in thek.kh-norm.
Table 5.24: Error rates for Example 2(a) with d= 2,p= 1, andν1 = 1 and ν2 = 100.
For the highest difference in ν1 and ν2, the convergence behaviour is again different than ford= 1. Here, we reach almost optimal rates, i.e., quadratic convergence in the L2-norm and linear convergence in thek.kh-norm (c.f. Table 5.25).
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 133 6.48×10−5 0 6.53×10−5 0 841 4.09×10−5 0.66 4.10×10−5 0.67 5969 1.36×10−5 1.59 1.36×10−5 1.59 44 961 3.94×10−6 1.78 3.94×10−6 1.79 348 993 1.03×10−6 1.94 1.03×10−6 1.94 2 750 081 2.45×10−7 2.07 2.45×10−7 2.07
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 133 9.89×10−3 0 9.94×10−3 0 841 1.16×10−2 −0.23 1.16×10−2 −0.23 5969 6.12×10−3 0.92 6.12×10−3 0.92 44 961 3.12×10−3 0.97 3.12×10−3 0.97 348 993 1.59×10−3 0.97 1.59×10−3 0.97 2 750 081 8.84×10−4 0.85 8.84×10−4 0.85
Error rates in thek.kh-norm.
Table 5.25: Error rates for Example 2(a) with d= 2, p= 1, andν1 = 795774 andν2 = 165.
Next we consider Example 2(b), where the coefficient ν depends on space and time.
Here,u(f ms) was calculated on a mesh with 25 363 073 dofs. Forν1 = 100 andν2 = 1, we observe that the convergence rates in both norms are worse than ford = 1.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 155 5.68×10−3 0 7.19×10−3 0 981 7.51×10−3 −0.40 9.29×10−3 −0.37 6953 2.87×10−3 1.39 3.72×10−3 1.32 52 305 1.02×10−3 1.50 1.27×10−3 1.55 405 665 3.96×10−4 1.36 4.61×10−4 1.47 3 195 201 1.33×10−4 1.57 1.47×10−4 1.65
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 155 9.35×10−2 0 1.21×10−1 0 981 1.83×10−1 −0.97 2.03×10−1 −0.74 6953 1.06×10−1 0.80 1.09×10−1 0.90 52 305 6.39×10−2 0.72 6.44×10−2 0.76 405 665 4.47×10−2 0.52 4.47×10−2 0.53 3 195 201 3.60×10−2 0.31 3.59×10−2 0.32
Error rates in thek.kh-norm.
Table 5.26: Error rates for Example 2(b) with d= 2, p= 1, andν1 = 100 andν2= 1.
For interchangedν1 and ν2, the rates are again better than for the previous case. We obtain linear convergence rates in the k.kh-norm, but for the L2-norm, the rates are worse than linear.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 155 5.47×10−2 0 7.83×10−2 0 981 2.61×10−2 1.07 3.40×10−2 1.20 6953 1.46×10−2 0.84 1.49×10−2 1.18 52 305 9.86×10−3 0.56 9.67×10−3 0.63 405 665 5.35×10−3 0.88 5.31×10−3 0.86 3 195 201 1.87×10−3 1.51 1.87×10−3 1.51
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate
155 1.14 0 1.20 0
981 8.97×10−1 0.35 9.07×10−1 0.40 6953 5.01×10−1 0.84 5.02×10−1 0.86 52 305 2.64×10−1 0.92 2.64×10−1 0.93 405 665 1.33×10−1 0.99 1.33×10−1 0.99 3 195 201 6.61×10−2 1.01 6.61×10−2 1.01
Error rates in thek.kh-norm.
Table 5.27: Error rates for Example 2(b) with d= 2, p= 1, andν1 = 1 andν2 = 100.
For the last case, i.e.,ν1 = 795774 and ν2 = 165, we notice a similar behaviour as for d = 1. Once a certain number of dofs is reached, the error in the k.kh-norm stops decreasing, while theL2-norm is still decreasing.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 155 3.41×10−5 0 3.40×10−5 0 981 4.65×10−5 −0.45 4.65×10−5 −0.45 6953 1.78×10−5 1.39 1.78×10−5 1.38 52 305 6.92×10−6 1.36 6.92×10−6 1.36 405 665 3.96×10−6 0.81 3.96×10−6 0.81 3 195 201 3.22×10−6 0.30 3.22×10−6 0.30
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 155 6.78×10−3 0 3.40×10−5 0 981 1.44×10−2 −1.08 4.65×10−5 −0.45 6953 8.30×10−3 0.79 1.78×10−5 1.38 52 305 5.50×10−3 0.59 6.92×10−6 1.36 405 665 5.71×10−3 −0.05 3.96×10−6 0.81 3 195 201 8.98×10−3 −0.65 3.22×10−6 0.30
Error rates in thek.kh-norm.
Table 5.28: Error rates for Example 2(b) with d= 2,p= 1, andν1 = 795774 and ν2= 165.
For a coefficient with non-smooth space-time interface, as in Example 2(c), we used a reference solution with 26 428 097 dofs. We have no convergence in thek.kh-norm for ν1 = 100 and ν2 = 1. The L2-norm is fine, however we only get reduced rates. This is most likely due to the loss in regularity of the analytical solution. Another reason could be the that we are too close to the reference solutionu(f ms).
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 164 2.46×10−3 0 6.52×10−3 0 1031 2.20×10−3 0.16 2.89×10−3 1.18 7277 7.18×10−4 1.62 9.09×10−4 1.67 54 617 2.51×10−4 1.52 2.90×10−4 1.65 423 089 1.31×10−4 0.93 1.37×10−4 1.08 3 330 401 6.64×10−5 0.98 6.70×10−5 1.03
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 164 5.57×10−2 0 1.12×10−1 0 1031 7.57×10−2 −0.44 8.13×10−2 0.46 7277 4.50×10−2 0.75 4.57×10−2 0.83 54 617 3.12×10−2 0.53 3.12×10−2 0.55 423 089 3.23×10−2 −0.05 3.23×10−2 −0.05 3 330 401 4.40×10−2 −0.45 4.40×10−2 −0.45
Error rates in thek.kh-norm.
Table 5.29: Error rates for Example 2(c) withd= 2, p= 1, andν1 = 100 andν2= 1.
However, for interchanged values of ν1 and ν2, the convergence rates in the L2-norm as well as in thek.kh-norm are almost optimal, i.e., O(h2) and O(h), respectively.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 164 3.91×10−2 0 6.91×10−2 0 1031 1.74×10−2 1.17 2.82×10−2 1.29 7277 5.62×10−3 1.63 8.37×10−3 1.75 54 617 1.65×10−3 1.77 2.26×10−3 1.89 423 089 4.84×10−4 1.76 6.04×10−4 1.90 3 330 401 1.29×10−4 1.91 1.49×10−4 2.02
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 164 2.92×10−1 0 4.59×10−1 0 1031 4.52×10−1 −0.63 4.71×10−1 −0.04 7277 2.62×10−1 0.79 2.64×10−1 0.83 54 617 1.38×10−1 0.92 1.39×10−1 0.93 423 089 7.18×10−2 0.95 7.18×10−2 0.95 3 330 401 3.90×10−2 0.88 3.90×10−2 0.88
Error rates in thek.kh-norm.
Table 5.30: Error rates for Example 2(c) withd= 2, p= 1, andν1 = 1 andν2 = 100.
And for the last coefficient choice, we again notice stagnating convergence rates in the k.kh-norm and reduced rates in the L2-norm.
dofs θE= 0 θE=hE/νE
ku−uhk0 Rate ku−uhk0 Rate 164 1.49×10−5 0 1.49×10−5 0 1031 1.37×10−5 0.12 1.37×10−5 0.12 7277 4.46×10−6 1.62 4.46×10−6 1.62 54 617 1.47×10−6 1.60 1.47×10−6 1.61 423 089 7.15×10−7 1.04 7.15×10−7 1.04 3 330 401 6.52×10−7 0.13 6.52×10−7 0.13
Error rates in theL2-norm.
dofs θE= 0 θE=hE/νE
ku−uhkh Rate ku−uhkh Rate 164 3.49×10−3 0 3.50×10−3 0 1031 5.60×10−3 −0.68 5.60×10−3 −0.68 7277 3.18×10−3 0.82 3.18×10−3 0.82 54 617 1.86×10−3 0.77 1.86×10−3 0.77 423 089 1.57×10−3 0.25 1.57×10−3 0.25 3 330 401 2.61×10−3 −0.73 2.61×10−3 −0.73
Error rates in thek.kh-norm.
Table 5.31: Error rates for Example 2(c) withd= 2,p= 1, andν1= 795774 and ν2 = 165.
As for solving times, we note that the case ν1 = 1 and ν = 100 performed far worse than the other two choices ofν1 and ν2. If we compare θE = 0 and θE =hE, then the solving times are very similar.
dofs θE= 0 θE=hE cores 133 0.01 0.01 128 841 0.01 0.01 128 5969 0.03 0.03 128 44 961 0.11 0.11 128 348 993 0.65 0.65 128 2 750 081 7.94 8.09 128
(a)
dofs θE= 0 θE=hE cores 155 0.01 0.01 128 981 0.02 0.02 128 6953 0.04 0.04 128 52 305 0.13 0.12 128 405 665 0.82 0.79 128 3 195 201 11.22 11.16 128
(b)
dofs θE= 0 θE=hE cores 164 0.01 0.01 128 1031 0.01 0.01 128 7277 0.04 0.04 128 54 617 0.14 0.14 128 423 089 0.99 0.98 128 3 330 401 13.78 13.79 128
(c)
Table 5.32: Time needed to solve the linear system for Example 2, with d= 2, p = 1, and ν1= 100 andν2 = 1.
dofs θE= 0 θE=hE cores 133 0.02 0.01 128 841 0.02 0.02 128 5969 0.05 0.05 128 44 961 0.22 0.21 128 348 993 1.66 1.62 128 2 750 081 18.73 18.69 128
(a)
dofs θE= 0 θE=hE cores 155 0.01 0.01 128 981 0.02 0.02 128 6953 0.05 0.05 128 52 305 0.27 0.27 128 405 665 2.41 2.76 128 3 195 201 35.27 33.26 128
(b)
dofs θE= 0 θE=hE cores 164 0.02 0.02 128 1031 0.03 0.03 128 7277 0.07 0.07 128 54 617 0.40 0.39 128 423 089 7.18 6.90 128 3 330 401 322.35 314.60 128
(c)
Table 5.33: Time needed to solve the linear system for Example 2, with d= 2, p = 1, and ν1= 1 and ν2 = 100.
dofs θE= 0 θE=hE cores 133 0.01 0.01 128 841 0.01 0.01 128 5969 0.03 0.03 128 44 961 0.10 0.11 128 348 993 0.62 0.61 128 2 750 081 6.93 6.84 128
(a)
dofs θE= 0 θE=hE cores 155 0.01 0.01 128 981 0.01 0.02 128 6953 0.04 0.04 128 52 305 0.12 0.13 128 405 665 0.81 0.80 128 3 195 201 10.53 10.50 128
(b)
dofs θE= 0 θE=hE cores 164 0.01 0.01 128 1031 0.01 0.01 128 7277 0.04 0.04 128 54 617 0.13 0.14 128 423 089 0.97 0.96 128 3 330 401 13.13 12.96 128
(c)
Table 5.34: Time needed to solve the linear system for Example 2, with d= 2, p = 1, and ν1= 795774 and ν2 = 165.
Example 3
For this example, we choose λ exactly as in the case d= 1, i.e. λ = 1.45. Moreover, forp= 2 andp= 3, we chosec= 0.01. For piecewise linear basis functions, i.e. p= 1, the convergence rates are optimal in the k.kh-norm. However, in contrast to d = 1, the L2-norm only shows a reduced convergence rate, and the reduction is more severe for θE = 0 (c.f. Table 5.16 and Table 5.35). This behaviour is due to the nature of our problem. When we assemble the right hand side of our linear system, we have to approximate the integral
Z
E
f(vh+θEhE∂tvh) d(x, t),
where f contains the term |t − 0.5|0.45. This expression is difficult to numerically integrate and introduces the reduction in the convergence rates.
dofs θE= 0 θE=hE
ku−uhk0 Rate ku−uhk0 Rate 84 3.42×10−2 0 5.70×10−2 0 503 1.76×10−2 0.96 2.98×10−2 0.94 3469 5.24×10−3 1.74 1.26×10−2 1.25 25 753 1.52×10−3 1.78 4.01×10−3 1.65 198 449 4.78×10−4 1.67 1.10×10−3 1.87 1 558 113 1.71×10−4 1.49 3.02×10−4 1.86 12 348 609 6.64×10−5 1.36 9.11×10−5 1.73
Error rates in theL2-norm.
dofs θE= 0 θE=hE
ku−uhkh Rate ku−uhkh Rate 84 2.65×10−1 0 6.29×10−1 0 503 1.79×10−1 0.57 3.19×10−1 0.98 3469 9.31×10−2 0.94 1.31×10−1 1.28 25 753 4.73×10−2 0.98 5.51×10−2 1.25 198 449 2.38×10−2 0.99 2.50×10−2 1.14 1 558 113 1.19×10−2 0.99 1.21×10−2 1.05 12 348 609 5.99×10−3 0.99 6.01×10−3 1.01
Error rates in thek.kh-norm.
Table 5.35: Error rates for Example 3 withd= 2 and p= 1 and λ= 1.45.
In case of piecewise quadratic basis functions, the reduced rates now occur in both norms. TheL2-norm behaves as forp= 1 and thek.kh-norm shows similar rates, but the difference betweenθE = 0 andθE =hE is now less.