In this section, we present the results of our numerical experiments for the schemes described in the previous section, with fourth order in temporal and fifth order in spatial accuracy. In some examples we also compare with the Lax-Wendroff type finite difference WENO scheme in [14], also with fourth order in temporal and fifth order in spatial accuracy. In all our examples, uniform meshes are used, and the CFL number is taken as 0.3. For one dimensional problems, the time step is chosen as
∆tn = 0.3∆x/α, where α= max
u |f′(u)| for the scalar cases and α= max
j=1,...,mmax
u |λj(u)|, where λj(u) are the eigenvalues of the Jacobianf′(u), for the systems.
Example 1. We first test the accuracy of the scheme on the linear scalar problem
ut+ux = 0 (34)
with the initial conditionu(x,0) = sin(πx) and 2-periodic boundary condition. The exact solution is u(x, t) = sin(π(x−t)). In Table 1, we show the errors at time t = 2. We can observe fifth order accuracy until quite refined meshes, despite of the formal fourth order temporal accuracy, supporting our claim before that spatial errors are usually dominating for modest meshes.
Example 2. We test our scheme on the nonlinear Burgers equation ut+
u2 2
x
= 0 (35)
Table 1: Accuracy on ut+ux = 0 with u(x,0) = sin(πx) at t= 2.
Nx L1 errors order L∞ error order
10 2.38E-02 – 3.67E-02 –
20 1.12E-03 4.41 1.99E-03 4.20 40 3.45E-05 5.02 6.64E-05 4.91 80 1.07E-06 5.00 2.16E-06 4.94 160 3.35E-08 5.00 6.51E-08 5.05 320 1.05E-09 5.00 1.95E-09 5.06 640 3.25E-11 5.01 5.71E-11 5.10
with the initial condition u(x,0) = 0.5 + sin(πx) and a 2-periodic boundary condition.
All the three monotone fluxes mentioned in Section 3.1, namely the Godunov flux, the Engquist-Osher flux and the Lax-Friedrichs flux, are tested.
When t= 0.5/π the solution is still smooth. The errors for all three monotone fluxes are shown in Tables 2 and 3 respectively. For comparison, we also show the errors by the Lax-Wendroff type finite difference WENO scheme in [14] with the same fourth order in temporal and fifth order in spatial accuracy in Table 2. We observe that all schemes achieve fifth order accuracy, despite of the formal fourth order temporal accuracy, once again supporting our claim before that spatial errors are usually dominating for modest meshes. For this test case, there is very little difference in the errors corresponding to the three different monotone fluxes. We do observe that the errors of the WENO scheme in [14] are significantly bigger than those for the schemes designed in this paper in Tables 2 and 3 when compared on the same mesh, indicating that the narrower effective stencil and flexibility in choosing numerical fluxes are helping to improve the magnitude of the errors. However, a more meaningful comparison is the CPU cost versus errors achieved. In Figure 1, we provide such a comparison for the current scheme with the scheme in [14] for the Lax-Friedrichs flux. Though our method has smaller errors when compared on the same mesh, which is shown in Table 2, it seems the two methods have similar efficiency, because of the complex calculation when computing high order time derivatives. This is also a favorable comparison for our method. For systems and
multi-dimensional cases, heavier algebra has a major influence, resulting from matrix calculus and mixed derivatives, and our method has a lower efficiency compared with the scheme in [14]. Our method, when coupled with the Lax-Wendroff time discretization, is therefore of advantage in allowing general monotone fluxes and maintenance of free-stream, rather than in efficiency.
+ +
+ +
+ +
+
Log(Cost)
Log(Error)
-10 -8 -6
-20 -15 -10 -5
L-F
+ [14]
Figure 1: Comparison of CPU time against L1 error for one-dimensional Burgers equa-tion between our scheme with the Lax-Friedrichs flux and the scheme in [14].
Next, we plot the solution with 80 grid points at time t= 1.5/π in Figure 2, when a shock has already appeared in the solution. We can see that for all the three monotone fluxes, the schemes simulate the solution with good resolution and without noticeable oscillations near the shock.
Example 3. We now consider the one-dimensional Euler system of compressible gas dynamics:
ξt+f(ξ)x = 0 (36)
where
ξ = (ρ, ρu, E)T, f(ξ) = (ρu, ρu2+p, u(E+p))T. (37)
Table 2: Accuracy on the Burgers equation with the Lax-Friedrichs flux and the WENO scheme in [14]. u(x,0) = 0.5 + sin(πx) at t= 0.5/π.
Lax-Friedrichs flux WENO scheme in [14]
Nx L1 errors order L∞ error order L1 errors order L∞ error order
10 4.57E-03 – 1.42E-02 – 3.65E-02 – 6.85E-02 –
20 4.85E-04 3.24 2.33E-03 2.61 4.44E-03 3.04 1.26E-02 2.45 40 2.59E-05 4.23 2.38E-04 3.29 2.55E-04 4.12 1.06E-03 3.57 80 1.37E-06 4.23 1.14E-05 4.38 9.99E-06 4.67 5.11E-05 4.38 160 5.97E-08 4.52 9.90E-07 3.53 3.76E-07 4.73 1.69E-06 4.92 320 1.80E-09 5.05 3.69E-08 4.74 1.15E-08 5.03 7.62E-08 4.47 640 4.07E-11 5.47 3.97E-10 6.54 2.96E-10 5.28 1.46E-09 5.70
Table 3: Accuracy on the Burgers equation with the Godunov flux and Engquist-Osher flux. u(x,0) = 0.5 + sin(πx) at t= 0.5/π.
Godunov flux Engquist-Osher flux
Nx L1 errors order L∞ error order L1 errors order L∞ error order
10 4.59E-03 – 1.41E-02 – 4.59E-03 – 1.41E-02 –
20 4.84E-04 3.24 2.32E-03 2.60 4.84E-04 3.24 2.32E-03 2.60 40 2.59E-05 4.23 2.38E-04 3.29 2.59E-05 4.23 2.38E-04 3.29 80 1.37E-06 4.23 1.14E-05 4.38 1.37E-06 4.23 1.14E-05 4.38 160 5.97E-08 4.52 9.90E-07 3.53 5.97E-08 4.52 9.90E-07 3.53 320 1.80E-09 5.05 3.69E-08 4.74 1.80E-09 5.05 3.69E-08 4.74 640 4.07E-11 5.47 3.97E-10 6.54 4.07E-11 5.47 3.97E-10 6.54
Here ρ is the density, u is the velocity, E is the total energy, and p is the pressure, which is related to the total energy by E = γ−1p + 12ρu2 with γ = 1.4 for air. The initial condition is set to be ρ(x,0) = 1 + 0.2 sin(πx), u(x,0) = 0.7, p(x,0) = 1, with a 2-periodic boundary condition. The exact solution is ρ(x, t) = 1 + 0.2 sin(π(x−ut)), u(x, t) = 0.7 andp(x, t) = 1. The errors and numerical orders of accuracy for the density ρ are shown in Table 4. We can see that the schemes with both the Lax-Friedrichs flux and the HLLC flux can achieve the designed order of accuracy, while the error magnitude is smaller for the HLLC flux than for the Lax-Friedrichs flux on the same mesh.
Example 4. Next, we solve the same one-dimensional Euler equations with a Riemann initial condition for the Sod’s problem
(ρ, u, p) =
(1,0,1), if x≤0
(0.125,0,0.1), if x >0 (38)
+++++++ +
+
+ ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
X
U
-1 -0.5 0 0.5 1
-0.5 0 0.5 1
1.5 Exact
Godunov E-O L-F
+
Figure 2: The Burgers equation with u(x,0) = 0.5 + sin(πx). t = 1.5/π. N = 80 grid points.
Table 4: Errors and orders of accuracy for the density ρ on the one-dimension Euler equation at t= 2. Lax-Friedrichs flux and HLLC flux.
Lax-Friedrichs flux HLLC flux
Nx L1 errors order L∞ error order L1 errors order L∞ error order
10 9.20E-03 – 1.30E-02 – 3.46E-03 – 5.58E-03 –
20 4.84E-04 4.25 7.74E-04 4.07 1.55E-04 4.48 2.93E-04 4.25 40 1.53E-05 4.98 2.80E-05 4.79 4.82E-06 5.01 1.00E-05 4.87 80 4.77E-07 5.00 8.89E-07 4.98 1.50E-07 5.01 3.06E-07 5.04 160 1.47E-08 5.02 2.67E-08 5.06 4.63E-09 5.02 8.38E-09 5.19 320 4.42E-10 5.06 7.52E-10 5.15 1.39E-10 5.06 2.37E-10 5.14 640 1.19E-11 5.21 1.99E-11 5.24 3.72E-12 5.22 6.44E-12 5.20
and the Lax’s problem (ρ, u, p) =
(0.445,0.698,3.528), if x≤0
(0.5,0,0.571), if x >0. (39)
We plot the computed density at t = 1.3 for both problems with 200 grid points. We can see the results of both problems are well resolved and non-oscillatory with both the Lax-Friedrichs flux and the HLLC flux.
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Figure 3: The Sod’s and Lax’s problems. t= 1.3. 200 grid points with the Lax-Friedrichs flux and the HLLC flux.
Example 5. We plot the numerical solution of the blast wave problem. We are still solving the Euler Equations (36)-(37), with the initial condition
ρ(x,0) = 1.0, 0≤x≤1 We compare numerical solution with a reference solution, which is obtained with the regular WENO scheme in [8] with 16000 grid points and hence can be considered as the exact solution. We use both the Lax-Friedrichs flux and the HLLC flux for the problem.
800 grid points are used. Our numerical results seem to agree with the reference solution very well.
Example 6. In this example, we solve the LeBlanc shock tube problem, still for the
+
Figure 4: Blast wave problem. Lines are from the reference solution. Circles are the numerical solution with the Lax-Friedrichs flux, and pluses are the numerical solution with the HLLC flux: (a) density; (b) pressure; (c) velocity.
Euler Equations (36)-(37), with the initial condition ρ(x,0) =
The Lax-Friedrichs flux is used. We plot the numerical solutions with 2400 and 19200 grid points in Figure 5, to compare with the results in [24], in which a third order DG method with 800 cells and 6400 cells are used and each cell contains three degrees of
freedom. We can see that our results are well resolved and agree well with the solutions in [24]. This is an extreme test case with very strong discontinuities and is used to test robustness of our scheme.
Example 7. We now consider the following two-dimensional linear equation
ut+aux+buy = 0 (42)
with the initial conditionu(x, y,0) = sin(π(x+y)) and a 2-periodic boundary condition, where a and b are constants. Here, we take a= 1, b =−2, and the final time t= 2. To avoid possible error cancellations due to symmetry, we use different mesh sizes in the x and y directions. Errors are shown in Table 5. We can clearly observe that the scheme achieves the designed fifth order accuracy.
Table 5: Accuracy on the linear equation ut+ux−2uy = 0 withu(x, y,0) = sin(π(x+y)) at t= 2.
Nx×Ny L1 errors order L∞ error order 8×12 4.97E-02 – 7.23E-02 – 16×24 5.70E-03 3.12 1.02E-02 2.83 32×48 1.70E-04 5.07 3.42E-04 4.90 64×96 4.72E-06 5.17 1.01E-05 5.08 128×192 1.41E-07 5.07 3.07E-07 5.04 256×384 4.32E-09 5.02 9.16E-09 5.06
Example 8. Next, we solve the two-dimensional nonlinear Burgers equation ut+
u2 2
x
+ u2
2
y
= 0 (43)
with the initial condition u(x,0) = 0.5 + sin(π(x+y)/2) and a 4-periodic boundary condition. When t = 0.5/π the solution is still smooth, and we give the errors and order of accuracy in Table 6 with the Lax-Friedrichs flux. Clearly, the scheme delivers the designed order of accuracy. For comparison, we also show the errors by the Lax-Wendroff type finite difference WENO scheme in [14] with the same fourth order in temporal and fifth order in spatial accuracy in Table 6. We again observe that the errors
of the WENO scheme in [14] are significantly bigger than those for the schemes designed in this paper in Table 6 when compared on the same mesh, indicating that the narrower effective stencil is helping in improving the magnitude of the errors (at the price of higher computational cost).
We plot the results att = 1.5/πwhen shocks have already appeared in Figure 6 with 160×160 grid points. On the left we are showing a one-dimensional cut at x=yof the solution, and on the right we are showing the surface of the computed solution. Clearly, the shocks are well resolved without spurious oscillations.
Table 6: Accuracy on the Burgers equation ut+ (u22)x + (u22)y = 0 with Lax-Friedrichs flux and the WENO scheme in [14]. u(x, y,0) = 0.5 + sin(π(x+y)/2) at t= 0.5/π.
Lax-Friedrichs flux WENO scheme in [14]
Nx×Ny L1 errors order L∞ error order L1 errors order L∞error order
8×12 7.51E-03 – 2.35E-02 – 2.23E-02 – 6.44E-02 –
16×24 9.86E-04 2.93 6.96E-03 1.75 2.98E-03 2.90 1.45E-02 2.15 32×48 8.23E-05 3.58 7.30E-04 3.25 2.90E-04 3.36 1.65E-03 3.13 64×96 4.26E-06 4.27 4.09E-05 4.16 8.58E-06 5.08 8.08E-05 4.36 128×192 1.84E-07 4.53 1.62E-06 4.65 3.27E-07 4.71 2.57E-06 4.97 256×384 6.24E-09 4.88 7.20E-08 4.49 1.01E-08 5.01 1.33E-07 4.28
X
U
0 1 2 3 4
-0.5 0 0.5 1 1.5
(a) (b)
Figure 6: The two-dimensional Burgers equation. u(x,0) = 0.5 + sin(π(x+y)/2). t = 1.5/π. 160×160 grid points. Left: a cut of the solution at x = y, where the solid line is the exact solution and the circles are the computed solution; right: the surface of the solution.
Example 9. We now consider two-dimensional Euler systems of compressible gas dy-namics:
ξt+f(ξ)x+g(ξ)y = 0 (44)
with
ξ= (ρ, ρu, ρv, E)T,
f(ξ) = (ρu, ρu2+p, ρuv, u(E+p))T, g(ξ) = (ρv, ρuv, ρv2+p, v(E+p))T.
Here, ρ is the density, (u, v) is velocity, E is the total energy, and p is the pressure, related to the total energy E by E = γ−1p + 12ρ(u2 +v2), with γ = 1.4. The initial condition is set to be ρ(x, y,0) = 1 + 0.2 sin(π(x+y)), u(x, y,0) = 0.7, v(x, y,0) = 0.3, p(x, y,0) = 1, with a 2-periodic boundary condition. The exact solution is ρ(x, y, t) = 1 + 0.2 sin(π(x+y−(u+v)t)),u= 0.7,v = 0.3,p= 1. The errors and order of accuracy for the density are shown in Table 7. We can see that the scheme achieves the designed order of accuracy.
Table 7: Accuracy on the two-dimensional Euler equation att = 2. Errors of the density ρ.
Nx×Ny L1 errors order L∞ error order 8×12 2.29E-02 – 3.28E-02 – 16×24 1.47E-03 3.97 2.22E-03 3.89 32×48 5.15E-05 4.83 9.03E-05 4.62 64×96 1.61E-06 5.00 2.99E-06 4.92 128×192 4.98E-08 5.01 9.20E-08 5.02 256×384 1.51E-09 5.04 2.58E-09 5.15
Example 10. Next, we use the following problem to illustrate further the high order accuracy of the method. Consider the following idealized problem for the Euler equation in 2D: The mean flow is ρ = 1, p = 1, and (u, v) = (1,1) (diagonal flow). We add, to this mean flow, an isentropic vortex (perturbation in (u, v) and the temperatureT = pρ,
no perturbation in the entropy S= ρpγ):
(δu, δv) = ǫ
2πe0.5(1−r2)(−¯y,x)¯ δT =− (γ−1)ǫ2
8γπ2 e1−r2 δS =0
where (¯x,y) = (x¯ −5, y −5), r2 = ¯x2+ ¯y2, and the vortex strength ǫ = 5 [17]. Since the mean flow is in the diagonal direction, the vortex movement is not aligned with the mash direction. It is clear that the exact solutions is just the passive convection of the vortex with the mean velocity. The computational domain is taken as [0,10]×[0,10], extended periodically in both directions. The simulation is performed until t= 0.2. The errors and orders of accuracy are shown in Table 8.
Table 8: Accuracy on the Vortex evolution at t= 0.2. Errors of the density ρ.
Nx×Ny L1 errors order L∞ error order 40×40 5.52E-05 – 1.98E-03 – 80×80 2.90E-06 4.25 1.41E-04 3.81 160×160 6.84E-08 5.41 2.57E-06 5.78 320×320 1.96E-09 5.13 5.53E-08 5.54
Example 11. Finally, we consider the double Mach reflection problem. The computa-tional domain for this problem is chosen to be [0,4]×[0,1]. The reflection wall lies at the bottom of the computational domain starting from x = 16. Initially a right-moving Mach 10 shock is positioned atx= 16, y= 0 and makes a 60◦ angle with thex-axis. For the bottom boundary, the exact post-shock condition is imposed for the part fromx= 0 to x= 16 and a reflective boundary condition is used for the rest. At the top boundary of the computational domain, the flow values are set to describe the exact motion of the Mach 10 shock. The initial pre-shock condition is
(ρ, p, u, v) = (8, 116.5, 8.25 cos(30◦), −8.25 sin(30◦)) (45)
and the post-shock condition is
(ρ, p, u, v) = (1.4, 1, 0, 0) (46)
We compute the solution up to t= 0.2. Uniform meshes with grid points 960×239 are used. In Figure 7 we show 30 equally spaced density contours from 1.5 to 22.7, and a
“zoomed-in” graph is provided in Figure 8. We can see that our results agree well with the results in [14].