The last test we run is using the arbitrary initial condition u0(x)= sin(πx)e−x2,
defined in (3.27), which we solve on [−5,5] for time T = 2.
So far the Semi-Implicit method has performed best, and we use this method to find a reference solution to which we can compare our methods. A drawback with the Semi-Implicit method is that it requires initial conditions at two time steps.
However, the others do not, so we can use one of these to obtain the second initial condition given the first at t= 0. We choose the Taylor Expansion method to find the solution at t= k, using a time step of kI= 10−5k. Then we have the conditions at t= 0 and t = k. The reference solution was found using N = 500, and a time step k= 10−8, such that kI= 10−13. The reference solution was found in 6 hours.
We also found a solution with k= 10−7, and the maximal point wise discrepancy between this and the better one was 5.9× 10−7. To make sure the reference solu-tion does not affect the result, we also did a solusolu-tion with the Taylor Expansion method. With N= 500 and k = 10−8, this took 16 hours, and the maximal point wise difference to the Semi-Implicit solution was 4.5× 10−9. Thus, the reference solution is deemed good enough for the following test.
We require that the maximal error in the|| · ||-norm is 1.0 × 10−4. To calculate the error in the nodal points chosen for each method, we find the continuous solution by the Fourier series
The time step for the Lax-Friedrichs method is chosen to be k= 5h. For the Semi-Implicit and Taylor Expansion methods, we chose time steps k= T/62000 and k= T/63000 respectively. That way we get an integer number of time steps up to T= 2, and can compare our methods at the same point in time. The results are found in Table 5.5.
# Method N k Time (s) ||e||∞ ||e||
1) T&VM 2000 2.50e-4 10.2 2.55e-4 1.0e-4
2) Semi-Implicit 24 3.22e-5 2.6 1.97e-4 1.0e-4 3) Taylor Expansion 25 3.17e-5 4.2 2.05e-4 1.0e-4 4) Lax-Friedrichs 110000 2.27e-4 318.2 2.42e-4 1.0e-4
Table 5.5: Comparison of the methods for the initial condition in (3.27) at T= 2.
The reference solution is calculated using the Semi-Implicit method with N= 500 and k= 10−8.
Yet again the Semi-Implicit method is best, but this time it is a much closer race.
The Operator Splitting with Taylor Expansion method requires about the same time and spatial step sizes, and takes less than twice the time. The Taylor Ex-pansion method solves the linear equation (4.10) with high accuracy. So, for test problems with many oscillations, where the uxxwill be large compared to the other terms, the Taylor Expansion method might have an advantage. Figure 5.13 shows the comparison of the uux and Huxx terms. For the arbitrary initial condition, the Hilbert term is clearly much larger compared to the uux term than in the two other test problems. The Lax-Friedrichs method solves the linear equation in the same way as the Taylor Expansion method, but the LxF step requires way too many points in space to be any effective. The method of T&VM needs more steps in space as expected, but can cope with a much larger time step than the Semi-Implicit and Taylor Expansion.
Figure 5.14 shows the solution at different times. We can see small waves breaking out at t= 0.2. It is hard to tell because of wave interactions, but it looks like we get a total of 7-8 waves.
(a) 1-Periodic solution at t=0. (b) 2-Periodic solution at t=0.
(c) Arbitrary initial condition at t=0
Figure 5.13: Plot of the uuxand Huxxterms in the BO equation for different initial conditions. Black is u(x, 0), blue is uux(x, 0) and red is Huxx(x, 0).
(a) t=0 (b) t=0.2
(c) t=0.3 (d) t=0.5
(e) t=1.0 (f) t=2.0
Figure 5.14: The solution with the arbitrary initial condition (3.27) at different times.
Chapter 6
Zero Dispersion Limit
In this chapter, we will study numerically the zero dispersion limit of the Burg-ers, Korteweg-de Vries and Benjamin-Ono equation. To do this, we introduce a parameter²> 0 in front of the dispersion term in the three equations,
ut+ uux− ²Huxx= 0, Benjamin-Ono, (6.1) ut+ uux+ ²uxxx= 0, Korteweg-de Vries, (6.2)
ut+ uux− ²uxx= 0, Burgers, (6.3)
for x∈ R and t > 0. Letting ² ↓ 0, with initial data independent of ², is what we refer to as the zero dispersion limit. If we let²= 0, we have the inviscid Burgers equation,
ut+ uux= 0. (6.4)
Formally, the solutions to the three equations in (6.1)-(6.3) should give us the so-lution to the inviscid Burgers equation as²↓ 0. However, it has been shown both for the KdV equation [LL05a, LL05b, LL05c, DVZ97], and for the BO equation [Mat98, MX11, Xu10], that this is not the case when the solution develops shocks.
The inviscid Burgers equation will develop shocks and/or rarefactions in finite time with certain initial conditions. For the Burgers equation, it is known that the solution will approach the solution to the inviscid Burgers equation.
We will now investigate the solutions to the three equations numerically, with smaller and smaller². As initial condition will we use an exact solution to the inviscid Burgers equation that immediately develops a shock and a rarefaction.
We can find the exact weak solution to the inviscid Burgers equation by the method of characteristics.
6.1 Weak Solution to the Inviscid Burgers Equation
We have the following initial value problem for the inviscid Burgers equation, ut+ uux= 0, u(x, 0)= u0(x)=
{1, −1 < x < 1,
0, otherwise, x∈ R, t > 0. (6.5) Immediately we get a shock at x= 1, and a rarefaction at x = −1. The shock will have a speed of 1/2, and the rarefaction will catch up with the shock after t= 4. So for t = (0,4) we have the following weak solution, found by the method of characteristics,
u(x, t)=
0, x< −1,
x+1
t , −1 < x < t − 1, 1, t− 1 < x <12t+ 1, 0, x>12t+ 1.
(6.6)
(a) t=0 (b) t=1
(c) t=2 (d) t=4
Figure 6.1: The weak solution to the initial value problem in (6.5).
We have chosen the initial condition such that it is reasonable to approximate the solution numerically with a spatially periodic initial value problem. The reason for doing this, is that all our methods are periodic.