The Solution of Burgers’ Equation for high Re’s

As the Reynolds number is ratcheted up to values of Re=1000, the equation becomes highly stiff. A 60-element MOL approach yields a solution that looks like the one shown below. The solution was plotted using the Rosenbrock method with the Shampine coefficients.

Severe discontinuities appear in the solution as one marches along the time domain. The number of elements along the spatial domain are clearly not adequate to capture the essence of the boundary value PDE. (How does one know what the “desired” solution is? In a case like this, one can cheat and refer to the literature to obtain it, as in the case of Blom et al’s paper).

Plotted below is the solution to the Burgers’ equation with Re=1000, but with 160 elements. The discontinuities appear to have quieted down with the increased number of elements. 

 

A Solution of the Burgers’ Equation

As intimidating as it looks, the Burgers’ equation does have closed form solutions for a handful of cases. Blom et al present a few such examples of initial value problems that do offer closed form solutions.

Presented here is a solution of the Burgers’ equation, for a Reynolds number of 100. The initial values at t=0 follow the solution:

 

A 60-element MOL approach was implemented with the Rosenbrock-Shampine method. Plotted below is the 3-D solution.

The real challenge, however, is to obtain solutions for Re's >= 1000! 

Rosenbrock – Shampine’s parameters

Shampine (Implementation of Rosenbrock Methods) derived a set of coefficients for the Rosenbrock methods, with a gamma value of 0.5. Presented below are the coefficients for Shampine’s method, with the scaling accounted for direct implementation in the generalized method (ref: here). The generalized Rosenbrock algorithm can be easily implemented with these coefficients.

         A21=2.
         A31=48./25.
         A32=6./25.
         C21=-8.
         C31=372./25.

         C32=12./5.
         C41=-112./125.
         C42=-54./125.
         C43=-2./5.
         B1=19./9.
         B2=1./2.
         B3=25./108.
         B4=125./108.
         E1=17./54.
         E2=7./36.
         E3=0.
         E4=125./108.
         GAMMA=.5
         C2= 0.1000000000000000E01
         C3= 0.6000000000000000
         D1= 0.5000000000000000
         D2=-0.1500000000000000E01
         D3= 0.2420000000000000E01
         D4= 0.1160000000000000