Numerical Tests With The MEBDFSO Code

Here we present numerical results obtained when the MEBDFSO code was applied to some ODEs that arise in the numerical method of lines solution of time dependent partial differential equations.

In each case NAME.F will denote the drivers and NAME.RES will denote the numerical results obtained. The file NAME.DAT specifies the PDE and should be in the same working directory as the three Fortran files. The file DISCRT.F contains the spatial approximation we used. The runs were done on an IBM RS6000. We used the Fortran 77 compiler with optimization: f77 -O3 MEBDFSO.F YALE.F NAME.F INITIAL.F DISCRT.F.

In what follows the subscripts in T and X denote partial derivatives with respect to these independent variables. To solve each of the following PDEs we used a uniform mesh spacing, DX, of N points, so that DX =(XU-XL)/(N-1), where XU, XL are the upper and lower limits of X respectively. We then applied finite difference approximations to the spatial derivatives. This reduces the problem to a linear system of ODEs with N equations. The Method of lines and some of the spatial approximations we used are explained in detail in [8].

Burgers´ equation in 1 dimension.

     U  = Vis * U  -U*U                 0< X < 1
      T          XX    X

The exact solution is :
      
 U(X,T)=PHI(X,T)=(0.1D+00*EA+0.5D+00*EB+EC)/(EA+EB+EC),

where
 EA=(0.05D+00/VIS)*(X-0.5D+00+4.95D+00*T)
 EB=(0.25D+00/VIS)*(X-0.5D+00+0.75D+00*T)
 EC=( 0.5D+00/VIS)*(X-0.375D+00).

The initial and boundary conditions are:

 U(X,0) = PHI(X,0) 
 U(0,T) = PHI(0,T)
 U(1,T) = PHI(1,T).

Here Vis denotes the viscosity. This parameter significantly affects the character of the equation. For small Vis the equation is hyperbolic, while for large Vis it is parabolic.

For our test we used Vis = 0.003 and N=201. The spatial derivatives

 U  and U   are approximated by a cubic spline polynomial.
  X      XX

To run this problem it is necessary to compile the first three Fortran programs together and the results are in the BURGER1D.RES file.

Burgers´ equation in 2 dimensions.

     U  = -U*U  + Vis*U   - U*U + Vis*U
      T       X        XX      Y       YY

where 0 < X < 1, 0 < Y< 1 and 0 < T < 2.

The exact solution is :

 U(X,Y,T)=UA(X,Y,T)=1/(1+EXP(X/(2*Vis)+Y/(2*Vis)-2*T/(4*Vis)))

The initial and boundary conditions are:

 U(X,Y,0) = UA(X,Y,0) 
 U(0,Y,T) = UA(0,Y,T)
 U(1,Y,T) = UA(1,Y,T)
 U(X,0,T) = UA(X,0,T)
 U(X,1,t) = UA(X,1,T).

We ran the problem with Vis=0.1 and 31 spatial grid points in both the X and Y direction. This gives 961 ordinary differential equations

We used the following spatial derivative approximations:


U  is approximated by a five point centered approximation
 X      

and U   is found by differentiating U  using the same 
     XX                              X

five point centered approximation.

 U  is approximated by a five point biased upwind approximation.
  X

      
Similarly for the spatial derivatives with respect to Y.

For more details about the spatial approximation see [8]. The error in the numerical solution is computed at the point (X=(NX/2),Y=(NY/2)).

Again the results are obtained by compiling the first three Fortran programs together and the results are in the BURGER2D.RES file.

Chemical flooding equation [4].

     U  = D*U   - U         -2 < X < 2, 0 < T < 2
      T      XX    X

The initial and boundary conditions are:

 U(X,0) = EXP(-X^2)
 U(-2,T)= [1/SQRT(1+4*D*T)]*EXP[-(-2-T)^2/(1+4*D*T)]
 U( 2,T)= [1/SQRT(1+4*D*T)]*EXP[-(2-T)^2/(1+4*D*T)]

The analytical solution: 

 U(X,T) = [1/SQRT(1+4*D*T)]*EXP[-(X-T)^2/(1+4*D*T)]

We ran the problem with N=2001 and D=2D-5. The spatial derivatives

 U  and U   are approximated by a cubic spline polynomial.
  X      XX

To run this problem it is necessary to compile the first four Fortran programs together where the file EXACT.F contains the reference solution and the results are in the FLOOD.RES file.

An expensive PDE.

                 2                  2
     U  = U   + U  - 0.5*U*S -(.5)*S     
      T    XX    
where
 S = SUM[ COS(I*X)*EXP(-I^2*T)], I=1,...,P
 0 < T < 5.0 and 0 < X < 1.0

The initial and boundary conditions are:

 U(X,0)= SUM[ COS(I*X) ],                 I,1,..,P
 U(0,T)= SUM[ EXP(-I^2*T) ]               I=1,..,P
 U(1,T)= SUM[ COS(I)*EXP(-I^2*T) ],       I=1,..,P

The analytical solution: 

 U(X,T) = SUM[ COS(I*X)*EXP(-I^2*T)], I=1,...,P

It is important to note that the equation we are solving is really the simple heat equation. We wish to test the performance of our algorithm as function evaluations become more expensive. To do this we give the results with P=1, P=10 and P=25. We used N=201 with the following spatial approximation :

U  : By five point, fourth order finite difference approximation.
 X

U  : By second order finite difference approximation. 
 XX

To run this problem it is necessary to compile the first four Fortran programs together where the file EXACT.F contains the reference solution.

Korteweg-de Vries equation (KDV).
The KDV equation is a classical nonlinear PDE, originally formulated to model shallow water flow. It is given by
```
     U  = -6U*U  - U         -INF < X < INF
      T        X    XXX

The initial condition is:

    U(X,0) = (1/2)*c*sech^2{SQRT(c)*X/2}.

The analytical solution:
 
    U(X,T) = (1/2)*c*sech^2{SQRT(c)(X-c*T)/2}

which is the equation for a soliton traveling from left
 
to right with velocity c and height (1/2)*c.
```
For our test we approximate the space domain by -30 < X < 70 with N=400. We used a fifth degree spline approximation to approximate the spatial derivatives. The results given here are for one soliton (KDV1S.RES), two equal solitons (KDV2ES.RES) and two unequal solitons (KDV2NES.RES).
To run this problem it is necessary to compile the first four Fortran programs together where the file EXACT.F contains the reference solution.
- KDV.F
- INITIAL.F
- DISCRT.F
- EXACT.F
- KDV.DAT
- KDV1S.RES
- KDV2ES.RES
- KDV2NES.RES

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Last updated April 02, 2000.

Jeff Cash (j.cash@ma.ic.ac.uk)