# The Navier-Stokes solver is second-order accurate in time

Let us test that for a 1D viscous-flow problem, evaluated on a 2D grid.

## set-up

The set-up consists of a periodic (left-right) channel with no-slip boundary conditions (top-bottom). The initialized flow is:

\displaystyle \overrightarrow{u_0}(x,y)= (u_0,v_0) =(\mathrm{cos}(y),0),

for $(y) /2 . Due to the fluid's viscousity ($), momemtum will diffuse over time, according to the (Navier-)Stokes equation, that is solved by;

\displaystyle \overrightarrow{u}(t)= \overrightarrow{u_0}e^{-\nu t}.

This equation will help to determine the error due to the time integration.

#include "grid/multigrid.h"
#include "navier-stokes/centered.h"

const face vector muv[]={1.,1.};
double timestepp;
int j;

int main(){

The grid resolution is chosen so fine that the presented errors here are dominated by the time stepping and are not significantly affected by the spatial discretization.

  init_grid(1<<9);
periodic(left);
u.t[top]=dirichlet(0);
u.t[bottom]=dirichlet(0.);
mu=muv;
L0=M_PI;
X0=Y0=-L0/2.;

Six experiments are run, the zeroth one (j=0) is not used because the timestepper thinks it has made a timestep with \Delta t = 0 for the -1-th timestep.

  for (j=0;j<7;j++){
timestepp=2.*pow(0.5,(double)j);
run();
}
}

event init(t=0){
CFL=10000.;
foreach()
u.x[]=cos(y);
boundary(all);
DT=timestepp;
}

This event is used to let the run have an equidistant timestep for j>0.

event setDT(i++){
if (i==0){
DT=1.1*timestepp/0.1;
}else{
DT=timestepp;
}
}

## Output

After a single ‘1/e’ timescale, the error in the numerically obtained solution is evaluated.

event check(t=1.){
static FILE * fp = fopen("resultvisc.dat","w");
double err=0.;
foreach()
err+=fabs(u.x[]-cos(y)*exp(-t))*sq(Delta);
if (j>0)
fprintf(fp,"%d\t%g\n",i,err);
}

## Results

  set xr [0.5:40]
set yr [0.02:1]
set logscale y
set logscale x
set xlabel '{Used number of Timesteps}'
set ylabel 'Total Error'
set key box 1
set size square
plot    (1*x**-1) lw 3 lc rgb 'purple' title '∝{dt}^{-1}',\
'resultvisc.dat' using 1:2 pt 4 title 'Error at t=1' A first-order accurate (global) timestepping can be diagnosed? (script)