# Source of a river

In this example we impose a variable flow rate for a “source” located inside the computation domain.

``````#include "saint-venant.h"
#include "discharge.h"``````

The domain is 10 metres squared, centered on the origin. Time is in seconds.

``````#define LEVEL 8

int main()
{
size (10.);
origin (- L0/2., - L0/2.);
G = 9.81;
N = 1 << LEVEL;
run();
}``````

## Boundary conditions

We create a new boundary for the source, with a Neumann condition for the normal velocity (i.e. an inflow).

``````bid source;
u.n[source] = neumann(0);``````

The flow rate varies in time and is set by computing the the elevation ${\eta }_{s}$ of the water surface necessary to match this flow rate.

``````double etas;

event inflow (i++) {
etas = eta_b (0.1*(1.1 - cos(4.*π*t)), source);
h[source] = max (etas - zb[], 0);
η[source] = max (etas - zb[], 0) + zb[];
}``````

## Initial conditions

The river bed is a single valley. The source is created by masking and is a narrow slot located at the head of the valley.

``````event init (i = 0)
{
mask (fabs(x) < 0.5 && fabs(y - 3.5) < Δ/2. ? source : none);``````

We start with a dry riverbed, so that the problem does not have a natural timescale the Saint-Venant solver can use. We set a maximum timestep to set this timescale.

``````  DT = 1e-2;

foreach()
zb[] = (- cos(x) + y)/2.;
boundary ({zb});
}``````

## Outputs

We compute the time-derivative of the total water volume (i.e. the net flow rate), and make a GIF movie.

``````event logfile (i++; t <= 2.) {
static double volo = 0., to = 0.;
double vol = statsf(h).sum;
if (i > 0)
fprintf (ferr, "%g %.6f %g %g\n", t, (vol - volo)/(t - to), vol, etas);
volo = vol, to = t;
}

event output (i += 5) {
static FILE * fp = popen ("ppm2gif > source.gif", "w");
output_ppm (h, fp, min = 0, max = 0.05);
}``````