I want to simulate the following programme on pdepe solver with third type inlet boundary of Heaviside nature with t0 = 140 days and a zero flux ourlet boundary

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Thomas TJOCK-MBAGA
评论： Sam Chak 2022-4-5
fonction [p1, q1, pr, qr]= pdex1bc (x1, u1, xr, ur) P1 =1; q1 = 1; pr = 0.71*(1+0.2*1)*Ur; qr = 1;
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Torsten
Torsten 2022-4-5
编辑：Torsten 2022-4-5
pl = ul - C0 * ( 1 + exp(-2*k*t0) )/( 1 + exp(2*k*(t-t0)) );
ql = 0.0;
I suggest you plot the function
f(t) = C0 * ( 1 + exp(-2*k*t0) )/( 1 + exp(2*k*(t-t0)) )
to deduce a suitable value for k.
Are you sure about your setting for f in pdex1pde ?
Note that the flow velocity will be
v = 0.71*(1+0.2*x)
, but that this setting will generate an artificial source term
s = -0.71*0.2*u
in your model.
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Torsten
Torsten 2022-4-5
@Thomas TJOCK-MBAGA Comment moved here:
Yes I'm sure about the expression of f. I have computed many numericals solution with pdepde solver usingba first type inlet boundary with this expression of f and other in the samedi form. They match well with analytical solution using GITT ans other analytical solutions. Now i wanted to plot the same problem but using third type boundary and using also a pulse boundary (Heaviside source).
Torsten
Torsten 2022-4-5
编辑：Torsten 2022-4-5
Ok. Then in your code variables, which boundary condition at the inlet do you want to set ?
0.6*dC/dx = 0.71*(C-f(t))
with f(t) defined as above ?

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Thomas TJOCK-MBAGA

Yes this is the boundary condition in the inlet that i wanted to set but written as follow -0.6*dC/dx + 0.71*C = C0 And un the ourlet dC/dx = 0.

Now i also wanted to set thr inlet boundary written like this: 0.6*dC/dx + 0.71*C = C0 for 0 < t =< t0 and 0.6*dC/dx + 0.71*C = 0 for t >t0 That is the definition of the pulse boundary.

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Torsten
Torsten 2022-4-5
If the diffusive flux is given by
f = 0.6*(1+0.2*x)^(1.5)*DuDx-0.71*(1+0.2*x)*u;
the settings for the left boundary are
pl = 0.71 * C0 * ( 1 + exp(-k*t0) )/( 1 + exp(k*(t-t0)) );
ql = 1.0;
(graph the function f(t) = C0 * ( 1 + exp(-k*t0) )/( 1 + exp(k*(t-t0)) ) with your values for C0 and t0 to get a suitable k)
and for the right boundary
pr = 0.71*1.2*ur;
qr = 1.0;
Sam Chak
Sam Chak 2022-4-5
@Torsten
Got to forgive him. If the View PDF and Download buttons can be missed, then it is not surprising that the "Comment" field was overlooked too. However, it is believed that the replies were genuine.

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