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

fonction [p1, q1, pr, qr]= pdex1bc (x1, u1, xr, ur) P1 =1; q1 = 1; pr = 0.71*(1+0.2*1)*Ur; qr = 1;

[Torsten]

What is "third type inlet boundary of Heaviside nature" ?

[Torsten]

@Thomas TJOCK-MBAGA Comment moved here:

-D dC(x=a, t) + u*C(x=a, t) = C0 0<t=<t0 and 0 else (t>t0)

OR -D dC(x=a, t) + u*C(x=a, t) = C0 H(t0-t) Where H is the heaviside step function For my problem a =0

[Sam Chak]

@Torsten

third -type inlet source boundary

Found the keyword here. The article is open access.

https://www.sciencedirect.com/science/article/pii/S0309170821001731

See Fig.1 for the idea of inlet boundary.

(Under a Creative Commons license)

Reference:

Liao, Z.-Y., Suk, H., Liu, C.-W., Liang, C.-P., & Chen, J.-S. (2021). Exact analytical solutions with great computational efficiency to three-dimensional multispecies advection-dispersion equations coupled with a sequential first-order reaction network. Advances in Water Resources, 155, 104018. doi:10.1016/j.advwatres.2021.104018

[Thomas TJOCK-MBAGA]

Can i have this paper ???

[Sam Chak]

Hi @Thomas TJOCK-MBAGA

I mentioned it clearly that it is an open-access paper. Click into link and download the PDF as you wish.

More importantly, please check if @Torsten's suggested solution works for you.

[Thomas TJOCK-MBAGA]

can't get hold of the pdf version. When I click on the link in any browser even in search sites I just see the HTML version

[Sam Chak]

Step 1: Click View PDF.

Step 2: Click the Download button to save the PDF.

Answer 1

[Torsten]

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.

[Thomas TJOCK-MBAGA]

In your proposition you used the first type boundary condition of a continuons source. How about the third type ???

[Torsten]

How about my question ?

If you have to modify your f, I have to modify the settings for the boundary condition.

[Torsten]

@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]

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 ?

Answer 2

[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.

[Torsten]

The usual condition at the inlet is

D*dC/dx = u*(C-C0)

Why in your case

D*dC/dx = u*C - C0

?

And please stop creating answers when you make a comment.

[Torsten]

@Thomas TJOCK-MBAGA Comment moved here

Please i have forgeten this the conditions as you stated

[Torsten]

@Thomas TJOCK-MBAGA Comment moved here

Thanks you, that is the rigth condition, it result the first problem that i have. I wrote -0.6*dC/dC + 0.71*C = C0(t) but it is 0.71*C0(t). Thanks. Now how Can i treat 0.6*dC/dC + 0.71*C = 0.71*C0(t) H(t0-t) ??? Where H(t0-t) is the Heaviside function. Meaning that C0(t) is only defined for values of t range for0 to t0 and is equal to zéro for t>t0. That is my stronghly problème..

[Torsten]

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]

@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.