Message Boards Message Boards

Solving equation on boundaries

Posted 3 months ago

Hello colleagues, I have a problem that is interesting.

I have the following equations:

1) diffusion equation

D[cb[t, x], {x, 2}] == D[cb[t, x], t]/d

d is the diffusion coefficient,

2) Adsorption on the boundary (x=0)

D[s[t], t] == k1*cb[t, 0]*s[t, x] -sat) -  k2*s[t]

The BC and IC:

D[s[t], t]==d*D[cb[t, x] at x=0

cb[t,a]== c0 (a and c0 constats)
cb[0,x] ==c0
s[0]==0

I found very akward to solve it. Any way to solve it? Or referentes about coupling ODE and PDE in Wolfram?

POSTED BY: Mnele Wolfram
6 Replies
Posted 3 months ago

Dear Mariusz Iwaniuk,

Greetings from Brazil.

Thanks for the solution. I particularly enjoyed the analytical one.

Please note that the ODE and PDE are coupled at the boundary:

D[s[t], t] == d*D[cb[t, x]] at x=0

The other BCs. are:

cb[t,a]== c0 (a and c0 constats)
cb[0,x] ==c0
s[0]==0
POSTED BY: Mnele Wolfram

I wrote earlier that: D[cb[t, x] have a syntax errors ?

what do you mean by: D[cb[t, x] ?

D[s[t], t] == d*D[cb[t, x],t] (* ? *)
D[s[t], t] == d*D[cb[t, x],x] (* ? *)
D[s[t], t] == d*D*cb[t, x]    (* ? *)
POSTED BY: Mariusz Iwaniuk
Posted 3 months ago

Apologize,

my bad.

D[cb[t, x], {x, 2}] == D[cb[t, x], t]]/d

D[s[t], t] == k1*cb[t, 0] * (s[t] -sat)  (* this is a BC*)

Double checking the equations in Wolfram now. No more mistakes.

POSTED BY: Mnele Wolfram
Posted 3 months ago

Thanks for the reply.

Sorry, I simplified the equations and some typos appeared;

D[cb[t, x], {x, 2}] == D[cb[t, x], t]/d

D[s[t], t] == k1*cb[t, 0] * (s[t] -sat)

s is only function of t.

POSTED BY: Mnele Wolfram

POSTED BY: Mariusz Iwaniuk

if x=0 why you have s[t, x] in the equation ?

What is : sat) ? and D[cb[t, x] ? you have a syntax errors.

POSTED BY: Mariusz Iwaniuk
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract