3
$\begingroup$

I'm interested in applying Robin boundary condition to a convection-diffusion problem in 1D.

In the following system, $$\frac{\partial C}{\partial t} = D\frac{\partial ^2 C}{\partial x^2} - v\frac{\partial C}{\partial x}$$

$$\frac{\partial C}{\partial t} = \frac{\partial}{\partial x}( D\frac{\partial C}{\partial x} - vC)$$ To implement no-flux boundary condition ,flux $$ N = D\frac{\partial C}{\partial x} - vC $$ is set to zero at the left and right boundary.

I'd like to know whether the sign of terms in the flux will vary at the right and left boundary.

According to the following description given in wiki,

enter image description here

enter image description here

Could someone explain if the above-mentioned method is the right way to implement?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

I think that the second boundary condition equation is incorrect. The first one should be right for both ends. Following your notation, the flux of mass in the domain should be: $$N = vC - D \frac{\partial C}{\partial x}$$ everywhere including the boundary points.

Keep in mind that the sign of the boundary condition value is positive if the direction of mass flux is positive and vice versa.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Many thanks for the response . However, I don't understand why you have denoted flux to be vC-DdC/dx and not DdC/dx - vC as represented in my post.Could you please explain? $\endgroup$
    – Natasha
    Commented Oct 30, 2019 at 2:29
  • $\begingroup$ The total fiux consists of two fluxes: The first one is advective flux that carries mass by the flow. This can be expressed in $vC$, which means the mass $C$ is transported along the direction of the flow. The second one is dispersive (or diffusive) flux that moves mass from higher concentration area to lower one. The direction of this flux is against the concentration gradient, so $-D \partial C / \partial x$. $\endgroup$
    – KJ Nam
    Commented Oct 30, 2019 at 4:46
  • $\begingroup$ Thank you. Your explanation makes sense. But I saw another post here and here that states flux (DdC/dx -vC).n =g .Side note: My flow is in the positive direction $\endgroup$
    – Natasha
    Commented Oct 30, 2019 at 5:06
  • $\begingroup$ It is amazing and unfortunate to see how inaccurate the information on the internet can be. I would rather recommend this Wikipedia document, en.wikipedia.org/wiki/…, to you. No matter where you look at, the equation should not vary. $\endgroup$
    – KJ Nam
    Commented Oct 30, 2019 at 5:15
  • $\begingroup$ I had a look at the description in the link that you provided. Over there, all the terms are on the LHS and the definition of flux that you stated holds true. But in pdepe solver of MATLAB the expression itself has the form of time derivative on LHS and rest on RHS. I would think in such a case N would be DdC/dx - vC. Second thoughts? $\endgroup$
    – Natasha
    Commented Oct 30, 2019 at 5:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.