Neumann boundary condition in a first order PDE

I'm trying to solve the following equation using PDEPE:

dC/dt + v * dC/dx = constant

With the boundary conditions:

C(t,0)=Cin dC(t,L)/dx=0

My question is how can I incorporate the second BC in the PDEPE syntax, if I should define f = [-v]. Is there any posibility to call the penultime value and make it equal to ur, so dC/dx=0? u(x_n) = u(x_(n-1))

Thanks for your cooperation!

Antonio

Answer 1

[Tom]

In this case you want to set

 pr = 0;
qr = -1/v; %to cancel out f

[Antonio]

Thanks for your promt reply. In this case, I would have:

p + q*f = 0.... 0 + (1/-v)*(-v) = 0... 1 = 0

What I need is to set dC/dx = 0. But I can´t since there is no derivative in the flux term (f).

[Tom]

The derivative is included in the q part, see the right hand boundary condition of example 1 of the pdepe help

[Tom]

You don't need to- the gradient is included. Can you provide some example code?

[Antonio]

In that case f = dudx...that's why q is just equal to 1. But in my case the PDE is a first order and f equals a constant.

[Antonio]

% dC/dt= -v * dC/dx - constant

function [c,f,s]=pde(u,t,x)

c=1;

f=-v;

s=-rp*(1-e)/e*.21;

%Boundary conditions

function [pl,ql,pr,qr] = pdebc(rl,ul,rr,ur,x,t)

global Cin

pl=[ul-Cin];

ql=[0];

pr=[?];

qr=[?];

[Tom]

I just realised- all the times I've used PDEPE my f term has included dudx in it... What is the equation you are solving?

[Antonio]

A mass balance in an adsorption column.

[Tom]

Can you post a link to the equation?

[Antonio]

https://www.dropbox.com/sh/3vp4f4ea1l1346f/9yjY_UKgCI/Dibujo.jpg

Thanks for your help, Tom. For now, im taking the last term as constant

[Tom]

I'm not sure how PDEPE can deal with two time terms, maybe it would be better to solve numerically in a for loop.

[Antonio]

About the last term (dq/dt), ill solve it differently. My problem now is this bc!