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shubiaohewan July 13, 2013 10:35

Boundary condition
 
Hi all,

I have equations to solve, which are
\frac{\partial A}{\partial t}=\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y}
\frac{\partial B}{\partial t} = \frac{\partial B}{\partial x}
where A=A(x,y,t), B=B(x,y,t)

The initial conditions are zero. The boundary condition are A(x=0)=a, and B(x=0)=b. My question is do I additionally have to impose boundary condition for B(y=0)=? It seems I don't have to, is this right?

Shu

Aeronautics El. K. July 13, 2013 11:37

You have to set boundary conditions on all the boundaries of your domain.

FMDenaro July 13, 2013 12:21

Quote:

Originally Posted by shubiaohewan (Post 439532)
Hi all,

I have equations to solve, which are
\frac{\partial A}{\partial t}=\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y}
\frac{\partial B}{\partial t} = \frac{\partial B}{\partial x}
where A=A(x,y,t), B=B(x,y,t)

The initial conditions are zero. The boundary condition are A(x=0)=a, and B(x=0)=b. My question is do I additionally have to impose boundary condition for B(y=0)=? It seems I don't have to, is this right?

Shu


First, the boundary conditions must be prescribed according to the mathematical character of the system, hyperbolic and parabolic equations require BCs on open domains, only elliptic equations require to prescribe BCs on the whole frontier.

Second, I see that this is not exactly a coupled system, you can solve the second equation indipendently. Also, the equation for B does not prescribe a dependence on y.

shubiaohewan July 15, 2013 04:33

Quote:

Originally Posted by FMDenaro (Post 439548)
First, the boundary conditions must be prescribed according to the mathematical character of the system, hyperbolic and parabolic equations require BCs on open domains, only elliptic equations require to prescribe BCs on the whole frontier.

Second, I see that this is not exactly a coupled system, you can solve the second equation indipendently. Also, the equation for B does not prescribe a dependence on y.

Hi FMDenaro,

Thank you for your reply. I agree with your first point. I think I don't have to specify the y boundary condition for B.

The second equation will evolve by itself as you point out. But, is coupling a problem?
What about \frac{\partial A}{\partial t}=\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y}
\frac{\partial B}{\partial t} = A\frac{\partial B}{\partial x}

I guess I still don't have to specify the y boundary condition in B, is that? Because there is no y-direction propagation in the second equation.

FMDenaro July 15, 2013 04:51

Quote:

Originally Posted by shubiaohewan (Post 439749)
Hi FMDenaro,

Thank you for your reply. I agree with your first point. I think I don't have to specify the y boundary condition for B.

The second equation will evolve by itself as you point out. But, is coupling a problem?
What about \frac{\partial A}{\partial t}=\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y}
\frac{\partial B}{\partial t} = A\frac{\partial B}{\partial x}

I guess I still don't have to specify the y boundary condition in B, is that? Because there is no y-direction propagation in the second equation.


But this second equation is now different ...it contains A and is coupled to the first. In any case it is valid for any y...
I suggest to classify the system....

Then, at a steady state, if A <> 0 you see that B must be a function only of y, then from the first equation you have dA/dx = f(y) -> A(x,y) = f(y) (x-x0) + A0

leflix August 18, 2013 11:42

Quote:

Originally Posted by Aeronautics El. K. (Post 439546)
You have to set boundary conditions on all the boundaries of your domain.


I would say that I agree with Aeronautics.

When you will discretize the first equation for A, you will need to know B at the boundary since you have the operator dB/dy involved in the equation. However if you use left sided FD scheme instead of central scheme you could avoid to specify B on the boundary.
Indeed you don't have y propagation of B in the B equation, but you have it on the A equation. So everything will depend on the scheme you use to discretize dB/dy in the A equation.


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