# Boundary condition

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 July 13, 2013, 10:35 Boundary condition #1 New Member   Join Date: Apr 2013 Posts: 15 Rep Power: 4 Hi all, I have equations to solve, which are where 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

 July 13, 2013, 11:37 #2 Senior Member   Lefteris Join Date: Oct 2011 Location: UK, Greece Posts: 187 Rep Power: 5 You have to set boundary conditions on all the boundaries of your domain. __________________ Lefteris

July 13, 2013, 12:21
#3
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Filippo Maria Denaro
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Quote:
 Originally Posted by shubiaohewan Hi all, I have equations to solve, which are where 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.

July 15, 2013, 04:33
#4
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Quote:
 Originally Posted by FMDenaro 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?

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.

July 15, 2013, 04:51
#5
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Filippo Maria Denaro
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Quote:
 Originally Posted by shubiaohewan 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 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

Last edited by FMDenaro; July 15, 2013 at 06:48.

August 18, 2013, 11:42
#6
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Quote:
 Originally Posted by Aeronautics El. K. 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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