Implicit boundary conditions
 

Dear sir, I have read from one paper refer about implicit boundary conditions. What does it mean? Is there the explicit boundary conditions? How can we apply it to our code? Thank you very much sir.

Atit Koonsrisuk.

Re: Implicit boundary conditions
 

In explicit schemes, dependent variables are computed one after another and then the boundary conditions are applied.

Implicit schemes do not give explicit expressions are dependent variables at each grid point. They produce linear (algebraic) equations that relate a dependent variable(s) at a grid point to the dependent variable(s) at other grid points. There are two types of linear equations. [1] Linear equations resulting from implicit discretization of the governing equations in the interior of the domain & [2] Linear equations that result from discretization of the boundary conditions.

These linear systems of equations should be solved (together) to simultaneously obtain the values of the dependent variables in the interior of the computational domain and on the boundaries. So, there is no separate boundary condition routine in implicit schemes. If you use the linear equations [1] (obtained as described above) to get the dependent variables in the interior AND THEN use system of equations [2] to obtain boundary values, the scheme may not converge. [1] and [2] should be clubbed together and solved simultaneously.

Hope this helps.

Re: Implicit boundary conditions
 

Dear sir, Thank you very much for your kind explaination. I am sorry but I don't understand exactly. Could you please give some examples? Thanks.

Atit Koonsrisuk

Re: Implicit boundary conditions
 

Here is a simple example of a 1-D diffusion problem with an implicit scheme/B.C.

u_t = u_xx ( "-" denote derivatives).

You need an initial condition and two boundary conditions for this problem.

Initial condition : u(t=0,x) = uo(x)

Boundary conditions : i) u(t,x=0) = 0 at x=0 (Dirchlet BC)

ii) u_x(t,x=0) = 0 at x=1 (Neumann BC)

Implicit scheme: "n+1" : new time step values

"n" : old time step values

"i" : spatial location.

Let spatial discretization be done on a grid with spacing = 1 (i.e. Dx = 1).

Discrete first order backward Euler implicit equation:

a){u(n+1,i) - u(n,i)}/dt = {u(n+1,i+1)-2u(n+1,i)+u(n+1,i-1)}

Discrete BC : i) u(n+1,0) = 0

ii) u(n+1,imax+1) = u(n+1,imax) (first order one-sided approximation for the spatial derivative = 0, which is a discrete Nuemann BC).

When you solve the equations by inverting a matrix, you have "imax" number of equations resulting from the discretization of the equations at all interior points (as shown in (a) above) AND two discrete equations that give you the boundary values as shown in (i) and (ii) above.

You have "imax" number of interior points, 2 boundary points. (a) gives you "imax" equations and (i) & (ii) give you two more equations. So you can find the values of u(n+1,i) for all "i" from 0 to "imax+1".

You have to solve (a) and (i) & (ii) simultaneously. If you solve (a) and then (i) & (ii), and then go back to (a) and iterate, the scheme will not converge.

Hope this helps.