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Old   March 22, 2000, 05:17
Default boundary conditions and convergence
Jongtae Kim
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Currently I am solving Laplace or Poisson equations such as heat conduction equation based on FVM.

For heat conduction on a square domain, Dirichlet BC makes the problem solved very quickly. But when I used Neumann-type BCs like heat-flux or convection condition q_b = h*(T_b - T_inf), the convergence slows down seriously.

I just think that the Neumann BCs destory the diagonal dominance of the discrete eq. of the boundary cells. Is there any good methods to make them diagonal-dominant? If there is any other reason for the convergence problem could you give me some teachings?

Sincerely yours,

Jongtae Kim
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Old   March 22, 2000, 10:06
Default Re: boundary conditions and convergence
John C. Chien
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(1). The slow convergence is due to the fact that the solution in the computational domain in T is changing all the time, trying to satisfy the gradient boundary conditions. Everything is floating except for the boundary gradient condition itself. (2). For the fixed boundary temperature type, the variable around the boundary is fixed and known, therefore, the convergence rate is a function of the distance of the unknown temperature from the boundary. For the variables near the boundary, the extent of the fluctuation is limited by the fixed temperature already. (3). So, the boundary condition itself changes the nature of the problem, and it affects the convergence of the process. The convergence of iterative methods is a different story, which deals with the numerical information propagation inside the computational domain.
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Old   March 22, 2000, 10:19
Default Re: boundary conditions and convergence
franck bertagnolio
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You have to check first that you fulfill the compatibility condition required for the Neumann problem, i.e. that the sum of the source term inside the domain is equal to the sum of the boundary fluxes (Gauss theorem). Moreover, this should be satisfied in the discrete sense!!

To increase the diagonal dominance, any iterative method can be written more or less as: dx/dt + f(x) = 0 where x is your unknow, f(x)=0 your actual equation, t being a pseudo-time step. Appropriately discretizing this pseudo-time derivative allows to recover a certain degree of diagonal dominance, at least when solving the problem at each iteration step.

Hope this helps.

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