One alternate to this is that on the first step the initial value of X_B
is not
taken as the know value but rather some prior known value. Then for the
first
iteration Y_B turns out to be the known increment; in subsequent
iterations it
is take as 0. This can sometimes be helpful with convergence in tough
problems.
-sg
On 2/18/14 5:56 PM, Barry Smith wrote:
On Feb 18, 2014, at 5:39 PM, Fande Kong <[email protected]> wrote:
Hi all,
I am just trying to solve a nonlinear system resulted from discretizating a
hyperelasticity problem by finite element method. When I solve a linear PDE, I
never put boundary solution either in a solution vector or a matrix, but
instead, I put boundary condition to the right hand size (load).
You adjust the right hand side to have zero as the boundary conditions.
This can be written as
(A_II A_IB ) ( X_I ) (F_I)
(A_BI A_BB)(X_B) = (F_B)
Which is equivalent to
(A_I A_B) (X_I) (F_I) - (A_B)*(X_B)
(0) =
A_I X_I = F_I - A_B*X_B
In the nonlinear case you have
F_I(X_I,X_B) = ( 0 )
F_B(X_I,X_B) ( 0)
where you know X_B with Jacobian
(J_II J_IB)
(J_BI J_BB)
Newtons’ method on all variables gives
(X_I)^{n+1} = (X_I)^{n} + (Y_I)
(X_B) (X_B) (Y_B)
where JY = F which written out in terms of I and B is
(J_II J_IB) (Y_I) = F_I( X_I,X_B)
(J_BI J_BB) (Y_B) F_B(X_I,X_B)
Now since X_B is the solution on the boundary the updates on the boundary
at zero so Y_B is zero so this system reduces to
J_II Y_I = F_I(X_I,X_B) so Newton reduces to just the interior
with
(X_I)^{n+1} = (X_I)^{n} + J_II^{-1} F_I(X_I,X_B)
Another way to look at it is you are simply solving F_I(X_I,X_B) = 0 with
given X_B so Newton’s method only uses the Jacobian of F_I with respect to X_I
Barry
How can I do a similar thing when solving a nonlinear system using a newton
method?
Thanks,
Fande,
