[petsc-users] Dirichlet boundary condition for a nonlinear system
Barry Smith
bsmith at mcs.anl.gov
Tue Feb 18 22:38:03 CST 2014
So you could solve a sequence of problems where you move the boundary condition from being “easy” to the final value you want as a kind of continuation method?
Hmm, wonder how you would do that cleanly in PETSc.
Barry
On Feb 18, 2014, at 10:31 PM, Sanjay Govindjee <s_g at berkeley.edu> wrote:
> 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 <fd.kong at siat.ac.cn> 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,
>
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