[firedrake] Pass approximate Jacobian to derivative() call

[Lawrence Mitchell]

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On 11/11/15 12:04, Buesing, Henrik wrote:
>>> is it possible to just pass the approximate Jacobian to the 
>>> derivative
>>>>> call? Since it's so complicated in my case to get the exact
>>>>> Jacobian I would like to do sth. like
>>>>> 
>>>>> 
>>>>> 
>>>>> du = [(F(u + h) - F(u - h))/(2*h)]
>>> 
>>> So you can't do this by passing something to derivative (which
>>>  instead computes the jacobian for you given a residual).
> 
> [Buesing, Henrik] Yes, true. I mixed this up. I want to do
> 
> solve(F == 0, u, J=approximate symbolic Jacobian)
> 
>>> 
>>> You can instead of course build an approximate, symbolic, 
>>> Jacobian "by hand" and provide that to the solver: you'll then
>>>  end up doing quasi- newton, rather than newton.
> [Buesing, Henrik] That's what I would like to do. In my 
> multi-physics case (see attachement) the Jacobian would have a 2x2 
> block structure.
> 
> 
> I'm unsure how to provide this... I tried to write sth. down for 
> the blocks (see test.py), but this is obviously wrong.

It looks like you getting mixed up between symbolic and actual
computation.  You look like you want to do the finite differencing
sort of by hand?

For example here:

delta = Constant(1e-8)
tmp = pw
eps = delta + delta*tmp
eps = assemble(eps)
pw = tmp + eps
val1 = F
pw = tmp - eps
val2 = F
pw = tmp
tmp00 = Constant(0.5)*(val1-val2)/eps

it looks like you want val1 to be F evaluated at tmp + eps and val2 to
be F evaluated a tmp - eps, but in fact, this is just symbolic
assignment so both val1 and val2 are just different names for F.

To write your approximate symbolic jacobian, you need to write J in
terms of functions of the coefficients and test and trial functions
(just like you did F).  Then it will have the appropriate block
structure, I think.


I don't quite understand what all the bits are doing, but it looks
like, for example the dF/dpw block is just zero (since F doesn't
depend on pw?)

Lawrence
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