I tested the modified code with uniform refinement and
exact_sol.extra_quadrature_order(1).  I printed out both l2 error and h1
error and got

time = 0.025, refinement step = 0, elements =    10240, l2_error =
0.002885, h1_error = 0.239394
Time = 0.025, refinement step = 1, elements =    40960, l2_error =
0.002643, h1_error = 0.207076
Time = 0.025, refinement step = 2, elements =   163840, l2_error =
0.002596, h1_error = 0.210480
Time = 0.025, refinement step = 3, elements =   655360, l2_error =
0.002575, h1_error = 0.214309
Time = 0.025, refinement step = 4, elements =  2621440, l2_error =
0.002569, h1_error = 0.215482

We can see l2_error monotonically decreases, however h1_error does not.
What is the possible reason?
Thanks.


--Junchao Zhang

On Thu, Jan 28, 2016 at 10:53 PM, Junchao Zhang <junchao.zh...@gmail.com>
wrote:

> I found an example of exact_sol.extra_quadrature_order()
> in adaptivity_ex3.C.
>
> Also, I found if I refined the mesh upfront (though the -n_refinements
> argument, e.g., set it to 5), L2 error would monotonically decrease with
> more refinements. Previously, I set n_refinements to 0 to get a small
> 10-elements-11-nodes mesh at the first time step for debugging purpose. Now
> I know that is questionable in view of error analysis as Roy explained.
>
> --Junchao Zhang
>
> On Thu, Jan 28, 2016 at 6:50 PM, Derek Gaston <fried...@gmail.com> wrote:
>
>> Also: it can depend on integration error in the integration of the L2
>> Error. We're still using quadrature to integrate... so if that quadrature
>> is poor you can "miss" solution features on a coarse grid that then show up
>> as you refine the mesh (which refines the quadrature too) leading to higher
>> error.
>>
>> One way to combat that is to use the "extra quadrature order" capability
>> when computing the L2 Norm of the error... it allows you to "fine up" your
>> quadrature to give you a better integral on the coarse mesh. I don't
>> remember exactly how to do that on the moment (currently on my phone on the
>> subway)... so if you snoop around a bit and don't find it... write back in.
>>
>> Derek
>> On Thu, Jan 28, 2016 at 1:36 PM Roy Stogner <royst...@ices.utexas.edu>
>> wrote:
>>
>>>
>>> On Wed, 27 Jan 2016, Junchao Zhang wrote:
>>>
>>> > Time = 0.025, refinement step = 0, elements =       10, l2_error =
>>> 0.443873
>>> > Time = 0.025, refinement step = 1, elements =       40, l2_error =
>>> 0.045196
>>> > Time = 0.025, refinement step = 2, elements =      160, l2_error =
>>> 0.131169
>>> > Time = 0.025, refinement step = 3, elements =      640, l2_error =
>>> 0.116789
>>> > Time = 0.025, refinement step = 4, elements =     2560, l2_error =
>>> 0.118175
>>> >
>>> >
>>> > I am curious why sometimes L2 error gets bigger, e.g., from r_step 1 to
>>> > r_step 2.  Don't more refinements give smaller errors?
>>>
>>> If your solve's discretization error is your only source of error, and
>>> if you're solving a self-adjoint problem, then more refinements should
>>> *always* give you smaller errors.
>>>
>>> This problem isn't self-adjoint, so the convection term can cause
>>> convergence to be more erratic, but I don't think that's the problem
>>> here.
>>>
>>> The problem might be that you've got two sources of discretization
>>> error here: the discretization for the solve, and the discretization
>>> for the initial conditions.  If you project the initial conditions and
>>> then refine, rather than refine and then project, you won't actually
>>> have improved your approximation of the initial conditions.  So you
>>> won't converge to the exact solution you want, you'll converge to the
>>> solution of the PDE with the wrong initial conditions.
>>> ---
>>> Roy
>>>
>>>
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>>
>
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