Kyusik,
Am Donnerstag, 3. November 2016 10:51:20 UTC+1 schrieb hank...@gmail.com:
>
> Hi all,
>
> I'm trying to calculate absolute values of (exact_solution -
> numeric_solution) and visualize it using Visit.
>
> So, I added the following in my code.
>
> error = solution;
> error -= exact_solution;
>
> for(unsigned it i=0;i<error.size(),++i)
> error(i)=std::fabs(error(i));
>
> ...
> data_out.add_data_vector(solution,"solution");
> data_out.add_data_vector(exact_solution,"exact_solution");
> data_out.add_data_vector(error,"error");
>
Yes, this gives you a representation of the discretization error.
And, I also calculated the {l2_norm of (exact_solution - numeric_solution)
> / sqrt(#cells)} (I divided l2norm by sqrt(#cell) Because I want to know
> kind of average value of difference between two solution.)
>
> i.e. I want to calculate sqrt{sum_i(exact_sol_i - sol_i)^2 / (#cell)} (I
> think it is kind of average value of difference between two solutions)
>
What do you mean by "average value" here? You probably just wnat to compute
the error in the L2-norm, isn't it?
> So, I added the following in my code....
>
> Vector<float> difference_per_cell (triangulation.n_active_cells());
> VectorTools::integrate_difference(dof_handler,
> solution,
> ExactSolution<dim>(),
> difference_per_cell,
> QGauss<dim>(2),
>
> VectorTools::L2_norm);
> const double L2_error = difference_per_cell.l2_norm();
>
> std::cout<<"l2norm(error)="<<L2_error/triangulation.n_active_cells()<<std::endl;
>
> So, l2_norm(error) is 1.14461e-09
>
> But the almost all values in the plot of error(Error_fe1.png) are about
> 10^-4 that is 10^5 times higher than l2norm.
>
If the error is O(1.e-4) you expect the L^2 error to be approximately
\sqrt{\int_\Omega 1.e-8} = |\Omega|^{1/2} 1.e-4.
Assuming |\Omega|~1, this would suggest that you use approximately
1.e-4/1.e-9=1.e5 cells.
What I want to say: Consider just the L2-norm! Due to |\Omega|~1, this
should give you an error in the magnitude of what you are observing in the
plot.
Best,
Daniel
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