For correction, the right code for the second method is
for (; cell != endc; ++cell) {
local_rhs = 0;
fe_values.reinit(cell);
fe_values.get_function_values(old_solution, old_sol_values);
for (unsigned int q = 0; q < n_q_points; ++q ){
for (unsigned int i = 0; i < dofs_per_cell; ++i) {
const double phi_i = fe_values.shape_value(i, q);
local_rhs(i) += old_sol_values[q] * phi_i * fe_values.JxW(q);
}
}
cell->get_dof_indices(local_dof_indices);
for (unsigned int i = 0; i < dofs_per_cell; ++i) {
system_rhs(local_dof_indices[i]) += local_rhs(i);
}
}
On Saturday, April 4, 2020 at 11:18:12 PM UTC-7, Xiang Sun wrote:
>
> Hi, I am implementing the backward Euler method to solve a time-dependent
> problem like u(n+1)= h*f(n+1)+u(n). n is the nth time step. h is the time
> step length. f is a function of time. I tried two ways to assemble the
> system right-hand side(rhs).
>
> One is to form a mass matrix first, then time the old value of u at the
> node. This way is similar to Step-26. The code is like
> mass_matrix.vmult(system_rhs, old_solution);
>
> The other way is to get the value of u at quadrature points by
> interpolation, then do the integral in cells to get the local rhs, finally
> assemble the system rhs by the local rhs. This method is from Step-31 and
> shown in the following code.
>
> for (; cell != endc; ++cell) {
> local_rhs = 0;
> fe_values.reinit(cell);
> fe_values.get_function_values(old_solution, old_sol_values);
> for (unsigned int q = 0; q < n_q_points; ++q ){
> for (unsigned int i = 0; i < dofs_per_cell; ++i) {
> const double phi_i = fe_values.shape_value(i, q);
> local_rhs(i) += old_sol_values[q] * phi_i * fe_values.JxW(q);
> }
>
> }
> cell->get_dof_indices(local_dof_indices);
> for (unsigned int i = 0; i < dofs_per_cell; ++i) {
> system_rhs(local_dof_indices[i]) += local_rhs(i)*0;
> }
> }
>
> I thought these two methods should give the same answer, but they can't. I
> don't know where I made a mistake. Thank you very much.
>
>
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