Non-linear resolution

We will now attempt a second approach. In reality, the parameters \(\sigma\) and \(k\) depend on the temperature in the following way:

\[\begin{align} \sigma(T) = \frac{\sigma_0}{1+\alpha(T-T_0)} k(T)=\frac{k_0}{1+\alpha(T-T_0)}\frac{T}{T_0} \end{align}\]

It is therefore necessary to update them every step of the time. To do this we will proceed as follows :

At time 0, \(\sigma_0\) and \(k_0\) are fixed and then the equations mqs are solved. Then, a fixed point method, here picard, is applied to approach the non-linear terms of the heat equation until a satisfactory relative error is obtained. Finaly \(\sigma\) and \(k\) are updated for the following time step.

1. Implementation under feelpp

The code is relatively close to that of the previous part, there are only a few small changes :

We define the tolerance for picard iterations, as well as the maximum number of iterations tolerated.

double dt = doption(_name = "ts.time-step");
double tmax = doption(_name = "ts.time-final");

We define \(\sigma\) in another way: We update the value of T at each time step with the previous value.

auto sigma = material.getScalar("sigma","T",idv(prevT));

where prevT is stem[T] at previous time.

We also do the same for k in heat part :

auto k = material.getScalar("k","T",idv(prevT));

Then, after solving the two mqs equations and calculating J, we start the picard loop for the heat equation

picardIter = 0;
while(incrT > tol )
    Feel::cout << "Picard iter num = " << picardIter+1 << std::endl;
    if (picardIter > maxiter)
        Feel::cout << "Picard maxiter reach" << std::endl;

After calculating T, the error L2 is calculated:

\[\begin{equation*} err(T) = \frac{||T-prevT||_2}{||prevT||_2} \end{equation*}\]
    incrT = normL2(_range=markedelements(mesh,range),_expr=idv(T)-idv(prevT));
    incrT /= normL2(_range=markedelements(mesh,range),_expr=idv(prevT));

And we make this loop until the error is smaller than the set tolerance. Note that \(\sigma\) and \(k\) are well updated at each iteration.

After that, we go to the time t+1 and start again.

2. Results