Control of the heat equation

We describe here a simple example for the one dimensional heat equation, over the domain $\Omega=[0,1]$ . We set $Q=\Omega\times [0,T]$ , where the final time is fixed. The control $u(t)$ is either (i) over a part of the domain, with Dirichlet conditions, or (ii) at the boundary by the Neumann condition. So the state equation is in case (i)

$\frac{d}{dt} y(x,t)- c_0\ y_{xx}(x,t) = \chi_{[0,a]} c_1\ u(t), \quad (x,t) \in Q,$

$y(\cdot,0)=y_0(x); \quad y(0,t)=y(1,t) = 0, \quad t\in [0,T],$

where $0<a\leq 1$ , and $\chi_{[0,a]}$ is the characteristic function of $[0,a]$ , and in case (ii)

$\frac{d}{dt} y(x,t) - c_0\ y_{xx}(x,t) = 0, \quad (x,t) \in Q,$

$y(\cdot,0) =y_0(x); \quad y_x(0,t)= -c_1 u(t); \quad y_x(1,t) = 0, \quad t\in [0,T].$

The cost function is, for $\gamma\geq 0$ and $\delta\geq 0$ :

$\frac{1}{2} \int_Q y(x,t)^2 dx dt + \int_0^T \left( \gamma u(t) + \delta u(t)^2\right) dt.$

We discretize in space by standard finite difference approximations.

Numerical simulations: problem heat
As an example, we take 50 space variables, with $c_0=0.02$ , $c_1=20$ , and a final time T=20.
The discretization method is implicit Euler with 200 steps.
We set here $\gamma = \delta=0$ , which gives a singular arc for the control.
We display on Fig. 1 the results in the case of the Dirichlet boundary condition $(a=0)$ .
Fig. 2 shows the Neumann case, this time with $c_0=0.2$ .
We can clearly see the differences between the boundary conditions $y(1,t)=0$ and $y_x(1,t)=0$ .

Figure 1: Heat equation, Dirichlet condition, $u(t)$ and $y(.,t)$ .

Figure 2: Heat equation, Neumann condition, $u(t)$ and $y(.,t)$ .

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