Neumann boundary conditions

Example 5.5: The first problem has a linear partial differential equation and nonlinear Neumann boundary conditions with data:

$$\begin{eqnarray*} \hspace*{-10mm} \begin{array}{lll} \mbox{on} \;\;\Omega \,: & ... ...eq 4.5 , \; u_d(x) \equiv 0 \,, \; \alpha = 0.01 . \end{array}\end{eqnarray*}$$

The following results were obtained:

Cost functional : $F(\bar{y},\bar{u}) = 0.553324$, CPU seconds : 494

The optimal control and state are shown in Figures 8 and 9. Note that the sign condition (2.5) holds since $\,b_y(x,y(x),u(x)) = -2 y(x) \leq 0 \,$. The optimal control is continuous and has two boundary arcs with $\,\bar{u}(x) = 3.7 \,$ and one boundary arc with $\,\bar{u}(x) = 4.5 \,$. The junction points with the boundary are the points $\,x_1=(.17,0), \,x_2=(.34,0) , \,x_3=(.66,0), \, x_4=(.83,0)\,$ on the bottom edge of $\,\Gamma\,$. The control law (2.16) was derived under the assumption $\,b_u \equiv 1 \,$ which obviously holds in this example. By looking at the adjoint variable in Figure 8, the reader may verify that the control law (2.16) is satisfied:

$$\begin{eqnarray*} \bar{u}(x) = \left \{ \begin{array}{llllll} \bar{q}(x) * 100 &... ...\mbox{if} & \bar{q}(x) * 100 \geq 4.5 \,. \end{array}\right \} \end{eqnarray*}$$

The active set for the state constraint $\, y(x) \leq 2.071 \,$ are the midpoints of the edges. The dual variable for this active inequality constraint is $\, \mu_{ij} = .00045274\,$. It can be checked that the adjoint equations (2.7) and (2.8) hold observing the scaling (4.13). As an example, let us test the Neumann boundary condition (2.8) at the active point $\,\bar{x}=x_{i0}=(0.5,0.)\,$. Hence, we have to verify the relation $\, q_{i0} - q_{i1} + 2 q_{i0}*h*y_{i0} + \mu_{i0} = 0 \,$ which corresponds to the equation $\, \partial_{\nu}\bar{q}(\bar{x}) + 2*\bar{q}(\bar{x})*\bar{y}(\bar{x})+ \bar{\mu} = 0 \,$. We find $\,q_{i0}= -.046506, \,q_{i1}= -.048885, \, y_{i0}= 2.071, \, \mu_{i0}= -.00045274\,$ und can check indeed the Neumann condition with $\, q_{i0} - q_{i1} + 2q_{i0}*h*y_{i0} = 0.0004528\,$ which agrees with $\, -\mu_{i0}= .00045274 \,$.

Figure 8: Optimal control and switching function for Example 5.5

Figure 9: Optimal state and adjoint variable for Example 5.5

Example 5.6: The data for the second Neumann problem are:

$$\begin{eqnarray*} \hspace*{-10mm} \begin{array}{lll} \mbox{on} \;\;\Omega \,: & ... ...x) \leq 9 , \; u_d(x) \equiv 0 \,, \; \alpha = 0 . \end{array}\end{eqnarray*}$$

Figure 10: Optimal control and switching function for Example 5.6

Figure 11: Optimal state and adjoint variable for Example 5.6

According to the optimal control law (2.17) we can expect either a bang-bang or a singular control. We get the following results:

Cost functional : $F(\bar{y},\bar{u}) = .015078$, CPU seconds : 864

The optimal control in Figure 10 is indeed bang-bang. The switching function $\,\bar{q}\,$ on $\,\Gamma\,$ as shown in Figure 10 obeys the optimal control law (2.17):

$$\begin{eqnarray*} \bar{u}(x) = \left \{ \begin{array}{llllll} 6 & , & \mbox{if} ... ... 9 & , & \mbox{if} & \bar{q}(x) > 0 \,. \end{array}\right \} \end{eqnarray*}$$

The switching points are approximately $\, x_1 = (.33,0),\, x_2= (.67,0) \,$. Again, the optimal state displayed in Figure 11 is active at the midpoints of the edges. The dual variable for this active inequality constraint is $\, \mu_{ij} = .00002929\,$.

Example 5.7: In this example we choose a nonlinear partial differential equation and linear Neumann boundary conditions:

$$\begin{eqnarray*} \hspace*{-11mm} \begin{array}{lll} \mbox{on} \;\;\Omega \,: & ... ...eq 2.5 , \; u_d(x) \equiv 0 \,, \; \alpha = 0.01 . \end{array}\end{eqnarray*}$$

These equations represent a simplified Ginzburg-Landau model for super-conductivity in the absence of internal magnetic fields with $\,y\,$ the wave function; cf. Ito, Kunisch [16] and Kunisch, Volkwein [18] where tracking functions and control or state constraints have been considered that are different from those used here. LOQO and AMPL provide the results:

Cost functional : $F(\bar{y},\bar{u}) = .264163$, CPU seconds : 317

The optimal control and state are shown in Figures 12 and 13. The optimal control is continuous and has two boundary arcs with $\,\bar{u}(x) = 1.8 \,$ and one boundary arc with $\,\bar{u}(x) = 2.5 \,$. The junction points with the boundary are the points $\,x_1=(.15,0), \,x_2=(.28,0), \,x_3=(.72,0), \,x_4=(.85,0)$ on the bottom edge of $\,\Gamma\,$. The adjoint variable in Figure 12 shows that the control law (2.16) is satisfied:

$$\begin{eqnarray*} \bar{u}(x) = \left \{ \begin{array}{llllll} \bar{q}(x) * 100 &... ...x{if} & \bar{q}(x) * 100 \,\geq \,2.5 \,. \end{array}\right \} \end{eqnarray*}$$

The active set for the state constraint $\, y(x) \leq 2.7 \,$ comprises the points adjacent to the corners of the domain. The dual variable for this active inequality constraint is $\, \mu_{ij} = .0034574\,$. For this problem the QP-code BPMPD is not applicaple. Both MINOS and SNOPT return with error messages as in Example 5.1, while LANCELOT solves the example on grids of $n = 50, 70, 90$ and needs $2$ respectively $8.4$ respectively $7.2$ times as long as LOQO.

Figure 12: Optimal control and switching function for Example 5.7

Figure 13: Optimal state and adjoint variable for Example 5.7

Example 5.8: The data are those from Example 5.7 but now we choose $\,\alpha=0.0\,$ expecting to obtain a bang-bang control. The numerical results are:

Cost functional : $F(\bar{y},\bar{u}) = .165531$, CPU seconds : 570

The optimal control shown in Figure 14 is indeed bang-bang. The switching function $\,\bar{q}\,$ on $\,\Gamma\,$ as shown in Figure 14 yields the optimal control law (2.17):

$$\begin{eqnarray*} \bar{u}(x) = \left \{ \begin{array}{llllll} 1.8 & , & \mbox{if... ...2.5 & , & \mbox{if} & \bar{q}(x) > 0 \,. \end{array}\right \} \end{eqnarray*}$$

The switching points are approximately $\, x_1 = (.21,0), \, x_2= (.79,0) \,$. Again, the optimal state displayed in Figure 15 is active at the points adjacent to the corners of the domain. The dual variable for this active inequality constraint is $\, \mu_{ij} = .030118\,$.