The test example

We fix here the following quantities in (P):

$$\begin{array}{rcl} T &=& 1, ~ l = \pi, ~u_a = 0,~u_b = 1, ~\... ...ght.\\ \\ a_u(t) &=&\nu + 1 - (1+ 2\nu) t,\\ \\ \end{array}$$

$$\begin{array}{rcl} e_Q(t) &=&\left\{ \begin{array}{ll} 0, & ... .../2)^4- (2t-1), & t \in (1/2,1]. \end{array} \right. \end{array}$$

THEOREM 1   The quantities

$$\begin{array}{rcl} \bar{u}&=&max\{0,2t-1\}\\ \bar{y}&=& \le... ...ght.\\ \bar{p}&=& (1-t) cosx\\ \bar{\lambda} &=&1 \end{array}$$

satisfy the system of first order necessary conditions.

Proof      Insert the definitions above in the state equation and adjoint equation. Then it is easy to check by elementary calculations that $\bar{u},~\bar{y}$ fulfill (1.2) and that $\bar{p}$ satisfies the adjoint equation (2.2). Moreover, the state-inequality constraint (1.4) holds as an equality, since the integral of $cosx$ over $[0,\pi]$ vanishes. Clearly, $\bar{u}$ is an admissible control. The variational inequality (2.3) is easy to verify by (2.5): We find

$$-1/\nu(\bar{p}(\pi,t) + a_u(t))= 2t-1=\left\{ \begin{array}{... ... 0, & t \in [0,1/2)\\ >0, & t \in (1/2,1]. \end{array}\right.$$

Therefore,

$$\Pi_{[0,1]}\left\{ -1/\nu(\bar{p}(\pi,t) + a_u(t)) \right\} = max \{0,2t-1\} = \bar{u}(t).$$

Next we consider the second order sufficient condition (SSC) for the example analytically. What conditions must be checked to verify them? Thanks to our construction, $\bar{u}$ is strongly active on $[0,1/2)$ and $\bar{u}=0$ holds there. If $b<1/2$ is given, then

$$\nu \bar{u}(t) + \bar{p}(\pi,t) + a_u(t) = \bar{p}(\pi,t) + a_u(t) = -\nu (2t -1) > \nu (2b-1)$$

holds for $t \in [0,b]$. Therefore, $t \in A^-(\tau)$ for $\tau = \vert\nu (2b-1)\vert$. To verify the second order sufficient conditions, it suffices to confirm the coercivity condition (2.7) for all pairs $(y,u)$ coupled through the linearized equation (2.8) and satisfying $u = 0$ on $[0,b]$ ($0<b<1/2$ being arbitrary but fixed). Assume that

$$\alpha (x,t) = \left\{ \begin{array}{rl} \alpha_0,& \, 0 \le t \le b \\ 1,& \, b < t \le 1. \end{array}\right.$$ (3.1)

THEOREM 2   Let $\alpha$ have the form (3.1), where $b \in [0,1/2)$. Then the second order sufficient conditions (SSC) are satisfied by $(\bar{y},\bar{u},\bar{p},\bar{\lambda})$ for arbitrary $\alpha_0 \in \mbox{\piz R}$.

Proof      Let $u$ vanish on $[0,b]$ and let $y$ solve (2.8). Then $y(x,t) = 0$ on $[0,b]$. For ${\cal L''}$ we get

$$\begin{array}{rcl} {\cal L''}(\bar{y},\bar{u},\bar{p},\bar{\... ...^2(\pi,t)\, dt\\ &\ge& \nu \int\limits_0^1 u^2 dt. \end{array}$$ (3.2)

Hence the coercivity condition (2.7) is satisfied.

Notice that $\alpha_0$ was not assumed to be positive. If $\alpha_0\ge 0$, then ${\cal L''}$ is obviously coercive on the whole space $W(0,1) \times L^2(0,1)$, and (SSC) is satisfied in a very strong sense. However, we might find negative values for $\alpha_0$ such that ${\cal L''}$ is partially indefinite.

THEOREM 3   If $\alpha_0 < 0$ is sufficiently small, then a pair $(y,u)$ exists, such that $u \ge 0$, $y$ solves the linearized equation (2.8), and

$${\cal L''}(\bar{y},\bar{u},\bar{p},\bar{\lambda}) [y,u]^2 < 0.$$ (3.3)

Proof      We take an arbitrary but fixed $b<1/2$ and set

$$u(t) = \left\{ \begin{array}{rl} 1 & \mbox{ on }[0,b]\\ 0& \mbox{ on }(b,1]. \end{array}\right.$$

Then $\int\limits_0^\pi\int\limits_0^b y^2\, dxdt$ is positive. Hence

$$\alpha_0 \int\limits_0^\pi\int\limits_0^b y^2\, dxdt \rightarrow - \infty$$

as $\alpha_0 \rightarrow - \infty$. Therefore, the expression (3.2) becomes negative for sufficiently small $\alpha_0$, if $y^2$ is substituted there for $0$ in the first integral.

For the numerical verification we need a rough estimate on how small $\alpha_0$ should be chosen. To get a negative value of ${\cal L''}(\bar{y},\bar{u},\bar{p},\bar{\lambda}) [y,u]^2$, we must have

$$\alpha_0 \int\limits_0^\pi \int\limits_0^b y^2 \, dxdt + \in... ...ts_0^1 2(1-t)y^2(\pi,t)\, dt + \nu \int\limits_0^1 u^2 dt < 0,$$

hence

$$\vert\alpha_0\vert > \displaystyle \frac{\int\limits_0^\pi \... ...limits_0^b y^2 \, dxdt} =\displaystyle \frac{I_1+I_2+I_3}{I_0}$$ (3.4)

must hold. Here, $b \in [0,1/2)$ can be chosen arbitrarily. We take the value $b = 1/4$. Thus, we evaluate the integrals $I_j$ for

$$u(t)= \left\{ \begin{array}{rl} 1& \, \mbox{ on }[0,1/4]\\ 0& \, \mbox{ on }(1/4,1] \end{array} \right.$$

and the associated state $y$. The state $y$ solves the homogeneous heat equation subject to homogeneous initial condition, homogeneous boundary condition at $x = 0$ and

$$y_x(\pi,t)= \left\{ \begin{array}{ll} 1& \, \mbox{ on }[0,1/... ... -2 \bar{y}(\pi,t)& \, \mbox{ on }(1/4,1]. \end{array} \right.$$

To avoid tedious estimates which might be performed by means of a Fourier series representation of $y$, we have evaluated the integrals $I_j$, $j = 0,..,3$ numerically. The result is

$I_0 = .0103271,\quad I_1 = .0401844,\quad I_2 = .0708107,\quad I_3 = .001$

$$\frac{I_1+I_2+I_3}{I_0} = 10.845 .$$ (3.5)