The test example

We fix here the following quantities in (P):

**Proof**
Insert the definitions above in the state equation and
adjoint equation. Then it is easy to check by elementary calculations that
fulfill (1.2) and that satisfies the
adjoint equation (2.2).
Moreover, the state-inequality constraint
(1.4) holds as an equality, since the integral of over
vanishes. Clearly, is an admissible control.
The variational inequality (2.3) is easy to verify by (2.5):
We find

Therefore,

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, is strongly active on and holds there.
If is given, then

holds for . Therefore, for . To verify the second order sufficient conditions, it suffices to confirm the coercivity condition (2.7) for all pairs coupled through the linearized equation (2.8) and satisfying on ( being arbitrary but fixed). Assume that

**Proof**
Let vanish on and let solve (2.8). Then
on . For we get

Notice that was not assumed to be positive. If , then is obviously coercive on the whole space , and (SSC) is satisfied in a very strong sense. However, we might find negative values for such that is partially indefinite.

**Proof**
We take an arbitrary but fixed and set

Then is positive. Hence

as . Therefore, the expression (3.2) becomes negative for sufficiently small , if is substituted there for in the first integral.

For the numerical verification we need a rough estimate on how small
should be chosen. To get a negative value of
, we must have

hence

must hold. Here, can be chosen arbitrarily. We take the value . Thus, we evaluate the integrals for

and the associated state . The state solves the homogeneous heat equation subject to homogeneous initial condition, homogeneous boundary condition at and

To avoid tedious estimates which might be performed by means of a Fourier series representation of , we have evaluated the integrals , numerically. The result is

2003-01-25