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# The test example

We fix here the following quantities in (P):

THEOREM 1   The quantities

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 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
 (3.1)

THEOREM 2   Let have the form (3.1), where . Then the second order sufficient conditions (SSC) are satisfied by for arbitrary .

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

 (3.2)

Hence the coercivity condition (2.7) is satisfied.

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.

THEOREM 3   If is sufficiently small, then a pair exists, such that , solves the linearized equation (2.8), and
 (3.3)

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
 (3.4)

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

 (3.5)

Next: Numerical Verification Up: paper94 Previous: Second order sufficient optimality
Hans D. Mittelmann
2003-01-25