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By Maksimov V. I.

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Extra resources for A Boundary Control Problem for a Nonlinear Parabolic Equation

Example text

Then the angle ∠x 0 y 0 y˜ is less than π/2, and, besides, since K¯ 2 is convex, it follows that y 0 , y˜ ∈ K¯ 2 . Let us take the point y˜ ∈ y 0 , y˜ such that x 0 , y˜ ⊥ y 0 , y˜ ˜ and y˜ and show that y˜ is not a point of K¯ 2 close to x 0 . Indeed, the points y 0 , y, belong to the same line and y˜ ∈ int H . 3 Appendix 39 (the shortest distance is one smaller than any other one). At the same time, y˜ ∈ ˜ so (y 0 , y), x 0 − y˜ < x 0 − y 0 . Also we have 0 ∈ H since, if this were not so, the line [0, ∞), crossing y 0 and belonging to K¯ 2 , should necessarily have points in common with int H .

L). (D) The multipliers νl∗ (l = 1, . . 106). 3 Appendix 41 is trivial. Suppose that gl0 x ∗ (T ) < 0. Then for δ > 0 the point δ, 0, . . , 0, gl0 x ∗ (T ) , 0, . . 120) l0 belongs to the set C. 118). 118) implies νl∗0 gl0 x ∗ (T ) ≥ −μ∗ δ. 121) Letting δ go to zero we obtain νl∗0 gl0 x ∗ (T ) ≥ 0 and since gl0 x ∗ (T ) < 0 it follows that νl∗0 ≤ 0. But in (C) it has been proven that νl∗0 ≥ 0. Thus, νl∗0 = 0, and, hence, νl∗0 gl0 x ∗ (T ) = 0. (E) Minimality condition for the Lagrange function.

L). (D) The multipliers νl∗ (l = 1, . . 106). 3 Appendix 41 is trivial. Suppose that gl0 x ∗ (T ) < 0. Then for δ > 0 the point δ, 0, . . , 0, gl0 x ∗ (T ) , 0, . . 120) l0 belongs to the set C. 118). 118) implies νl∗0 gl0 x ∗ (T ) ≥ −μ∗ δ. 121) Letting δ go to zero we obtain νl∗0 gl0 x ∗ (T ) ≥ 0 and since gl0 x ∗ (T ) < 0 it follows that νl∗0 ≤ 0. But in (C) it has been proven that νl∗0 ≥ 0. Thus, νl∗0 = 0, and, hence, νl∗0 gl0 x ∗ (T ) = 0. (E) Minimality condition for the Lagrange function.

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A Boundary Control Problem for a Nonlinear Parabolic Equation by Maksimov V. I.


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