By Maksimov V. I.
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Extra resources for A Boundary Control Problem for a Nonlinear Parabolic Equation
Ulbrich Obviously, g is linear. Furthermore, for all v ∈ H01 (Ω), there holds n (g 0 , v)L2 + (g j , vxj )L2 j =1 n ≤ g0 L2 v L2 + gj vxj L2 L2 j =1 1/2 n ≤ gj 2 L2 1/2 n 2 L2 v j =0 + vxj j =1 L2 1/2 n = gj 2 L2 v H1. 12) . j =0 To show the formula for g H −1 let g 0 , . . , g n ∈ L2 (Ω) be an arbitrary representation of g. Moreover let u be the Riesz representation of g and choose (g¯ 0 , . . , g¯ n ) := (u, ux1 , . . , uxn ) as above. 12). This shows that g¯ 0 , . . , g¯ n is the representation with minimum norm and yields g H −1 .
Similarly as above, existence can be shown under the following assumptions. 44 1. Uad ⊂ U is convex, bounded and closed. 2. 78) has a feasible point. 3. The state equation e(y, u) = 0 has a bounded solution operator u ∈ Uad → y(u) ∈ Y . 4. (y, u) ∈ Y × U → e(y, u) ∈ Z is continuous under weak convergence. 5. J is sequentially weakly lower semicontinuous. 44 hold. 78) has an optimal solution (y, ¯ u). 43. 78) by Fad . , 3. ensure the existence of a bounded minimizing sequence (yk , uk ) ⊂ Fad . Since U, Y are reflexive, we can extract a weakly convergent subsequence (yki , uki ) − (y, ¯ u).
T ∈ [0, T ], where H = L2 (Ω), V = H01 (Ω). This yields a(y(t), w; t) = − yt (t), w (H01 )∗ ,H01 + (f (t), w)L2 = (−yt (t) + f (t), w)L2 ∀w ∈ H01 (Ω). 28 y(t) H 2 (Ω ) ≤ C( yt L∞ (0,T ;L2 ) + f L∞ (0,T ;L2 ) + y either for Ω ⊂⊂ Ω or for Ω = Ω if Ω has C 2 -boundary. 34) y(0, ·) = y0 , where the operator L is given by n Ly := − (aij yxi )xj , i,j =1 and L is assumed to be uniformly elliptic in the sense that there is a constant θ > 0 such that n aij (x)ξi ξj ≥ θ ξ 2 for almost all x ∈ Ω and all ξ ∈ Rn .