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Calculus of variations by Miersemann E.

25 February 2017 adminCalculus

By Miersemann E.

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Example text

Suppose that x is a local minimizer of f in V and that x is a regular point. Then there exists λj ∈ R, λj ≥ 0 if j ∈ I0 , such that f (x) + λj gj (x) = 0. j∈E∪I0 Proof. 3) with K = {z ∈ Rn : gj (x), z ≥ 0 and −gj (x), z ≥ 0 for each j ∈ E, and −gj (x), z ≥ 0 for each j ∈ I0 } (1) is satisfied, there exists nonnegative real numbers µj if j ∈ I0 , µj and (2) µj if j ∈ E such that f (x) = j∈I0 = − µj −gj (x) + (1) j∈E j∈E µj j∈E (1) µj gj (x) + j∈I0 (2) µj gj (x) + (2) µj − µj gj (x). 2. 6: let A be a real matrix with m rows and n columns and let b ∈ Rn , then b, y ≥ 0 ∀y ∈ Rn with Ay ≥ 0 if and only if there exists an x ∈ Rm , such that x ≥ 0 and AT x = b.

Assertion (i) follows from the maximal rank condition and the implicit function theorem. Since f (y) = L(y, λ) for all y ∈ V0 , we obtain f (y) − f (x) = = L (x, λ), y − x + 1 L (x, λ)(y − x), y − x + o(||x − y||2 ) 2 1 L (x, λ)(y − x), y − x + o(||x − y||2 ). 2 On the other hand f (y) − f (x) = f (x), y − x + 1 f (x)(y − x), y − x + o(||y − x||2 ). 5. 3 Supplement In the above considerations we have focused our attention on the necessary condition that x ∈ V is a local minimizer: f (x), w ≥ 0 for all w ∈ C(V, x).

Let x be a local minimizer of f in V which is regular and λj denote Lagrange multipliers such that λj gj (x) = 0, f (x) + j∈E∪I0 where λj ≥ 0 if j ∈ I0 . Let I0+ = {j ∈ I0 ; λj > 0}, V0 = {y ∈ V ; gj (y) = 0 for each j ∈ I0+ }, Z = {y ∈ Rn : gj (x), y = 0 for each j ∈ E ∪ I0+ } and L(y, λ) ≡ f (y) + j∈E∪I0 λj gj (y). Then (i) (ii) (iii) T (V0 , x) = Z, L (x, λ)z, z ≥ 0 lim tk f (x), wk k→∞ for all 1 = 2 z ∈ T (V0 , x) (≡ Z), j∈E∪I0 λj gj (x)w, w , wk := tk (xk − x), for all w ∈ T (V0 , x) and for all associated sequences xk , tk to w ∈ T (V0 , x) (≡ Z).

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