Ñ 1. Introduction - Issues of Analysis

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The theorem on existence of singular solutions 29We conclude the discussion of the regular case by the following example,which is very simple, but serves to illustrate the basic idea of Theorem 1.Example. Consider the problem (1) with (x 0 , y 0 ) = (0, 0) andF : R 2 → R,where F (x, y) = x 3 + x − y + 11100. Moreover, ‖F (0, 0)‖ = η =100 ,√ ∥ [F ′ (0, 0)] −1∥ ∥ 2 = δ =2 , sup ‖F ′′ (x, y)‖ ≤ C = 6 · ε,(x,y)∈U ε (0,0)where ε = 0, 1. The assumptions of Theorem 1 are fullled, so there existssuch point (x ∗ , y ∗ ), that F (x ∗ , y ∗ ) = 0, and for instance (x ∗ , y ∗ ) = (0,1100 ).Ÿ 4. p-regular (singular) caseLet us mention one of the consequences of the Mean Value Theorem,which is important for our further investigations. For the proof we referthe reader to [7, 8].Lemma 4. Let F : U → Y, where U ⊆ X, such that [a, b] ⊂ U, then‖F (b) − F (a) − Λ(b − a)‖ ≤ sup ‖F ′ (ξ) − Λ‖ · ‖a − b‖,ξ∈[a,b]for any Λ ∈ L(X, Y ).Next lemma follows from homogeneity properties of p-form (see [4]).Lemma 5. [4] Let U(x 0 ) neighborhood of x 0 in X, F : U(x 0 ) → Y,F ∈ C p+1 (U(x 0 )), and F (i) (x 0 ) = 0, for i = 1, . . . , p − 1. Then for every(R = max{1, 2 ɛ · 1(p−1)!∥∥F (p) (x 0 ) ∥ })ɛ > 0 there exist δ > 0 and R > 0such that for any x ∈ X, ‖x‖ ≤ δ and for any x 1 , x 2 ∈ X, ‖x j ‖ ≤‖x‖/R, j = 1, 2, the following condition∥ F (x 10 + x + x 1 ) − F (x 0 + x + x 2 ) −(p − 1)! F (p) (x 0 )[x] p−1 (x 1 − x 2 )∥ ≤≤ ɛ · ‖x‖ p−1 · ‖x 1 − x 2 ‖ holds.Definition 4. (Banach condition) Let F : X → Y, and let F ∈C p (X). Then for any y ∈ Y, ‖y‖ = 1 there exists x ∈ X, such thatF (p) (x 0 )[x] p = y, ‖x‖ ≤ c,
30 A. Prusinska, A. Tret'yakovwhere c > 0 is a constant independent of y.Let us introduce following additional notationsĥ =δ = ‖F (x 0 )‖ ̸= 0,{ [ h‖h‖ , where ] −1 1 h ∈ p! F (p) (x 0 ) (−F (x 0 ))}, h ≠ 0,as a generalization of (2) we have[η =∥F (p) (x 0 )] −1∥ ∥∥∥= sup‖y‖=1C ={}inf ‖x‖ : F (p) (x 0 )[x] p = y, x ∈ X ,supx∈U ε (x 0 )∥∥F (p+1) (x) ∥ ,[∥C 1 =∥ F (p) (x 0 )[ĥ]p−1] −1 ∥∥and∥C 2 = ∥F (p) (x 0 ) ∥ .We can now formulate our main result which is a generalization of Theorem 1in the degenerate case.Theorem 2. Let F : X → Y and assume that for F ∈ C p+1 (U ε (x 0 ))the Banach condition holds and F is p-regular mapping at the point x 0along ĥ and F (i) (x 0 ) = 0, for i = 1, . . . , p − 1. Moreover assume thefollowing inequalities1) p! · η · δ 1 p ≤ε2 ,2) 4 p · (p − 1)! · C · C 1 · ε ≤ 1 2 ,3) ε < 1.Then the equation F (x) = 0 has a solution x ∗ ∈ U ε (x 0 ).Äîêàçàòåëüñòâî.As in the proof of Theorem 1 consider a multivalued mappingΨ : U ε (x 0 ) → 2 X , U ε (x 0 ) ⊂ X,
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