By Alexander J.C.

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**Example text**

Xn−1 ) ∈ Rn , put fm t, x0 , . . , xn−1 = hm t, σ0∗ 1 , x0 , σ1 x1 , . . , σn−1 xn−1 m . 4), fm ∈ Car([0, T] × Rn ) and the inequalities ψ(t) ≤ fm t, x0 , . . e. t ∈ [0, T] and each x0 , . . 45) and 0 < fm t, x0 , . . e. t ∈ [0, T] and each x0 , . . , xn−1 ∈ R \ {0} n −1 ×R hold for m ≥ m0 ≥ 1/K. 46) follows from the fact that σi xi σ0∗ 1 , x0 m ωi σi xi for 1 ≤ i ≤ n − 1, ≤ xi ≤ 1 + x0 , ≤ ωi xi σ0∗ + ωi (1), 1 , x0 ≥ σ0 x0 , m 0 ≤ i ≤ n − 2. Consider auxiliary regular equation −u(n) = fm t, u, .

N−1 , x n −1 m m . 36) Consider the sequence of regular diﬀerential equations: (−1)n− p u(n) = fm t, u, . . , u(n−1) depending on m ∈ N. 5. 4. 15) has a solution um ∈ B(r, a) and (n−1) um ∞ < S. 38) Proof . Fix an arbitrary m ∈ N. 36) yield fm ∈ Car([0, T] × Rn ). Put gm (t) = sup f ∗ t, x0 , . . 34). Then gm ∈ L1 [0, T] and ≤ gm (t) fm t, x0 , . . e. t ∈ [0, T] and all (x0 , . . , xn−1 ) ∈ Rn . 15) has a solution um ∈ AC n−1 [0, T]. 17), we see that the inequalities a(T − t)r ≤ fm t, x0 , .

Xn−1 ) may be singular at the value 0 of its space variables x0 , . . , xn−2 . Notice that the (n, 0) problem is simultaneously the (1, n − 1) conjugate problem discussed in Chapter 4. 2) have singular points of type I at t = 0, T and also singular points of type II. 1): n −2 f ∈ Car [0, T] × D , where D = (0, ∞) × R \ {0} ×R and there exist a positive function ψ ∈ L1 [0, T] and K > 0 such that ψ(t) ≤ f t, x0 , . . e. t ∈ [0, T] and each x0 , . . , xn−1 ∈ (0, K] × R \ {0} n −1 0 < f t, x0 , .

### Dynamical Systems by Alexander J.C.

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