characteristic numbers of non-autonomous emden-fowler type ...

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406 A. Gritsans, F. SadyrbaevThen the functionx(t) ={x1 (t), if −1 ≤ t ≤ 0,x 2 (t), if 0 ≤ t ≤ 1is a C 2 -solution of the problem (2.1), (2.2) if additionally the smoothnessconditionx ′ 1(0) = x ′ 2(0)is satisfied. The problem (2.3) can be explicitly resolved as( )x 1 (t, β 1 ) = β 1 21 (ht + η) · sl β 1 t + 121 ,ht + ηwhereandβ 1 = x ′ 1 (−1) > 0x 1 (0; β 1 ) = τ.The derivative is given by( )( )x ′ 1(t; β 1 ) = β 1 21 h · sl β 1 t + 1 −h + η21 + β 1ht + η ht + η · sl′ β 1 t + 121 .ht + ηSimilar formulas are valid for x 2 (t). Notice that x ′ 2 (1) = −β 2 < 0. In orderto get an explicit formula for a solution of the BVP (2.1), (2.2) one have tosolve a system of two equations with respect to (β 1 , β 2 )x 1 (0; β 1 ) = x 2 (0; β 2 ),x ′ 1(0; β 1 ) = x ′ 2(0; β 2 ).This system after replacements and simplifications looks as⎧ ( ) ( )⎪⎨ β 1 β 1 221 · sl 1= β 1 β 1 22η 2 · sl 2,η( ))( )⎪⎩ β 1 β 1 221 h · sl 1+ β 1(β 1 2η η · 1 sl′ = −β 1 β 1 22η2 h · sl 2− β 2(β 1 2η η · 2 sl′ η),where 0 < β 1 2 1η, β 2 1 2η< 2A. In new variables u := β 2 1 1η, v := β 1 2 2ηthe systemtakes the form{ u sl u = v sl v, 0 < u, v < 2A,hu sl u + u 2 sl ′ u = −hv sl v − v 2 sl ′ v, h > 0.(2.4)Notice that if a solution (ū, ¯v) of the system (2.4) exists, then a solution x(t)of the BVP (2.1), (2.2) can be constructed such thatx ′ (−1) = β 1 = ū 2 (h + 1) 2 , x ′ (1) = −β 2 = −¯v 2 (h + 1) 2 .
Characteristic Numbers of Emden – Fowler Equations 407Proposition 1. For h large the system (2.4) has exactly three solutions, whichhave the following characteristics.1. There exists a unique symmetric solution (u 0 , v 0 ), that is, u 0 = v 0 . Onehas that (u 0 , v 0 ) → (2A, 2A) as h → +∞.2. There exists a unique solution (u 1 , v 1 ) in the triangle {0 < u, v < 2A, v >u)} for h large. Moreover, (u 1 , v 1 ) → (0, 2A) as h → +∞.3. There exists a unique solution (u 2 , v 2 ) in the triangle {0 < u, v < 2A, v 1are depicted in the Figure 1. Notice that a set of zeros of Φ consists of thediagonal u = v and two symmetric branches.2A2.521.510.50.5 1 1.5 2 2.52AFigure 1. Zeros of Φ(u, v) (solid line) and Ψ(u, v) (dashed line), h = 2.Nehari’s numbers. The respective solutions of the boundary value problem(2.1), (2.2) look as shown in the picture.30252015105-1 -0.5 0.5 1Let H = 1 21∫−1x ′2 (t)dt. Denote by H sym and H asym the respective valuesof H for a symmetric solution (which is depicted by solid line), and for asymmetricsolutions (depicted by dashed lines). Notice that H(x) is the same forboth asymmetric solutions. The values of H sym and H asym are
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