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Let (x t)) be a negative solution of the boundary value problem (1.10 - 1.11) in ((a, b)).
Let (x t)) be a negative solution of equation (1.10) satisfying the boundary conditions (1.11) in ((a, b)).
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She show that (1.2) has at least two solutions, one of which is a negative solution.
Then {v n } is a negative solution of inequality (2.25), a contradiction.
Since and are even, and are odd, then is a negative solution of.
If, satisfies (3.4) on and, then (3.19). is a positive solution of (3.2), and is a negative solution to (3.3).
for s ≥ s 2, where v 1 = s ϕ 1 a − λ 1 is a negative solution of (1.1).
When j is even, (alpha_{j}uvert _{Omega (rho_{j}^{n}, rho^{n}_{j+1})}) is a positive solution of (1.2), and when j is odd, (alpha_{j}uvert _{Omega (rho _{j}^{n}, rho^{n}_{j+1})}) is a negative solution.
Lemma 2.5 Assume that − ∞ < a < λ 1 < b < λ 2 and s > 0. Then there exists a small open neighborhood B of v 0 in V such that v 0 is a strict local point maximum of I ˜ ( v, s ) and there exists a small open neighborhood D of v 1 in V such that v 1 is a strict point of minimum of I ˜ ( v, s ), where v 0 = s ϕ 1 b − λ 1 is a positive solution and v 1 = s ϕ 1 a − λ 1 is a negative solution of (1.1).
Meanwhile m (r, u) is called a negative solution of (1.1) and (1.2), if (r, -u) is a positive solution of (1.1) and (1.2).
A sequence { x ( 0 ), x ( 1 ), …, x ( N + 1 ) } is said to be a positive (negative) solution of (BP) if it satisfies (BP) and x ( k ) > 0 (<0) for k ∈ Z [ 1, N ].
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com