The Distribution of Zeroes and Critical Points of Solutions of a Second Order Half-Linear Differential Equation

and Applied Analysis 3 given the linear character ofy(a)+(y(b)−y(a))((s−a)/(b− a󸀠)) and the fact that y(x), y󸀠(x) ≥ 0 on [a, b]. Equation (14) gives (7). Let us focus now on (8). Again, for the sake of simplicity we will only consider the case y(x) ≥ 0 on [a, b]. From the concavity of y(x) (Lemma 1) one has y (s) ≥ y (b 󸀠 ) + y (a󸀠) − y (b󸀠) b − a (b 󸀠 − s) , s ∈ [a 󸀠 , b 󸀠 ] , (15) that is, Φ(y (s)) ≥ [y (b 󸀠 ) + y (a󸀠) − y (b󸀠) b − a (b 󸀠 − s)] r−1