Boundary Value Problems for a Super-Sublinear Asymmetric Oscillator: The Exact Number of Solutions

Figure 2

Layout of envelopes depends on sign(𝑝𝑞−1).

(a) the case 𝑝=5 and 𝑞=1/3. Since 𝑝𝑞−1>0, the envelopes are ordered as ℰ^±₁≺ℰ⁺₂≺ℰ⁻₂≺ℰ^±₃ < ⋯

(b) the case 𝑝=3 and 𝑞=1/3. Since 𝑝𝑞−1=0, the envelopes are ordered as ℰ^±₁≺ℰ^±₂≺ℰ^±₃ < ⋯

(c) the case 𝑝=2 and 𝑞=1/3. Since 𝑝𝑞−1<0, the envelopes are ordered as ℰ^±₁≺ℰ⁻₂≺ℰ⁺₂≺ℰ^±₃ < ⋯