Potential Vector Fields in R^3 and α-Meridional Mappings of the Second Kind (α ∈ R)

The paper’s Theorem 5 asserts that for any α, every equilibrium of the gradient system ẋ = ∇h with α-axial-hyperbolic harmonic, axially symmetric potential h is degenerate and has both index and degree of instability equal to one. The setup is: h solves (x1^2 + x2^2)Δh − α(x1 h_{x1} + x2 h_{x2}) = 0 (eq. (4)) and axial symmetry is x2 h_{x1} = x1 h_{x2} ⇔ ∂θ h = 0, equivalently the cylindrical form ρ^2(h_{x0x0} + h_{ρρ}) − (α − 1)ρ h_ρ + h_{θθ} = 0 reduces to ρ^2(h_{x0x0} + h_{ρρ}) − (α − 1)ρ h_ρ = 0 under ∂θ h = 0 (Proposition 3 and eqs. (33)–(36)) . The proof of Theorem 5 identifies one zero eigenvalue and claims the other two are ±√((∂Vρ/∂x0)^2 + (∂Vρ/∂ρ)^2), concluding index = degree = 1 for any α . However, the same formulas show that those two eigenvalues are zero whenever ∂Vρ/∂x0 = ∂Vρ/∂ρ = 0 at an equilibrium, which the paper does not rule out; indeed, the earlier Jacobian/eigenvalue formulas admit this possibility . The candidate solution constructs an explicit counterexample for α = 1 (where the reduced equation is h_{x0x0} + h_{ρρ} = 0), taking h(x0,ρ) = Re[(ρ − ρ0 + i(x0 − x0,0))^3], which has h_ρ = h_{x0} = 0 and h_{ρρ} = h_{ρx0} = h_{x0x0} = 0 at (x0,ρ) = (x0,0, ρ0), so the Jacobian (Hessian) vanishes and the index/degree are 0, not 1. Thus the theorem, as stated “for any α,” is false. With an added nondegeneracy assumption (e.g., (∂Vρ/∂x0, ∂Vρ/∂ρ) ≠ (0,0) or equivalently B ≠ 0 in the meridional 2×2 Hessian), the paper’s intended conclusion follows from its own eigenvalue formula. Therefore, the paper’s claim is wrong as stated; the model’s analysis and counterexample are correct.