Higher-order superintegrability of a Holt related potential (Q2857975)

Consider a natural Hamiltonian system with two degrees of freedom and Hamiltonian \(H=\frac{1}{2}(p_x^2+p_y^2) + U(x,y)\). Suppose that the potential \(U\) is integrable, that is, there exists a second (independent on \(H\)) first integral \(K\). Then \(U\) is named separable, if \(K\) is quadratic in the momenta, or nonseparable if \(K\) has order \(\geq 3\) in the momenta. On the other hand, we consider \(U\) to be super-integrable if it is a part of two different (integrable in different ways) families of integrable potentials. The authors prove that the potential \(U=(k_2x+k_3){y^{-2/3}}\) is super-integrable with two integrals of motion NEWLINE\[NEWLINE K_1 = 2 p_x^3 + 3 p_x p_y^2 + [6p_x(k_2x+k_3)+9k_2yp_y]{y^{-2/3}}, NEWLINE\]NEWLINE NEWLINE\[NEWLINE K_2 = p_x^4 + 2 p_x^2 p_y^2 + 4 p_x^2(k_2x+k_3)y^{-2/3} + 12 k_2 p_x p_y y^{1/3} + 18 k_2^2 y^{2/3} NEWLINE\]NEWLINE for the corresponding so-called Holt potentials \( V_1= k_1 (4x^2+3y^2)y^{-2/3}+U \) and \( V_2= k_1 (2x^2+9y^2)y^{-2/3}+U \) (cf. [\textit{C. R. Holt}, J. Math. Phys. 23, 1037--1046 (1982; Zbl 0501.58020)]). Originally, the Holt's potentials corresponds to \(k_1=1, k_2=k_3=0\). Thus, generalizing two original Holt's potentials, the authors prove the super-integrability of \(U\) in a very simple way.NEWLINENEWLINEEditorial note: According to the authors, the results were obtained by a different approach in [\textit{S. Post} and \textit{P. Winternitz}, J. Phys. A, Math. Theor. 44, No. 16, Article ID 162001, 8 p. (2011; Zbl 1214.81107)].