New classes of quadratically integrable systems with velocity dependent potentials: non-subgroup type cases

Abstract

We study quadratic integrability of systems with velocity dependent potentials in three-dimensional Euclidean space. Unlike in the case with only scalar potential, quadratic integrability with velocity dependent potentials does not imply separability in the configuration space. The leading order terms in the pairs of commuting integrals can either generalize or have no relation to the forms leading to separation in the absence of a vector potential. We call such pairs of integrals generalized, to distinguish them from the standard ones, which would correspond to separation. Here we focus on three cases of generalized non-subgroup type integrals, namely elliptic cylindrical, prolate/oblate spheroidal and circular parabolic integrals, together with one case not related to any coordinate system. We find two new integrable systems, non-separable in the configuration space, both with generalized elliptic cylindrical integrals. In the other cases, all systems found were already known and possess standard pairs of integrals. In the limit of vanishing vector potential, both systems reduce to free motion and therefore separate in every orthogonal coordinate system.

Graphical abstract

Appendix A: Solution of the determining equations for the generalized elliptic cylindrical case

Let us show in detail the solution of the determining equations that lead to the integrable systems (3.22)–(3.23) and (3.31)–(3.32) in the following Sects. Appendix A.1 and Appendix A.2, respectively.

1.1 Appendix A.1: $a\ne 0$

If $a\ne 0$, Eq. (3.21) implies $\partial _z S_{22}(z)=0$, therefore $S_{22}(z)=S_{22}$, $S_{22}\in {\mathbb {R}}$. By substituting into (3.14) we conclude that necessarily

$$\begin{aligned} S_{21}^2(z)= -\frac{1}{2} a \partial _z S_{21}^1(z). \end{aligned}$$

(A.1)

Equation (3.15) then implies

$$\begin{aligned} \frac{1}{2} a^2 \partial _z^2 S_{21}^1(z)+2 S_{21}^1(z)=0, \end{aligned}$$

solved by

$$\begin{aligned} S_{21}^1(z)=\alpha _1 \sin \left( \frac{2 z}{a}\right) +\alpha _2\cos \left( \frac{2 z}{a}\right) ,\;\; \alpha _i\in {\mathbb {R}}. \end{aligned}$$

(A.2)

By substituting (A.1) and (A.2) into (3.14), (3.15) and (3.17) we obtain

$$\begin{aligned}{} & {} c \left( \alpha _1 \cos \left( \frac{2 z}{a}\right) - \alpha _2 \sin \left( \frac{2 z}{a}\right) \right) =0, \end{aligned}$$

(A.3)

$$\begin{aligned}{} & {} d\left( \alpha _1 \sin \left( \frac{2 z}{a}\right) +\alpha _2 \cos \left( \frac{2 z}{a}\right) \right) =0, \end{aligned}$$

(A.4)

$$\begin{aligned}{} & {} c \left( \alpha _1 \sin \left( \frac{2 z}{a}\right) +\alpha _2\cos \left( \frac{2 z}{a}\right) \right) +d \left( \alpha _1\cos \left( \frac{2 z}{a}\right) -\alpha _2 \sin \left( \frac{2 z}{a}\right) \right) - a b S_{22}=0. \end{aligned}$$

(A.5)

We are thus lead to a second level splitting, depending on the values of the constants c and d.

1.1.1 Appendix A.1.1: $c^2+ d^2 \ne 0$

If c or d is not zero, we see that (A.3)–(A.5) can be satisfied only if $\alpha _1=\alpha _2=S_{22}=0$ (both a and b cannot be zero). This implies that

$$\begin{aligned} s_2^1(x,y,z)&= {} s_2^2(x,y,z)=0,\;\;s_2^3(x,y,z)=S_2^3(x,y), \end{aligned}$$

(A.6)

$$\begin{aligned} B_1(x,y,z)&= {} -\frac{1}{2} \partial _y S_2^3(x,y),\;\;B_2(x,y,z)= \frac{1}{2} \partial _x S_2^3(x,y), \end{aligned}$$

(A.7)

while the remaining second order Eqs. (3.18)–(3.19) simplify to

$$\begin{aligned} \partial _z s_1^1(x,y,z)=\partial _z s_1^2(x,y,z)=0 \end{aligned}$$

and are therefore solved by

$$\begin{aligned} s_1^1(x,y,z)= S_1^1(x,y),\;\; s_1^2(x,y,z)= S_1^2(x,y), \end{aligned}$$

(A.8)

where $S_1^1(x,y)$, $S_1^2(x,y)$ and $S_2^3(x,y)$ are arbitrary functions of x and y. At this point, both second order conditions for the integral $X_2$ and the commutativity condition (2.7) are solved.

From (3.2) we get

$$\begin{aligned} B_3(x,y,z)=\frac{(a x+d) \partial _y S_2^3(x,y)-2\partial _y S_1^2(x,y)}{4 x y}. \end{aligned}$$

(A.9)

The remaining second order equations for $X_1$ are quite complicated. Let us continue with the simpler first order conditions coming from (2.6) for the integral $X_2$, namely

$$\begin{aligned}{} & {} S_2^3(x,y) \partial _x S_2^3(x,y)-2\partial _x m_2(x,y,z)=0, \end{aligned}$$

(A.10)

$$\begin{aligned}{} & {} S_2^3(x,y) \partial _ yS_2^3(x,y)-2\partial _y m_2(x,y,z)=0, \end{aligned}$$

(A.11)

$$\begin{aligned}{} & {} 2 \partial _z W(x,y,z)- \partial _z m_2(x,y,z)=0, \end{aligned}$$

(A.12)

where we substituted (A.6), (A.7) and (A.9).

Since $\partial _{xz} m_2=\partial _{z x} m_2$ and $\partial _{yz} m_2=\partial _{z y} m_2$, the above equations imply $\partial _{xz} W= \partial _{yz} W=0$ and therefore

$$\begin{aligned} W(x,y,z)=W_1(x,y)+ W_2(z). \end{aligned}$$

(A.13)

Thus, (A.10)–(A.12) are solved by

$$\begin{aligned} m_2(x,y,z)=\frac{1}{4} s_2^3(x,y)^2+2 W_2(z). \end{aligned}$$

(A.14)

After substituting (A.6), and (A.14), the first order equations for the involutivity condition (2.7) simplify to

$$\begin{aligned} -2 (c-a y) W_2'(z)=0,\;\;-2 (a x+d) W_2'(z)=0,\;\; \partial _z m_1(x,y,z)=0. \end{aligned}$$

(A.15)

Therefore, $m_1$ does not depend on z,

$$\begin{aligned} m_1(x,y,z)= M_1(x,y) \end{aligned}$$

(A.16)

and since $a\ne 0$, necessarily $W_2(z)=0$.

By using all we know so far about the solutions for $s_2^j$ and $m_j$, we see that the zero order equations coming from the integral $X_2$ and from the involutivity relation (2.7) are identically zero. In particular, $X_2$ takes the form

$$\begin{aligned} X_2= { \left( p_3^A+ \frac{1}{2} S_2^3(x,y) \right) ^2} \end{aligned}$$

(A.17)

and if the gauge is chosen so that $A_3(x,y,z)=- { \frac{1}{2} S_2^3(x,y) }$ it simplifies to $X_2=p_3$. This reflects the fact that the z coordinate is cyclic and indeed

$$\begin{aligned} W(x,y,z)=W_1(x,y). \end{aligned}$$

(A.18)

However, we still have to solve the determining equations for the integral $X_1$, except (3.2). Let us start by the remaining second order conditions from (3.1)–(3.6). We find it convenient to change the coordinate system to a shifted and scaled one, namely

$$\begin{aligned} x=\frac{X-d}{a},\;\;y= \frac{c+Y}{a}, \end{aligned}$$

(A.19)

so that (3.1)–(3.6) reduce to (where (3.2) was already solved by (A.9))

$$\begin{aligned}{} & {} X \partial _Y S_{2}^3(X,Y)-Y \partial _X S_{2}^3(X,Y)=0, \end{aligned}$$

(A.20)

$$\begin{aligned}{} & {} 2(\partial _X S_{1}^1(X,Y)+\partial _Y S_{1}^2(X,Y))- X \partial _Y S_{2}^3(X,Y)+ Y\partial _X S_{2}^3(X,Y)=0, \end{aligned}$$

(A.21)

$$\begin{aligned}{} & {} -2 a^4 Y\partial _y S_{1}^2(X,Y)-4 (c+Y) (d-X) \left( a^2 \partial _y S_{1}^3(X,Y) +(c+Y) (d-X) \partial _x S_{2}^3(X,Y)\right) \nonumber \\ {} & \qquad {} +\left( a^4 X Y-4 (d-X)^3 (c+ Y)\right) \partial _Y S_{2}^3(X,Y)=0, \end{aligned}$$

(A.22)

$$\begin{aligned}{} & {} 4 (c+Y) (X-d) \left( a^2\partial _X S_{1}^3(X,Y)+ \partial _X S_{2}^3(X,Y)\left( (c+Y)^2+a^2 b\right) \right) \nonumber \\ {} & \qquad {} -2 a^4 X \partial _Y S_{1}^2(X,Y)+\left( a^4 X^2-4 (c+Y)^2 (d-X)^2\right) \partial _Y S_{2}^3(X,Y)=0, \end{aligned}$$

(A.23)

$$\begin{aligned}{} & {} 2\partial _Y S_{1}^2(X,Y) \left( a^2 b+(c+Y)^2- (d-X)^2\right) \nonumber \\ {} & \qquad {} +2 (c+Y) (X-d)\left( \partial _Y S_{1}^1(X,Y) +\partial _X S_{1}^2(X,Y)-\frac{1}{2} X\partial _X S_{2}^3(X,Y)\right) \nonumber \\ {} & \qquad {} +\left( -a^2 b X-c Y (d+X)-X (c+d-X) (c-d+X)-d Y^2\right) \partial _Y S_{2}^3(X,Y)=0 . \end{aligned}$$

(A.24)

The first equation above implies

$$\begin{aligned} S_{2}^3(X,Y)= F(r), \end{aligned}$$

(A.25)

if polar coordinates

$$\begin{aligned} X= r \cos (\varphi ),\;\; Y=r \sin (\varphi ),\;\;r\ge 0,\;\; 0\le \varphi <2\pi , \end{aligned}$$

(A.26)

are introduced. Though it seems more convenient to work in polar coordinates, the remaining equations do not simplify in polar coordinates. However, if we multiply the third equation above by X and subtract from it the fourth equation multiplied by Y and then transform the expression so obtained into polar coordinates we arrive at

$$\begin{aligned}{} & {} r \partial _r F(r) \left( a^2 b \sin (2 \varphi )+c^2 \sin (2 \varphi )-2 c d \cos (2 \varphi )+2 c r \cos (\varphi )-d^2 \sin (2 \varphi )+2 d r \sin (\varphi )\right) \nonumber \\ {} &\qquad {} -2 a^2 \partial _{\varphi } S_1^3(r,\varphi )=0, \end{aligned}$$

(A.27)

solved by

$$\begin{aligned} S_1^3(r,\varphi )=G(r)-\frac{r \partial _r F(r) \left( \cos (2 \varphi ) \left( a^2 b+c^2-d^2\right) +4 d \cos (\varphi ) (c \sin (\varphi )+r)-4 c r \sin (\varphi )\right) }{4 a^2}. \end{aligned}$$

(A.28)

Now let us continue in the (X, Y) coordinates. We solve (A.20) for $\partial _X S_2^3(X,Y)$, namely $\partial _X S_2^3(X,Y)=\frac{X}{Y} \partial _Y S_2^3(X,Y)$ and substitute it into the remaining Eqs. (A.21)–(A.24), for simplicity let these be in the following $Q_j=0$, $j=1,\ldots 4$, respectively. We solve $Q_2=0$ for $\partial _Y S_1^2$ and together with $\partial _X S_2^3(X,Y)=\frac{X}{Y} \partial _Y S_2^3(X,Y)$ we substitute it into

$$\begin{aligned} \partial _Y(\partial _Y Q_1- \partial _X (((-d + X) (c + Y))^{-1}Q_4). \end{aligned}$$

(A.29)

In this way we manage to arrive at an equation involving only $S_1^3$ and $S_2^3$. Once expressed in polar coordinates, this gives us an equation for the functions F(r) and G(r), cf. (A.25) and (A.28). We then expand the trigonometric expressions in (A.29) to obtain a polynomial in $\sin (\varphi )$, $\sin (\varphi )\cos (\varphi )$, $\sin (\varphi )\cos ^2(\varphi )$, $\cos ^3(\varphi )$ and $\cos ^4(\varphi )$ with coefficients depending on r. Equation (A.29) is satisfied if all such coefficients are equal to zero. This corresponds to a system of equations that the functions F, G have to satisfy. If we look at the coefficient of $\cos ^4(\varphi )$, we get

$$\begin{aligned} c d \left( a^2 b+c^2-d^2\right) \left( 15 F'(r)-r \left( 15 F''(r)+r \left( r F^{(4)}(r)-6 F'''(r)\right) \right) \right) =0. \end{aligned}$$

(A.30)

If $c d \left( a^2 b+c^2-d^2\right) \ne 0$, Eq. (A.30) implies

$$\begin{aligned} F(r)=\frac{1}{6} r^6 S^3_{23}+\frac{1}{4} r^4 S^3_{22}+\frac{1}{2} r^2 S^3_{21}+S^3_{20},\;\; S^3_{2j}\in {\mathbb {R}}. \end{aligned}$$

(A.31)

By substituting (A.31) into (A.29) and looking at the coefficient of $\sin (\varphi )\cos (\varphi )$ we obtain the condition

$$\begin{aligned}{} & {} -16 \left( a^2 \left( a^2 b+c^2-d^2\right) \left( 3 G'(r)+r \left( r G^{(3)}(r)-3 G''(r)\right) \right) +2 r^5 \left( 2 S^3_{23} \left( -7 r^2 \left( a^2 b-5 c^2+5 d^2\right. \right. \right. \right. )\nonumber \\ {} &\qquad +\left. \left. \left. \left( a^2 b+c^2\right) ^2-d^4\right) -5S^3_{22} \left( a^2 b-3 c^2+3 d^2\right) \right) \right) =0, \end{aligned}$$

(A.32)

solved by

$$\begin{aligned} G(r)&= {} \frac{S^{3}_{11}}{4} r^4 + \frac{S^{3}_{10}}{2} r^2 + S^3_{12} + 5 S^{3}_{22} \frac{a^2 b - 3 c^2 + 3 d^2}{24 a^2 (a^2 b + c^2 - d^2)} r^6 \nonumber \\ {} & \qquad {} + S^{3}_{23} r^6 \left( 7 \frac{a^2 b - 5 c^2 + 5 d^2}{48 a^2 (a^2 b + c^2 - d^2)} r^2 -\frac{a^2 b + c^2 + d^2}{12 a^2} \right) , \end{aligned}$$

(A.33)

where $S^3_{ij}\in {\mathbb {R}}$. By substituting also (A.33) into (A.29), we obtain that the remaining coefficients of trigonometric expressions vanish only if

$$\begin{aligned} S^3_{21}=\frac{2}{3} a^2 S^3_{11},\; S^3_{22}= 0,\; S^3_{23}=0. \end{aligned}$$

(A.34)

Transforming back to the (X, Y) coordinates, we therefore arrive at

$$\begin{aligned} S_2^3(X,Y)&= {} \frac{1}{3} a^2 S^3_{11} \left( X^2+Y^2\right) + S^3_{20}, \end{aligned}$$

(A.35)

$$\begin{aligned} S_1^3(X,Y)&= {} \frac{1}{12} S^3_{11} \left( 2 Y^2 \left( a^2 b+c^2-d^2-4 d X+3 X^2\right) +X^2 \left( -2 a^2 b-2 c^2+2 d^2-8 d X+3 X^2\right) \right. \nonumber \\ {} & \qquad {} +\left. 8 c X Y (X-d)+8 c Y^3+3 Y^4\right) +\frac{1}{2} S^3_{10}\left( X^2+Y^2\right) +S^3_{12} \end{aligned}$$

(A.36)

with $S^3_{1j}$, $S^3_{2 i}$ $\in {\mathbb {R}}$. Since $X_2$ is given by (A.17), we can set $S^3_{20}=S^3_{12}=0$ in the following without loss of generality (i.e. subtracting the first order integral $p_3^A+ S_2^3(X,Y)$ from $X_1$). By substituting into (3.1)–(3.6), we are finally left with equations for $S_1^1$ and $S_1^2$ that can be solved. Namely, we find

$$\begin{aligned} S_1^1(X,Y)&= {} S_{11}^1-\frac{1}{6 a^2}\left( 6 a^2 Y (2 b S^3_{10}+S_{12}^2)+S^3_{11} \left( 6 c Y^2 \left( 3 a^2 b+3 c^2+d^2-6 d X+3 X^2\right) \right. \right. \nonumber \\ {} & \qquad {} - c X (2 d-X) \left( 2 \left( a^2 b+c^2+d^2\right) - 6 d X+3 X^2\right) +Y^3 \left( a^4+6 a^2 b+2 \left( 13 c^2+d^2-6 d X+3 X^2\right) \right) \nonumber \\ {} & \qquad {} +Y \left( -4 d X \left( a^2 b+7 c^2+d^2\right) +4 \left( a^2 b+c^2\right) ^2+X^2 \left( a^4+2 a^2 b+14 \left( c^2+d^2\right) \right) \right. \nonumber \\ {} & \qquad {} - \left. \left. 4 d^4-12 d X^3+3 X^4\right) +15 c Y^4+3 Y^5\right) +6 S^3_{10} Y \left( (c+Y) (2 c+Y)-2 d^2-2 d X+X^2\right) \nonumber \\ {} & \qquad {} +\left. 6 c S^3_{10} X (X-2 d)\right) \end{aligned}$$

(A.37)

and

$$\begin{aligned} S_1^2(X,Y)&= {} \frac{S_{10}^3 \left( -d \left( Y (2 c+Y)+3 X^2\right) +X Y (2 c+Y)+X^3\right) }{a^2}+S_{12}^2 X+S_{11}^2\nonumber \\ {} & \qquad {} +\frac{1}{6 a^2}\left( X^2 \left( a^4 X+a^2 b (6 d-2 X)+2 c^2 (X-3 d)-18 d^3+26 d^2 X-15 d X^2+3 X^3\right) \right. \nonumber \\ {} & \qquad {} +4 c Y (X-d) \left( a^2 b+c^2+d^2-6 d X+3 X^2\right) \nonumber \\ {} & \qquad {} +\left. Y^2 \left( a^4 X+2 a^2 b (X-d)-2 (d-X) \left( 7 c^2+d^2-6 d X+3 X^2\right) \right) +12 c Y^3 (X-d)\right. \nonumber \\ {} & \qquad {} + \left. 3 Y^4 (X-d)\right) S_{11}^3, \end{aligned}$$

(A.38)

where $S_{1j}^i\in {\mathbb {R}}$.

All second order equations for the integral $X_1$ are now solved. After substituting the solution found for W and $m_1$, cf. (A.18), (A.16), we have

$$\begin{aligned}{} & {} \frac{2 \left( a^2 b+(c+Y)^2\right) \partial _X W_1(X,Y)}{a^2}+\frac{a^2 S_1^2(X,Y) \left( 2 \partial _X S_1^1(X,Y)+X \partial _Y S_2^3(X,Y)\right) }{4 (c+Y) (d-X)}\nonumber \\ {} &\qquad {} +\frac{2 (c+Y) (d-X) \partial _Y W_1 (X,Y)}{a^2}+\frac{X \partial _Y S_2^3(X,Y) S_1^3(X,Y)}{2 Y}-\partial _x M_1(X,Y)=0, \end{aligned}$$

(A.39)

$$\begin{aligned}{} & {} \frac{1}{4} \left( \frac{a^2 S_1^1(X,Y) \left( 2\partial _X S_1^1 (X,Y)+X \partial _Y S_2^3(X,Y)\right) }{(c+Y) (d-X)} -2 \partial _Y S_2^3(X,Y) S_1^3(X,Y)\right. \nonumber \\ {} &\qquad {} \left. -\frac{8 (d-X) \left( (c+Y) \partial _X W_1(X,Y)+(d-X) \partial _Y W_1(X,Y)\right) }{a^2}\right) +\partial _Y M_1(X,Y)=0, \end{aligned}$$

(A.40)

$$\begin{aligned}{} & {} \frac{1}{2} \left( \frac{X \partial _YS_2^3(X,Y)S_1^1(X,Y)}{Y}+\partial _Y S_2^3(X,Y) S_1^2(X,Y)\right. \nonumber \\ {} &\qquad {} \left. -2 X\partial _Y W_1(X,Y)+2 Y \partial _X W_1(X,Y)\right) =0 \end{aligned}$$

(A.41)

and

$$\begin{aligned} S_1^1(X,Y)\partial _X W_1(X,Y)+S_1^2(X,Y) \partial _Y W_1(X,Y)=0 \end{aligned}$$

(A.42)

for the first and zero order equations, respectively. For consistency with Eqs. (A.20)–(A.24), we express the above equations in the (X, Y) coordinates, though it is easier to solve them in the polar coordinates (A.26). The functions $S_i^j$ are known from (A.37)–(A.38), therefore the only unknowns are $M_1$ and the potential $W_1$. It is best to start with Eq. (A.41), that expressed in the polar coordinates (A.26) would involve only $\partial _{\varphi } W_1$ and known functions, thus it can be easily integrated. In this way we find W up to an arbitrary function of r that is determined by (A.42). Finally, the function $m_1$ can be found by solving the remaining Eqs. (A.39)–(A.40). By omitting constant terms in the potential W and in $m_1$, we arrive at the solution given in Sect. 3.1, where the constants have been renamed according to

$$\begin{aligned} \beta _1&= {} \frac{1}{3} S_{11}^3 \left( 2 \left( c^2+d^2\right) -a^2 b\right) -S_{10}^3,\;\;\beta _2=\frac{1}{3} a^3S_{11}^3,\\ \omega _1&= {} \frac{6 \beta _2 d^3 \left( c^2+d^2\right) }{a^5}-\frac{4 b \beta _2 d^3}{a^3}-\frac{2 \beta _1 d^3}{a^2}-\frac{\beta _2 d \left( c^2+d^2\right) }{2 a}-d S_{12}^2-S_{11}^2,\\ \omega _2&= {} -\frac{6\beta _2 c d^2 \left( c^2+d^2\right) }{a^5}+2 b c \left( \frac{\beta _2 \left( c^2+3 d^2\right) }{a^3}-\beta _1\right) +\frac{2 \beta _1 c d^2}{a^2}+\frac{\beta _2 c \left( c^2+d^2\right) }{2 a}+c S_{12}^2+S_{11}^1,\\ \omega _3&= {} \frac{a^5 (4 b \beta _1-2 S_{12}^2)-a^4 \beta _2\left( 3 c^2+d^2\right) -2 a^3 \beta _1 \left( c^2+3 d^2\right) -12 a^2 b \beta _2 d^2+3 \beta _2\left( c^2+d^2\right) \left( c^2+5 d^2\right) }{4 a^4}. \end{aligned}$$

If $c d \left( a^2 b+c^2-d^2\right) =0$, Eq. (A.30) vanishes identically. We look at the coefficient of $\sin (\varphi ) \cos (\varphi )^3$ in (A.29) which implies

$$\begin{aligned} \left( (a^2 b + c^2 - d^2)^2 - 4 c^2 d^2\right) \left( 15 F'(r)-r \left( 15 F''(r)+r \left( r F^{(4)}(r)-6 F^{(3)}(r)\right) \right) \right) =0. \end{aligned}$$

(A.43)

Thus unless we have also $(a^2 b + c^2 - d^2)^2 - 4 c^2 d^2=0$ the Eq. (A.31) determining F(r) still holds. Similarly, the remaining conditions coming from (A.29) after substituting the solution for F(r), cf. (A.31), imply that if $(a^2 b + c^2 - d^2)^2 - 4 c^2 d^2\ne 0$, we also must have

$$\begin{aligned} G(r) = \frac{ S^{3}_{11} }{4} r^4 + \frac{ S^{3}_{10} }{2} r^2 + S^3_{12} \end{aligned}$$

(A.44)

and

$$\begin{aligned} S^3_{21}=\frac{2}{3} a^2 S^3_{11},\; S^3_{22}= 0,\; S^3_{23}=0; \end{aligned}$$

(A.45)

i.e. the computation in this case proceeds exactly as above.

The equations characterizing the case when both (A.30) and (A.43) identically vanish, namely

$$\begin{aligned} c d \left( a^2 b+c^2-d^2\right) =0,\qquad (a^2 b + c^2 - d^2)^2 - 4 c^2 d^2=0, \end{aligned}$$

(A.46)

are equivalent to the equations

$$\begin{aligned} c d=0, \qquad a^2 b+c^2-d^2=0, \end{aligned}$$

(A.47)

which have two solutions, $c=0$, $b=\frac{c^2}{a^2}$ or $d=0$, $b=-\frac{c^2}{a^2}$. However, they are related by a rotation of the coordinate frame around the z–axis by the angle $\frac{\pi }{2}$ accompanied by an appropriate linear combination of the integrals H, $X_1$ and $X_2$. Thus we proceed considering only the case $d=0$, $b=-\frac{c^2}{a^2}$.

In this case, the Eq. (A.29) significantly simplifies and implies only two independent conditions on the functions F(r) and G(r), namely

$$\begin{aligned} r^3 F^{(4)}(r) + 5 r^2 F'''(r) + 2 r F''(r) - 2 F'(r)&= {} 0, \nonumber \\ a^2 \left( r^2 G'''(r)+ r G''(r) - G'(r) \right) -r^4 F'''(r) - 6 r^3 F''(r) - 6 r^2 F'(r)&= {} 0. \end{aligned}$$

(A.48)

Their general solution reads

$$\begin{aligned} F(r)&= {} S^3_{21} \ln (r) + S^3_{22} r^2 + \frac{S^3_{23}}{r}+ S^{3}_{20}, \nonumber \\ G(r)&= {} \frac{S^3_{10}}{2} r^2 + S^3_{11} \ln (r) + S^3_{12}+\frac{3 S^3_{22}}{4 a^2}r^4 + \frac{S^3_{21}}{2 a^2} r^2 \left( \ln (r) - 1 \right) . \end{aligned}$$

(A.49)

Thus we have determined the functions $S^3_2(r,\varphi )$ and $S^3_1(r,\varphi )$ in terms of seven integration constants. Continuing our analysis of the Eqs.  (A.20)–(A.24) in polar coordinates (A.26), we determine also $S^1_1(r,\varphi )$ and $S^2_1(r,\varphi )$, introducing three more integration constants and fully solving the Eqs.  (A.20)–(A.24).

The consistency of the lower order conditions requires

$$\begin{aligned} S^3_{11}=S^3_{21}=0 \end{aligned}$$

(A.50)

and leads to the splitting into two cases: either

$$\begin{aligned} S^3_{23}=0, \end{aligned}$$

(A.51)

which after solution of all the lower order equations implies the solution of the form introduced in Sect. 3.1 for the particular values $d=0$, $b=-\frac{c^2}{a^2}$; or

$$\begin{aligned} S^3_{22}=0. \end{aligned}$$

(A.52)

The system corresponding to the case (A.52) is characterized by

$$\begin{aligned} B_x(x, y, z)&= {} \frac{\beta \left( y - \frac{c}{a} \right) }{\sqrt{x^2 + \left( y - \frac{c}{a} \right) ^2 }^3}, \qquad B_y(x, y, z) = -\frac{\beta x}{\sqrt{ x^2 + \left( y - \frac{c}{a} \right) ^2 }^{3}}, \nonumber \\ B_z(x, y, z)&= {} 0, \qquad W(x, y, z) = \frac{\omega }{\sqrt{x^2 + \left( y - \frac{c}{a} \right) ^2}} - \frac{\beta ^2}{4 \left(x^2 + \left( y - \frac{c}{a} \right) ^2\right) }; \end{aligned}$$

(A.53)

however, it possesses two commuting first order integrals

$$\begin{aligned} X_3={ \ell _3^A}, \quad \tilde{X}_2=p_3^A+\frac{\beta }{2 \sqrt{x^2 + \left( y - \frac{c}{a} \right) ^2}} \end{aligned}$$

(A.54)

of standard cylindrical type together with a pair of standard pair of parabolic cylindrical integrals, and therefore it is contained in [28], only expressed in a shifted coordinate system. Thus, the generalized elliptical integral $X_1$ can be expressed as a function of the standard integrals.

1.1.2 Appendix A.1.2: $c=d=0$

If $c=d=0$ then the structure of the leading order terms hints that the equations are easiest to investigate in the cylindrical coordinates (4.1). Expressing the conditions (3.1)–(3.10) and (3.14)–(3.19) in these coordinates, it is rather straightforward if somewhat tedious, to solve them. The easiest approach is to solve first the conditions coming from the leading order terms of $\{ H,X_2 \}=0$, next the involutivity condition $\{ X_1,X_2 \}=0$ and finally use the leading order terms $\{ H,X_1 \}=0$, under the assumption $a\ne 0$ and $b\ne 0$. We arrive at a general solution of the form

$$\begin{aligned} B_r\left( r, \varphi , Z\right)&= {} \frac{r}{a} \left( S^r_{212} \sin \left( \frac{2 Z}{a} - \varphi \right) -S^r_{211} \cos \left( \frac{2 Z}{a} - \varphi \right) \right) , \end{aligned}$$

(A.55)

$$\begin{aligned} B_\varphi \left( r, \varphi , Z\right)&= {} 3 S^Z_{21} r^5 + 2 S^Z_{22} r^3 + S^Z_{23} r - \frac{S^r_{211}}{a} \sin \left( \frac{2 Z}{a} - \varphi \right) - \frac{S^r_{212}}{a} \cos \left( \frac{2 Z}{a} - \varphi \right) , \nonumber \\ B_Z\left( r, \varphi , Z\right)&= {} \frac{1}{a} \left[ -S^Z_{101} r^3 - \frac{S^Z_{102}}{a} r + S^Z_{21} r^5 \left( 6 b \cos \left( 2 \varphi \right) - 7 r^2\right) + \right. \nonumber \\ {}{} &\qquad {} + \left. S^Z_{22} r^3 \left( 2 b \cos \left( 2 \varphi \right) - 5 r^2 - 2 b\right) - S^Z_{23} b r\right] , \nonumber \\ s_1^r\left( r, \varphi , Z\right)&= {} -\frac{b S^Z_{101}}{3 a} r^3 \sin \left( 2 \varphi \right) - \frac{b S^Z_{102}}{a} r \sin \left( 2 \varphi \right) + S^r_{110} \cos \left( \varphi \right) + S^r_{120} \sin \left( \varphi \right) \nonumber \\ {}{} & \qquad {} + \frac{b S^Z_{21}}{5 a} r^5 \left( -5 r^2 \sin \left( 2 \varphi \right) + 3 b \sin \left( 4 \varphi \right) \right) + \frac{b S^Z_{22}}{3a} r^3 \left( 2 b \cos \left( 2 \varphi \right) - 3 r^2 - 2 b\right) \sin \left( 2 \varphi \right) \nonumber \\ {} & \qquad {} - \frac{b^2 S^Z_{23}}{a} r \sin \left( 2 \varphi \right) - \frac{S^r_{211}}{4} \left( \left( a^2 + 2 b\right) \cos \left( \frac{2 Z}{a}-\varphi \right) + 2 b \cos \left( \frac{2 Z}{a}+\varphi \right) \right) \nonumber \\ {} & \qquad {} + \frac{S^r_{212}}{4} \left( \left( a^2 + 2 b\right) \sin \left( \frac{2 Z}{a}-\varphi \right) + 2 b \sin \left( \frac{2 Z}{a}+\varphi \right) \right) , \nonumber \\ s_1^\varphi \left( r, \varphi , Z\right)&= {} -2 \frac{S^Z_{101}}{3a} r^2 \left( b \cos \left( 2 \varphi \right) - \frac{3}{4} r^2\right) - \frac{S^Z_{102}}{a} \left( b \cos \left( 2 \varphi \right) r^2\right) + S^\varphi _{10} \nonumber \\ {} & \qquad {} - S^r_{110} \frac{\sin \left( \varphi \right) }{r} + S^r_{120} \frac{\cos \left( \varphi \right) }{r} + \frac{S^Z_{21}}{20 a} r^4 \left( 35 r^4 + 10 a^2 r^2+ 36 b^2 \cos \left( 2 \varphi \right) ^2 \right. \nonumber \\ {}{} &\qquad {} - \left. 80 b r^2 \cos \left( 2 \varphi \right) + 12 b^2\right) + \frac{S^Z_{22}}{6a} r^2 \left[ 10 r^4+\left( 3 a^2 + 6 b \right) r^2 + 4 b^2 \cos \left( 2 \varphi \right) ^2 \right. \nonumber \\ {}{} & \qquad{} \left. - 18 b r^2 \cos \left( 2 \varphi \right) - 8 b^2 \cos \left( 2 \varphi \right) + 4 b^2\right] + \frac{S^Z_{23}}{2 a} \left( \left( a^2 + 2 b\right) r^2 -2 b^2 \cos \left( 2 \varphi \right) \right) \nonumber \\ {}{} &\qquad {} - \frac{S^r_{211}}{4 r} \left( \left( a^2 + 4 r^2 + 2 b\right) \sin \left( \frac{2 Z}{a}-\varphi \right) - 2 b \sin \left( \frac{2 Z}{a}+\varphi \right) \right) \nonumber \\ {}{} &\qquad {} - \frac{S^r_{212}}{4 r} \left( \left( a^2 + 4 r^2 + 2 b\right) \cos \left( \frac{2 Z}{a} - \varphi \right) - 2 b \cos \left( \frac{2Z}{a}+\varphi \right) \right) ,\\ s_1^Z\left( r, \varphi , Z\right)&= {} \frac{S^Z_{101}}{4} r^4 + \frac{S^Z_{102}}{2} r^2 + S^Z_{103} - \frac{S^Z_{21}}{8} r^6 \left( 12 b \cos \left( 2 \varphi \right) - 7 r^2 + 4 b\right) \nonumber \\ {}{} &\qquad {} - S^Z_{22} r^4 \left( b \cos \left( 2 \varphi \right) - \frac{5}{6} r^2\right) - \frac{b S^Z_{23}}{2} r^2 \cos \left( 2 \varphi \right) \nonumber \\ {}{} &\qquad {} - \frac{a S^r_{211}}{2} r \sin \left( \frac{2 Z}{a} - \varphi \right) - \frac{a S^r_{212}}{2} r \cos \left( \frac{2 Z}{a} - \varphi \right) , \nonumber \\ s_2^r\left( r, \varphi , Z\right)&= {} S^r_{211} \cos \left( \frac{2 Z}{a} - \varphi \right) - S^r_{212} \sin \left( \frac{2 Z}{a} - \varphi \right) , \nonumber \\ s_2^\varphi \left( r, \varphi , Z\right)&= {} \frac{1}{r} \left( S^r_{211} \sin \left( \frac{2 Z}{a} - \varphi \right) + S^r_{212} \cos \left( \frac{2 Z}{a} - \varphi \right) \right) , \nonumber \\ s_2^Z\left( r, \varphi , Z\right)&= {} S^Z_{21} r^6 + S^Z_{22} r^4 + S^Z_{23} r^2 + S^Z_{24}.\nonumber \end{aligned}$$

(A.56)

However, substituting the solution (A.55) and (A.56) in the remaining, lower order determining equations and proceeding with their solution, the compatibility of the first order equations for the derivatives of the function $m_2$, namely $\partial _\varphi \left( \partial _r m_2(r, \varphi , Z) \right) = \partial _r \left( \partial _\varphi m_2(r, \varphi , Z)\right)$, immediately implies splitting into two cases:

• $S^r_{211} = S^r_{212} = 0$, or

• $S^Z_{101} = S^Z_{23}$ and $S^Z_{21} = S^Z_{22} = 0$.

In the first case, solving the remaining conditions, we find four systems, all already known:

• three systems for which other choice of integrals of the standard cylindrical type is possible: the constant magnetic field and vanishing electrostatic potential, the constant magnetic field and isotropic harmonic oscillator in the plane perpendicular to the magnetic field [7] and the superintegrable system (4.15);

• the system with nonreducible generalized integrals (2.8) defined in (3.22) and (3.23), restricted to the particular values $c=d=0$.

In the second case, there is only one solution, namely the helical undulator of [12], for which again other choices of commuting integrals exist, e.g. of standard Cartesian type.

1.2 Appendix A.2: $a=0$

If $a=0$, Eq. (3.21) is identically satisfied and (3.14),(3.15) and (3.17) imply

$$\begin{aligned} S_{21}^2(z)= -\frac{1}{2} d \partial _z S_{22}(z), \;\;S_{21}^{1}(z)= \frac{1}{2} c \partial _z S_{22}(z) \end{aligned}$$

(A.57)

and

$$\begin{aligned}{} & {} c d \partial ^2_{z}S_{22}(z)=0, \nonumber \\ {} & {} 4 b S_{22}(z)- c^2 \partial ^2_{z}S_{22}(z)+ d^2 \partial ^2_{z}S_{22}(z)=0. \end{aligned}$$

(A.58)

This leads to a second splitting depending on whether $cd=0$ or not.

1.2.1 Appendix A.2.1: c or d equals zero

Without loss of generality (i.e. up to a rotation and redefinition of the integrals) we can consider only the case c not zero and $d=0$, therefore from (A.57) we have $S_{21}^2(z)=0$ and the real solutions of Eq. (A.58) are given by

$$\begin{aligned} S_{22}(z)=\alpha _1 \cosh \left( \frac{2 \gamma z}{c}\right) + \alpha _2 \sinh \left( \frac{2 \gamma z}{c}\right) , \;\; \alpha _i\in {\mathbb {R}} \end{aligned}$$

(A.59)

for $\gamma ^2=b$, and

$$\begin{aligned} S_{22}(z)=\alpha _1 \cos \left( \frac{2 \gamma z}{c}\right) + \alpha _2 \sin \left( \frac{2 \gamma z}{c}\right) , \;\; \alpha _i\in {\mathbb {R}} \end{aligned}$$

(A.60)

for $\gamma ^2=- b$. Let us continue with this last case, i.e. $S_{22}$ as in (A.60), for (A.59) the computation is analogous.

The last two Eqs. (3.18) and (3.19) in the involutivity condition are solved by

$$\begin{aligned} s_1^1(x,y,z)&= {} S_1^1(x,y) -y\left( \sin \left( \frac{2 \gamma z}{c}\right) \left( \alpha _1\gamma x+\alpha _2\left( x^2+y^2- \gamma ^2\right) \right) -\cos \left( \frac{2 \gamma z}{c}\right) \left( \gamma \alpha _2 x-\alpha _1 \left( x^2+y^2-\gamma ^2\right) \right) \right) ,\nonumber \\ s_2^1(x,y,z)&= {} S_1^2(x,y) +\frac{1}{4 \gamma }\left( \sin \left( \frac{2 \gamma z}{c}\right) \left( 4\gamma \alpha _2 x (x^2+y^2)+\alpha _1(\gamma ^2 x^2-c^2)\right) \right. \nonumber \\ {} & \qquad{} \left. +\gamma \cos \left( \frac{2 \gamma z}{c}\right) \left( 4\gamma \alpha _1 x(x^2+y^2)-\alpha _2(\gamma ^2 x^2-c^2)\right) \right) , \end{aligned}$$

(A.61)

where again $S_1^1(x,y)$ and $S_1^2(x,y)$ are arbitrary functions of x and y.

We substitute the solution we have so far for $s_j^2$, $s_3^1$ and $B_1$, $B_2$, i.e. (3.11), (3.20) and (3.12)–(3.13) respectively, together with (A.60) and (A.61) into the second order Eqs. (3.1)–(3.6) for the integral $X_1$. These give equations for the yet unknown functions $S_2^3$, $B_3$, $S_j^1$, $j=1,2,3$ that can be solved without much difficulty in this case. We obtain

$$\begin{aligned} S_2^3(x,y)&= {} \frac{S_{21}^3}{20 y^4}+\frac{S_{22}^3}{6 y^2}+S_{24}^3 y+ S_{23}^3,\end{aligned}$$

(A.62)

$$\begin{aligned} S_1^1(x,y)&= {} \frac{1}{120 c y^4}\left( -40 \eta _2 y^2 \left( \gamma ^2+x^2\right) +4 S_{21}^3 \left( 3 \left( \gamma ^2+x^2\right) ^2-4 \gamma ^2 y^2\right) \right. \nonumber \\ {} & \qquad {} +30S_{24}^3 y^5 \left( 3 \left( x^2+y^2\right) ^2-4 \gamma ^2 y^2\right) +c^2 \left( 10 S_{22}^3 y^2+3 S_{21}^3\right) \nonumber \\ {} &\qquad {} \left. +120 c y^4 (\omega _2 y+\omega _0)+120 \eta _1 y^5 \left( x^2+y^2\right) \right) , \end{aligned}$$

(A.63)

$$\begin{aligned} S_2^1(x,y)&= {} \omega _1-\omega _2 x+ \frac{1}{2} c S_{24}^3 x+ \frac{x}{60 c y^3}\left( 20 y^2 \left( -3 \eta _1 y \left( 2 \gamma ^2+x^2+y^2\right) -2 \eta _2\right) \right. \nonumber \\ {} & {} \left. +8 S_{21}^3 \left( \gamma ^2+x^2\right) -15 S_{24}^3 y^3 \left( 4 \gamma ^2 x^2+3 \left( x^2+y^2\right) ^2\right) \right) , \end{aligned}$$

(A.64)

and

$$\begin{aligned} B_3(x,y,z)&= {} \frac{6 \gamma ^2 S_{21}^3+30 \eta _1 y^5-10 \eta _2 y^2+45 S_{24}^3 x^2 y^5+6 S_{21}^3 x^2+45 S_{24}^3 y^7}{30 c y^5}\nonumber \\ {} & \qquad {} -\left( \alpha _1 \left( \cos \left( \frac{2 \gamma z}{c}\right) \right) + \alpha _2 \sin \left( \frac{2 \gamma z}{c}\right) \right) , \end{aligned}$$

(A.65)

where $\eta _1,\eta _2, S^3_{21},S^{3}_{22},S^{3}_{23}$ and $S^{3}_{24}$ are constants of integration.

We continue with the solution of the first order equations. They read

$$\begin{aligned}{} & {} -2 b^2 \partial _x W(x,y,z)-B_2(x,y,z) S_1^3(x,y,z)+B_3(x,y,z) S_1^2(x,y,z)\nonumber \\ {} & \qquad {} -c\partial _z W(x,y,z)+\partial _x m_1(x,y,z)-2 y^2 \partial _x W(x,y,z)+2 x y \partial _y W(x,y,z)=0, \end{aligned}$$

(A.66)

$$\begin{aligned}{} & {} B_1(x,y,z) S_1^3(x,y,z)-B_3(x,y,z)S_1^1(x,y,z)\nonumber \\ {} &\qquad {} +\partial _y m_1(x,y,z)-2 x^2\partial _y W(x,y,z)+2 x y\partial _x W(x,y,z)=0, \end{aligned}$$

(A.67)

$$\begin{aligned}{} & {} -B_1(x,y,z)S_1^2(x,y,z)+B_2(x,y,z) S_1^1(x,y,z)\nonumber \\ {} &\qquad {} -c \partial _x W(x,y,z)+\partial _z m_1(x,y,z)=0 \end{aligned}$$

(A.68)

and

$$\begin{aligned}{} & {} B_2(x,y,z) S_2^3(x,y,z)-B_3(x,y,z) S_2^2(x,y,z)-\partial _x m_2(x,y,z)=0, \end{aligned}$$

(A.69)

$$\begin{aligned}{} & {} B_3(x,y,z) S_2^1(x,y,z)-B_1(x,y,z) S_2^3(x,y,z)-\partial _y m_2(x,y,z)=0, \end{aligned}$$

(A.70)

$$\begin{aligned}{} & {} B_1(x,y,z) S_2^2(x,y,z)-B_2(x,y,z) S_2^1(x,y,z)-\partial _z m_2(x,y,z)+2 \partial _z W(x,y,z)=0, \end{aligned}$$

(A.71)

for $X_1$ and $X_2$, respectively. The magnetic field $\vec {B}$ and the functions $S_i^j$ are already constrained by the previous steps.

We use compatibility conditions for the above equations, consequence of the fact that the second order mixed derivatives of each $m_i$ must be equal. In particular, by imposing that $\partial _ {x y} m_2=\partial _{y x} m_2$ we get (after substituting all we know about $B_j$ and $S_i^j$)

$$\begin{aligned}{} & {} -\frac{1}{10 c y^6}\left( \cos \left( \frac{2 \gamma z}{c}\right) \left( 10 \alpha _1 S_{21}^3 x^3-10 \gamma \alpha _2 S_{21}^3 x^2\right. \right. \nonumber \\ {} & {} \left. +2 \alpha _1 x \left( 5 \gamma ^2 S_{21}^3+y^2 (2 S_{21}^3-5 \eta _2)\right) +\gamma \alpha _2 \left( y^2 \left( 10 \eta _2-20 S_{23}^3 y^4\right. \right. \right. \left. \left. \left. +S_{21}^3\right) -10 \gamma ^2 S_{21}^3\right) \right) \nonumber \\ {} & \qquad {} +\left( 10 \gamma ^3 \alpha _1 S_{21}^3+10 \gamma ^2 \alpha _2 S_{21}^3 x+2 \alpha _2 x \left( y^2 (2 S_{21}^3-5 \eta _2)+5 S_{21}^3 x^2\right) \right. \nonumber \\ {} & \qquad {} +\left. \left. \gamma \alpha _1 \left( -y^2 (10 \eta _2+S_{21}^3)+10 S_{21}^3 x^2+20 S_{23}^3 y^6\right) \right) \sin \left( \frac{2 \gamma z}{c}\right) \right) =0. \end{aligned}$$

(A.72)

The above equation is polynomial in x and y. By collecting the coefficients of different powers of x and y, we see that necessarily

$$\begin{aligned} 10 S_{21}^3 \left( \alpha _2 \sin \left( \frac{2 \gamma z}{c}\right) + \alpha _1 \cos \left( \frac{2 \gamma z}{c}\right) \right) =0. \end{aligned}$$

(A.73)

This leads to two subcases depending on $S_{21}^3$ vanishing or not. We find that there can be a solution corresponding to a new integrable system only for $S_{21}^3=S_{22}^3=S_{23}^3=\eta _2=\omega _0=0$. Namely, we obtain the integrable system given in Sect. 3.2. The constants are for simplicity renamed according to $\gamma = \frac{c \delta }{2}$, $\eta _1=\beta _1$, $S_{24}^3=\beta _2$.

1.2.2 Appendix A.2.2: $cd\ne 0$

If $cd\ne 0$ then the Eq. (A.58) imply that $S_{22}(z)$ vanishes and thus also $S_{21}^2(z)$ and $S_{22}^1(z)$ due to Eq. (A.57).

Consequently, the Eqs. (3.11)–(3.13) simplify to

$$\begin{aligned} s^1_2(x, y, z)&= {} 0, \qquad s^2_2(x, y, z) = 0, \qquad s^3_2(x, y, z) = S_2^3(x, y), \nonumber \\ B_1(x, y, z)&= {} -\frac{1}{2} \partial _y S_2^3(x, y), \qquad B_2(x, y, z) = \frac{1}{2} \partial _x S_2^3(x, y). \end{aligned}$$

(A.74)

The remaining equations in (3.14)–(3.19) imply

$$\begin{aligned} s^1_1(x, y, z) = S^1_1(x,y), \qquad s^2_1(x, y, z) = S^2_1(x,y), \qquad s^3_1(x, y, z) = S_3^1(x, y). \end{aligned}$$

(A.75)

Proceeding to solve the Eqs. (3.1)–(3.6) we determine the third component of the magnetic field using (3.2),

$$\begin{aligned} B_3(x, y, z) = \frac{1}{4 xy} \left( d \partial _y S_2^3(x, y) - 2 \partial _y S_1^2(x, y) \right) . \end{aligned}$$

(A.76)

The Eq. (3.3) then implies

$$\begin{aligned} c \partial _x S_2^3(x, y) + d\partial _y S_2^3(x, y)=0. \end{aligned}$$

(A.77)

We can solve the remaining equations in (3.1)–(3.6) with respect to the derivatives $\partial _y S_1^2(x, y)$, $\partial S_1^2(x, y)$, $\partial _y S_1^3(x, y)$, $\partial _x S_1^3(x, y)$ and consider their compatibility, namely

$$\begin{aligned} \partial _y \left( \partial _x S_1^2(x, y) \right) =\partial _x \left( \partial _y S_1^2(x, y) \right) , \qquad \partial _y \left( \partial _x S_1^3(x, y) \right) =\partial _x \left( \partial _y S_1^3(x, y) \right) . \end{aligned}$$

(A.78)

These together with Eq. (A.77) define a system of linear partial differential equations for $S_2^3(x, y)$ and $S_1^1(x, y)$. Considering its differential consequences and their compatibility we find that we must have

$$\begin{aligned} S_1^1(x, y)&= {} S^1_{11} y + S^1_{12} + \frac{S^1_{13}}{2} y (x^2 + y^2) + \frac{S^3_{21}}{4 c} \left( 3 y (x^2 + y^2)^2 + 4 b y^3 - 2 c d x \right) , \nonumber \\ S_2^3(x, y)&= {} S^3_{21} \left( y - \frac{d}{c} x \right) + S^3_{20}. \end{aligned}$$

(A.79)

Next, integrating the equations determining the first order derivatives $\partial _y S_1^2(x, y)$, $\partial S_1^2(x, y)$, $\partial _y S_1^3(x, y)$, $\partial _x S_1^3(x, y)$ we find the complete solution of the second order conditions (3.1)–(3.10) and (3.14)–(3.19) in the form

$$\begin{aligned} B_1(x, y, z)&= {} - \frac{S^3_{21}}{2}, \quad B_2(x, y, z) = - \frac{d S^3_{21}}{2c}, \quad B_3(x, y, z) = \frac{3 S^3_{21}}{2c} ( x^2 + y^2 ) + \frac{S^1_{13}}{2}, \nonumber \\ s_1^1(x, y, z)&= {} \frac{S^3_{21}}{4c} \left( 3 y \left( x^2 + y^2 \right) ^2 + 2 (2 b y^3 - c d x) \right) +\frac{S^1_{13}}{2} y \left( x^2 + y^2 \right) +S^1_{11} y +S^1_{12}, \nonumber \\ s_1^2(x, y, z)&= {} - \frac{S^3_{21}}{4c} \left( 3 x \left( x^2+y^2 \right) ^2 + 2 \left( \left( d^2 - c^2 \right) x - 2 b x^3 - c d y \right) \right) \\ {} & \qquad {} - \frac{S^1_{13}}{2} x \left( x^2 + y^2 \right) + \left( S^1_{13} b - S^1_{11} \right) x + S^2_{1}, \nonumber \\ s_1^3(x, y, z)&= {} \frac{S^3_{21}}{2} \left( \left( y - \frac{d}{c} x \right) \left( x^2 + y^2\right) + 2 b \frac{d}{c} x \right) + \frac{S^1_{13}}{2} c \left( y - \frac{d}{c} x \right) +S_1^3, \nonumber \\ s_2^1(x, y, z)&= {} 0, \qquad s_2^2(x, y, z) = 0, \qquad s_2^3(x, y, z) = S^3_{21} \left( y - \frac{d}{c} x \right) + S^3_{20}.\nonumber \end{aligned}$$

(A.80)

Proceeding to solve the lower order conditions, we immediately find from the involutivity condition $\{ X_1,X_2 \}=0$ that W(x, y, z) can depend only on the xy–coordinates and express $\partial _x W(x, y)$ in terms of $\partial _y W(x, y)$ and the integration constants $S^3_{21}, S_{13}^1,\ldots$ introduced in (A.80). Substituting it into the compatibility conditions $\partial _u \left( \partial _v m_a\right) = \partial _v \left( \partial _u m_a\right)$, $a=1,2$, $u,v=x,y,z$ derived from the first order conditions coming from $\{H,X_1\}=0$ and $\{H,X_2\}=0$, and easily integrating them, we find W(x, y) fully determined up to an irrelevant additive constant. However, comparing it with the previously derived results, we see that this system is the limit of the system (3.22)–(3.23) as $a\rightarrow 0$, $a \beta _1 \rightarrow \frac{S^1_{13}}{2}$, $\beta _2=-\frac{S^3_{21}}{2}$, $\omega _1=-\frac{S^2_1}{c}$, $\omega _2=\frac{S^1_{12}}{c}$, $\omega _3=\frac{S^1_{11}}{2 c}$.