First, we study some special (x,y)-linear Weingarten canal surface with Bishop frame in E3 including developable canal surfaces, minimal canal surfaces, and the canal surfaces with vanishing second Gaussian curvature.
Remark 1
The (x,y)-linear Weingarten canal surfaces are considered as a natural generalization of canal surfaces with constant Gaussian curvature, constant mean curvature, or constant second Gaussian curvature.
Theorem 4
A canal surface Υ=f(u,v) according to Bishop frame in E3 is developable if and only if it is a circular cylinder or a circular cone.
Proof
Υ=f(u,v) is developable if and only if its Gaussian curvature K=0. From Eq. (14), we have q=0. Also, from Eq. (13), we get
$$ q=\sin{\theta}(-k_{1}\cos{\nu}+k_{2}\sin{\nu}+\theta')=\sin{\theta}(-k_{1}\cos{\nu}+k_{2}\sin{\nu})+r^{\prime\prime}=0. $$
(36)
□
From Eq. (36), we have r′′=0 and k1=k2=0 (i.e., κ=0). Therefore, r(u)=c1u+c2, where c1 and c2 are constants and c1≠±1 or (else sinθ, a contradiction). Then, Υ is a circular cylinder if c1=0 or a circular cone if c1≠0 and c1≠±1.
Theorem 5
A canal surface Υ=f(u,v) according to Bishop frame in E3 is minimal if and only if it is a catenoid.
Proof
Since Υ is minimal if and only if its mean curvature H=0, then Eq. (15) implies
$$\begin{array}{@{}rcl@{}} &2p-\sin^{2}{\theta}=0,\\ &\sin^{2}{\theta}+2r\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-2rr^{\prime\prime}=0. \end{array} $$
(37)
□
Then, we have sinθ−2rr′′=0 and 2r sinθ(k1 cosν−k2 sinν). Since r≠0 and sinθ, then k1 cosν−k2 sinν=0 implies that κ=0 and Υ is a surface of revolution. It is well known that the only minimal surface of revolution is the catenoid.
It is recognized that a minimal surface satisfied that KII=0. However, a surface with vanishing second Gaussian curvature is not necessary to be minimal [6].
Theorem 6
A non-developable canal surface Υ=f(u,v) according to Bishop frame in E3 with vanishing second Gaussian curvature KII=0 is a surface of revolution which satisfies
$$\begin{array}{@{}rcl@{}} &-2r{r^{\prime\prime}}^{2} (-1+rr^{\prime\prime})(-1+2rr^{\prime\prime})+{r'}^{6} (r^{\prime\prime}-r{r^{\prime\prime}}^{2})+{r'}^{2} (r^{\prime\prime}+5r{r^{\prime\prime}}^{2}-6r^{2} {r^{\prime\prime}}^{3}-r{r^{\prime\prime}}^{2})\\ &+{r'}^{4} (-r^{\prime\prime} (2+3rr^{\prime\prime})+2r{r^{\prime\prime}}^{2})+(-rr'+2r{r'}^{3}-r{r'}^{5})r^{\prime\prime\prime}=0. \end{array} $$
(38)
Proof
When KII=0, we have from Eq. (19)
$$\begin{array}{@{}rcl@{}} &H=-\frac{B}{2r^{2}p^{2}q^{2}}, B=-q^{3}(2p-\sin^{2}{\theta}),\\& B=\frac{-1}{2048}\sum\nolimits^{6}_{i=0,j=-6}\sin{\theta}\Big(\gamma_{i,j}\,\cos{(i\nu+j\theta)+\delta_{i,j}\,\sin{(i\nu+j\theta)}}\Big). \end{array} $$
□
All the coefficients of Eq. (19), γi,j and δi,j, are vanished exept γ1,0 and δ1,0, then we have k1=k2=0 which implies that κ=0. Then, the canal surface is a surface of revolution. Furthermore, by Eqs. (27) and (35), we have
$$\begin{array}{@{}rcl@{}} &\frac{1}{2}\Big(-2r{r^{\prime\prime}}^{2} (-1+rr^{\prime\prime})(-1+2rr^{\prime\prime})+{r'}^{6} (r^{\prime\prime}-r{r^{\prime\prime}}^{2})+{r'}^{2} (r^{\prime\prime}+5r{r^{\prime\prime}}^{2}-6r^{2} {r^{\prime\prime}}^{3}-r{r^{\prime\prime}}^{2})\\ &+{r'}^{4} (-r^{\prime\prime} (2+3rr^{\prime\prime})+2r{r^{\prime\prime}}^{2}) +(-rr'+2r{r'}^{3}-r{r'}^{5})r^{\prime\prime\prime}\Big)=0. \end{array} $$
Now, we study some properties of (x;y)-linear Weingarten canal surfaces. Without losing generality, we may assume that c=1 in ax+by=c.
Theorem 7
A canal surface Υ=f(u,v) according to Bishop frame in E3 is a (K,H)-linear Weingarten canal surface if and only if it is one of the following surfaces:
-
i. Υ tube with radius \(r=\frac {b}{a}\),
-
i.. Υ surface of revolution such as
$$f(u,v)=\Big(-r(u)\cos{\nu}\sin{\theta},\sin{\nu}\sin{\theta},u+ \cos{\theta}\Big), $$
where r(u) is given by Eq. (41).
Proof
A (K,H)-linear Weingarten canal surface satisfied
where a,b∈R, and (a,b)≠(0,0). From Eqs. (14) and (15), we have
$$\begin{array}{c} (-b+ar)\big(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})\big) \\ \left(\begin{array}{c} 2r\big(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-{r^{\prime\prime}}\big)^{2}+\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})\\ \left(-2r^{2}\big(1-{r'}^{2}+r(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-{r^{\prime\prime}}^{2})\big)^{2}+(-b+2ar)\right.\\ \left(\begin{array}{c} (1-{r'}^{2})(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-{r^{\prime\prime}}^{2})-(-b+2ar)\\ \left.(1-{r'}^{2})r^{\prime\prime}+2r^{2}(1-{r'}^{2}+rr^{\prime\prime})^{2}\right) \end{array} \right) \end{array} \right) \\ -(-b-2ar)(-1+{r'}^{2}){r^{\prime\prime}}+2r(-b+ar){r^{\prime\prime}}^{2}-2r^{2}(-1+{r'}^{2}+r{r^{\prime\prime}})^{2} \end{array}=0. $$
Therefore, we get
$$\begin{array}{c} (-b+ar)\big(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})\big) \\ \left(\begin{array}{c} 2r\big(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-{r^{\prime\prime}}\big)^{2}+\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})\\ \left(-2r^{2}\big(1-{r'}^{2}+r(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-{r^{\prime\prime}}^{2})\big)^{2}+(-b+2ar)\right.\\ \left(\begin{array}{c} (1-{r'}^{2})(\sin{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu})-{r^{\prime\prime}}^{2})-(-b+2ar)\\ \left.(1-{r'}^{2})r^{\prime\prime}+2r^{2}(1-{r'}^{2}+rr^{\prime\prime})^{2}\right) \end{array} \right) \end{array} \right) \end{array}=0, $$
(39)
and
$$ -(-b-2 a r)(-1+{r'}^{2}){r^{\prime\prime}}+2r(-b+a r){r^{\prime\prime}}^{2}-2r^{2}(-1+{r'}^{2}+r{r^{\prime\prime}})^{2}=0. $$
(40)
□
Case 1
From Eq. (39), if k1=k2=0, this mean that κ=0. Thus, Υ is a surface of revolution and its radial function satisfies Eq. (40)
$$-(b-2ar)(-1+{r'}^{2}) r^{\prime\prime}+2r(-b+ar) {r^{\prime\prime}}^{2}-2r^{2} (-1+{r'}^{2}+rr^{\prime\prime})^{2}=0. $$
Solving the above equation, we get
$$ u=c_{2}\pm\int\sqrt{\frac{c_{1}+r(b-a r+r^{3})}{r(b-a r+r^{3})}} dr, $$
(41)
where c1 and c2 are constants [23]. Since κ=0, we assume that the spine curve can be taken as the following form α(u)=(0,0,u) and T(u)=(0,0,1),P(u)=(1,0,0), and M(u)=(0,1,0). Then, f(u,v) can be expressed by
$$f(u,v)=\Big(-r(u)\cos{\nu}\sin{\theta},\sin{\nu}\sin{\theta},u+ \cos{\theta}\Big), $$
where r(u) is given by Eq. (
41
).
Case 2
If κ≠0, then −b+ar=0. Hence, \(r=\frac {b}{a}\) is a non-zero constant, f(u,v) is a tube. Note that Υ=f(u,v) is a circular cylinder if k1=k2=κ=0 and −b+ar=0.
Corollary 1
The canal surface Υ=f(u,v) according to Bishop frame in E3 which has non-zero constant Gaussian curvature is a surface of revolution such as
$$f(u,v)=\Big(-r(u)\cos{\nu}\sin{\theta},\sin{\nu}\sin{\theta},u+ \cos{\theta}\Big), $$
where r(u) is given by Eq. (
42
).
Proof
By Remark 1 and Theorem 7 with b=0, Υ has non-zero constant Gaussian curvature \(K=\frac {1}{a}\). Then, Υ cannot be a tube and it is a surface of revolution. Additionally, by a similar development as was given in Theorem 7, Υ can be expressed by
$$f(u,v)=\Big(-r(u)\cos{\nu}\sin{\theta},\sin{\nu}\sin{\theta},u+ \cos{\theta}\Big), $$
in which r(u) is given by
$$ u=c_{2}\pm \int\sqrt{\frac{c_{1}+r(-ar+r^{3})}{r(-ar+r^{3})}}dr, $$
(42)
where c1andc2 are constants [23]. □
Corollary 2
The canal surface Υ=f(u,v) according to Bishop frame in E3 which has non-zero constant mean curvature is a surface of revolution such as
$$f(u,v)=\Big(-r(u)\cos{\nu}\sin{\theta},\sin{\nu}\sin{\theta},u+ \cos{\theta}\Big), $$
where r(u) is given by Eq. (
43
).
Proof
By Remark 1 and Theorem 7 with a=0, f(u,v) has non-zero constant mean curvature \(H=\frac {1}{b}\). Similarly, as Corollary 1, f(u,v) is a surface of revolution and it can be expressed by
$$f(u,v)=\Big(-r(u)\cos{\nu}\sin{\theta},\sin{\nu}\sin{\theta},u+ \cos{\theta}\Big), $$
in which, r(u) is given by
$$ u=c_{2}\pm \int\sqrt{\frac{c_{1}+r(b+r^{3})}{r(b+r^{3})}}\,dr, $$
(43)
where c1andc2 are constants [23]. □
Theorem 8
A non-developable canal surface Υ=f(u,v) according to Bishop frame in E3 is a (H,KII)-linear Weingarten canal surface if and only if it is an open part of a surface of revolution satisfies
$$\begin{array}{@{}rcl@{}} &r^{\prime\prime} \big(-a{r'}^{2} (-1+2{r'}^{2}+{r'}^{4})+r(-2(a+2r)+4(a+2r){r'}^{2}-(a+4r){r'}^{4}\\ &+a{r'}^{6})r^{\prime\prime}-2(b+r^{2}(3a+4r))(-1+{r'}^{2}){r^{\prime\prime}}^{2}-4r(b+r^{2} (a+r)){r^{\prime\prime}}^{3}\big)\\ &-arr' (-1+{r'}^{2})^{2}r^{(3)}=0, \end{array} $$
(44)
where a,b∈R, and (a,b)≠(0,0).
Proof
Suppose f(u,v) is a (H,KII)-linear Weingarten canal surface. It satisfies
where a,b∈R, and (a,b)≠(0,0). From Eqs. (15) and (16), we have
$$\begin{array}{@{}rcl@{}} \frac{-a}{2048(r^{2}p^{2}q^{2})}&\sum\nolimits^{6}_{i=0,j=-6}\sin{\theta}\Big(\xi_{i,j}\,\cos{(i\nu+j\theta)+\eta_{i,j}\,\sin{(i\nu+j\theta)}}\Big)\\ &+b\frac{q(2p-\sin^{2}{\theta})}{2r^{2}p^{2}}=1. \end{array} $$
Then
$$\begin{array}{@{}rcl@{}} -a\sum\limits^{6}_{i=0,j=-6}\sin{\theta}&\Big(\xi_{i,j} \cos{(i\nu+j\theta)+\eta_{i,j} \sin{(i\nu+j\theta)}}\Big)\\ &=2048\left(r^{2}p^{2}q^{2}\right)\left(1-b\frac{q(2p-\sin^{2}{\theta})}{2r^{2}p^{2}}\right), \end{array} $$
$$\begin{array}{@{}rcl@{}} &-a\sum\nolimits^{6}_{i=0,j=-6}\sin{\theta}\Big(\xi_{i,j}\,\cos{(i\nu+j\theta)+\eta_{i,j}\sin{(i\nu+j\theta)}}\Big)\\ &=-512\left(\sin^{4}{\theta}\left(k_{1}\cos{\nu}-k_{2}\sin{\nu}-\theta'\right)^{2} \left(-2r^{2}+2r^{2}\cos{2\theta}-2brk_{1}^{2}-2r^{4}k_{1}^{2}\right.\right.\\ &+\sin{(\nu+\theta)}(-bk_{1}-4r^{3} k_{1})+\sin{(\nu-\theta)}\left(bk_{1}+4r^{3} k_{1}\right)-2brk_{2}^{2}-2r^{4}k_{2}^{2}\\ &+\cos{(\nu+\theta)}\left(-bk_{2}-4r^{3}k_{2}\right)+\cos{(\nu-\theta)}\left(bk_{2}+4r^{3} k_{2}\right)+\sin{2\nu}(4brk_{1}k_{2}\\ &+4r^{4} k_{1}k_{2}) +\cos{2\nu}(-2brk_{1}^{2}-2r^{4} k_{1}^{2}+2brk_{2}^{2}+2r^{4} k_{2}^{2})-4br{\theta'}^{2}-4r^{4}{\theta'}^{2}\\ &\left.\left.+\sin{\theta}(2b{\theta'}+8r^{3}{\theta'})+\cos{\nu}\left(8brk_{1}{\theta'}+8r^{4}{\theta'}k_{1}\right)\sin{\nu}(-8br{\theta'}k_{2} -8r^{4}{\theta'} k_{2})\right)\right). \end{array} $$
(45)
By comparing the coefficient of cos5ν, sin5ν in Eq. (45), we have
$$\begin{array}{@{}rcl@{}} &a \sin{\theta}\,\xi_{5}=0,\\ & a \sin{\theta}\big(-92r^{2}k^{5}_{1}+920r^{2}k^{3}_{1}k^{2}_{2}-460r^{2}k_{1}k^{4}_{2}\big)=0,\\ & a \sin{\theta} \eta_{5}=0,\\ & a \sin{\theta}\big(460r^{2}k^{4}_{1}k_{2}-920r^{2}r^{2}k^{2}_{1}k^{3}_{2}+92k^{5}_{2}\big)=0. \end{array} $$
It is follows that k1=k2=0, this means κ=0. Thus, the surface is a surface of revolution. From Eqs. (27) and (37), we have
$$\begin{array}{@{}rcl@{}} &r^{\prime\prime} \left(-a{r'}^{2} (-1+2{r'}^{2}+{r'}^{4})+r(-2(a+2r)+4(a+2r){r'}^{2}-(a+4r){r'}^{4}+a{r'}^{6})r^{\prime\prime}\right.\\ &\left.-2(b+r^{2}(3a+4r))(-1+{r'}^{2}){r^{\prime\prime}}^{2}-4r(b+r^{2} (a+r)){r^{\prime\prime}}^{3}\right)-arr' (-1+{r'}^{2})^{2}r^{(3)}=0, \end{array} $$
□
Theorem 9
A non-developable canal surface Υ=f(u,v) according to Bishop frame in E3 is a (K,KII)-linear Weingarten canal surface if and only if it is an open part of a surface of revolution satisfies
$$\begin{array}{@{}rcl@{}} &2r^{\prime\prime} \left(a({r'}-{r'}^{3})^{2}-r(-1+{r'}^{2})(-2(a+2r)+2(a+2r){r'}^{2}+a{r'}^{4})r^{\prime\prime}\right.\\ &\left.-2r(-2b+3ar+4r^{2})(-1+{r'}^{2}){r^{\prime\prime}}^{2}-4r^{2}(-b+r(a+r)){r^{\prime\prime}}^{3}\right)\\ &-2arr' (-1+{r'}^{2})^{2} r^{(3)}=0, \end{array} $$
(46)
where a,b∈R, and (a,b)≠(0,0).
Proof
Suppose f(u,v) is a (K,KII)-linear Weingarten canal surface. It satisfies
where a,b∈R, and (a,b)≠(0,0). From Eqs. (15) and (16), we have
$$\begin{array}{@{}rcl@{}} &\frac{-a}{2048(r^{2}p^{2}q^{2})}\sum\nolimits^{6}_{i=0,j=-6}\sin{\theta}\Big(\xi_{i,j}\,\cos{(i\nu+j\theta)+\eta_{i,j}\,\sin{(i\nu+j\theta)}}\Big) +b\left(\frac{-q}{rp}\right)=1\\ &-a\sum\nolimits^{6}_{i=0,j=-6}\sin{\theta}\Big(\xi_{i,j}\,\cos{(i\nu+j\theta)+\eta_{i,j}\,\sin{(i\nu+j\theta)}}\Big)\\ &=2048(r^{2}p^{2}q^{2})\left(1-b\left(\frac{-q}{rp}\right)\right), \end{array} $$
$$ \begin{aligned} &-a\sum\limits^{6}_{i=0,j=-6}\sin{\theta}\Big(\xi_{i,j}\,\cos{(i\nu+j\theta)+\eta_{i,j}\,\sin{(i\nu+j\theta)}}\Big)\\ &=2048\left(r\sin^{4}{\theta}(k_{1}\cos{\nu}-k_{2}\sin{\nu}-\theta')^{2}\left(\sin{\theta}+rk_{1}\cos{\nu}-rk_{2}\sin{\nu}\right.\right.\\ &\left.\left.-r\theta'\right)\big(r\sin{\theta}+\cos{\nu}(-bk_{1}+r^{2}k_{1}) +\sin{\nu} (bk_{2}-r^{2} k_{2})+b\theta'-r^{2} \theta'\big)\right). \end{aligned} $$
(47)
By comparing the coefficient of cos5ν, sin5ν in Eq. (47), we have
$$\begin{array}{@{}rcl@{}} &a\sin{\theta}\,\xi_{5}=0,\\ & a\,\sin{\theta}\big(-92r^{2}k^{5}_{1}+920r^{2}k^{3}_{1}k^{2}_{2}-460r^{2}k_{1}k^{4}_{2}\big)=0,\\ & a\sin{\theta}\,\eta_{5}=0,\\ & a\,\sin{\theta}\big(460r^{2}k^{4}_{1}k_{2}-920r^{2}k^{2}_{1}k^{3}_{2}+92k^{5}_{2}\big)=0. \end{array} $$
From the above equations, we have k1=k2=0; this means κ=0. Thus, the surface is a surface of revolution. From Eqs. (27) and (37), we have
$$\begin{array}{@{}rcl@{}} &2r^{\prime\prime} \left(a({r'}-{r'}^{3})^{2}-r(-1+{r'}^{2})(-2(a+2r)+2(a+2r){r'}^{2}+a{r'}^{4})r^{\prime\prime}\right.\\ &\left.-2r(-2b+3ar+4r^{2})(-1+{r'}^{2}){r^{\prime\prime}}^{2}-4r^{2}(-b+r(a+r)){r^{\prime\prime}}^{3}\right)\\ &-2arr' (-1+{r'}^{2})^{2} r^{(3)}=0. \end{array} $$
□
Corollary 3
The canal surface Υ=f(u,v) according to Bishop frame in E3 which has non-zero second Gaussian curvature is an open part of a surface of revolution satisfies
$$\begin{array}{@{}rcl@{}} &2r^{\prime\prime} \left(a({r'}-{r'}^{3})^{2}-r(-1+{r'}^{2})(-2(a+2r)+2(a+2r){r'}^{2}+a{r'}^{4})r^{\prime\prime}\right.\\ &\left.-2r(-2b+3ar+4r^{2})(-1+{r'}^{2}){r^{\prime\prime}}^{2}-4r^{2}(-b+r(a+r)){r^{\prime\prime}}^{3}\right)\\ & -2arr' (-1+{r'}^{2})^{2} r^{(3)}=0. \end{array} $$
At last, we study the (x,y)-linear Weingarten canal surface f(u,v) according to Bishop frame in E3 which satisfies KII=K and KII=H, respectively.
Theorem 10
A non-developable canal surface Υ=f(u,v) according to Bishop frame in E3 satisfying KII=K a surface of revolution which satisfies
$$\begin{array}{@{}rcl@{}} &{r'}^{6} r^{\prime\prime} (-1+rr^{\prime\prime})+{r'}^{4} r^{\prime\prime} (2+rr^{\prime\prime})+{r'}^{2} r^{\prime\prime} \big(-1+4r(-1+r^{\prime\prime})r^{\prime\prime}+6r^{2}{r^{\prime\prime}}^{2}\big)\\ & +2r{r^{\prime\prime}}^{2} \big(1-(2+3r)r^{\prime\prime}+2r(1+r){r^{\prime\prime}}^{2}\big)+rr' r^{(3)}-2r{r'}^{3} r^{(3)}+r{r'}^{5} r^{(3)}=0, \end{array} $$
where a,b∈R, and (a,b)≠(0,0).
Proof
When KII=K, we have by Eqs. (14) and (16)
$$ \frac{1}{2048(r^{2}p^{2}q^{2})}\sum\limits^{6}_{i=0,j=-6}\sin{\theta}\Big(\xi_{i,j}\,\cos{(i\nu+j\theta)+\eta_{i,j}\,\sin{(i\nu+j\theta)}}\Big) =\frac{-q}{rp}. $$
(48)
Comparing the coefficient of cos5ν, sin5ν in Eq. (48), we have
$$\begin{array}{@{}rcl@{}} &a \sin{\theta}\,\xi_{5}=0,\\ & a\,\sin{\theta}\big(-92r^{2}k^{5}_{1}+920r^{2}k^{3}_{1}k^{2}_{2}-460r^{2}k_{1}k^{4}_{2}\big)=0,\\& a\sin{\theta}\,\eta_{5}=0,\\ & a\,\sin{\theta}\big(460r^{2}k^{4}_{1}k_{2}-920r^{2}k^{2}_{1}k^{3}_{2}+92k^{5}_{2}\big)=0. \end{array} $$
From the above equations, we have k1=k2=0; this means κ=0. Then, the canal surface Υ is a surface of revolution. From Eqs. (27) and (37), we have
$$\begin{array}{@{}rcl@{}} &{r'}^{6} r^{\prime\prime} (-1+rr^{\prime\prime})+{r'}^{4} r^{\prime\prime} (2+rr^{\prime\prime})+{r'}^{2} r^{\prime\prime} \big(-1+4r(-1+r^{\prime\prime})r^{\prime\prime}+6r^{2}{r^{\prime\prime}}^{2}\big)\\ &+2r{r^{\prime\prime}}^{2} \big(1-(2+3r)r^{\prime\prime}+2r(1+r){r^{\prime\prime}}^{2}\big)+rr' r^{(3)}-2r{r'}^{3} r^{(3)}+r{r'}^{5} r^{(3)}=0, \end{array} $$
□
Theorem 11
For a non-developable canal surface Υ=f(u,v) according to Bishop frame in E3 satisfying KII=H, then the canal surface is a surface of revolution which satisfies
$$\begin{array}{@{}rcl@{}} {r'}^{6} r^{\prime\prime} \big(-1+rr^{\prime\prime}\big)&+{r'}^{4} r^{\prime\prime} \big(2+rr^{\prime\prime}\big)+2{r^{\prime\prime}}^{2}\big(r+r^{\prime\prime}-3r^{2} r^{\prime\prime}-2r{r^{\prime\prime}}^{2}+2r^{3} {r^{\prime\prime}}^{2}\big)\\ & -{r'}^{2} \big(r^{\prime\prime}+4r{r^{\prime\prime}}^{2}+{r^{\prime\prime}}^{3}(2-6r^{2})\big)+r{r'} r^{(3)} -2r{r'}^{3} r^{(3)}+r{r'}^{5} r^{(3)}=0. \end{array} $$
Proof
When KII=H, from Eq. (19), we have
$$B=\frac{-1}{2048}\sum\limits^{6}_{i=0,j=-6}\sin{\theta}\Big(\gamma_{i,j}\,\cos{(i\nu+j\theta)+\delta_{i,j}\,\sin{(i\nu+j\theta)}}\Big), $$
$$\begin{array}{@{}rcl@{}} &\gamma_{5,6}=\gamma_{5,-6}=r^{2}k^{5}_{1}-10r^{2}k^{3}_{1}k^{2}_{2}+5r^{2}k_{1}k^{4}_{2}=0,\\ &\delta_{2,5}=-\delta_{2,-5}=8rk^{2}_{1}+8rk^{4}_{1}+2r^{3}k^{6}_{1}-8rk^{2}_{2}+2r^{3}k^{4}_{1}k^{2}_{2}-8rk^{4}_{2}-2r^{3}k^{2}_{1}k^{4}_{2}-2r^{3}k^{6}_{2}=0. \end{array} $$
Considering the coefficient of cos(5ν+6θ), sin(2ν+5θ) in B, we get k1=k2=0; this means κ=0. Thus, the surface is a surface of revolution; we have from Eqs. (27) and (37)
$$\begin{array}{@{}rcl@{}} &{r'}^{6} r^{\prime\prime} \big(-1+rr^{\prime\prime}\big)+{r'}^{4} r^{\prime\prime} \big(2+rr^{\prime\prime}\big)+2{r^{\prime\prime}}^{2}\big(r+r^{\prime\prime}-3r^{2} r^{\prime\prime}-2r{r^{\prime\prime}}^{2}+2r^{3} {r^{\prime\prime}}^{2}\big)\\ &-{r'}^{2} \big(r^{\prime\prime}+4r{r^{\prime\prime}}^{2}+{r^{\prime\prime}}^{3}(2-6r^{2})\big)+r{r'} r^{(3)} -2r{r'}^{3} r^{(3)}+r{r'}^{5} r^{(3)}=0. \end{array} $$
□