Surfaces with a canonical principal direction and prescribed mean curvature

Abstract

A surface in Euclidean space has a canonical principal direction with respect to a fixed direction ${{\mathbf {d}}}$ if its tangent part ${{\mathbf {d}}}^\top$ is a principal direction along the surface. In this paper, we classify all such surfaces with prescribed mean curvature given as an affine function of one of the following three functions: the height function, the angle function and the support function.

1 Introduction and statement of the result

A constant angle surface in the three-dimensional Euclidean space ${\mathbb {R}}^3$ is an orientable surface whose Gauss map makes a constant angle with a fixed vector direction ${{\mathbf {d}}}$ ( [1]). In particular, the projection of ${{\mathbf {d}}}$ on the tangent plane of the surface is a principal direction and the corresponding principal curvature is 0. Dillen et al. generalized, in [4], the notion of the constant angle surfaces to those surfaces where the tangent projection of ${{\mathbf {d}}}$ is a principal direction without necessarily corresponding with zero principal curvature. See also [9] for the case of hypersurfaces.

Definition 1.1

Let ${{\mathbf {d}}}\in {\mathbb {R}}^3$ be a unitary vector. We say that a surface M in ${\mathbb {R}}^3$ has a canonical principal direction with respect to ${{\mathbf {d}}}$ if the tangent part ${{\mathbf {d}}}^\top$ along M is a principal direction.

We abbreviate by saying a CPD surface where the direction ${{\mathbf {d}}}$ is understood. The study of CPD surfaces in different ambient spaces has been of interest recently: See [4, 5, 8, 17].

The purpose of this paper is to classify all CPD surfaces whose mean curvature is a linear function of the next three functions on M:

$$\begin{aligned} \text{ height } \text{ function }&f(p)=\langle p,{\mathbf {v}}\rangle ,\\ \text{ angle } \text{ function }&f(p)=\langle N(p),{\mathbf {v}}\rangle ,\\ \text{ support } \text{ function }&f(p)=\langle N(p) , p \rangle . \end{aligned}$$

Here, ${\mathbf {v}}\in {\mathbb {R}}^3$ is a fixed unit vector and N is the Gauss map of M. The interest for these three types of mean curvatures is the following:

• 1. Height function. A surface M whose mean curvature H satisfies

  $$\begin{aligned} H(p)=\lambda \langle p,{\mathbf {v}}\rangle +\mu ,\ \lambda ,\mu \in {\mathbb {R}}, \end{aligned}$$

  (1)

  is a model of a liquid drop where the direction of the gravity is indicated by the vector ${\mathbf {v}}$, the constant $\lambda$ depends only on the physical properties of the drop and $\mu$ is a volume constraint: See [7] as a general reference. A surface whose mean curvature satisfies (1) is called a capillary surface.

• 2. Angle function. A surface M whose mean curvature H satisfies

  $$\begin{aligned} H(p)=\lambda \langle N(p),{\mathbf {v}}\rangle +\mu ,\ \lambda ,\mu \in {\mathbb {R}}, \end{aligned}$$

  (2)

  is called a $\lambda$-translating soliton. This notion generalizes the translating soliton ($\mu =0$). Translating solitons appear in the theory of the mean curvature flow being the equation of the limit flow by a proper blowup procedure near type II singular points ( [12, 13, 20]).

• 3. Support function. A surface M whose mean curvature H satisfies

  $$\begin{aligned} H(p)=\lambda \langle N(p),p\rangle +\mu ,\ \lambda ,\mu \in {\mathbb {R}}, \end{aligned}$$

  (3)

  is called a $\lambda$-shrinker surface. This generalizes the concept of self-expander ($\lambda >0, \mu =0)$ and self-shrinker ($\lambda <0, \mu =0$). Both types of surfaces appear in the context of self-similar solutions to the mean curvature flow as models of the flow behavior near a singularity, where M does not change the shape but is contracted (self-shrinkers) or dilated (self-expanders) by the flow: See [6, 11, 13].

We observe that if $\lambda =0$ in the three above equations, then the mean curvature H is constant on the surface. This particular case was treated previously in [9] (see Proposition 2.2 below).

Although in the literature, Eqs. (2) and (3) have been studied when $\mu =0$, the case $\mu \not =0$ makes sense in the context of manifolds with density [10, 16]. Indeed, consider ${\mathbb {R}}^3$ with a positive smooth density function $e^\phi$, $\phi \in C^\infty ({\mathbb {R}}^3)$, which serves as a weight for the volume and the surface area. For a variation $\{M_t:t\in (-\epsilon ,\epsilon )\}$ of M, let $A_\phi (t)$ and $V_\phi (t)$ be the weighted area and the enclosed weighted volume of $M_t$, respectively. Then, the first variation of $A_\phi (t)$ and $V_\phi (t)$ is

$$\begin{aligned} A'_\phi (0)=-2\int _M H_\phi u\, \mathrm{d}A_\phi ,\quad V'_\phi (0)=\int _M u\, \mathrm{d}A_\phi , \end{aligned}$$

where u is the normal component of the variation and $H_\phi =H- \langle \nabla \phi ,N\rangle /2$ is called the weighted mean curvature  [19]. Then, it is immediate that M is a critical point of $A_\phi$ for a given weighted volume if and only if $H_\phi =\mu$ is a constant function. Equations (2) and (3) appear with suitable choices of the density function $\phi$; namely, for Eq. (2), we take $\phi (p)=2\lambda \langle p,{\mathbf {v}}\rangle$, and for Eq. (3), the function if $\phi (p)=\lambda |p|^2$.

Surfaces with a canonical principal direction have suitable parametrization. It was proved in [9] that a CPD surface with respect to ${{\mathbf {d}}}$ admits the following parametrization

$$\begin{aligned} X(s,t) = \gamma (s)+ (f(t) \eta (s) + g(t) {{\mathbf {d}}}), \end{aligned}$$

(4)

where $\gamma =\gamma (s)$, $s\in I\subset {\mathbb {R}}$ is a curve parametrized by the arc length and contained in a plane orthogonal to ${{\mathbf {d}}}$, $\eta =\eta (s)$ is a unit vector field along $\gamma$ and orthogonal to both $\gamma$ and ${{\mathbf {d}}}$. Finally, $f=f(t)$, $g=g(t)$, $t\in J\subset {\mathbb {R}}$, are smooth functions that satisfy $f'(t)^2 +g'(t)^2=1$ for all $t\in J$. The curve $\gamma$ is named the directrix of M, and the curve (f(t), g(t)) is the profile curve of M. As a consequence, we may take $\eta$ to be the principal normal vector to $\gamma$ and thus $\gamma ''(s)=\kappa (s)\eta (s)$, where $\kappa$ is the curvature of $\gamma$.

Once introduced the above notation, we give the main result of this paper.

Theorem 1.1

Let M be a CPD surface parametrized as (4). If M satisfies (1), (2) or (3), then M is one of the following surfaces:

• 1. A cylindrical surface where the base curve is $\alpha (t)=f(t)\eta +g(t){{\mathbf {d}}}$, $\eta$ is a constant vector, the rulings are orthogonal to ${{\mathbf {d}}}$ and $\alpha$ satisfies the one-dimensional case of (1), (2) or (3).

• 2. A surface of revolution whose generating curve is $\alpha$ and the rotational axis is parallel to ${{\mathbf {d}}}$. The curve $\alpha$ satisfies a specific second-order ODE depending on condition (1), (2) or (3).

• 3. A cylindrical surface where the base curve is $\gamma _a(s)=\gamma (s)+a\eta (s)$, $a\in {\mathbb {R}}$, $\gamma _a$ satisfies the one-dimensional case of (1), (2) or (3) and the rulings are parallel to ${{\mathbf {d}}}$.

By a cylindrical surface, we mean a ruled surface where all the rulings are parallel. Then, the base curve of the surface is contained in a plane orthogonal to the rulings. In Theorem 1.1, items (1) and (3), the base curve of the cylindrical surface satisfies that its curvature is a linear function of the height, angle or support function depending on the case. In such a case, we will say then that the curve satisfies the one-dimensional case of (1), (2) or (3).

The proof of Theorem 1.1 is separated in each one of the three equations: See Sects. 3, 4 and 5, respectively.

2 Preliminaries

In this section, we give an expression of the mean curvature H for a CPD surface and we classify the surfaces with constant mean curvature. Firstly, we need to give some observations about parametrization (4) in the following proposition.

Proposition 2.1

Let M be a CPD surface parametrized by (4).

• 1. We have $1-f(t)\kappa (s)\not =0$ for all $s\in I$, $t\in J$.

• 2. If the function g is constant, then M is a plane orthogonal to ${{\mathbf {d}}}$.

• 3. If $\gamma$ is a straight line, then M is a cylindrical surface whose rulings are orthogonal to ${{\mathbf {d}}}$.

• 4. If $\gamma$ is a circle, then M is a surface of revolution whose rotational axis is parallel to ${{\mathbf {d}}}$.

Proof

• 1. Since $\partial _sX=(1-\kappa (s)f(t))\gamma '(s)\not =0$ by regularity of M, we conclude that $1-\kappa (s)f(t)\not =0$.

• 2. If $g(t)=a$ for all $s\in I$, then $f(t)= \epsilon t + b$, with $b\in {\mathbb {R}}$ and $\epsilon \in \{-1,1\}$. Then, (4) gives

  $$\begin{aligned} X(s,t) = \gamma (s)+ (\epsilon t+b) \eta (s) + a {{\mathbf {d}}}= (\gamma (s)+ a {{\mathbf {d}}}) + (\epsilon t+b) \eta (s). \end{aligned}$$

  It follows that $\gamma (s)+a{{\mathbf {d}}}$ is a planar curve and the curves $t\mapsto (\epsilon t+b)\eta (s)$ are segments of straight lines coplanar to $\gamma$. This proves that M is a plane orthogonal to ${{\mathbf {d}}}$.

• 3. If $\gamma$ is a straight line, then $\eta$ is a unit constant vector and from (4), the surface M is a cylindrical surface whose base curve is $\alpha (t)= f(t)\eta +g(t){{\mathbf {d}}}$ and the rulings are orthogonal to ${{\mathbf {d}}}$.

• 4. After a change of coordinates, we suppose $\gamma (s)=(r\cos (s),r\sin (s),0)$ and ${{\mathbf {d}}}=(0,0,1)$. By (4), we find

  $$\begin{aligned} X(s,t)=(r(1-f(t))\cos (s), r(1-f(t))\sin (s),g(t)), \end{aligned}$$

  proving that M is a surface of revolution about the ${{\mathbf {d}}}$-axis.

$\square$

We need to have an expression of the mean curvature H of a CPD surface. The following result is proved in [9].

Proposition 2.2

Let M be a CPD surface. If M is not a plane orthogonal to ${{\mathbf {d}}}$, then the principal curvatures $\kappa _1$ and $\kappa _2$ with respect to the Gauss map $N(s,t)=- g'(t) \eta (s) + f'(t) {{\mathbf {d}}}$ are

$$\begin{aligned} \kappa _1(X(s,t))= \dfrac{f''(t)}{g'(t)} , \ \ \kappa _2(X(s,t)) = \dfrac{g'(t) \kappa (s)}{1- f(t) \kappa (s)}. \end{aligned}$$

The mean curvature H with respect to N is

$$\begin{aligned} H(X(s,t))= \dfrac{1}{2} \left( \dfrac{f''(t)}{g'(t)} + \dfrac{g'(t) \kappa (s)}{1- f(t) \kappa (s)} \right) . \end{aligned}$$

(5)

In Theorem 1.1, we exclude the case $\lambda =0$, that is, that the surface has constant mean curvature. The CPD surfaces with constant mean curvature were obtained in [9, Cor. 14]. By completeness, we give its proof.

Proposition 2.3

If M is a CPD surface with constant mean curvature H, then M is a plane, a circular cylinder with axis orthogonal to ${{\mathbf {d}}}$ or a surface of revolution whose axis is parallel to ${{\mathbf {d}}}$.

Proof

It is immediate that a plane orthogonal to ${{\mathbf {d}}}$ is a CPD surface with respect to ${{\mathbf {d}}}$ and with zero mean curvature. Suppose now that M is not the above surface, in particular, the function g(t) in (4) is not constant by item 2 of Proposition 2.1. By Proposition 2.2, we have

$$\begin{aligned} \dfrac{f''(t)}{g'(t)} + \dfrac{g'(t) \kappa (s)}{1- f(t) \kappa (s)}=2H, \end{aligned}$$

where the right-hand side is a constant. Differentiating with respect to s, we obtain

$$\begin{aligned} \frac{g'(t)\kappa '(s)}{(1-f(t)\kappa (s))^2}=0. \end{aligned}$$

Since $g'(t)\not =0$, we conclude $\kappa '(s)=0$ for every $s\in I$. This means that $\gamma$ is a straight line or a circle.

• 1. Case $\gamma$ is a straight line. By Proposition 2.1, the surface M is a cylindrical surface whose rulings are parallel to $\gamma$. Since H is constant, then M is a plane parallel to $\gamma$ ($H=0$) or M is a circular cylinder ($H\not =0$) whose axis is the straight line, being $\gamma$ orthogonal to ${{\mathbf {d}}}$.

• 2. Case $\gamma$ is a circle. By Proposition 2.1 again, M is a surface of revolution whose axis is parallel to ${{\mathbf {d}}}$.

$\square$

3 CPD capillary surfaces

In this section, we prove Theorem 1.1 for capillary surfaces. By the expression of H in (5), Eq. (1) is written as

$$\begin{aligned} \dfrac{1}{2} \left( \dfrac{f''(t)}{g'(t)} + \dfrac{g'(t) \kappa (s)}{1- f(t) \kappa (s)} \right) =\lambda \left( \langle \gamma (s),{\mathbf {v}}\rangle +(f(t)\langle \eta (s),{\mathbf {v}}\rangle +g(t)\langle {{\mathbf {d}}},{\mathbf {v}}\rangle ) \right) +\mu , \end{aligned}$$

for all $s\in I$, $t\in J$. We differentiate with respect to s, obtaining

$$\begin{aligned} \frac{g'(t)\kappa '(s)}{(1-f(t)\kappa (s))^2}=2\lambda (1-f(t)\kappa (s))\langle \gamma '(s),{\mathbf {v}}\rangle , \end{aligned}$$

(6)

where we have used $\gamma ''(s)=\kappa (s)\eta (s)$. We discuss according to $\kappa$.

• 1. Case $\kappa '(s)=0$ for every $s\in I$. Then, the curvature function $\kappa$ is constant. Moreover, Eq. (6) implies $\langle \gamma '(s),{\mathbf {v}}\rangle = 0$, that is, $\langle \gamma (s),{\mathbf {v}}\rangle$ is a constant function. From Proposition 2.1, we know that M is a cylindrical surface whose rulings are orthogonal to ${{\mathbf {d}}}$ or M is a surface of revolution whose axis is parallel to ${{\mathbf {d}}}$.

  • (a) Case $\kappa =0$. Then, $\eta$ is a constant vector and Eq. (1) becomes

    $$\begin{aligned} \dfrac{1}{2} \dfrac{f''(t)}{g'(t)} = \lambda \langle f(t)\eta +g(t){{\mathbf {d}}},{\mathbf {v}}\rangle +\lambda \langle \gamma (s),{\mathbf {v}}\rangle +\mu . \end{aligned}$$

    Define $\alpha (t)=f(t)\eta +g(t){{\mathbf {d}}}$ and ${\tilde{\mu }}= \lambda \langle \gamma (s),{\mathbf {v}}\rangle +\mu$. Let us observe that ${\tilde{\mu }}$ is constant. On the other hand, $\alpha$ is parametrized by the arc length and its curvature ${\tilde{\kappa }}$ satisfies

    $$\begin{aligned} \frac{{\tilde{\kappa }}(t)}{2}=\lambda \langle \alpha (t) , {\mathbf {v}}\rangle + {\tilde{\mu }}. \end{aligned}$$

    Consequently, the planar curve $\alpha$ satisfies the one-dimensional case of Eq. (1) as a linear combination of its height function.

  • (b) Case $\kappa \not =0$. We know that M is a rotational surface whose axis is parallel to ${{\mathbf {d}}}$. Since $\langle \gamma '(s),{\mathbf {v}}\rangle = 0$, a differentiation with respect to s yields $\kappa (s) \langle \eta (s), {\mathbf {v}}\rangle = 0$. We deduce that $\langle \eta (s), {\mathbf {v}}\rangle = 0$, and it follows that ${\mathbf {v}}$ is orthogonal to both $\gamma '$ and $\eta$. This says that ${{\mathbf {d}}}$ is parallel to ${\mathbf {v}}$. By Eq. (1), the generating curve $\alpha (t)=f(t)\eta +g(t){{\mathbf {d}}}$ satisfies the ODE

    $$\begin{aligned} \dfrac{1}{2} \left( \dfrac{f''(t)}{g'(t)} + \dfrac{g'(t) \kappa }{1- f(t) \kappa } \right) = \epsilon \lambda g(t) + b \end{aligned}$$

    where $\epsilon = \langle {{\mathbf {d}}}, {\mathbf {v}}\rangle = \pm 1$ and $b=\lambda \langle \gamma (s),{\mathbf {v}}\rangle + \mu$ is a constant.

• 2. Case $\kappa '(s_0)\not =0$ at some $s_0\in I$. In particular, $\kappa '(s)\not =0$ in a subinterval $I'$ of I around $s=s_0$. By simplicity, we assume $I'=I$. We now write (6) as

  $$\begin{aligned} 2 \frac{\lambda \langle \gamma '(s),{\mathbf {v}}\rangle }{\kappa '(s)}= \frac{g'(t)}{(1-f(t)\kappa (s))^3}. \end{aligned}$$

  By differentiating with respect to t, we have

  $$\begin{aligned} g''(t) (1 - f(t) \kappa (s)) + 3 g'(t) f'(t) \kappa (s) = 0. \end{aligned}$$

  (7)

  We distinguish two possibilities:

  • (a) Subcase $g''(t) \ne 0$ in some $t_0\in J$. Then, $g''(t)\not =0$ in an interval $K\subset J$ around $t=t_0$. In $I\times K$, we can write (7) as

    $$\begin{aligned} \frac{1}{\kappa (s)}= f(t)- 3 \frac{g'(t) f'(t)}{g''(t)}. \end{aligned}$$

    Since the left-hand side depends only on s and the right-hand side on t, then they are constant functions; in particular, $1/\kappa (s)$ is constant, a contradiction because $\kappa '(s)\not =0$.

  • (b) Subcase $g''(t) = 0$ for every t. From (7), we have $f'(t)\kappa (s)=0$ for every $s\in I$, $t\in J$. Since $\kappa '(s)\not =0$, then $\kappa (s)\not =0$ for some $s\in I$, and thus $f'(t)=0$ for every $t\in J$. Hence, $g'(t)= \epsilon \in \{-1,1\}$, f is a constant function $f(t)=a$, $a\in {\mathbb {R}}$, and $g(t)=\epsilon t + b$, $b\in {\mathbb {R}}$. Parametrization (4) of M becomes

    $$\begin{aligned} X(s,t) = \gamma (s)+ a \eta (s) + (\epsilon t + b) {{\mathbf {d}}}. \end{aligned}$$

    This proves that M is a cylindrical surface whose base curve is the planar curve $\gamma _a(s) = \gamma (s)+a\eta (s)$ and the rulings are parallel to ${{\mathbf {d}}}$. Under this situation, Eq. (1) becomes

    $$\begin{aligned} \dfrac{\epsilon \kappa (s)}{2(1- a \kappa (s))} = \lambda \left( \langle \gamma _a(s) ,{\mathbf {v}}\rangle +(\epsilon t + b)\langle {{\mathbf {d}}},{\mathbf {v}}\rangle \right) +\mu . \end{aligned}$$

    The derivative with respect to t gives $\lambda \epsilon \langle {{\mathbf {d}}},{\mathbf {v}}\rangle =0$, and we deduce that ${{\mathbf {d}}}$ is orthogonal to ${\mathbf {v}}$. Eq. (1) simplifies into

    $$\begin{aligned} \dfrac{ \kappa (s)}{1- a \kappa (s)} = 2 \epsilon \lambda \langle \gamma _a(s),{\mathbf {v}}\rangle + 2 \epsilon \mu . \end{aligned}$$

    (8)

    Finally, a simple computation gives that the curvature of $\gamma _a$ is $\kappa _{\gamma _a}=\dfrac{\epsilon \kappa (s)}{1- a \kappa (s)}$. Thus, from (8), we deduce that $\gamma _a$ satisfies the one-dimensional case of (1).

We point out that the one-dimensional case of capillary Eq. (1) and the rotational capillary surfaces are well known. For the first one, we refer [18, pp. 1130–40], also [14]. A detailed description of the rotational capillary surfaces lies in [7].

4 CPD $\lambda$-translating solitons

We now give the proof of Theorem 1.1 for CPD $\lambda$-translating solitons. In order to compute (2), we need to know $\langle N(p),{\mathbf {v}}\rangle$. By Proposition 2.2, we find

$$\begin{aligned} \langle N(s,t), {\mathbf {v}}\rangle = - g'(t) \langle \eta (s) , {\mathbf {v}}\rangle + f'(t) \langle {{\mathbf {d}}}, {\mathbf {v}}\rangle . \end{aligned}$$

By combining (2) and (5), we obtain

$$\begin{aligned} \frac{f''(t)}{g'(t)} + \frac{g'(t) \kappa (s)}{1- f(t) \kappa (s)} = - 2 \lambda g'(t) \langle \eta (s) , {\mathbf {v}}\rangle + 2 \lambda f'(t) \langle {{\mathbf {d}}}, {\mathbf {v}}\rangle +2 \mu , \end{aligned}$$

for all $s\in I$, $t\in J$. By taking the derivative with respect to s, and using that $g'(t)\not =0$, we get

$$\begin{aligned} \frac{\kappa '(s)}{(1- f(t) \kappa (s) )^2} = - 2 \lambda \langle \eta '(s) , {\mathbf {v}}\rangle = 2 \lambda \kappa (s) \langle \gamma '(s) , {\mathbf {v}}\rangle . \end{aligned}$$

We differentiate this expression with respect to t, obtaining now

$$\begin{aligned} \frac{2 \kappa (s) \kappa '(s)f'(t)}{(1- f(t) \kappa (s) )^3} =0. \end{aligned}$$

Then, $(\kappa ^2(s))'f'(t)=0$. We discuss two cases.

• 1. Suppose there exists $t_0\in J$ such that $f'(t_0)\not =0$. Then, $\kappa$ is a constant function. In such a case, by Proposition 2.1 we conclude that M is either a cylindrical surface whose rulings are orthogonal to ${{\mathbf {d}}}$ or M is a surface of revolution with axis parallel to ${{\mathbf {d}}}$.

  • (a) Case $\kappa =0$. Now $\eta$ is a constant vector. Equation (2) is now

    $$\begin{aligned} \frac{f''(t)}{g'(t)} = - 2 \lambda g'(t) \langle \eta (s) , {\mathbf {v}}\rangle + 2 \lambda f'(t) \langle {{\mathbf {d}}}, {\mathbf {v}}\rangle +2 \mu . \end{aligned}$$

    (9)

    Define the planar curve $\alpha (t)= f(t) \eta + g(t) {{\mathbf {d}}}$ is a planar curve. A computation of its principal normal vector $N_\alpha$ gives $N_\alpha (s)= - g'(t) \eta + f'(t) {{\mathbf {d}}}$ and its curvature is $\kappa _\alpha (t)= \frac{f''(t)}{g'(t)}$. Thus, (9) is written as

    $$\begin{aligned} \kappa _\alpha (t) = 2 \lambda \langle N_\alpha (t),{\mathbf {v}}\rangle + 2 \mu , \end{aligned}$$

    proving that $\alpha$ satisfies the one-dimensional case of (2).

  • (b) Case $\kappa \not =0$. Then, $\langle \gamma '(s) , {\mathbf {v}}\rangle =0$, that is, $\langle \gamma (s) , {\mathbf {v}}\rangle$ is a constant function. Moreover, taking another derivative with respect to s, we find $\kappa \langle \eta (s) , {\mathbf {v}}\rangle = 0$, so ${\mathbf {v}}$ is orthogonal to both $\gamma '(s)$ and $\eta (s)$ which proves that ${\mathbf {v}}$ is parallel to ${{\mathbf {d}}}$. Now Eq. (2) becomes

    $$\begin{aligned} \frac{f''(t)}{g'(t)} + \frac{g'(t) \kappa }{1- f(t) \kappa } = 2 \epsilon \lambda f'(t) +2 \mu , \end{aligned}$$

    with $\epsilon = \langle {{\mathbf {d}}}, {\mathbf {v}}\rangle = \pm 1$.

• 2. Case $f'(t)=0$ for every $t\in J$. Then, f is a constant function $f(t)=a$, $a\in {\mathbb {R}}$. From $f'(t)^2+g'(t)^2=1$, we deduce that $g(t)= \epsilon t +b$, $\epsilon \in \{-1,1\}$, $b\in {\mathbb {R}}$. The parametrization X in (4) is now

  $$\begin{aligned} X(s,t) = (\gamma (s)+ a \eta (s)) + (\epsilon t +b) {{\mathbf {d}}}. \end{aligned}$$

  This proves that M is a cylindrical surface over the planar curve $\gamma _a(s)=\gamma (s)+a\eta (s)$ and the rulings are parallel to ${{\mathbf {d}}}$. Now Eq. (2) becomes

  $$\begin{aligned} \frac{ \kappa (s)}{1- a \kappa (s)} = - 2 \lambda \langle \eta (s) , {\mathbf {v}}\rangle +2 \mu \epsilon . \end{aligned}$$

  (10)

  Again the left-hand side of (10) is the curvature $\kappa _{\gamma _a}$ of $\gamma _a$, and consequently (10) asserts that $\gamma _a$ satisfies the one-dimensional case of (2).

We point out that the rotational surfaces that are $\lambda$-translating solitons are classified in [3] if $\mu =0$ and in [15] when $\mu \not =0$.

5 CPD $\lambda$-shrinker surfaces

We now give the proof of Theorem 1.1 for CPD $\lambda$-shrinker surface. From the expression of N in Proposition 2.2, we have

$$\begin{aligned} \begin{array}{ccl} \langle N(p),p \rangle &{}=&{} \langle - g'(t) \eta (s) + f'(t) d , \gamma (s)+ (f(t) \eta (s) + g(t) d) \rangle \\ &{}=&{}-g'(t) \langle \eta (s) , \gamma (s) \rangle - f(t) g'(t) + f'(t) g(t). \end{array} \end{aligned}$$

By the expression of H in (5), Eq. (3) is written as

$$\begin{aligned} \frac{f''(t)}{g'(t)} + \frac{g'(t) \kappa (s)}{1- f(t) \kappa (s)} = -2 \lambda g'(t) \langle \eta (s) , \gamma (s) \rangle -2 \lambda f(t) g'(t) + 2 \lambda f'(t) g(t) + 2\mu , \end{aligned}$$

for all $s\in I$, $t\in J$. Since $g'(t)\not =0$, we divide the above expression by $g'(t)$, obtaining

$$\begin{aligned} \frac{f''(t)}{g'(t)^2} + \frac{ \kappa (s)}{1- f(t) \kappa (s)} = -2 \lambda \langle \eta (s) , \gamma (s) \rangle -2 \lambda f(t) + 2 \lambda \frac{f'(t) g(t)}{g'(t)} + 2\frac{\mu }{g'(t)}. \end{aligned}$$

We observe that the right-hand side in the above equation is a sum of a function depending only on s and other function depending only on t. Thus, if we differentiate with respect to s and then with respect to t, we obtain 0. Therefore, looking at the left-hand side, we conclude

$$\begin{aligned} 0=\frac{2\kappa (s)\kappa '(s)f(t)}{(1-f(t)\kappa (s))^3}= \frac{(\kappa ^2(s))' f(t)}{(1-f(t)\kappa (s))^3}. \end{aligned}$$

Then, $(\kappa ^2(s))' f'(t) =0$. Now the discussion follows the next two cases.

• 1. Suppose there exists $t_0\in J$ such that $f'(t_0)\not =0$. Then, $\kappa$ is a constant function. In such a case, by Proposition 2.1, we conclude that M is either a cylindrical surface whose rulings are orthogonal to ${{\mathbf {d}}}$ or M is a surface of revolution with axis parallel to ${{\mathbf {d}}}$. By the above equations, $\kappa$ constant implies that the function $\langle \gamma (s) , \eta (s) \rangle$ is constant.

  • (a) Case $\kappa =0$. In particular, $\gamma '$ and $\eta$ are constant coplanar vectors and mutually orthogonal. Define the curve $\alpha (t)=f(t)\eta +g(t){{\mathbf {d}}}$. The principal normal vector of $\alpha$ is $N_\alpha (t)=-g'(t)\eta +f'(t){{\mathbf {d}}}$. Equation (3) is written as

    $$\begin{aligned} \frac{f''(t)}{g'(t)} = 2 \lambda \langle N_\alpha (t) , \gamma (s)+ \alpha (t) \rangle +2 \mu . \end{aligned}$$

    Let us observe that $\langle N_\alpha (t) , \gamma (s) \rangle$ does not depend on s and $\kappa _\alpha (t)= f''(t)/g'(t)$ is the curvature of $\alpha$. Consequently, $\alpha$ satisfies the one-dimensional case of (3).

  • (b) Case $\kappa \ne 0$. Then, $\gamma$ is a circle. Furthermore, we have

    $$\begin{aligned} - \kappa \langle \gamma (s) , \gamma '(s) \rangle = \langle \gamma (s) , \eta '(s) \rangle = \langle \gamma (s) , \eta (s) \rangle ' = 0 , \end{aligned}$$

    which implies that $\langle \gamma (s) , \gamma '(s) \rangle = 0$. This shows that $\gamma$ is a circle with center in the origin. Taking another derivative, we find

    $$\begin{aligned} \langle \gamma (s) , \gamma '(s) \rangle ' = \langle \gamma (s) , \gamma ''(s) \rangle + \langle \gamma '(s) , \gamma '(s) \rangle = \kappa \langle \gamma (s) , \eta (s) \rangle +1=0 . \end{aligned}$$

    Therefore, $\langle \gamma (s) , \eta (s) \rangle = -1/\kappa$. Thus, M is a rotational surface whose axis of revolution passes through the origin. Moreover, the generating curve satisfies

    $$\begin{aligned} \frac{f''(t)}{g'(t)} + \frac{g'(t) \kappa }{1- f(t) \kappa } = (2/ \kappa ) \lambda g'(t) -2 \lambda f(t) g'(t) + 2 \lambda f'(t) g(t) + 2\mu . \end{aligned}$$

• 2. Case $f'(t)=0$ for every $t\in J$. Then, f is a constant function $f(t)=a$, $a\in {\mathbb {R}}$. From $f'(t)^2+g'(t)^2=1$, we deduce that $g(t)= \epsilon t +b$, $\epsilon \in \{-1,1\}$, $b\in {\mathbb {R}}$. From (4), the parametrization X is written as

  $$\begin{aligned} X(s,t) = (\gamma (s)+ a \eta (s)) + (\epsilon t +b) {{\mathbf {d}}}. \end{aligned}$$

  This proves that M is a cylindrical surface over the planar curve $\gamma _a = \gamma (s)+a\eta (s)$ and the rulings are parallel to ${{\mathbf {d}}}$. Finally, Eq. (3) becomes

  $$\begin{aligned} \frac{ \kappa (s)}{1- a \kappa (s)} = -2 \lambda \langle \eta (s) , \gamma (s) \rangle -2 \lambda a + 2 \epsilon \mu = 2 \lambda \langle \eta (s) , \gamma _a(s) \rangle + 2 \epsilon \mu . \end{aligned}$$

  Because the curvature of $\gamma _a$ is $\kappa _{\gamma _a}= \kappa (s)/(1- a \kappa (s))$, then $\gamma _a$ satisfies the one-dimensional case of (3).

The cylindrical shrinker surfaces are classified in [2].