Classification of separable surfaces with constant Gaussian curvature
CCLASSIFICATION OF SEPARABLE SURFACES WITHCONSTANT GAUSSIAN CURVATURE
THOMAS HASANIS AND RAFAEL L ´OPEZ
Abstract.
We classify all surfaces with constant Gaussian curvature K in Eu-clidean 3-space that can be expressed as an implicit equation of type f ( x ) + g ( y ) + h ( z ) = 0, where f , g and h are real functions of one variable. If K = 0, we provethat the surface is a surface of revolution, a cylindrical surface or a conical surface,obtaining explicit parametrizations of such surfaces. If K (cid:54) = 0, we prove that thesurface is a surface of revolution. Introduction and statement of the results
The objective of our investigation is the classification of all surfaces with constantGaussian curvature in Euclidean 3-space that can be expressed by an implicit equa-tion of type f ( x ) + g ( y ) + h ( z ) = 0, where f , g and h are real functions of onevariable. Our motivation arises from the classical theory of minimal surfaces. For ex-ample, historically the first two minimal surfaces are separable, namely, the catenoidcosh( z ) = x + y by Euler in 1744, and the helicoid tan( z ) = y/x by Meusnier1776.In 1835, Scherk discovered all minimal surfaces of type z = φ ( x ) + ψ ( y ), where φ and ψ are two real functions ([8]). Later, Weingarten addressed the classificationproblem of all minimal surfaces of type f ( x ) + g ( y ) + h ( z ) = 0, realizing that forma rich and large family of minimal surfaces ([10]). For example, this family containsa variety of minimal surfaces given in term of elliptic integrals as well as periodicminimal surfaces such as the Schwarz surfaces of type P and D . In the middle of theabove century, Fr´echet gave a deep study of these surfaces obtaining examples withexplicit parametrizations [2, 3]. The reader can see a description of these surfacesin [7, II-5.2].We introduce the following terminology. Let R denote the Euclidean 3-dimensionalspace, that is, the real vector 3-space R endowed with the Euclidean metric (cid:104) , (cid:105) = dx + dy + dz , where ( x, y, z ) stand for the canonical coordinates of R . Locally,any surface of R is the zero level set F ( x, y, z ) = 0 of a function F defined in an Mathematics Subject Classification.
Primary 53A10; Secondary 53C42.
Key words and phrases.
Gaussian curvature, separable surface, cylindrical surface, conical sur-face, rotational surface.Rafael L´opez has partially supported by the grant no. MTM2017-89677-P,MINECO/AEI/FEDER, UE.. a r X i v : . [ m a t h . DG ] D ec THOMAS HASANIS AND RAFAEL L ´OPEZ open set O ⊂ R . Our interest are those surfaces where the function F is a separablefunction of its variables x , y and z . Definition 1.1.
A (regular) surface S in R is said to be separable if can be expressedas S = { ( x, y, z ) ∈ R : f ( x ) + g ( y ) + h ( z ) = 0 } . (1) Here f , g and h are smooth functions defined in some intervals I , I and I of R ,respectively. By the regularity of S , f (cid:48) ( x ) + g (cid:48) ( y ) + h (cid:48) ( z ) (cid:54) = 0 for every x ∈ I , y ∈ I and z ∈ I .In this paper, we consider the following question: What are the separable surfaces in Euclidean space R with constantGaussian curvature? There are three particular examples of separable surfaces that deserve be pointedout because they are obtained by simples choices of the functions f , g and h in theequation (1).(1) Right cylinders . A right cylinder is formed by all the lines that are orthogonalto a given planar curve C . If C is contained in one coordinate plane, thenthe surface is separable where one of the functions f , g or h is constant. So,if C is contained in the xy -plane, then C writes as f ( x ) + g ( y ) = 0. Thecorresponding right cylinder is the surface { ( x, y, z ) ∈ R : f ( x ) + g ( y ) = 0 } .(2) Translation surfaces . A translation surface is a surface that, after renamingthe the coordinates, can be expressed as z = φ ( x ) + ψ ( y ), where φ and ψ aresmooth functions. This surface is the sum of the plane curves x (cid:55)→ ( x, , φ ( x ))and z (cid:55)→ (0 , y, ψ ( y )), so the surface is generated when we move one of thesecurves by means of the translations along the other ones. A separable surfaceis a translation surface if and only one of the three functions in (1) is linear,say, h ( z ) = az + b , with a, b ∈ R and a (cid:54) = 0 (the case a = 0 corresponds witha right cylinder).(3) Rotational surfaces . A surface of revolution with respect to the z -line writesas h ( z ) = x + y , hence that it is a separable surface. In general, if therotation axis is parallel to the z -line, the implicit equation of the surface is h ( z ) = x + y + ax + by + c , with a, b, c ∈ R .Among the above surfaces, our question has a known answer. Indeed, any rightcylinder has zero constant Gaussian curvature. On the other hand, translationsurfaces z = φ ( x ) + ψ ( y ) with constant Gaussian curvature were classified by Liu,proving that K = 0 and one of the functions φ or ψ is linear ([5]). In such acase, and if φ ( x ) = ax + b , a, b ∈ R , a (cid:54) = 0, the implicit equation of the surfaceis z = ax + b + φ ( y ), hence that the surface is a non-right cylindrical surface.Finally, rotational surfaces with constant Gaussian curvature are well known in theliterature; see for example [4]. LASSIFICATION OF SEPARABLE SURFACES WITH CONSTANT GAUSSIAN CURVATURE 3
A first approach to the study of separable surfaces with constant Gaussian curvaturewas done by the second author and Moruz in [6], where were classified all surfacesof type z = φ ( x ) ψ ( y ) with zero constant Gaussian curvature: these surfaces willappear as a particular case of Theorem 1.2 below.The purpose of this paper is to provide a complete classification of all separablesurfaces with constant Gaussian curvature, obtaining explicit parametrizations ofthese surfaces. The classification depends on if the Gaussian curvature is zero or isnot zero. Theorem 1.2.
The only separable surfaces with zero constant Gaussian curvatureare congruent to: (1)
A right cylinder over a planar curve contained in one of the coordinate planes. (2)
A translation surface z = ax + g ( y ) , where a (cid:54) = 0 and g is any smoothfunction. (3) A surface of revolution with zero constant Gaussian curvature. (4)
A cylindrical surface and a conical surface with parametrizations (9) , (11) and (12) ; see Section 3. In case that K is a nonzero constant, the classification is the following. Theorem 1.3.
The only separable surfaces with nonzero constant Gaussian curva-ture are the rotational surfaces of constant curvature whose rotational axis is parallelto one of the coordinate axes.
The organization of this paper is as follows. In Section 2 we obtain the expressionof the Gaussian curvature K of a separable surface and we distinguish the abovethree special cases of separable surfaces. In Section 3 we give the proof of Theorem1.2 and in Section 4 we prove Theorem 1.3.2. Preliminaries
In our study, we need the expression of the Gaussian curvature K for a surfacedefined by the implicit equation F ( x, y, z ) = 0 for a smooth function F defined in adomain of R . Although this calculation may be seen as a mere exercise, as far as theauthors know, the first reference where appears such a computation is [1]. In fact,Dombrowski obtains the expression of the Gauss-Kronecker curvature K for a hyper-surface in Euclidean space R n +1 given by an implicit function F ( x , . . . , x n +1 ) = 0;see also [9]. In [1], it was derived the following formula for K : K = 1 |∇ F | n +2 ∇ F t · co(Hess)( F ) · ∇ F, where ∇ F is the gradient of F and co(Hess)( F ) is the matrix formed by the cofactorsof Hess( F ). In the Euclidean space R , the above formula reduces into THOMAS HASANIS AND RAFAEL L ´OPEZ K |∇ F | = F x (cid:12)(cid:12)(cid:12)(cid:12) F yy F yz F yz F zz (cid:12)(cid:12)(cid:12)(cid:12) + F y (cid:12)(cid:12)(cid:12)(cid:12) F zz F xz F xz F xx (cid:12)(cid:12)(cid:12)(cid:12) + F z (cid:12)(cid:12)(cid:12)(cid:12) F xx F xy F xy F yy (cid:12)(cid:12)(cid:12)(cid:12) − F x F y (cid:12)(cid:12)(cid:12)(cid:12) F xy F yz F xz F zz (cid:12)(cid:12)(cid:12)(cid:12) − F y F z (cid:12)(cid:12)(cid:12)(cid:12) F yz F xz F xy F xx (cid:12)(cid:12)(cid:12)(cid:12) − F x F z (cid:12)(cid:12)(cid:12)(cid:12) F xz F xy F yz F yy (cid:12)(cid:12)(cid:12)(cid:12) . If the surface is defined by the implicit equation f ( x ) + g ( y ) + h ( z ) = 0, then theabove expression of K simplifies in f (cid:48) g (cid:48)(cid:48) h (cid:48)(cid:48) + g (cid:48) f (cid:48)(cid:48) h (cid:48)(cid:48) + h (cid:48) f (cid:48)(cid:48) g (cid:48)(cid:48) = K ( f (cid:48) + g (cid:48) + h (cid:48) ) . (2)A first case to distinguish is when one of the functions f , g or h is constant. Withoutloss of generality, we suppose that h is constant, so h ( z ) = a , z ∈ I , for some a ∈ R .Then the equation of the surface is f ( x ) + g ( y ) + a = 0, that is, the surface is aright cylinder over the plane curve C = { ( x, y ) ∈ R : f ( x ) + g ( y ) + a = 0 } and itsGaussian curvature is K = 0. This case is the item 1 in Theorem 1.2.In what follows, we suppose that S is not a right cylinder over a plane curve containedin one of the three coordinate planes. This is equivalent to f (cid:48) ( x ) g (cid:48) ( y ) h (cid:48) ( z ) (cid:54) = 0everywhere in I × I × I . Thus we can introduce the new variables u = f ( x ) , v = g ( y ) , w = h ( z ) , (3)which are related by the equation u + v + w = 0 thanks to (1). Define the functions X ( u ) = f (cid:48) ( x ) , Y ( v ) = g (cid:48) ( y ) , Z ( w ) = h (cid:48) ( z ) . Then X (cid:48) ( u ) = 2 f (cid:48)(cid:48) ( x ) , Y (cid:48) ( v ) = 2 g (cid:48)(cid:48) ( y ) , Z (cid:48) ( w ) = 2 h (cid:48)(cid:48) ( z ) . With this change of variables, equation (2) becomes XY (cid:48) Z (cid:48) + Y X (cid:48) Z (cid:48) + ZX (cid:48) Y (cid:48) = 4 K ( X + Y + Z ) , (4)for all values u , v and w under the condition u + v + w = 0.Throughout this paper we need to differentiate equations similar to (4) involvingfunctions depending on u , v and w . Since these variables are not independentbecause u + v + w = 0, the following result will be useful in our computations. Lemma 2.1.
Let Q = Q ( u, v, w ) be a smooth function defined in a domain Ω ⊂ R .If Q ( u, v, w ) = 0 for any triple of the section Ω ∩ Π , where Π is the plane of equation u + v + w = 0 , then on the section we have Q u = Q v = Q w , where Q u , Q v and Q v are the derivatives of Q with respect to u , v and w , respectively.Proof. Since w = − u − v , then Q ( u, v, − u − v ) = 0. Differentiating with respectto u , we deduce Q u − Q w = 0. Changing the roles of u , v and w , we conclude theresult. (cid:3) (cid:3) We need to distinguish the three special cases of separable surfaces described in theIntroduction in terms of the functions X , Y and Z . LASSIFICATION OF SEPARABLE SURFACES WITH CONSTANT GAUSSIAN CURVATURE 5
Proposition 2.2.
With the above notation, we have: (1)
If one of the functions X , Y or Z vanishes, then the surface is a right cylinderover a planar curve contained in one of the three coordinates planes. (2) If one of the functions X (cid:48) , Y (cid:48) or Z (cid:48) vanishes, then the surface is a rightcylinder over a planar curve contained in one of the three coordinates planesor it is a translation surface. (3) If one of the functions X (cid:48) − Y (cid:48) , X (cid:48) − Z (cid:48) or Y (cid:48) − Z (cid:48) vanishes, then the surfaceis one of the type studied in the previous item, or it is a surface of revolutionwhose rotation axis is parallel to one of the three coordinate axes.Proof. We discuss case-by-case.(1) When we introduced the variables u , v and w in (3), we showed that if oneof the functions f (cid:48) , g (cid:48) or h (cid:48) vanishes, then the surface is a right cylinder overa planar curve contained in one of the three coordinate planes.(2) Without loss of generality, we suppose that Z (cid:48) = 0. Because Z = h (cid:48) , then h ( z ) = az + b with a, b ∈ R . This proves that the surface is a right cylinder( a = 0) or a translation surface ( a (cid:54) = 0).(3) Without loss of generality, we suppose that X (cid:48) − Y (cid:48) = 0. Then there is a ∈ R such that X (cid:48) = Y (cid:48) = a . The case a = 0 has been studied in the previousitem. Assume now a (cid:54) = 0. Solving the equations X (cid:48) ( u ) = a , Y (cid:48) ( v ) = a , wefind f (cid:48) ( x ) = af ( x ) + b , g (cid:48) ( y ) = ag ( y ) + b , for some constants b , b ∈ R . The solutions of these ODEs are f ( x ) = ( ax + c ) a − b a , g ( y ) = ( ay + c ) a − b a , where c , c ∈ R . Thus the surface is a surface of revolution with respect toa straight-line parallel to the z -axis. (cid:3) (cid:3) Case K = 0 : proof of Theorem 1.2 In this section we will study separable surfaces with zero constant Gaussian curva-ture and we will prove Theorem 1.2. If the Gauss curvature K is constantly zero,then equation (4) becomes XY (cid:48) Z (cid:48) + Y X (cid:48) Z (cid:48) + ZX (cid:48) Y (cid:48) = 0 , for all u + v + w = 0 . (5)By Proposition 2.2, the cases that one of the functions X , Y or Z is constant, or X (cid:48) , Y (cid:48) or Z (cid:48) is 0, or X (cid:48) − Y (cid:48) , Y (cid:48) − Z (cid:48) or X (cid:48) − Z (cid:48) is 0 corresponds with the items 1,2 and 3 of Theorem 1.2, respectively. THOMAS HASANIS AND RAFAEL L ´OPEZ
From now on, we will suppose that the surface is not of the above three cases. Inparticular, X (cid:48) , Y (cid:48) , Z (cid:48) (cid:54) = 0, so equation (5) can be expressed as X (cid:48) Y (cid:48) Z (cid:48) (cid:18) XX (cid:48) + YY (cid:48) + ZZ (cid:48) (cid:19) = 0 , or equivalently, XX (cid:48) + YY (cid:48) + ZZ (cid:48) = 0 . (6)By using Lemma 2.1, we differentiate with respect to u , v and w , obtaining (cid:18) XX (cid:48) (cid:19) (cid:48) = (cid:18) YY (cid:48) (cid:19) (cid:48) = (cid:18) ZZ (cid:48) (cid:19) (cid:48) . Because we have three functions depending in the variables u , v and w , there is k ∈ R such that (cid:18) XX (cid:48) (cid:19) (cid:48) = (cid:18) YY (cid:48) (cid:19) (cid:48) = (cid:18) ZZ (cid:48) (cid:19) (cid:48) = k. (7)We distinguish two cases.(1) Case k = 0. Because XY Z (cid:54) = 0, we deduce that there are a, b, c ∈ R , a, b, c (cid:54) = 0, such that XX (cid:48) = a , YY (cid:48) = b , ZZ (cid:48) = c , with a + b + c = 0 because of (6). The integration of these equations leadsto f ( x ) = − a log( m x + n ) g ( y ) = − b log( m y + n ) h ( z ) = − c log( m z + n ) , where m i , n i ∈ R and m i (cid:54) = 0, 1 ≤ i ≤
3. Thus the implicit equation of thesurface f ( x ) + g ( y ) + h ( z ) = 0 becomes( m x + n ) a ( m y + n ) b ( m z + n ) c = 1 , (8)or equivalently (cid:18) x + n m (cid:19) a (cid:18) y + n m (cid:19) b = m − a m − b m a + b (cid:18) z + n m (cid:19) a + b . This surface is the generalized cone with apex ( − n /m , − n /m , − n /m )and the directrix is the planar curve (cid:18) x + n m (cid:19) a (cid:18) y + n m (cid:19) b = m − a m − b m a + b (cid:18) d + n m (cid:19) a + b z = d (cid:54) = − n m . This case is included in the item (4) of Theorem 1.2. Equation (8) can beexpressed as m z + n = ( m x + n ) p ( m y + n ) q , p + q = 1 . (9) LASSIFICATION OF SEPARABLE SURFACES WITH CONSTANT GAUSSIAN CURVATURE 7
This surface is the graph of the product of two functions on the variables x and y : see [6]. The surface defined by the equation (9) appeared in [6, Th.1.3].(2) Case k (cid:54) = 0. Replacing k by 1 / (2 k ) in (7), and integrating, we deduce thatthere are a, b, c ∈ R such that X (cid:48) X = 2 ku + a , YY (cid:48) = 2 kv + b , ZZ (cid:48) = 2 kw + c . (10)Substituting in (6), u + v + w + a + b + c = a + b + c = 0 , because of u + v + w = 0. Thus a + b + c = 0. Integrating (10), we find f (cid:48) ( x ) = m ( f ( x ) + a ) k g (cid:48) ( y ) = m ( g ( y ) + b ) k h (cid:48) ( z ) = m ( h ( z ) + c ) k , for some nonzero real numbers m i , 1 ≤ i ≤
3. The solutions of the aboveequations depend if k = 1 or k (cid:54) = 1.(a) Case k = 1. There are n i ∈ R such that f ( x ) = n e m x − ag ( y ) = n e m y − bh ( z ) = n e m z − c, and thus the equation of the surface S is n e m x + n e m y + n e m z = 0 . (11)This surface is a generalized cylinder whose generators are parallel to(1 /m , /m , /m ) and the directrix curve is (cid:26) xm + ym + zm = 0 n e m x + n e m y + n e m z = 0 . (b) Case k (cid:54) = 1. There are n i ∈ R , such that f ( x ) = ((1 − k )( m x + n )) − k − ag ( y ) = ((1 − k )( m y + n )) − k − bh ( z ) = ((1 − k )( m z + n )) − k − c. Then the implicit equation of the surface is( m x + n ) − k + ( m y + n ) − k + ( m z + n ) − k = 0 . (12)This surface is conical with apex the point ( − n /m , − n /m , − n /m )and the directrix curve is (cid:26) z = d (cid:54) = − n m ( m x + n ) − k + ( m y + n ) − k = − ( m d + n ) − k . THOMAS HASANIS AND RAFAEL L ´OPEZ
It deserves to point out that for some values of k , as k = (2 n − / (2 n ), n ∈ N , equation (12) represents only a point.The previous cases k = 1 and k (cid:54) = 1 correspond with the item 4 of Theorem 1.2 andthis completes the proof.We finish this section showing some explicit parametrizations of separable surfaceswith zero constant Gaussian curvature for the case 4 of Theorem 1.2. Example 3.1.
Case k = 0 in equation (9). We choose p = 2 , q = − , m i = 1 and n i = 0 . Then the surface is x = yz . See figure 1, left. Example 3.2.
Case k = 1 in equation (11). We choose n = − , n = n = 1 and m i = 1 . The implicit equation of the surface is − e x + e y + e z = 0 . See figure 1,middle. Example 3.3.
Case k = 2 in equation (12). We choose m i = 1 and n i = 0 ,obtaining /x + 1 /y + 1 /z = 0 , or equivalently, z = xy/ ( x + y ) with x + y (cid:54) = 0 . Seefigure 1, right. Figure 1.
Left: the surface x = yz . Middle: the surface − e x + e y + e z = 0. Right: the surface z = xy/ ( x + y ).4. Case K (cid:54) = 0 : proof of Theorem 1.3 In this section we prove Theorem 1.3. Consider a separable surface defined by (1)and suppose that the Gaussian curvature K is a nonzero constant. In this setting,the particular cases that one of the functions X , Y or Z is constant, or that X (cid:48) , Y (cid:48) or Z (cid:48) is zero and described in Proposition 2.2, can not appear because in such a casethe Gaussian curvature should be zero. On the other hand, the case that one of thefunctions X (cid:48) − Y (cid:48) , X (cid:48) − Z (cid:48) or Y (cid:48) − Z (cid:48) vanishes proves that the surface is rotationalabout an axis parallel to one of the coordinate axes, proving just the statement ofTheorem 1.3.Therefore, and besides the rotational surfaces, it remains to prove that there are notmore surfaces of separable surfaces with nonzero constant Gaussian curvature. Theproof of Theorem 1.3 is by contradiction. Assume on the contrary that the surface LASSIFICATION OF SEPARABLE SURFACES WITH CONSTANT GAUSSIAN CURVATURE 9 is not a surface of revolution about a straight-line parallel to one of the coordinatesaxes. In particular, none of the functions X (cid:48) − Y (cid:48) , X (cid:48) − Z (cid:48) or Y (cid:48) − Z (cid:48) is identically0. We write down again equation (4) XY (cid:48) Z (cid:48) + Y X (cid:48) Z (cid:48) + ZX (cid:48) Y (cid:48) = 4 K ( X + Y + Z ) . (13)Since K (cid:54) = 0, the surface is not a cylindrical surface neither a translation surface.This implies that XY Z (cid:54) = 0 and X (cid:48) Y (cid:48) Z (cid:48) (cid:54) = 0. Then equation (13) is( X (cid:48) Y + XY (cid:48) ) Z (cid:48) + ( X (cid:48) Y (cid:48) − K ( X + Y )) Z − KZ − K ( X + Y ) = 0 . (14)We simplify the notation of this equation by setting P Z (cid:48) + QZ + R = 4 KZ , (15)where P ( u, v ) = X (cid:48) Y + XY (cid:48) Q ( u, v ) = X (cid:48) Y (cid:48) − K ( X + Y ) R ( u, v ) = − K ( X + Y ) . Before to indicate the arguments to prove Theorem 1.3, we need a lemma that saysus that the coefficient of Z (cid:48) in (14), namely, the function P , is not zero. Lemma 4.1.
The function P = X (cid:48) Y + XY (cid:48) in (14) can not vanish in any open set.Proof. The proof is by contradiction. If X (cid:48) Y + XY (cid:48) = 0 in an open set, then X (cid:48) X = a = − Y (cid:48) Y (16)where a ∈ R is a constant. Now equation (13) is − a XY Z = 4 K ( X + Y + Z ) , (17)in particular, a (cid:54) = 0. By applying Lemma 2.1 to this equation, and differentiatingwith respect to u and v , we obtain − a XY Z = 8 Ka ( X + Y )( X + Y + Z ) , (18)where we have used (16). By combining (17) and (18), we conclude 8 KaZ ( X + Y + Z ) = 0, arriving to a contradiction. (cid:3) (cid:3) Once proved Lemma 4.2, we return to the equation (15). We apply Lemma 2.1 bydifferentiating (15) with respect to u and v . Then we find( P u − P v ) Z (cid:48) + ( Q u − Q v ) Z + R u − R v = 0 , or equivalently( X (cid:48)(cid:48) Y − XY (cid:48)(cid:48) ) Z (cid:48) + ( X (cid:48)(cid:48) Y (cid:48) − X (cid:48) Y (cid:48)(cid:48) − K ( X (cid:48) − Y (cid:48) )) Z − K ( X + Y )( X (cid:48) − Y (cid:48) ) = 0 . (19) We set A ( u, v ) = P u − P v = X (cid:48)(cid:48) Y − XY (cid:48)(cid:48) B ( u, v ) = Q u − Q v = X (cid:48)(cid:48) Y (cid:48) − X (cid:48) Y (cid:48)(cid:48) − K ( X (cid:48) − Y (cid:48) ) C ( u, v ) = R u − R v = − K ( X + Y )( X (cid:48) − Y (cid:48) ) . (20)Then (19) is AZ (cid:48) + BZ + C = 0 , for all u + v + w = 0 . (21)We now present the steps of the proof of Theorem 1.3. We will prove two technicallemmas. First, we show that the functions A , B and C can not vanish (Lemma 4.2)and then we show that B/A and
C/A are functions on the variable u + v (Lemma 4.3).After that, the proof returns to the equation (15) and, after successive applicationsof Lemma 2.1, we arrive to the desired contradiction. Lemma 4.2.
The coefficients A or B or C in (21) can not be identically zero.Proof. It is clear that C = 0 implies X (cid:48) − Y (cid:48) = 0, which is not possible. Bycontradiction, we suppose A = 0 or B = 0 identically.(1) Case A = 0. Then X (cid:48)(cid:48) Y = XY (cid:48)(cid:48) , so X (cid:48)(cid:48) /X = Y (cid:48)(cid:48) /Y = a , where a ∈ R is aconstant. Then (19) is F := ( a ( XY (cid:48) − X (cid:48) Y ) − K ( X (cid:48) − Y (cid:48) )) Z − K ( X + Y )( X (cid:48) − Y (cid:48) ) = 0 . (22)By Lemma 2.1, the expression F u − F v = 0 is G := (cid:0) aX (cid:48) Y (cid:48) − a XY − aK ( X + Y ) (cid:1) Z − K (( X (cid:48) − Y (cid:48) ) + a ( X + Y ) ) = 0 . Using Lemma 2.1 again for the function G , the equation G u − G v = 0 becomes (cid:0) a ( XY (cid:48) − X (cid:48) Y ) − Ka ( X (cid:48) − Y (cid:48) ) (cid:1) Z − aK ( X + Y )( X (cid:48) − Y (cid:48) ) = 0 . (23)Subtracting (23) from (22), we deduce24 aK ( X (cid:48) − Y (cid:48) ) Z = 0 , which it is a contradiction.(2) Case B = 0. Then X (cid:48)(cid:48) X (cid:48) + 8 K X (cid:48) = a = Y (cid:48)(cid:48) Y (cid:48) + 8 K Y (cid:48) for some constant a ∈ R . Then X (cid:48)(cid:48) = aX (cid:48) − K, Y (cid:48)(cid:48) = aY (cid:48) − K. (24)Taking into account both expressions, equation (19) is F := ( a ( X (cid:48) Y − XY (cid:48) ) + 8 K ( X − Y )) Z (cid:48) − K ( X + Y )( X (cid:48) − Y (cid:48) ) = 0 . (25)We utilize Lemma 2.1 by differentiating with respect to u and v . Then F u − F v = 0 and using (24), we deduce G : = (cid:0) a ( X (cid:48) Y + XY (cid:48) ) − aK ( X + Y ) − aX (cid:48) Y (cid:48) + 8 K ( X (cid:48) + Y (cid:48) ) (cid:1) Z (cid:48) − K (cid:0) ( X (cid:48) − Y (cid:48) ) − aK ( X + Y )( X (cid:48) + Y (cid:48) ) + 128 K ( X + Y ) (cid:1) = 0 . LASSIFICATION OF SEPARABLE SURFACES WITH CONSTANT GAUSSIAN CURVATURE11
Again, we compute G u − G v = 0 and using (24), we have (cid:0) a ( X (cid:48) Y − XY (cid:48) ) − aK ( X (cid:48) − Y (cid:48) ) + 8 a K ( X − Y ) (cid:1) Z (cid:48) − aK ( X (cid:48) − Y (cid:48) )( X (cid:48) + Y (cid:48) ) − a K ( X + Y )( X (cid:48) − Y (cid:48) ) + 384 K ( X (cid:48) − Y (cid:48) ) = 0 . If a = 0, then X (cid:48) − Y (cid:48) = 0, which is not possible. So, a (cid:54) = 0. By combiningthis equation and (25), we deduce H := 2 aZ (cid:48) + 3 a ( X (cid:48) − Y (cid:48) ) − K = 0 . Finally, using Lemma 2.1 for the function H and (24), the equation H u − H v =0 yields 3 a ( X (cid:48)(cid:48) − Y (cid:48)(cid:48) ) = 3 a ( X (cid:48) − Y (cid:48) ) = 0 , which is a contradiction. (cid:3) (cid:3) After Lemma 4.2, we differentiate (21) with respect to u and v , obtaining( A u − A v ) Z (cid:48) + ( B u − B v ) Z + ( C u − C v ) = 0 . Lemma 4.3.
With the above notation, we have A u − A v A = B u − B v B = C u − C v C , or equivalently, (cid:18) BA (cid:19) u = (cid:18) BA (cid:19) v , (cid:18) CA (cid:19) u = (cid:18) CA (cid:19) v , (cid:18) CB (cid:19) u = (cid:18) CB (cid:19) v . Proof.
By eliminating Z (cid:48) from (14) and (21), we find4 KAZ + (cid:110) (8 K ( X + Y ) − X (cid:48) Y (cid:48) ) A + ( X (cid:48) Y + XY (cid:48) ) B (cid:111) Z +4 K ( X + Y ) A + ( X (cid:48) Y + XY (cid:48) ) C = 0 . We write this relation as LZ + M Z + N = 0 , (26)where L = 4 KAM = (8 K ( X + Y ) − X (cid:48) Y (cid:48) ) A + ( X (cid:48) Y + XY (cid:48) ) BN = 4 K ( X + Y ) A + ( X (cid:48) Y + XY (cid:48) ) C. Applying Lemma 2.1 to (26) differentiating with respect to u and v , we obtain( L u − L v ) Z + ( M u − M v ) Z + N u − N v = 0 . (27)Simple calculations give L u − L v = 4 K ( A u − A v ) M u − M v = (8 K ( X + Y ) − X (cid:48) Y (cid:48) )( A u − A v ) + ( X (cid:48) Y + XY (cid:48) )( B u − B v ) N u − N v = 4 K ( X + Y ) ( A u − A v ) + ( X (cid:48) Y + XY (cid:48) )( C u − C v ) . (28) We distinguish two cases in the proof of Lemma 4.3:(1) Case L u − L v (cid:54) = 0. Since equations (26) and (27) have the same solutions, L u − L v L = M u − M v M = N u − N v N , or equivalently, M ( L u − L v ) = L ( M u − M v ) N ( L u − L v ) = L ( N u − N v ) . (29)Using the expressions of L u − L v , M u − M v and N u − N v in (28), the twoequations of (29) are, respectively, B ( X (cid:48) Y + XY (cid:48) )( A u − A v ) = A ( X (cid:48) Y + XY (cid:48) )( B u − B v ) C ( X (cid:48) Y + XY (cid:48) )( A u − A v ) = A ( X (cid:48) Y + XY (cid:48) )( C u − C v ) . Since X (cid:48) Y + XY (cid:48) (cid:54) = 0 by Lemma 4.1, if follows the result of Lemma 4.3.(2) Case L u − L v = 0.(a) Subcase M u − M v = 0. Then (27) implies N u − N v = 0. From (28) wededuce A u − A v = B u − B v = C u − C v = 0, proving the result.(b) Subcase M u − M v (cid:54) = 0. From (27), we deduce that (26) has a uniquesolution, namely, Z = − M L .
Then (cid:18) ML (cid:19) u − (cid:18) ML (cid:19) v = 0 . This implies ( M u − M v ) L = 0, and we conclude M u − M v = 0, a con-tradiction. (cid:3) (cid:3) Once proved Lemmas 4.2 and 4.3, we are in position to complete the proof of The-orem 1.3. of Theorem 1.3.
From Lemma 4.3 we deduce that there exist functions Φ and Ψ ofone variable such thatΨ( u + v ) = B ( u, v ) A ( u, v ) , Φ( u + v ) = C ( u, v ) A ( u, v ) . By Lemma 4.2, A (cid:54) = 0 and we write (26) as Z + ML Z + NL = 0 , (30)This a polynomial equation on Z . Then the two roots Z ( w ) and Z ( w ) of thisequation satisfy Z ( w ) + Z ( w ) = − ML , Z ( w ) Z ( w ) = − NL .
LASSIFICATION OF SEPARABLE SURFACES WITH CONSTANT GAUSSIAN CURVATURE13
Since u + v + w = 0, the functions M/L and
N/L depend only on the variable u + v .Denote Ψ ( u + v ) = M ( u, v ) L ( u, v ) , Φ ( u + v ) = N ( u, v ) L ( u + v ) . Thus (30) is now Z + Ψ Z + Φ = 0 . Applying Lemma 2.1 to this equation with respect to w and u , we find(2 Z + Ψ ) Z (cid:48) − Ψ (cid:48) Z − Φ (cid:48) = 0 . From this equation and (15), we eliminate Z (cid:48) , obtaining((2 Z + Ψ ) Q + P Ψ (cid:48) ) Z + (2 Z + Ψ ) R + P Φ (cid:48) = 4 KZ (2 Z + Ψ ) , or equivalently, Z + (cid:18) Ψ − Q K (cid:19) Z − (cid:18) P Ψ (cid:48) + 2 R + Q Ψ K (cid:19) Z − P Φ (cid:48) + R Ψ K = 0 . (31)This is a polynomial equation on Z . Let Z ( w ), Z ( w ) and Z ( w ) denote the threeroots of this equation. Then the function T ( u + v ) = Z ( w ) + Z ( w ) + Z ( w )is the opposite of the coefficient of Z in (31), so T ( u + v ) = Q ( u, v )4 K − Ψ ( u + v )2 . We apply to this equation Lemma 2.1 differentiating with respect to the variables u and v , obtaining 0 = 14 K ( Q u − Q v ) = 14 K B by (20). This is a contradiction by Lemma 4.2, and this concludes the proof ofTheorem 1.3. (cid:3) (cid:3)
Acknowledgements
Rafael L´opez has been partially supported by the grant no. MTM2017-89677-P,MINECO/AEI/FEDER, UE.
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