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Authors requiring further information regarding Elsevier’s archiving and manuscript policies are encouraged to visit: http://www.elsevier.com/copyright Author's personal copy Computers and Mathematics with Applications 65 (2013) 673–681 Contents lists available at SciVerse ScienceDirect Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa Constructing PDE-based surfaces bounded by geodesics or lines of curvature Wei-Xian Huang, Hua-Jing-Ling Wu, Guo-Jin Wang ∗ Department of Mathematics, Zhejiang University, Hangzhou 310027, China State Key Laboratory of CAD&CG, Zhejiang University, Hangzhou 310027, China article info Article history: Received 1 July 2012 Received in revised form 14 October 2012 Accepted 20 November 2012 Keywords: Partial differential equations Geodesic Line of curvature Surface modeling Biharmonic surfaces abstract In order to explore a new approach to construct surfaces bounded by geodesics or lines of curvature, a method of surface modeling based on fourth-order partial differential equations (PDEs) is presented. Compared with the free-form surface modeling based on finding control points, PDE-based surface modeling has the following three advantages. First, the corresponding biharmonic surface can naturally be derived under some degenerative conditions; second, the parameters in the PDE implicate some physical meaning, such as elasticity or rigidity; third, there are only a few parameters that need to be evaluated, and hence the computation is simple. In addition, this paper constructs two adjacent surfaces with C1 continuity whose common boundary is the same given curve as well as respective geodesic (or line of curvature). Examples show that this method to construct PDE-based surfaces bounded by geodesics or lines of curvature is easy and effective. © 2012 Elsevier Ltd. All rights reserved. 1. Introduction Surface modeling is a central issue in computer-aided design, and surfaces are usually represented by implicit or parametric forms. For example, [1] transformed the construction of a surface into Poisson problems and then obtained implicit surfaces. Compared with implicit surfaces, parametric surfaces are used more widely, and the most usual modeling technique is to construct free-form parametric surfaces based on determining control points. [2] constructed parametric spline surfaces by using variable-degree polynomial splines. [3] constructed harmonic and homogenous biharmonic Bézier surfaces bounded by the given curves. [4] constructed nonhomogenous biharmonic and tetraharmonic Bézier surfaces according to given boundary curves and tangent conditions along them. In addition, for free-form parametric surfaces, there is another modeling technique that needs to be paid attention to, which is the parametric surface construction based on PDEs. [5] constructed parametric surfaces using a fourth-order PDE with three shape control parameters, and presented exact analytic solutions of the PDE in some cases. [6] constructed surfaces using a sixth-order PDE, which interpolate the given boundary curves, boundary tangents, and boundary curvatures. In recent years, people have become interested in two additional problems in surface modeling. One is to construct surfaces bounded by given geodesic curves, and the other is to construct surfaces bounded by lines of curvature. A geodesic curve is intrinsic to the geometric characterization of surfaces. Geodesics are used in many fields. For example, they are used in object segmentation [7,8] and multi-scale image analysis [9], in computer vision and image processing. In surface modeling, [10] constructed parametric surfaces bounded by the given geodesic curves; [11] constructed polynomial ruled surfaces with the given boundary as geodesic curves; similarly, [12] constructed cubic polynomial ruled patches with the given geodesic boundary. Recently, [13] constructed Bézier surfaces bounded by the four given geodesic curves, and [14] ∗ Corresponding author at: Department of Mathematics, Zhejiang University, Hangzhou 310027, China. E-mail address: wanggj@zju.edu.cn (G.-J. Wang). 0898-1221/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2012.11.018 Author's personal copy 674 W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 constructed triangular Bézier surfaces using the methods in [13]. As we know, the existed method of constructing parametric surfaces bounded by geodesic curves is usually based on finding corresponding control points, but a method based on PDEs has not yet been published in the literature. On the other hand, lines of curvature have important applications in computer graphics and product manufacturing. [15] studied an approach to compute lines of curvature near the umbilical points on the surfaces, and applied it to extracting the generic features of free-form parametric surfaces for shape interrogation. [16] presented a method to determine lines of curvature in point cloud models, and applied it to reconstruct meshes that interpolate the given lines of curvature. [17] showed an application of lines of curvature in plate-metal-based manufacturing. In the surface modeling literature, there are only a few articles concerning lines of curvature. Recently, [18] constructed Bézier parametric surfaces bounded by the four lines of curvature, but the construction of a PDE-based surface bounded by lines of curvature has not been studied in the literature. Hence, this paper aims to address the issue of constructing a PDE-based surface bounded by geodesics or lines of curvature. The work of this paper can be listed as follows. First, the method to generate a surface bounded by geodesics or lines of curvature based on a vector-valued fourth-order PDE is presented, and the exact solutions of the PDE for some given conditions are obtained by using the method in [5]. Second, the numerical solution of the PDE by the least-square method is given. Third, the process shows that a fourth-order PDE-based surface can be degenerated into a biharmonic surface by choosing appropriate coefficients. Fourth, two adjacent surfaces are constructed so that they have a common boundary which is the same given curve as well as the respective geodesic, and the continuity between them is also discussed. The rest of the paper is arranged as follows. Section 2 presents the construction of surfaces bounded by geodesic curves. Section 3 presents the construction of surfaces bounded by lines of curvature, and the conclusion is presented in the last section. 2. Construction of surfaces bounded by geodesics 2.1. Description of surfaces with geodesic boundary curves Suppose that the curves p0 (u), p1 (u), u ∈ [a, b] are given, where a, b are both arbitrarily real numbers. Then the surface p(u, v) = (x(u, v), y(u, v), z (u, v))T with the given curves as boundary curves should satisfy p(u, 0) = p0 (u), (1) p(u, 1) = p1 (u). (2) In order to make the given curves p0 (u), p1 (u) be geodesics on the surface p(u, v), the necessary condition is that the vectors b0 (u) = p′0 (u) × p′′0 (u) and b1 (u) = p′1 (u) × p′′1 (u) are all tangent to the surface. The vectors b0 (u) and b1 (u) are parallel to the binormals of the curves p0 (u) and p1 (u), respectively. Specifically, the partial derivative with respect to v of the surface should satisfy the following equations: pv (u, 0) = α0 (u)p′0 (u) + β0 (u)b0 (u), (3) pv (u, 1) = α1 (u)p′1 (u) + β1 (u)b1 (u), (4) where αi (u), βi (u) ̸= 0, i = 0, 1 are any functions defined on [a, b]. In addition, suppose that the surface satisfies the following fourth-order PDE: ∂4 ∂4 ∂4 a 4 +b 2 2 +c 4 ∂u ∂ u ∂v ∂v p(u, v) = 0, (5) where a = (ax , ay , az )T , b = (bx , by , bz )T , c = (cx , cy , cz )T are vectors with positive components. The PDE in Eq. (5) includes all forms of the existing fourth-order PDEs used for surface generation [5], and another advantage of this PDE is that the surface p(u, v) becomes a biharmonic surface if we choose the coefficients as a = (1, 1, 1)T , b = (2, 2, 2)T , c = (1, 1, 1)T . So the surface Eq. (5) is interesting. A biharmonic surface is a surface satisfying that satisfies the equation ∆2 p(u, v) = 0, where ∆ = ∂ ∂ u2 + ∂ ∂v 2 . Besides, the parameters in the PDE imply some physical meaning. For example, a fourth-order PDE can be derived from the theory of bending thin elastic plates; thus the coefficients of such a PDE are closely related to the physical properties of the surface that it represents [19]. The parametric surface determined by Eqs. (1)–(5) is the one we need, i.e., the surface with the common boundary as respective geodesic. Remark 1. The operations of the vectors defined in this paper are ab = (ax bx , ay by , ay by )T and a/b = (ax /bx , ay /by , ay /by )T , called vector multiplication or division, respectively, where a = (ax , ay , az )T , b = (bx , by , bz )T . In addition, a < b means ax < bx , ay < by , az < bz , 1 − a = (1 − ax , 1 − ay , 1 − az ), ξ a = (ξ ax , ξ ay , ξ az ), b a = bx ax , by ay , bz az . Author's personal copy W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 675 2.2. Solution of the fourth-order PDE For the fourth-order PDE, shown as in Eq. (5), with boundary conditions (1)–(4), explicit solutions do not always exist. [5] presented three cases in which explicit solutions do exist. Hence, we divide the process of solving the PDE into two parts. For cases where explicit solutions exist, we will simply present the explicit solutions [5]. For other cases, we will use the least-square method to obtain the numerical solution. We represent the boundary conditions written as Eqs. (1)–(4) as follows: p(u, 0) = J aj,1 gj (ξ u), pv (u, 0) = j =0 p(u, 1) = J J aj,2 gj (ξ u), (6) aj,4 gj (ξ u), (7) j =0 aj,3 gj (ξ u), pv (u, 1) = j =0 J j =0 where {gj (ξ u) = (gj,x (ξ u), gj,y (ξ u), gj,z (ξ u))T , j = 0, . . . , J } are independent functions with respect to the parameter u, the parameter ξ is a constant coefficient, and their corresponding coefficients are chosen as aj,i = (aj,i,x , aj,i,y , aj,i,z )T , i = 1, 2, 3, 4. The reason that we use Eqs. (6) and (7) to represent the boundary conditions is that the solution of the PDE with the above boundary conditions can be represented by p(u, v) = J Gj (v)gj (ξ u), j =0 where Gj (v) = (Gj,x (v), Gj,y (v), Gj,z (v))T ; that is, the solution separates the parameters u and v . Therefore, our goal has been transformed to deriving the functions Gj (v). If the basis functions satisfy one of the following three cases, then explicit solutions exist. Case 1. ∂∂u2 gj (ξ u) = ∂∂u4 gj (ξ u) = 0. The unknown functions Gj (v) can be expressed as 4 2 Gj (v) = 4 cj,k v k−1 , j = 0, 2, . . . , J , k=1 where cj,k are unknown coefficients, which can be determined by Eqs. (6) and (7) uniquely, and J is the number of independent functions in Eqs. (6) and (7). 2 Case 2. ∂∂u2 gj (ξ u) = −ξ 2 gj (ξ u). If the coefficients in Eq. (5) satisfy 4ac < b2 , then the unknown functions Gj (v) can be expressed as Gj (v) = cj,1 erj,1 v + cj,2 erj,2 v + cj,3 erj,3 v + cj,4 erj,4 v , where b 4ac rj,i = ±ξ 1± 1− 2 , erj,i v = (erj,i,x v , erj,i,z v , erj,i,z v )T , b 2a i = 1, 2, 3, 4. If the coefficients in Eq. (5) satisfy 4ac = b2 , then the unknown functions Gj (v) can be expressed as Gj (v) = cj,1 + cj,2 v erj,1 v + cj,3 + cj,4 v erj,2 v , where rj,i = ±ξ b 2a , i = 1, 2. It should be noticed that the computations of ac , b2 , ac /b2 , 1 − 4ac /b2 , rj,i , a, b, c are all vectors. √ b/(2a), etc. are all defined in Remark 1, and 2 Case 3. ∂∂u2 gj (ξ u) = ξ 2 gj (ξ u). If the coefficients in Eq. (5) satisfy 4ac < b2 , then the unknown functions Gj (v) can be expressed as Gj (v) = cj,1 cos rj,1 v + cj,2 sin rj,1 v + cj,3 cos rj,2 v + cj,4 sin rj,2 v , Author's personal copy 676 W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 where b 4ac rj,i = ξ 1± 1− 2 , cos rj,i v = cos rj,i,x v , cos rj,i,y v , cos rj,i,z v b 2a sin rj,i v = sin rj,i,x v , sin rj,i,y v , sin rj,i,z v T , T , i = 1, 2. If the coefficients in Eq. (5) satisfy 4ac = b2 , then the unknown functions Gj (v) can be expressed as Gj (v) = cj,1 + cj,2 v cos rj,1 v + cj,3 + cj,4 v sin rj,1 v , where rj,1 = ξ b 2a . The unknown coefficients cj,k in the expressions of the above function Gj (v) can be determined by the boundary conditions written as Eqs. (6) and (7) uniquely. If the boundary conditions do not satisfy the above three cases, then we can use the following least-square method to obtain the numerical solution. In order to satisfy the boundary conditions expressed as (6) and (7), we represent the numerical solution as follows: p(u, v) = J m j =0 pj,i Bm i (v) gj (ξ u), (8) i=0 where {gj (ξ u) = (gj,x (ξ u), gj,y (ξ u), gj,z (ξ u))T , j = 0, . . . , J } are independent functions with respect to the parameter u, the parameter ξ is a constant coefficient, Bm i (v) are Bernstein polynomials, and pj,i are unknown coefficients. It should be noticed that the higher the degree m of the parameter v is, the smaller the error of the numerical solution is, which is similar to numerical solution of a sixth-order PDE in [6]. In this paper, we take m = 6. Because Eq. (8) satisfies Eqs. (6) and (7), the unknown coefficients pj,0 , pj,1 , pj,m−1 , pj,m can be derived as follows: pj,0 = aj,1 , pj,1 = aj,1 + aj,2 /m, pj,m−1 = aj,3 − aj,4 /m, pj,m = aj,3 . Substituting the above equations into Eq. (8), we can obtain that p(u, v) = J m−2 f (v, aj,i ) + j=0 pj,i Bm i (v) gj (ξ u), (9) i=2 where f (v, aj,i ) = aj,1 (1 − v)m + m aj,1 + aj,2 m aj,4 (1 − v)m−1 v + m aj,3 − (1 − v)v m−1 + aj,3 v m . m Substituting Eq. (9) into Eq. (5) and uniformly sampling n2 ≥ (J + 1)(m − 3) points in the parametric domain, denoted by {(uk , vk ), k = 1, . . . , n2 }, then we can obtain the following n2 equations: ∆(uk , vk ) = J m −2 A(uk , vk )j,i pj,i + B(uk , vk ), k = 1, . . . n2 , j=0 i=2 where ∂2 m ∂2 ∂4 m ∂4 g (ξ u ) + b B (v ) g (ξ u ) + cg (ξ u ) B (vk ), j k k j k k j ∂ u4 ∂v 2 i ∂ u2 ∂v 4 i J ∂4 ∂2 ∂2 ∂4 B(uk , vk ) = af (vk , aj,i ) 4 gj (ξ uk ) + b 2 f (vk , aj,i ) 2 gj (ξ uk ) + cg j (ξ uk ) 4 f (vk , aj,i ) . ∂u ∂v ∂u ∂v j=0 A(uk , vk )j,i = aBm i (vk ) For simplicity, Eq. (10) can be represented in matrix form as ∆ = AC − B, where the vectors are ∆ = (∆(u1 , v1 ), . . . , ∆(un2 , vn2 ))T , B = (−B(u1 , v1 ), . . . , −B(un2 , vn2 )), C = (p0,2 , p0,3 , . . . , p0,m−2 , p1,2 , . . . , pJ ,m−2 )T , (10) Author's personal copy W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 677 Fig. 1. The biharmonic surface with given geodesic boundaries; the boundaries are drawn in blue. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.) and the elements of matrix A are Ak,j(m−3)+i = A(uk , vk )j,i+1 , k = 1, . . . , n2 , j = 0, . . . , J , i = 1, . . . , m − 3. In order to minimize the error ∆T ∆, according to the least-square method, it can be derived that −1 C = AT A AT B. (11) Hence, the unknown coefficients in Eq. (9) can be obtained by Eq. (11). 2.3. Numerical examples Two examples are presented in this section. Example 1 constructs a biharmonic surface bounded by the given geodesics, and Example 2 constructs two adjacent surfaces that have a common geodesic boundary curve. Suppose that the two surfaces are s1 (u, v) and s2 (u, v), and that the common boundary is p(u); that is, s1 (u, 1) = p(u) = s2 (u, 0). Also, suppose that the surface s1 (u, v) chooses the functions α1 (u), β1 (u) in Eq. (4), and the surface s2 (u, v) chooses the functions α0 (u), β0 (u) in Eq. (3). If the two surfaces choose the same functions α0 (u) = α1 (u), β0 (u) = β1 (u) on the common boundary which is the given curve, then we can prove that they are C1 continuous along the common boundary, as follows. Since s1 (u, 1) = ∂ s (u,1) = ∂ p∂(uu) = p(u) = s2 (u, 0), the two surfaces are C 0 continuous. Denoting b(u) = p′ (u) × p′′ (u), it can be seen that 1∂ u ∂ s2 (u,0) ∂u ∂ s (u,1) ∂ s (u,0) and 1∂v == α1 (u)p′ (u) + β1 (u)b(u) = α0 (u)p′ (u) + β0 (u)b(u) = 2∂v , so the two surfaces are also C1 continuous. The proof is valid for lines of curvature too. If we need to make the two adjacent surfaces be C 2 continuous along the common boundary, then a sixth-order PDE is needed to construct the corresponding surfaces; see [20]. Example 1 (To Construct Biharmonic Surface Bounded by Geodesics). Two curves p0 (u) = (sin u, cos u, 1)T , p1 (u) = (2 sin u, 3 cos u, 0)T , u ∈ [0, 2π ] are given, the functions in Eqs. (3) and (4) are chosen as α0 (u) = α1 (u) ≡ 0, β0 (u) ≡ 0.5, β1 (u) ≡ 1, and the coefficients in Eq. (5) are chosen as a = (1, 1, 1)T , b = (2, 2, 2)T , c = (1, 1, 1)T (see Fig. 1). In this example, it can be seen that J = 3, and the three independent functions are {sin u, cos u, 1}. Example 2 (To Construct Two Adjacent Biharmonic Surfaces with Prescribed Geodesic Boundaries, One of them Common to Both Surfaces). For Surface 1, the given geodesic boundary curves are p0 (u) = (3 sin u, cos u, 4)T , p1 (u) = (2 sin u, 3 cos u, 0)T , u ∈ [0, 2π ], and the functions in Eqs. (3) and (4) are chosen as α0 (u) = α1 (u) ≡ 0, β0 (u) ≡ 0.1, β1 (u) ≡ 1. For Surface 2, the given geodesic boundary curves are p0 (u) = (2 sin u, 3 cos u, 0)T , p1 (u) = (3 sin u, 2 cos u, −1)T , u ∈ [0, 2π ], and the functions in Eqs. (3) and (4) are chosen as α0 (u) = α1 (u) ≡ 0, β0 (u) = β1 (u) ≡ 1. In addition, the coefficients in Eq. (5) are chosen as a = (1, 1, 1)T , b = (2, 2, 2)T , c = (1, 1, 1)T . The two biharmonic surfaces are C1 continuous along the common boundary (see Fig. 2). In this example, it can also be seen that J = 3, and the three independent functions are {sin u, cos u, 1}. Author's personal copy 678 W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 Fig. 2. Two biharmonic surfaces with the same given geodesic boundary. Surface 1 is drawn in red, and the other is drawn in green. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.) 3. Construction of surfaces bounded by lines of curvature 3.1. Rational rotation-minimizing frames for spatial Pythagorean-hodograph (PH) curves [21] presented the sufficient and necessary condition that quintic PH curves have rational rotation-minimizing frames. We simply describe it as follows. If the derivative of the r (t ) satisfies x′2 (t ) + y′2 (t ) + z ′2 (t ) = σ 2 (t ), where r ′ (t ) = (x′ (t ), y′ (t ), z ′ (t )) and σ (t ) is a polynomial, then the curve r (t ) is called a spatial Pythagorean-hodograph (PH) curve. What is more, there are polynomials u(t ), v(t ), p(t ), q(t ) such that they satisfy x′ (t ) = u2 (t ) + v 2 (t ) − p2 (t ) − q2 (t ), z (t ) = 2 [v(t )q(t ) − u(t )p(t )] , ′ y′ (t ) = 2 [u(t )q(t ) + v(t )p(t )] , σ (t ) = u2 (t ) + v 2 (t ) + p2 (t ) + q2 (t ). Meanwhile, the derivative of the PH curve r (t ) can also be represented by a quaternion, as follows: r ′ (t ) = A(t )iA∗ (t ), (12) where A(t ) = u(t ) + v(t )i + p(t )j + q(t )k , ii = jj = kk = −1, ij = k , A∗ (t ) = u(t ) − v(t )i − p(t )j − q(t )k , ji = −k , jk = i, kj = −i, ki = j , ik = −j . Suppose that the Frenet frame of the curve r (t ) is defined by T = r′ ∥r ′ ∥ , N = r ′ × r ′′ ∥r ′ × r ′′ ∥ × T, B= r ′ × r ′′ ∥r ′ × r ′′ ∥ ; then the curve’s angular velocity is given by w = kB + τ T , where k is the curvature, and τ is the torsion. The Frenet frame is not a rotation-minimizing frame, because the angular velocity contains the term τ T . A rotation of (N , B) through the angle ϕ(ξ ) = ϕ0 − ξ τ (u)σ (u)du 0 can change the Frenet frame into a rotation-minimizing frame, such that the angular velocity only contains the term kB, where ϕ0 is the free integration constant, σ (u) = ∥r ′ (u)∥. If ϕ(ξ ) is a rational function, then the Frenet frame is changed into a rational rotation-minimizing frame. [21] pointed out the sufficient and necessary condition that quintic spatial PH curves defined by Eq. (12) have rational rotation-minimizing frames, namely, if and only if the coefficients of the quaternion polynomial A(t ) = A0 (1 − t )2 + 2A1 (1 − t )t + A2 t 2 satisfy the expression A2 iA∗0 = A1 iA∗1 . (13) Author's personal copy W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 679 If we take A0 = 1, A1 = u1 + v1 i + p1 j + q1 k, A2 = u21 + v12 − p21 − q21 + 2 (u1 p1 − v1 q1 ) j + 2 (u1 q1 + v1 p1 ) k , where u1 , v1 , p1 , q1 are any constants, then A(t ) satisfies the requirement. The key polynomials u(t ), v(t ), p(t ), q(t ) can be obtained as follows by simple computation: u(t ) = (1 − t )2 + 2u1 (1 − t )t + u21 + v12 − p21 − q21 t 2 , v(t ) = 2v1 (1 − t )t , p(t ) = 2p1 (1 − t )t + 2(u1 p1 − v1 q1 )t 2 , q(t ) = 2q1 (1 − t )t + 2 (u1 q1 + v1 p1 ) t 2 . In the following, all quintic spatial PH curves are defined in this way. We denote a(t ) = (1 − t )2 + 2u1 (1 − t )t + u21 + v12 + p21 + q21 t 2 , (14) b(t ) = 2v1 (1 − t )t . (15) Then Euler–Rodrigues [21] frame is defined as follows: t = A(t )iA∗ (t ) ∥A(t )∥2 , n= A(t )jA∗ (t ) ∥A(t )∥2 , b= A(t )kA∗ (t ) ∥A(t )∥2 , where ∥A(t )∥2 = u2 (t ) + v 2 (t ) + p2 (t ) + q2 (t ). (16) Further, we can obtain the rational rotation-minimizing frame of the curve r (t ) as e1 = t , e2 = a2 (t ) − b2 (t ) a2 (t ) + b2 (t ) n− 2a(t )b(t ) a2 ( t ) + b 2 ( t ) b, e3 = 2a(t )b(t ) a2 ( t ) + b 2 ( t ) n+ a2 (t ) − b2 (t ) a2 (t ) + b2 (t ) b. (17) 3.2. Description of surfaces bounded by lines of curvature [18] presented a lemma stating that a curve on a smooth surface is a line of curvature of that surface if and only if the Darboux frame is the same as the rational rotation-minimizing frame with respect to the curve’s tangent. Suppose that the curve r (ξ ) = s(u(ξ ), v(ξ )) is a curve on the surface s(u, v); then the Darboux frame (t , n, h) along this curve is defined as follows: t is the tangent to the curve r (ξ ), n is the surface normal along r (ξ ), and h = n×t. Suppose that the curves p0 (u), p1 (u), u ∈ [0, 1] and their rotation-minimizing frames (ti , ni , hi ), i = 0, 1 are given, where (ti , ni , hi ) is defined by (e1 , e2 , e3 ) as Eq. (17). And suppose that the surface bounded by the given curves as lines of curvature is denoted by p(u, v) = (x(u, v), y(u, v), z (u, v))T . Then the surface should satisfy p(u, 0) = p0 (u), (18) p(u, 1) = p1 (u). (19) According to the lemma in [18], mentioned above, in order to make the curves p0 (u), p1 (u) be lines of curvature on the surface p(u, v), the partial derivative with respect to v of the surface should satisfy pv (u, 0) = α0 (u)t0 (u) + β0 (u)h0 (u), (20) pv (u, 1) = α1 (u)t1 (u) + β1 (u)h1 (u), (21) where αi (u), βi (u) ̸= 0, i = 0, 1 are any functions defined on [0, 1]. In addition, suppose that the surface satisfies the fourthorder PDE defined by Eq. (5). Then the surface determined by Eqs. (18)–(21) and (5) is just the one bounded by the given lines of curvature. Specially, the surface p(u, v) will be degenerated into a biharmonic surface if we choose the coefficients as a = (1, 1, 1)T , b = (2, 2, 2)T , c = (1, 1, 1)T . The solution of the PDE has been given in Section 2.2. 3.3. Numerical examples Example 3 (To Construct Biharmonic Surface Bounded by Lines of Curvature). Two quintic spatial PH curves, p0 (u) and p1 (u), are given as follows. The curve p0 (u) is defined as p0 (0) = (0, 0, 0), and the parameters in Eq. (13) are chosen as u1 = 1, v1 = 2, p1 = 3, q1 = 2. The functions in Eq. (20) are taken as α0 (u) ≡ 0, β0 (u) ≡ 0.0000017(a2 (u) + b2 (u))∥A(u)∥2 , where a(u), b(u), A(u) are defined by Eqs. (14), (15) and (16), respectively. The curve p1 (u) is defined as p1 (0) = (0, 0, −100), and the parameters in Eq. (13) are chosen as u1 = 1, v1 = 2, p1 = 3, q1 = 2. The functions in Eq. (21) are taken as α1 (u) ≡ 0, β1 (u) ≡ 0.0000017(a2 (u) + b2 (u))∥A(u)∥2 , where a(u), b(u), A(u) are defined by Eqs. (14), (15) and (16), respectively. Author's personal copy 680 W.-X. Huang et al. / Computers and Mathematics with Applications 65 (2013) 673–681 Fig. 3. The biharmonic surface is bounded by lines of curvature; the boundaries are drawn in blue. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.) Fig. 4. Two adjacent approximate biharmonic surfaces are bounded by lines of curvature. Surface 1 is drawn in red, and the other is drawn in green. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.) The coefficients in Eq. (5) are taken as a = (1, 1, 1)T , b = (2, 2, 2)T , c = (1, 1, 1)T , and m = 6, n = 20 in Eq. (10). In this example, we can see thatJ = 8, and the independent functions are {ui |i = 0, . . . , 7}. Hence, we need to solve the PDE by the least-square method. We can obtain the approximate biharmonic surface with given curves as boundary lines of curvature as shown in Fig. 3. Example 4 (To Construct Two Adjacent Biharmonic Surfaces Bounded by Lines of Curvature). For Surface 1, two quintic spatial PH curves, p0 (u) and p1 (u), are given as follows. The curve p0 (u) is defined as p0 (0) = (0, 0, 0), and the parameters in Eq. (13) are chosen as u1 = 1, v1 = 2, p1 = 3, q1 = 2. The functions in Eq. (20) are taken as α0 (u) ≡ 0, β0 (u) ≡ 0.000017(a2 (u) + b2 (u))∥A(u)∥2 , where a(u), b(u), A(u) are defined by Eqs. (14), (15) and (16), respectively. The curve p1 (u) is defined as p1 (0) = (0, 0, −100), and the parameters in Eq. (13) are chosen as u1 = 1, v1 = 3, p1 = 3, q1 = 2. The functions in Eq. (21) are taken as α1 (u) ≡ 0, β1 (u) ≡ 0.000017(a2 (u)+ b2 (u))∥A(u)∥2 , where a(u), b(u), A(u) are defined by Eqs. (14), (15) and (16), respectively. For Surface 2, two quintic spatial PH curves, p0 (u) and p1 (u), are given as follows. The curve p0 (u) is defined as p0 (0) = (0, 0, 100), and the parameters in Eq. (13) are chosen as u1 = 1, v1 = 2, p1 = 2, q1 = 3. The functions in Eq. (20) are taken as α0 (u) ≡ 0, β0 (u) ≡ 0.000017(a2 (u)+ b2 (u))∥A(u)∥2 , where a(u), b(u), A(u) are defined by Eqs. (14), (15) and (16), respectively. The curve p1 (u) is defined as p1 (0) = (0, 0, 0), and the parameters in Eq. (13) are chosen as u1 = 1, v1 = 2, p1 = 3, q1 = 2. The functions in Eq. (21) are taken as α1 (u) ≡ 0, β1 (u) ≡ 0.000017(a2 (u) + b2 (u))∥A(u)∥2 , where a(u), b(u), A(u) are defined by Eqs. (14), (15) and (16), respectively. The coefficients in Eq. (5) are taken as a = (1, 1, 1)T , b = (2, 2, 2)T , c = (1, 1, 1)T , and m = 6, n = 20 in Eq. (10). In this example, we can see that J = 8, and the independent functions are {ui |i = 0, . . . , 7}. Hence, we need to solve the PDE by the least-square method. We can obtain two approximate biharmonic surfaces with the given curves as boundary lines of curvature, see Fig. 4. 4. Conclusion This paper presents a method to construct PDE-based surfaces bounded by geodesics or lines of curvature. Examples show that this method is effective and easy. The method for constructing a fourth-order PDE-based surface has some advantages that the one for constructing a free-form surface does not have. It is simple to obtain a biharmonic surface, Author's personal copy W.-X. 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