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Optimal parameter values for approximating conic sections by the quartic Bézier curves
Journal of Computational and Applied Mathematics 322 (2017) 86–95
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Optimal parameter values for approximating conic sections by the quartic Bézier curves Xuli Han ∗ , Xiao Guo School of Mathematics and Statistics, Central South University, Changsha, 410083, PR China
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Article history: Received 20 June 2016 Received in revised form 24 March 2017 Keywords: Conic section Quartic Bézier curves Hausdorff distance Optimal parameters
abstract The previous approximation curves of conic section by quartic Bézier curves interpolate the conic section at the specified parameter values. In this paper, by solving the minimax problem, we present the optimal parameter values for approximating conic sections by the quartic Bézier curves. The upper bound on the Hausdorff distance between the conic section and the approximation curves is minimized. The method of solving the minimax problem is changed to solve a quartic algebraic equation by delicate reasoning. © 2017 Elsevier B.V. All rights reserved.
1. Introduction Approximation of conic section by Bézier curve is an important task in CAGD or CAD/CAM. In the recent twenty years, many works on the approximation of conic sections including circular arcs and quadric surfaces by Bézier curves or splines with high order approximation have been developed. The approximations of circular arcs by Bézier curves were presented in [1,2]. Ahn [3] provided two approximation methods of conic sections by quartic Bézier curves with G1 or G2 continuity at the end points. Fang proposed the approximations of conic sections by quintic polynomial curves in [4,5]. Floater constructed a quadratic spline having optimal convergence order 4 to approximate a conic section in [6] and found an approximation of a conic section by a Bézier curve of any odd degree n having optimal approximation order 2n in [7]. In [8], the approximation of the rational curve by Bézier curve with optimal approximation order 2n was presented. In [9], approximation of circular arcs by cubic polynomials was presented. Hu proposed algorithms for approximating conic sections by constrained Bézier curves in [10,11]. In [12], the cubic and quartic Bézier approximations of circular arcs that respectively satisfy G1 and G2 endpoint interpolation conditions were presented. The necessary and sufficient conditions for such approximations to be the best were identified. Kim and Ahn presented approximation of circular arcs by quartic Bézier curves in [13]. On polynomial approximation of circular arcs and helices was discussed in [14]. The present authors were mainly interested in the approximation orders of conic section. Few optimal upper bound or uniform approximations were presented. Generally, it is difficult to obtain a uniform approximation by high degree polynomial. Based on the Hausdorff distance, the purpose of this paper is to present the optimal parameter values for approximating conic sections by the quartic Bézier curves in [3,11]. Thus, the given approximation method has the smallest error bound than the previous approximation methods. The remainder of this paper is organized as follows. In Section 2, based on the Hausdorff distance, the optimal parameter values for approximating conic sections are presented. The proofs for the results of the optimal parameters are provided in Section 3.
∗
Corresponding author. E-mail addresses: [email protected] (X. Han), [email protected] (X. Guo).
http://dx.doi.org/10.1016/j.cam.2017.03.029 0377-0427/© 2017 Elsevier B.V. All rights reserved.
X. Han, X. Guo / Journal of Computational and Applied Mathematics 322 (2017) 86–95
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2. Quartic Bézier approximation of conic section A conic section is represented in the standard rational quadratic Bézier form as R(t ) =
(1 − t )2 P0 + 2w(1 − t )tP1 + t 2 P2 , (1 − t )2 + 2w(1 − t )t + t 2
t ∈ [0, 1],
where P0 , P1 , P2 ∈ R3 are the control points and w > 0 is the weight associated with P1 . It is well known that R(t ) is an ellipse when 0 < w < 1, a parabola when w = 1 and a hyperbola when w > 1, see [15]. Any point P ∈ △P0 P1 P2 can be written uniquely in terms of barycentric coordinates τ0 , τ1 , τ2 , where τ0 + τ1 + τ2 = 1, with respect to the triangle △P0 P1 P2 : P = τ0 P0 + τ1 P1 + τ2 P2 . Consequently any function on △P0 P1 P2 can be expressed as a function of τ0 , τ1 , τ2 . Let f : △P0 P1 P2 → R be a function given by f (P) = τ12 − 4w 2 τ0 τ2 .
(1)
It is well known that for t ∈ [0, 1], the curve R(t ) satisfies the equation f (R(t )) = 0 for all t. For any planar curve C(t ) inside △P0 P1 P2 , Floater presented a sharp analysis for the Hausdorff distance dH (C, R) between the curve C(t ) and the conic R(t ) as follows [7]. Lemma 1. Suppose that C(t ), t ∈ [0, 1], is any continuous curve which lies entirely inside the (closed) triangle △P0 P1 P2 and such that C(0) = P0 and C(1) = P2 . Then dH (C, R) ≤
1 4
max
1
w2
,1
max |f (C(t ))| |P0 − 2P1 + P2 |,
t ∈[0,1]
where dH (C, R) is the Hausdorff distance defined by
dH (C, R) = max max min |C(s) − R(t )|, max min |C(s) − R(t )| . t ∈[0,1] s∈[0,1]
s∈[0,1] t ∈[0,1]
Let Q(t ) be a quartic Bézier approximation curve of the conic R(t ), Q(t ) =
4
Qi Bi,4 (t ),
t ∈ [0, 1],
(2)
i =0
where Qi , 0 ≤ i ≤ 4, are the control points, and Bi,4 (t ) are the Bernstein polynomials of degree 4. In the papers [3,11], the control points of Q(t ) satisfy Q0 = P0 ,
Q1 = (1 − α)P0 + α P1 ,
1
(1 − β)(P0 + P2 ) + β P1 , 2 Q3 = α P1 + (1 − α)P2 , Q4 = P2 . Q2 =
For 0 < α, β < 1, the quartic Bézier curve Q(t ) is a G1 end-point interpolation of the conic R(t ) and contained in △P0 P1 P2 . When w = 1, we have Q(t ) = R(t ) with α = 1/2 and β = 2/3. Therefore, we consider w ̸= 1. Substituting (2) into (1) yields [3,11]
f (Q(t )) = 4(1 − t )2 t 2 (1 − w 2 )(4α − 3β)2 (t − 0.5)4 + 8(1 + w 2 )α 2 +
+
1
9 2
(w 2 − 1)β 2 − 16w2 α + 4w 2 (t − 0.5)2
16
[(4α + 3β)(1 + w) − 8w][(4α + 3β)(1 − w) + 8w] .
From this we can see that f (Q(t )) has zeros at t = 0, 1 of order two, and f (Q(t )) has zero at t = 0.5 of order at least two if and only if
β=
8w
4
3(w ± 1)
− α. 3
In order to avoid that β goes to infinity as w approaches 1, we take
β=
8w
4
3(w + 1)
− α. 3
(3)
Thus we obtain f (Q(t )) =
16
(1 − t )2 t 2 (t − 0.5)2 16(1 − w)[(w + 1)α − w]2 (t − 0.5)2 (w + 1) + 4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7) .
(4)
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If t ∗ is a zero point of f (Q(t )), then 1 − t ∗ is also a zero point of f (Q(t )). The methods presented in [3] are that f (Q(t )) has factor (1 − t )3 (t − 0.5)2 t 3 or (1 − t )2 (t − 0.5)4 t 2 . The method given in [11] is that f (Q(t )) has factor (1 − t )2 (t − 0.25)(t − 0.5)2 (t − 0.75)t 2 . The quartic Bézier approximations of the conic section in [3] are the extensions of the quartic Bézier approximations of the circular presented in [1,13]. The quartic Bézier approximation of the conic section in [11] is an extension to the conic case of the quartic Bézier approximation of a circular arc proposed in [16]. It was shown that the approximation method given in [11] has a smaller error bound than the approximation method given in [3]. The values of α and β are obtained by appointing the zeros of f (Q(t )) in [3,11]. Different zero points of f (Q(t )) correspond to the different α , and then maxt ∈[0,1] |f (Q(t ))| depends on α . Thus there is a minimax problem min max |f (Q(t ))|.
(5)
α∈[0,1] t ∈[0,1]
Since the zero point t ∗ ̸∈ [0, 1] for some α and w , we need consider not only t ∗ ∈ [0, 1] but also t ∗ ̸∈ [0, 1]. The discussion on maxt ∈[0,1] |f (Q(t ))| is also influenced by the parameter w . By minimizing maxt ∈[0,1] |f (Q(t ))|, we give the optimal values of α and β for the quartic Bézier approximation curve. Theorem 1. The solution of (5) is
α=
√ w (w + 1)[(3 − 2λ)w + 2λ − 2] − (w − 1) λw2 + 4w + 4 − λ 2(w + 1)[(1 − λ)w 2 + λ]
,
(6)
where
λ=
2 1+
√ ≈ 0.362596.
(7)
10 + 6 3
By (3) and (6), we have
√
β=
2w (1 − 2λ)w 2 − w + 2λ + 2 + (w − 1) λw 2 + 4w + 4 − λ 3(w + 1)[(1 − λ)w 2 + λ]
.
(8)
Theorem 2. The values (6) and (8) satisfy α > 0, 0 < β < 1 for all w > 0 and α < 1 for 0 < w < w0 ,
w0 =
1 3
7 + 72λ + 91 + i (12λ + 25)3 − (72λ + 91)2
1/3
1/3 + 72λ + 91 − i (12λ + 25)3 − (72λ + 91)2 ≈ 5.834912,
(9)
where i is the complex unit and the complex number [·]1/3 means the principal value. Theorems 1 and 2 will be proved by delicate reasoning. We will give the proofs in next section. With (6), for (4) we will have 16(1 − w)3 w 2
2 2 2 2 (10) 2 (1 − t ) (t − 0.5) t [(2t − 1) − λ]. √ (w + 1) w + 2 + λw2 + 4w + 4 − λ √ √ Therefore f (Q(t )) has other zeros at t = (1 − λ)/2 ≈ 0.198920, t = (1 + λ)/2 ≈ 0.801080. These are different from
f (Q(t )) =
the methods in [3,11]. Under the conditions described in Theorem 2, the quartic Bézier approximation curve Q(t ) lies entirely inside the triangle △P0 P1 P2 . Here the upper bound value w0 of w is a little larger than the upper bound value 5.753038 in [11]. With (6), we will obtain min max |f (Q(t ))| =
α∈[0,1] t ∈[0,1]
(2 − λ)(9λ2 − 4λ + 4)3/2 |1 − w|3 w2 2 . √ 211 (w + 1) w + 2 + λw 2 + 4w + 4 − λ
Let
ϕ(w) =
1 4
max
1
w
,1 2
min max |f (Q(t ))|,
α∈[0,1] t ∈[0,1]
with the optimal values of α and β , for curve (2) we have dH (Q, R) ≤ ϕ(w)|P0 − 2P1 + P2 |.
(11)
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Fig. 1. The curves of the error bound function and the corresponding functions in [3,11].
Fig. 2. The conic section and the quartic Bézier approximation curve with w = 0.5.
Fig. 1 shows the curves (solid lines) of the error bound function ϕ(w). For the numerical comparison, the curves (dash dot lines) corresponding to bβ2 (t ) in [3] and the curves (dashed lines) corresponding to Q2 (t ) in [11] are shown in the figure. For 0 ≤ w ≤ 1, the curves are shown on the left. For w ≥ 1, the curves are shown on the right. Clearly, the presented approximation method has a smaller error bound than the previous approximation methods. For the examples in [3,11], the conic section R(t ) is given with the control points P0 = (0, 0), P1 = (120, 150), and P2 = (100, 0). With w = 0.5, Fig. 2 shows the conic section on the left and the quartic Bézier approximation curve on the right. The quartic Bézier approximation curve Q(t ) has control points Q0 = (0, 0), Q1 = (37.9564, 47.4455), Q2 = (82.7006, 70.0727), Q3 = (106.3261, 47.4455), Q4 = (100, 0). The Hausdorff error bounds of the curve Q(t ), the curve Q2 (t ) in [11] and the curve bβ2 (t ) in [3] are dH (Q, R) ≤ 1.660922 × 10−3 ,
dH (Q2 , R) ≤ 2.413908 × 10−3 ,
dH (bβ2 , R) ≤ 4.399094 × 10−3
respectively. For the same control points, with w = 3, Fig. 3 shows the conic section on the left and the quartic Bézier approximation curve on the right. The quartic Bézier approximation curve Q(t ) has control points Q0 = (0, 0), Q1 = (99.6282, 124.5353), Q2 = (112.5114, 133.9529), Q3 = (116.6047, 124.5353), Q4 = (100, 0). The Hausdorff error bounds of the curve Q(t ), the curve Q2 (t ) in [11] and the curve bβ2 (t ) in [3] are dH (Q, R) ≤ 0.098315,
dH (Q2 , R) ≤ 0.147200,
dH (bβ2 , R) ≤ 0.287378
respectively. As stated as in [3,11], from (11) we can know that the approximation order of the approximation curve (2) with (6) and (8) is eight. If the bound on the Hausdorff error dH (Q, P) is larger than a user-specified error tolerance, the subdivision scheme can be considered for Q(t ), as stated as in [3,7,11].
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Fig. 3. The conic section and the quartic Bézier approximation curve with w = 3.
3. The proofs of Theorems 1 and 2 3.1. The properties of |f (Q(t ))| Let u = (2t − 1)2 , then u ∈ [0, 1], (1 − t )t = (1 − u)/4, and (4) is simplified to f (Q(t )) =
( 1 − u) 2 u 4(1 − w)[(w + 1)α − w]2 u + 4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7) . 4(w + 1)
When α = w/(w + 1), we have f (Q(t )) =
w 2 (w − 1)2 w 2 (w − 1)2 2 ( 1 − u ) u ≤ . 4(w + 1)2 27(w + 1)2
This result does not provide the optimal value of |f (Q(t ))| comparing with (11). Hence we set α ̸= w/(w + 1) in the following. For w ̸= 1, let a = 4(1 − w)[(w + 1)α − w]2 , 1
λ = − [4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7)], a
then f (Q(t )) =
a 4(w + 1)
(1 − u)2 u(u − λ).
From 4(1 − w)[(w + 1)α − w]2 λ + 4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7) = 0 we obtain
α=
√ w (w + 1)[(3 − 2λ)w + 2λ − 2] ± (w − 1) w 2 λ − λ + 4w + 4 2(w + 1)[(1 − λ)w 2 + λ]
,
and then a=
2 (1 − w)3 w2 2 λ − λ + 4w + 4 w + 2 ± w . [(1 − λ)w2 + λ]2
For any given λ, there are two values of a. In order to minimize |a| in |f (Q(t ))|, we set
α=
√ w (w + 1)[(3 − 2λ)w + 2λ − 2] − (w − 1) w 2 λ − λ + 4w + 4 2(w + 1)[(1 − λ)w 2 + λ]
,
(12)
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and then
2 (1 − w)3 w 2 w + 2 − w2 λ − λ + 4w + 4 2 2 [(1 − λ)w + λ] (1 − w)3 w2 = 2 √ w + 2 + w 2 λ − λ + 4w + 4
a=
provides a less |a|. From this a we have f (Q(t )) = F (u, λ) :=
(1 − w)3 w2 (1 − u)2 u(u − λ) 2 . √ 4(w + 1) w + 2 + w 2 λ − λ + 4w + 4
(13)
For (5), now we need to consider the minimax problem min max |F (u, λ)|. λ
(14)
u∈[0,1]
Here we consider not only λ ∈ [0, 1] but also λ ∈ R. Let d du
[(1 − u)2 u(u − λ)] = (u − 1)[4u2 − (3λ + 2)u + λ] = 0,
we have the extreme points u1 (λ) =
1 8
(3λ + 2 −
9λ2 − 4λ + 4),
u2 (λ) =
1 8
(3λ + 2 +
9λ2 − 4λ + 4).
(15)
Let A1 (λ) = 16 − 32λ − 56λ2 + 72λ3 − 27λ4 + (λ − 2)(9λ2 − 4λ + 4)3/2 , A2 (λ) = 16 − 32λ − 56λ2 + 72λ3 − 27λ4 − (λ − 2)(9λ2 − 4λ + 4)3/2 ,
B(λ) = w + 2 +
2 w 2 λ − λ + 4w + 4 ,
then the nonzero extreme values of F (u, λ) are F (u1 , λ) =
(1 − w)3 w 2 A1 (λ) , 211 (w + 1)B(λ)
F (u2 , λ) =
(1 − w)3 w 2 A2 (λ) . 211 (w + 1)B(λ)
(16)
In order to obtain the optimal values of λ and F (u, λ), we need discuss the extreme values F (u1 , λ) and F (u2 , λ) for the cases λ ≤ 0, 0 ≤ λ ≤ 1 and λ ≥ 1. Lemma 2. A1 (λ) ≤ 0. When λ ≤ 0, A′1 (λ) ≥ 0. When λ ≥ 0, A′1 (λ) ≤ 0. A′2 (λ) ≤ 0. When λ ≤ 1, A2 (λ) ≥ 0. When λ ≥ 1, A2 (λ) ≤ 0. Proof. We have A′1 (λ) = 4(b0 − a0 ),
A′2 (λ) = −4(a0 + b0 ),
where a0 = 27λ3 − 54λ2 + 28λ + 8,
b0 = (9λ2 − 16λ + 4) 9λ2 − 4λ + 4.
A straightforward computation gives that a20 − b20 = 1024λ(λ − 1)2 . When λ ≤ 0, |a0 | ≤ |b0 |. Since b0 ≥ 0, we have A′1 (λ) ≥ 0,
A′2 (λ) ≤ 0.
When λ ≥ 0, |a0 | ≥ |b0 |. Since
√
a0 ≥ 6( 84 − 9)λ2 + 8 > 0, we have A′1 (λ) ≤ 0, A′2 (λ) ≤ 0. Since A1 (0) = 0, A2 (1) = 0, we have A1 (λ) ≤ 0, A2 (λ) ≥ 0 for λ ≤ 1, and A2 (λ) ≤ 0 for λ ≥ 1.
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Lemma 3. For λ ≥ 0, F (u1 , λ) is decreasing when 0 < w < 1 and F (u1 , λ) is increasing when w > 1. Proof. We have
√ (w 2 − 1)[w + 2 + w2 λ − λ + 4w + 4] . √ w 2 λ − λ + 4w + 4 When 0 < w < 1, we have A1 (λ) ≤ 0, A′1 (λ) ≤ 0, B(λ) ≥ 0, B′ (λ) ≤ 0. Therefore, A′1 (λ)B(λ) − A1 (λ)B′ (λ) ≤ 0 and then F (u1 , λ) is decreasing. When w > 1, we have B(λ) > (w 2 − 1)λ and then (w2 λ − λ + 4w + 4)B(λ) > (w2 − 1)λ. B′ (λ) =
Therefore,
√ A1 (λ)B(λ) − A1 (λ)B (λ) = √ ′
′
B(λ)
(w 2 λ − λ + 4w + 4)B(λ)A′1 (λ) − (w 2 − 1)A1 (λ)
w 2 λ − λ + 4w + 4 √ B(λ) (w 2 − 1)(λA′1 (λ) − A1 (λ)) ≤ √ 2 w λ − λ + 4w + 4 = B′ (λ)(λA′1 (λ) − A1 (λ)).
It is easy to get
λA′1 (λ) − A1 (λ) = b1 − a1 where
b1 = (27λ3 − 42λ2 + 4λ + 8) 9λ2 − 4λ + 4.
a1 = 81λ4 − 144λ3 + 56λ2 + 16,
The zero points of a′1 = 4λ(81λ2 − 108λ + 28) are λ = 0, 2(3 ±
√
√
2)/9. a1 (0) = 16 > 0. When λ = 2(3 +
√
2)/9,
a1 > 81λ − 128λ + 56λ ≥ 2(9 56 − 64)λ > 0. 4
3
2
3
Therefore a1 > 0 for λ ≥ 0. Since a21 − b21 = 4096λ2 (λ − 1)2 ≥ 0, we have |a1 | ≥ |b1 | and then λA′1 (λ) − A1 (λ) ≤ 0. Therefore, F (u1 , λ) is increasing when w > 1. Lemma 4. For λ ≤ 1, F (u2 , λ) is decreasing when 0 < w < 1 and F (u2 , λ) is increasing when w > 1. Proof. When w > 1, we have A2 (λ) ≥ 0, A′2 (λ) ≤ 0, B(λ) ≥ 0, B′ (λ) ≥ 0. Therefore, A′2 (λ)B(λ) − A2 (λ)B′ (λ) ≤ 0 and then F (u2 , λ) is increasing. When 0 ≤ w ≤ 1, we prove the lemma by two cases. (1) The case for λ ≤ 0. Since
(w2 λ − λ + 4w + 4)B(λ) > (2 +
w 2 λ − λ + 4) w 2 λ − λ + 4
≥ 4 + (w 2 − 1)λ + 4, we have A′2 (λ)B(λ) − A2 (λ)B′ (λ) = √
√
B(λ)
(w 2 λ − λ + 4w + 4)B(λ)A′2 (λ) − (w 2 − 1)A2 (λ)
w 2 λ − λ + 4w + 4 √ 2 B(λ) ≤ √ (w λ − λ + 8)A′2 (λ) + (1 − w2 )A2 (λ) 2 w λ − λ + 4w + 4 √ ′ B(λ) = √ [λA2 (λ) − A2 (λ)]w 2 + (8 − λ)A′2 (λ) + A2 (λ) . 2 w λ − λ + 4w + 4 Since a1 > 0 for λ ≤ 0 and |a1 | ≥ |b1 |, we have λA′2 (λ) − A2 (λ) = −b1 − a1 ≤ 0. It is easy to know
(8 − λ)A′2 (λ) + A2 (λ) = a2 + b2 , where a2 = 81λ4 − 1008λ3 + 1784λ2 − 896λ − 240,
b2 = (27λ3 − 330λ2 + 516λ − 120) 9λ2 − 4λ + 4 < 0, b22 − a22 = 4096λ(λ − 1)2 (23λ − 240) ≥ 0. Thus (8 − λ)A′2 (λ) + A2 (λ) ≤ 0 and then A′2 (λ)B(λ) − A2 (λ)B′ (λ) ≤ 0. Therefore F (u2 , λ) is decreasing.
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(2) The case for 0 ≤ λ ≤ 1. Since
(w 2 λ − λ + 4w + 4)B(λ) ≥ w + 2 + w 2 + 4w + 3 w2 + 4w + 3 > 2(w 2 + 4w + 3) > 6,
we have
√ A2 (λ)B(λ) − A2 (λ)B (λ) ≤ √ ′
B(λ)
′
w λ − λ + 4w + 4 2
[6A′2 (λ) + A2 (λ)].
It is easy to know 6A′2 (λ) + A2 (λ) = −a3 − b3 , where a3 = 27λ4 + 576λ3 − 1240λ2 + 704λ + 176,
b3 = (9λ3 + 194λ2 − 372λ + 88) 9λ2 − 4λ + 4. Since a23 − b23 = 4096λ(λ − 1)2 (λ2 + 29λ + 132) ≥ 0,
√
a3 ≥ 27λ4 + (384 11 − 1240)λ2 + 176 > 0, we have 6A′2 (λ) + A2 (λ) ≤ 0. Therefore A′2 (λ)B(λ) − A2 (λ)B′ (λ) ≤ 0 and then F (u2 , λ) is decreasing.
3.2. The proof of Theorem 1 For λ ≤ 0, from (15) we have λ ≤ u1 ≤ 0 and 0 < u2 < 1. Therefore, by Lemma 2 max |F (u, λ)| =
u∈[0,1]
F (u2 , λ), −F (u2 , λ),
0 < w < 1, w > 1.
By Lemma 4, we have max |F (u, λ)| ≥ |F (u2 , 0)| =
u∈[0,1]
F (u2 , 0), −F (u2 , 0),
0 < w < 1, w > 1.
For λ ≥ 1, from (15) we have 0 < u1 < 1 and 1 ≤ u2 ≤ λ. Therefore, by Lemma 2 max |F (u, λ)| =
u∈[0,1]
−F (u1 , λ), F (u1 , λ),
0 < w < 1, w > 1.
By Lemma 3, we have max |F (u, λ)| ≥ |F (u1 , 1)| =
u∈[0,1]
−F (u1 , 1), F (u1 , 1),
0 < w < 1, w > 1.
As stated as above, for (14) we have min max |F (u, λ)| = min max |F (u, λ)|. λ∈R u∈[0,1]
λ∈[0,1] u∈[0,1]
(17)
For 0 ≤ λ ≤ 1, from (15) we have 0 ≤ u1 ≤ λ ≤ u2 ≤ 1. Therefore max |F (u, λ)| = max {|F (u1 , λ)|, |F (u2 , λ)|} .
u∈[0,1]
By Lemma 2, F (u1 , λ) and F (u2 , λ) are the negative and the positive amount of deviation of F (u, λ) respectively when 0 < w < 1, and F (u1 , λ) and F (u2 , λ) are the positive and the negative amount of deviation of F (u, λ) respectively when w > 1. By Lemmas 3 and 4, F (u1 , λ) and F (u2 , λ) are decreasing when 0 < w < 1, and F (u1 , λ) and F (u2 , λ) are increasing when w > 1. Thus the negative amount of deviation and the positive amount of deviation of F (u, λ) have the same monotonicity. Therefore, based on these properties of amount of deviation, the solution of minimax problem (17) must be obtained from F (u2 , λ) = −F (u1 , λ). From this and (16) we have 27λ4 − 72λ3 + 56λ2 + 32λ − 16 = 0.
(18)
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It can be shown that
√
27λ4 − 72λ3 + 56λ2 + 32λ − 16 = (6 3 − 9)λ2 − 4λ + 4
√ (6 3 − 9)λ2 + 4λ − 4 .
√
√
The equation (6 3 − 9)λ2 − 4λ + 4 = 0 has not a real root. The equation (6 3 − 9)λ2 + 4λ − 4 = 0 has two real roots:
λ¯ =
2 1−
√ < 0,
10 + 6 3
λ∗ =
2 1+
√ .
10 + 6 3
¯ is not the solution. Thus λ∗ is the optimal solution of (17) and we obtain (7). By (12), with λ = λ∗ , we Obviously, λ obtain (6). With λ = λ∗ , we obtain (10) from (13), and obtain (11) from (16) and (18). 3.3. The proof on the value α in Theorem 2
√ √ λw2 − λ + 4w + 4 ≥ 4 − λ > 1.9. Therefore (3 − 2λ)w 2 + w + 2λ − 2 − (w − 1) λw2 − λ + 4w + 4 > (3 − 2λ)w2 + w + 2λ − 2 − 1.9(w − 1)
When 0 < w < 1, we have
= (3 − 2λ)w2 − 0.9w + 2λ − 0.1 > 0. This means α > 0. When w > 1, since (λ − 1)2 w 2 + (10λ − 6)w + 4λ + 9 > 0, we have 4(λw 2 − λ + 4w + 4) < (λ + 1)2 w 2 + 10(λ + 1)w + 25 and then
2 λw 2 − λ + 4w + 4 < (λ + 1)w + 5. Therefore
(3 − 2λ)w 2 + w + 2λ − 2 − (w − 1) λw2 − λ + 4w + 4 1
> (3 − 2λ)w2 + w + 2λ − 2 − (w − 1)[(λ + 1)w + 5] 2
= >
1 2 1 2
5(1 − λ)w + (λ − 2)w + 1 2
[5(1 − λ)w + (λ − 2)w + 1] > 0.
This means α > 0. Now we prove α < 1 for 0 < w < w0 . By (6),
√ w 3 + (2λ − 1)w 2 − 2w − 2λ − w(w − 1) λw2 − λ + 4w + 4 α−1= . 2(w + 1)[(1 − λ)w 2 + λ] We have
w 3 + (2λ − 1)w 2 − 2w − 2λ − w(w − 1) λw2 − λ + 4w + 4 × w 3 + (2λ − 1)w 2 − 2w − 2λ + w(w − 1) λw2 − λ + 4w + 4 = (1 − λ)w 6 + 6(λ − 1)w 5 + (2w − 1)2 w 4 + (8 − 14λ)w3 + (5λ − 8λ2 )w 2 + 8λw + 4λ2 = (w + 1)[(1 − λ)w 2 + λ][w 3 − 7w 2 + (8 − 4λ)w + 4λ]. Therefore, the positive roots of the equation α − 1 = 0 must be the roots of the equation
w3 − 7w2 + (8 − 4λ)w + 4λ = 0.
(19)
This equation has three real roots w2 ≈ −0.2, w1 ≈ 1.3 and w0 ≈ 5.8. The root w0 is the unique positive root of the equation α − 1 = 0. Therefore α < 1 for 0 < w < w0 . From Eq. (19) we obtain (9).
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3.4. The proof on the value β in Theorem 2 (1) The proof of β > 0. When 0 < w < 1, since
λw2 − λ + 4w + 4 <
w 2 + 4w + 4 = w + 2,
we have
(1 − w) λw2 − λ + 4w + 4 ≤ (1 − w)(w + 2) < (1 − 2λ)w 2 − w + 2λ + 2. This means β > 0. When w > 1, since (1 − 2λ)w 2 − w + 2λ + 2 > 0, we have
(1 − 2λ)w2 − w + 2λ + 2 > (1 − w) λw2 − λ + 4w + 4 and then β > 0. (2) The proof of β < 1. √ √ When 0 < w < 1, since λw 2 − λ + 4w + 4 > λw > w/2, we have
2w(w − 1) λw 2 − λ + 4w + 4 ≤ w 2 (w − 1). Since λw 3 + 3(2 − λ)w 2 − (λ + 4)w + 3λ > λw 3 > 0, we have
(λ + 1)w 3 + (5 − 3λ)w2 − (λ + 4)w + 3λ > w2 (w − 1) ≥ 2w(w − 1) λw2 − λ + 4w + 4. This means β < 1. When w > 1, since (1 − 2λ)w 2 + (1 − λ)w + 3λ > 0, we have
(λ + 1)w 3 + (5 − 3λ)w2 − (λ + 4)w + 3λ > w(w − 1)[(λ + 1)w + 5] ≥ 2w(w − 1) λw2 − λ + 4w + 4. This means β < 1.
Acknowledgments The authors would like to thank the referees for their valuable comments and suggestions. The research is supported by the National Natural Science Foundation of China (Nos. 11271376, 61375063). References [1] Y.J. Ahn, H.O. Kim, Approximation of circular arcs by Bézier curves, J. Comput. Appl. Math. 81 (1997) 145–163. [2] Y.J. Ahn, Y.S. Kim, Y.S. Shin, Approximation of circular arcs and offset curves by Bézier curves of high degree, J. Comput. Appl. Math. 167 (2004) 405–416. [3] Y.J. Ahn, Approximation of conic sections by curvature continuous quartic Bézier curves, Comput. Math. Appl. 60 (7) (2010) 1986–1993. [4] L. Fang, Circular arc approximation by quintic polynomial curves, Comput. Aided Geom. Design 15 (1998) 843–861. [5] L. Fang, G3 approximation of conic sections by quintic polynomial curves, Comput. Aided Geom. Design 16 (1999) 755–766. [6] M.S. Floater, High order approximation of conic sections by quadratic splines, Comput. Aided Geom. Design 12 (1995) 617–637. [7] M.S. Floater, An O(h2n ) Hermite approximation for conic sections, Comput. Aided Geom. Design 14 (1997) 135–151. [8] M.S. Floater, High order approximation of rational curves by polynomial curves, Comput. Aided Geom. Design 23 (2006) 621–628. [9] M. Goldapp, Approximation of circular arcs by cubic polynomials, Comput. Aided Geom. Design 8 (1991) 227–238. [10] Q.Q. Hu, Approximating conic sections by constrained Bézier curves of arbitrary degree, J. Comput. Appl. Math. 236 (2012) 2813–2821. [11] Q.Q. Hu, G1 approximation of conic sections by quatic Bézier curves, Comput. Math. Appl. 68 (2014) 1882–1891. [12] S. Hur, T.W. Kim, The best G1 cubic and G2 quartic Bézier approximations of circular arcs, J. Comput. Appl. Math. 236 (2011) 1183–1192. [13] S.H. Kim, Y.J. Ahn, Approximation of circular arcs by quartic Bézier curves, Comput. Aided Geom. Design 39 (2007) 490–493. [14] L.Z. Lu, On polynomial approximation of circular arcs and helices, Comput. Math. Appl. 63 (2012) 1192–1196. [15] G. Farin, Curves and Surfaces for Computer Aided Geometric Design, Morgan-Kaufmann, San Francisco, 2002. [16] Z. Liu, J.Q. Tan, X.Y. Chen, L. Zhang, An approximation method to circular arcs, Appl. Math. Comput. 219 (2012) 1306–1311.
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Optimal parameter values for approximating conic sections by the quartic Bézier curves Xuli Han ∗ , Xiao Guo School of Mathematics and Statistics, Central South University, Changsha, 410083, PR China
article
info
Article history: Received 20 June 2016 Received in revised form 24 March 2017 Keywords: Conic section Quartic Bézier curves Hausdorff distance Optimal parameters
abstract The previous approximation curves of conic section by quartic Bézier curves interpolate the conic section at the specified parameter values. In this paper, by solving the minimax problem, we present the optimal parameter values for approximating conic sections by the quartic Bézier curves. The upper bound on the Hausdorff distance between the conic section and the approximation curves is minimized. The method of solving the minimax problem is changed to solve a quartic algebraic equation by delicate reasoning. © 2017 Elsevier B.V. All rights reserved.
1. Introduction Approximation of conic section by Bézier curve is an important task in CAGD or CAD/CAM. In the recent twenty years, many works on the approximation of conic sections including circular arcs and quadric surfaces by Bézier curves or splines with high order approximation have been developed. The approximations of circular arcs by Bézier curves were presented in [1,2]. Ahn [3] provided two approximation methods of conic sections by quartic Bézier curves with G1 or G2 continuity at the end points. Fang proposed the approximations of conic sections by quintic polynomial curves in [4,5]. Floater constructed a quadratic spline having optimal convergence order 4 to approximate a conic section in [6] and found an approximation of a conic section by a Bézier curve of any odd degree n having optimal approximation order 2n in [7]. In [8], the approximation of the rational curve by Bézier curve with optimal approximation order 2n was presented. In [9], approximation of circular arcs by cubic polynomials was presented. Hu proposed algorithms for approximating conic sections by constrained Bézier curves in [10,11]. In [12], the cubic and quartic Bézier approximations of circular arcs that respectively satisfy G1 and G2 endpoint interpolation conditions were presented. The necessary and sufficient conditions for such approximations to be the best were identified. Kim and Ahn presented approximation of circular arcs by quartic Bézier curves in [13]. On polynomial approximation of circular arcs and helices was discussed in [14]. The present authors were mainly interested in the approximation orders of conic section. Few optimal upper bound or uniform approximations were presented. Generally, it is difficult to obtain a uniform approximation by high degree polynomial. Based on the Hausdorff distance, the purpose of this paper is to present the optimal parameter values for approximating conic sections by the quartic Bézier curves in [3,11]. Thus, the given approximation method has the smallest error bound than the previous approximation methods. The remainder of this paper is organized as follows. In Section 2, based on the Hausdorff distance, the optimal parameter values for approximating conic sections are presented. The proofs for the results of the optimal parameters are provided in Section 3.
∗
Corresponding author. E-mail addresses: [email protected] (X. Han), [email protected] (X. Guo).
http://dx.doi.org/10.1016/j.cam.2017.03.029 0377-0427/© 2017 Elsevier B.V. All rights reserved.
X. Han, X. Guo / Journal of Computational and Applied Mathematics 322 (2017) 86–95
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2. Quartic Bézier approximation of conic section A conic section is represented in the standard rational quadratic Bézier form as R(t ) =
(1 − t )2 P0 + 2w(1 − t )tP1 + t 2 P2 , (1 − t )2 + 2w(1 − t )t + t 2
t ∈ [0, 1],
where P0 , P1 , P2 ∈ R3 are the control points and w > 0 is the weight associated with P1 . It is well known that R(t ) is an ellipse when 0 < w < 1, a parabola when w = 1 and a hyperbola when w > 1, see [15]. Any point P ∈ △P0 P1 P2 can be written uniquely in terms of barycentric coordinates τ0 , τ1 , τ2 , where τ0 + τ1 + τ2 = 1, with respect to the triangle △P0 P1 P2 : P = τ0 P0 + τ1 P1 + τ2 P2 . Consequently any function on △P0 P1 P2 can be expressed as a function of τ0 , τ1 , τ2 . Let f : △P0 P1 P2 → R be a function given by f (P) = τ12 − 4w 2 τ0 τ2 .
(1)
It is well known that for t ∈ [0, 1], the curve R(t ) satisfies the equation f (R(t )) = 0 for all t. For any planar curve C(t ) inside △P0 P1 P2 , Floater presented a sharp analysis for the Hausdorff distance dH (C, R) between the curve C(t ) and the conic R(t ) as follows [7]. Lemma 1. Suppose that C(t ), t ∈ [0, 1], is any continuous curve which lies entirely inside the (closed) triangle △P0 P1 P2 and such that C(0) = P0 and C(1) = P2 . Then dH (C, R) ≤
1 4
max
1
w2
,1
max |f (C(t ))| |P0 − 2P1 + P2 |,
t ∈[0,1]
where dH (C, R) is the Hausdorff distance defined by
dH (C, R) = max max min |C(s) − R(t )|, max min |C(s) − R(t )| . t ∈[0,1] s∈[0,1]
s∈[0,1] t ∈[0,1]
Let Q(t ) be a quartic Bézier approximation curve of the conic R(t ), Q(t ) =
4
Qi Bi,4 (t ),
t ∈ [0, 1],
(2)
i =0
where Qi , 0 ≤ i ≤ 4, are the control points, and Bi,4 (t ) are the Bernstein polynomials of degree 4. In the papers [3,11], the control points of Q(t ) satisfy Q0 = P0 ,
Q1 = (1 − α)P0 + α P1 ,
1
(1 − β)(P0 + P2 ) + β P1 , 2 Q3 = α P1 + (1 − α)P2 , Q4 = P2 . Q2 =
For 0 < α, β < 1, the quartic Bézier curve Q(t ) is a G1 end-point interpolation of the conic R(t ) and contained in △P0 P1 P2 . When w = 1, we have Q(t ) = R(t ) with α = 1/2 and β = 2/3. Therefore, we consider w ̸= 1. Substituting (2) into (1) yields [3,11]
f (Q(t )) = 4(1 − t )2 t 2 (1 − w 2 )(4α − 3β)2 (t − 0.5)4 + 8(1 + w 2 )α 2 +
+
1
9 2
(w 2 − 1)β 2 − 16w2 α + 4w 2 (t − 0.5)2
16
[(4α + 3β)(1 + w) − 8w][(4α + 3β)(1 − w) + 8w] .
From this we can see that f (Q(t )) has zeros at t = 0, 1 of order two, and f (Q(t )) has zero at t = 0.5 of order at least two if and only if
β=
8w
4
3(w ± 1)
− α. 3
In order to avoid that β goes to infinity as w approaches 1, we take
β=
8w
4
3(w + 1)
− α. 3
(3)
Thus we obtain f (Q(t )) =
16
(1 − t )2 t 2 (t − 0.5)2 16(1 − w)[(w + 1)α − w]2 (t − 0.5)2 (w + 1) + 4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7) .
(4)
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If t ∗ is a zero point of f (Q(t )), then 1 − t ∗ is also a zero point of f (Q(t )). The methods presented in [3] are that f (Q(t )) has factor (1 − t )3 (t − 0.5)2 t 3 or (1 − t )2 (t − 0.5)4 t 2 . The method given in [11] is that f (Q(t )) has factor (1 − t )2 (t − 0.25)(t − 0.5)2 (t − 0.75)t 2 . The quartic Bézier approximations of the conic section in [3] are the extensions of the quartic Bézier approximations of the circular presented in [1,13]. The quartic Bézier approximation of the conic section in [11] is an extension to the conic case of the quartic Bézier approximation of a circular arc proposed in [16]. It was shown that the approximation method given in [11] has a smaller error bound than the approximation method given in [3]. The values of α and β are obtained by appointing the zeros of f (Q(t )) in [3,11]. Different zero points of f (Q(t )) correspond to the different α , and then maxt ∈[0,1] |f (Q(t ))| depends on α . Thus there is a minimax problem min max |f (Q(t ))|.
(5)
α∈[0,1] t ∈[0,1]
Since the zero point t ∗ ̸∈ [0, 1] for some α and w , we need consider not only t ∗ ∈ [0, 1] but also t ∗ ̸∈ [0, 1]. The discussion on maxt ∈[0,1] |f (Q(t ))| is also influenced by the parameter w . By minimizing maxt ∈[0,1] |f (Q(t ))|, we give the optimal values of α and β for the quartic Bézier approximation curve. Theorem 1. The solution of (5) is
α=
√ w (w + 1)[(3 − 2λ)w + 2λ − 2] − (w − 1) λw2 + 4w + 4 − λ 2(w + 1)[(1 − λ)w 2 + λ]
,
(6)
where
λ=
2 1+
√ ≈ 0.362596.
(7)
10 + 6 3
By (3) and (6), we have
√
β=
2w (1 − 2λ)w 2 − w + 2λ + 2 + (w − 1) λw 2 + 4w + 4 − λ 3(w + 1)[(1 − λ)w 2 + λ]
.
(8)
Theorem 2. The values (6) and (8) satisfy α > 0, 0 < β < 1 for all w > 0 and α < 1 for 0 < w < w0 ,
w0 =
1 3
7 + 72λ + 91 + i (12λ + 25)3 − (72λ + 91)2
1/3
1/3 + 72λ + 91 − i (12λ + 25)3 − (72λ + 91)2 ≈ 5.834912,
(9)
where i is the complex unit and the complex number [·]1/3 means the principal value. Theorems 1 and 2 will be proved by delicate reasoning. We will give the proofs in next section. With (6), for (4) we will have 16(1 − w)3 w 2
2 2 2 2 (10) 2 (1 − t ) (t − 0.5) t [(2t − 1) − λ]. √ (w + 1) w + 2 + λw2 + 4w + 4 − λ √ √ Therefore f (Q(t )) has other zeros at t = (1 − λ)/2 ≈ 0.198920, t = (1 + λ)/2 ≈ 0.801080. These are different from
f (Q(t )) =
the methods in [3,11]. Under the conditions described in Theorem 2, the quartic Bézier approximation curve Q(t ) lies entirely inside the triangle △P0 P1 P2 . Here the upper bound value w0 of w is a little larger than the upper bound value 5.753038 in [11]. With (6), we will obtain min max |f (Q(t ))| =
α∈[0,1] t ∈[0,1]
(2 − λ)(9λ2 − 4λ + 4)3/2 |1 − w|3 w2 2 . √ 211 (w + 1) w + 2 + λw 2 + 4w + 4 − λ
Let
ϕ(w) =
1 4
max
1
w
,1 2
min max |f (Q(t ))|,
α∈[0,1] t ∈[0,1]
with the optimal values of α and β , for curve (2) we have dH (Q, R) ≤ ϕ(w)|P0 − 2P1 + P2 |.
(11)
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Fig. 1. The curves of the error bound function and the corresponding functions in [3,11].
Fig. 2. The conic section and the quartic Bézier approximation curve with w = 0.5.
Fig. 1 shows the curves (solid lines) of the error bound function ϕ(w). For the numerical comparison, the curves (dash dot lines) corresponding to bβ2 (t ) in [3] and the curves (dashed lines) corresponding to Q2 (t ) in [11] are shown in the figure. For 0 ≤ w ≤ 1, the curves are shown on the left. For w ≥ 1, the curves are shown on the right. Clearly, the presented approximation method has a smaller error bound than the previous approximation methods. For the examples in [3,11], the conic section R(t ) is given with the control points P0 = (0, 0), P1 = (120, 150), and P2 = (100, 0). With w = 0.5, Fig. 2 shows the conic section on the left and the quartic Bézier approximation curve on the right. The quartic Bézier approximation curve Q(t ) has control points Q0 = (0, 0), Q1 = (37.9564, 47.4455), Q2 = (82.7006, 70.0727), Q3 = (106.3261, 47.4455), Q4 = (100, 0). The Hausdorff error bounds of the curve Q(t ), the curve Q2 (t ) in [11] and the curve bβ2 (t ) in [3] are dH (Q, R) ≤ 1.660922 × 10−3 ,
dH (Q2 , R) ≤ 2.413908 × 10−3 ,
dH (bβ2 , R) ≤ 4.399094 × 10−3
respectively. For the same control points, with w = 3, Fig. 3 shows the conic section on the left and the quartic Bézier approximation curve on the right. The quartic Bézier approximation curve Q(t ) has control points Q0 = (0, 0), Q1 = (99.6282, 124.5353), Q2 = (112.5114, 133.9529), Q3 = (116.6047, 124.5353), Q4 = (100, 0). The Hausdorff error bounds of the curve Q(t ), the curve Q2 (t ) in [11] and the curve bβ2 (t ) in [3] are dH (Q, R) ≤ 0.098315,
dH (Q2 , R) ≤ 0.147200,
dH (bβ2 , R) ≤ 0.287378
respectively. As stated as in [3,11], from (11) we can know that the approximation order of the approximation curve (2) with (6) and (8) is eight. If the bound on the Hausdorff error dH (Q, P) is larger than a user-specified error tolerance, the subdivision scheme can be considered for Q(t ), as stated as in [3,7,11].
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Fig. 3. The conic section and the quartic Bézier approximation curve with w = 3.
3. The proofs of Theorems 1 and 2 3.1. The properties of |f (Q(t ))| Let u = (2t − 1)2 , then u ∈ [0, 1], (1 − t )t = (1 − u)/4, and (4) is simplified to f (Q(t )) =
( 1 − u) 2 u 4(1 − w)[(w + 1)α − w]2 u + 4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7) . 4(w + 1)
When α = w/(w + 1), we have f (Q(t )) =
w 2 (w − 1)2 w 2 (w − 1)2 2 ( 1 − u ) u ≤ . 4(w + 1)2 27(w + 1)2
This result does not provide the optimal value of |f (Q(t ))| comparing with (11). Hence we set α ̸= w/(w + 1) in the following. For w ̸= 1, let a = 4(1 − w)[(w + 1)α − w]2 , 1
λ = − [4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7)], a
then f (Q(t )) =
a 4(w + 1)
(1 − u)2 u(u − λ).
From 4(1 − w)[(w + 1)α − w]2 λ + 4w 2 (w + 1)α 2 − 4w(w + 1)(3w − 2)α + w 2 (9w − 7) = 0 we obtain
α=
√ w (w + 1)[(3 − 2λ)w + 2λ − 2] ± (w − 1) w 2 λ − λ + 4w + 4 2(w + 1)[(1 − λ)w 2 + λ]
,
and then a=
2 (1 − w)3 w2 2 λ − λ + 4w + 4 w + 2 ± w . [(1 − λ)w2 + λ]2
For any given λ, there are two values of a. In order to minimize |a| in |f (Q(t ))|, we set
α=
√ w (w + 1)[(3 − 2λ)w + 2λ − 2] − (w − 1) w 2 λ − λ + 4w + 4 2(w + 1)[(1 − λ)w 2 + λ]
,
(12)
X. Han, X. Guo / Journal of Computational and Applied Mathematics 322 (2017) 86–95
91
and then
2 (1 − w)3 w 2 w + 2 − w2 λ − λ + 4w + 4 2 2 [(1 − λ)w + λ] (1 − w)3 w2 = 2 √ w + 2 + w 2 λ − λ + 4w + 4
a=
provides a less |a|. From this a we have f (Q(t )) = F (u, λ) :=
(1 − w)3 w2 (1 − u)2 u(u − λ) 2 . √ 4(w + 1) w + 2 + w 2 λ − λ + 4w + 4
(13)
For (5), now we need to consider the minimax problem min max |F (u, λ)|. λ
(14)
u∈[0,1]
Here we consider not only λ ∈ [0, 1] but also λ ∈ R. Let d du
[(1 − u)2 u(u − λ)] = (u − 1)[4u2 − (3λ + 2)u + λ] = 0,
we have the extreme points u1 (λ) =
1 8
(3λ + 2 −
9λ2 − 4λ + 4),
u2 (λ) =
1 8
(3λ + 2 +
9λ2 − 4λ + 4).
(15)
Let A1 (λ) = 16 − 32λ − 56λ2 + 72λ3 − 27λ4 + (λ − 2)(9λ2 − 4λ + 4)3/2 , A2 (λ) = 16 − 32λ − 56λ2 + 72λ3 − 27λ4 − (λ − 2)(9λ2 − 4λ + 4)3/2 ,
B(λ) = w + 2 +
2 w 2 λ − λ + 4w + 4 ,
then the nonzero extreme values of F (u, λ) are F (u1 , λ) =
(1 − w)3 w 2 A1 (λ) , 211 (w + 1)B(λ)
F (u2 , λ) =
(1 − w)3 w 2 A2 (λ) . 211 (w + 1)B(λ)
(16)
In order to obtain the optimal values of λ and F (u, λ), we need discuss the extreme values F (u1 , λ) and F (u2 , λ) for the cases λ ≤ 0, 0 ≤ λ ≤ 1 and λ ≥ 1. Lemma 2. A1 (λ) ≤ 0. When λ ≤ 0, A′1 (λ) ≥ 0. When λ ≥ 0, A′1 (λ) ≤ 0. A′2 (λ) ≤ 0. When λ ≤ 1, A2 (λ) ≥ 0. When λ ≥ 1, A2 (λ) ≤ 0. Proof. We have A′1 (λ) = 4(b0 − a0 ),
A′2 (λ) = −4(a0 + b0 ),
where a0 = 27λ3 − 54λ2 + 28λ + 8,
b0 = (9λ2 − 16λ + 4) 9λ2 − 4λ + 4.
A straightforward computation gives that a20 − b20 = 1024λ(λ − 1)2 . When λ ≤ 0, |a0 | ≤ |b0 |. Since b0 ≥ 0, we have A′1 (λ) ≥ 0,
A′2 (λ) ≤ 0.
When λ ≥ 0, |a0 | ≥ |b0 |. Since
√
a0 ≥ 6( 84 − 9)λ2 + 8 > 0, we have A′1 (λ) ≤ 0, A′2 (λ) ≤ 0. Since A1 (0) = 0, A2 (1) = 0, we have A1 (λ) ≤ 0, A2 (λ) ≥ 0 for λ ≤ 1, and A2 (λ) ≤ 0 for λ ≥ 1.
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Lemma 3. For λ ≥ 0, F (u1 , λ) is decreasing when 0 < w < 1 and F (u1 , λ) is increasing when w > 1. Proof. We have
√ (w 2 − 1)[w + 2 + w2 λ − λ + 4w + 4] . √ w 2 λ − λ + 4w + 4 When 0 < w < 1, we have A1 (λ) ≤ 0, A′1 (λ) ≤ 0, B(λ) ≥ 0, B′ (λ) ≤ 0. Therefore, A′1 (λ)B(λ) − A1 (λ)B′ (λ) ≤ 0 and then F (u1 , λ) is decreasing. When w > 1, we have B(λ) > (w 2 − 1)λ and then (w2 λ − λ + 4w + 4)B(λ) > (w2 − 1)λ. B′ (λ) =
Therefore,
√ A1 (λ)B(λ) − A1 (λ)B (λ) = √ ′
′
B(λ)
(w 2 λ − λ + 4w + 4)B(λ)A′1 (λ) − (w 2 − 1)A1 (λ)
w 2 λ − λ + 4w + 4 √ B(λ) (w 2 − 1)(λA′1 (λ) − A1 (λ)) ≤ √ 2 w λ − λ + 4w + 4 = B′ (λ)(λA′1 (λ) − A1 (λ)).
It is easy to get
λA′1 (λ) − A1 (λ) = b1 − a1 where
b1 = (27λ3 − 42λ2 + 4λ + 8) 9λ2 − 4λ + 4.
a1 = 81λ4 − 144λ3 + 56λ2 + 16,
The zero points of a′1 = 4λ(81λ2 − 108λ + 28) are λ = 0, 2(3 ±
√
√
2)/9. a1 (0) = 16 > 0. When λ = 2(3 +
√
2)/9,
a1 > 81λ − 128λ + 56λ ≥ 2(9 56 − 64)λ > 0. 4
3
2
3
Therefore a1 > 0 for λ ≥ 0. Since a21 − b21 = 4096λ2 (λ − 1)2 ≥ 0, we have |a1 | ≥ |b1 | and then λA′1 (λ) − A1 (λ) ≤ 0. Therefore, F (u1 , λ) is increasing when w > 1. Lemma 4. For λ ≤ 1, F (u2 , λ) is decreasing when 0 < w < 1 and F (u2 , λ) is increasing when w > 1. Proof. When w > 1, we have A2 (λ) ≥ 0, A′2 (λ) ≤ 0, B(λ) ≥ 0, B′ (λ) ≥ 0. Therefore, A′2 (λ)B(λ) − A2 (λ)B′ (λ) ≤ 0 and then F (u2 , λ) is increasing. When 0 ≤ w ≤ 1, we prove the lemma by two cases. (1) The case for λ ≤ 0. Since
(w2 λ − λ + 4w + 4)B(λ) > (2 +
w 2 λ − λ + 4) w 2 λ − λ + 4
≥ 4 + (w 2 − 1)λ + 4, we have A′2 (λ)B(λ) − A2 (λ)B′ (λ) = √
√
B(λ)
(w 2 λ − λ + 4w + 4)B(λ)A′2 (λ) − (w 2 − 1)A2 (λ)
w 2 λ − λ + 4w + 4 √ 2 B(λ) ≤ √ (w λ − λ + 8)A′2 (λ) + (1 − w2 )A2 (λ) 2 w λ − λ + 4w + 4 √ ′ B(λ) = √ [λA2 (λ) − A2 (λ)]w 2 + (8 − λ)A′2 (λ) + A2 (λ) . 2 w λ − λ + 4w + 4 Since a1 > 0 for λ ≤ 0 and |a1 | ≥ |b1 |, we have λA′2 (λ) − A2 (λ) = −b1 − a1 ≤ 0. It is easy to know
(8 − λ)A′2 (λ) + A2 (λ) = a2 + b2 , where a2 = 81λ4 − 1008λ3 + 1784λ2 − 896λ − 240,
b2 = (27λ3 − 330λ2 + 516λ − 120) 9λ2 − 4λ + 4 < 0, b22 − a22 = 4096λ(λ − 1)2 (23λ − 240) ≥ 0. Thus (8 − λ)A′2 (λ) + A2 (λ) ≤ 0 and then A′2 (λ)B(λ) − A2 (λ)B′ (λ) ≤ 0. Therefore F (u2 , λ) is decreasing.
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(2) The case for 0 ≤ λ ≤ 1. Since
(w 2 λ − λ + 4w + 4)B(λ) ≥ w + 2 + w 2 + 4w + 3 w2 + 4w + 3 > 2(w 2 + 4w + 3) > 6,
we have
√ A2 (λ)B(λ) − A2 (λ)B (λ) ≤ √ ′
B(λ)
′
w λ − λ + 4w + 4 2
[6A′2 (λ) + A2 (λ)].
It is easy to know 6A′2 (λ) + A2 (λ) = −a3 − b3 , where a3 = 27λ4 + 576λ3 − 1240λ2 + 704λ + 176,
b3 = (9λ3 + 194λ2 − 372λ + 88) 9λ2 − 4λ + 4. Since a23 − b23 = 4096λ(λ − 1)2 (λ2 + 29λ + 132) ≥ 0,
√
a3 ≥ 27λ4 + (384 11 − 1240)λ2 + 176 > 0, we have 6A′2 (λ) + A2 (λ) ≤ 0. Therefore A′2 (λ)B(λ) − A2 (λ)B′ (λ) ≤ 0 and then F (u2 , λ) is decreasing.
3.2. The proof of Theorem 1 For λ ≤ 0, from (15) we have λ ≤ u1 ≤ 0 and 0 < u2 < 1. Therefore, by Lemma 2 max |F (u, λ)| =
u∈[0,1]
F (u2 , λ), −F (u2 , λ),
0 < w < 1, w > 1.
By Lemma 4, we have max |F (u, λ)| ≥ |F (u2 , 0)| =
u∈[0,1]
F (u2 , 0), −F (u2 , 0),
0 < w < 1, w > 1.
For λ ≥ 1, from (15) we have 0 < u1 < 1 and 1 ≤ u2 ≤ λ. Therefore, by Lemma 2 max |F (u, λ)| =
u∈[0,1]
−F (u1 , λ), F (u1 , λ),
0 < w < 1, w > 1.
By Lemma 3, we have max |F (u, λ)| ≥ |F (u1 , 1)| =
u∈[0,1]
−F (u1 , 1), F (u1 , 1),
0 < w < 1, w > 1.
As stated as above, for (14) we have min max |F (u, λ)| = min max |F (u, λ)|. λ∈R u∈[0,1]
λ∈[0,1] u∈[0,1]
(17)
For 0 ≤ λ ≤ 1, from (15) we have 0 ≤ u1 ≤ λ ≤ u2 ≤ 1. Therefore max |F (u, λ)| = max {|F (u1 , λ)|, |F (u2 , λ)|} .
u∈[0,1]
By Lemma 2, F (u1 , λ) and F (u2 , λ) are the negative and the positive amount of deviation of F (u, λ) respectively when 0 < w < 1, and F (u1 , λ) and F (u2 , λ) are the positive and the negative amount of deviation of F (u, λ) respectively when w > 1. By Lemmas 3 and 4, F (u1 , λ) and F (u2 , λ) are decreasing when 0 < w < 1, and F (u1 , λ) and F (u2 , λ) are increasing when w > 1. Thus the negative amount of deviation and the positive amount of deviation of F (u, λ) have the same monotonicity. Therefore, based on these properties of amount of deviation, the solution of minimax problem (17) must be obtained from F (u2 , λ) = −F (u1 , λ). From this and (16) we have 27λ4 − 72λ3 + 56λ2 + 32λ − 16 = 0.
(18)
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X. Han, X. Guo / Journal of Computational and Applied Mathematics 322 (2017) 86–95
It can be shown that
√
27λ4 − 72λ3 + 56λ2 + 32λ − 16 = (6 3 − 9)λ2 − 4λ + 4
√ (6 3 − 9)λ2 + 4λ − 4 .
√
√
The equation (6 3 − 9)λ2 − 4λ + 4 = 0 has not a real root. The equation (6 3 − 9)λ2 + 4λ − 4 = 0 has two real roots:
λ¯ =
2 1−
√ < 0,
10 + 6 3
λ∗ =
2 1+
√ .
10 + 6 3
¯ is not the solution. Thus λ∗ is the optimal solution of (17) and we obtain (7). By (12), with λ = λ∗ , we Obviously, λ obtain (6). With λ = λ∗ , we obtain (10) from (13), and obtain (11) from (16) and (18). 3.3. The proof on the value α in Theorem 2
√ √ λw2 − λ + 4w + 4 ≥ 4 − λ > 1.9. Therefore (3 − 2λ)w 2 + w + 2λ − 2 − (w − 1) λw2 − λ + 4w + 4 > (3 − 2λ)w2 + w + 2λ − 2 − 1.9(w − 1)
When 0 < w < 1, we have
= (3 − 2λ)w2 − 0.9w + 2λ − 0.1 > 0. This means α > 0. When w > 1, since (λ − 1)2 w 2 + (10λ − 6)w + 4λ + 9 > 0, we have 4(λw 2 − λ + 4w + 4) < (λ + 1)2 w 2 + 10(λ + 1)w + 25 and then
2 λw 2 − λ + 4w + 4 < (λ + 1)w + 5. Therefore
(3 − 2λ)w 2 + w + 2λ − 2 − (w − 1) λw2 − λ + 4w + 4 1
> (3 − 2λ)w2 + w + 2λ − 2 − (w − 1)[(λ + 1)w + 5] 2
= >
1 2 1 2
5(1 − λ)w + (λ − 2)w + 1 2
[5(1 − λ)w + (λ − 2)w + 1] > 0.
This means α > 0. Now we prove α < 1 for 0 < w < w0 . By (6),
√ w 3 + (2λ − 1)w 2 − 2w − 2λ − w(w − 1) λw2 − λ + 4w + 4 α−1= . 2(w + 1)[(1 − λ)w 2 + λ] We have
w 3 + (2λ − 1)w 2 − 2w − 2λ − w(w − 1) λw2 − λ + 4w + 4 × w 3 + (2λ − 1)w 2 − 2w − 2λ + w(w − 1) λw2 − λ + 4w + 4 = (1 − λ)w 6 + 6(λ − 1)w 5 + (2w − 1)2 w 4 + (8 − 14λ)w3 + (5λ − 8λ2 )w 2 + 8λw + 4λ2 = (w + 1)[(1 − λ)w 2 + λ][w 3 − 7w 2 + (8 − 4λ)w + 4λ]. Therefore, the positive roots of the equation α − 1 = 0 must be the roots of the equation
w3 − 7w2 + (8 − 4λ)w + 4λ = 0.
(19)
This equation has three real roots w2 ≈ −0.2, w1 ≈ 1.3 and w0 ≈ 5.8. The root w0 is the unique positive root of the equation α − 1 = 0. Therefore α < 1 for 0 < w < w0 . From Eq. (19) we obtain (9).
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95
3.4. The proof on the value β in Theorem 2 (1) The proof of β > 0. When 0 < w < 1, since
λw2 − λ + 4w + 4 <
w 2 + 4w + 4 = w + 2,
we have
(1 − w) λw2 − λ + 4w + 4 ≤ (1 − w)(w + 2) < (1 − 2λ)w 2 − w + 2λ + 2. This means β > 0. When w > 1, since (1 − 2λ)w 2 − w + 2λ + 2 > 0, we have
(1 − 2λ)w2 − w + 2λ + 2 > (1 − w) λw2 − λ + 4w + 4 and then β > 0. (2) The proof of β < 1. √ √ When 0 < w < 1, since λw 2 − λ + 4w + 4 > λw > w/2, we have
2w(w − 1) λw 2 − λ + 4w + 4 ≤ w 2 (w − 1). Since λw 3 + 3(2 − λ)w 2 − (λ + 4)w + 3λ > λw 3 > 0, we have
(λ + 1)w 3 + (5 − 3λ)w2 − (λ + 4)w + 3λ > w2 (w − 1) ≥ 2w(w − 1) λw2 − λ + 4w + 4. This means β < 1. When w > 1, since (1 − 2λ)w 2 + (1 − λ)w + 3λ > 0, we have
(λ + 1)w 3 + (5 − 3λ)w2 − (λ + 4)w + 3λ > w(w − 1)[(λ + 1)w + 5] ≥ 2w(w − 1) λw2 − λ + 4w + 4. This means β < 1.
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