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Solving nonlinear programming problems via a homotopy continuation method under three unbounded conditions
Nonlinear Analysis 70 (2009) 3099–3103
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Solving nonlinear programming problems via a homotopy continuation method under three unbounded conditions Menglong Su a,b,∗ , Qiuling Hua b a Mathematics and Information Science College, Luoyang Normal University, Luoyang 471022, PR China b College of Mathematics, Jilin University, Changchun 130012, PR China
article
a b s t r a c t
info
Article history: Received 14 October 2007 Accepted 15 April 2008
In this paper, we give three unbounded conditions under which we are able to solve the nonlinear programming problems in unbounded sets by a homotopy continuation method. In addition, we also discuss their relations. © 2008 Elsevier Ltd. All rights reserved.
Keywords: Three unbounded conditions Nonlinear programming problems Homotopy continuation method
1. Introduction Consider the following nonlinear programming problems: minn f (x) x ∈R
s.t. gi (x) ≤ 0,
i = 1, . . . , m,
(1.1)
where f : R → R , gi : R → R , i = 1, . . . , m, are three times continuously differentiable. We call a point x∗ a Karush–Kuhn–Tucker point of (1.1) and y∗ are the corresponding Lagrangian multiplier vectors if (x∗ , y∗ ) satisfies n
1
n
1
∇ f (x) + ∇ g(x)y = 0, Yg(x) = 0,
g(x) ≤ 0, y ≥ 0,
(1.2)
where y ∈ Rm , ∇ f (x) = (∂f (x)/∂x)T ∈ Rn , ∇ g(x) = (∇ g1 (x), . . . , ∇ gm (x)) ∈ Rn×m , g(x) = (g1 (x), . . . , gm (x))T ∈ Rm , and Y = diag(y) ∈ Rm×m . In [2], to solve problem (1.2), the authors constructed the following homotopy: ! (1 − µ)(∇ f (x) + ∇ g(x)y) + µ(x − x(0) ) (0) H(w, w , µ) = = 0, (1.3) (0) (0) Yg(x) − µY
g (x
)
(0) = diag(y(0) ), Rm and Rm represent the nonnegative and where w = (x, y) ∈ Rn+m , w(0) = (x(0) , y(0) ) ∈ Ω 0 × Rm ++ and Y + ++ positive orthants of Rm , respectively. But they obtained the global convergence of the homotopy continuation method requiring the boundedness of the feasible set. In this paper, we will attempt to remove the boundedness assumption. To this end, we give three unbounded
∗ Corresponding author at: Mathematics and Information Science College, Luoyang Normal University, Luoyang 471022, PR China. E-mail address: [email protected] (M. Su). 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.04.012
M. Su, Q. Hua / Nonlinear Analysis 70 (2009) 3099–3103
3100
conditions under which we also obtain the global convergence results for the homotopy continuation method. In addition, we discuss the relations of the three unbounded conditions. Throughout this paper, let Ω = {x ∈ Rn : gi (x) ≤ 0, i = 1, . . . , m}, Ω 0 = {x ∈ Rn : gi (x) < 0, i = 1, . . . , m} and ∂Ω = Ω \ Ω 0 be the boundary set of Ω . For any x ∈ ∂Ω , denote the active index set at x by B(x) = {i ∈ (1, . . . , m) : gi (x) = 0}. 2. Main results In this section, we make the first assumptions:
(A1 ) (A2 ) (A3 ) (A4 )
Ω 0 is nonempty. For any x ∈ ∂Ω , the matrix {∇ gi (x) : i ∈ B(x)} is of full column rank. gi (x), i = 1, . . . , m, are convex. There exists some ρ > 0 such that for any x ∈ Ω and d ∈ Rn , we have dT ∇ 2 f (x)d ≥
ρkdk22 .
Remark 2.1. It should be pointed out that under assumptions (A1 )–(A4 ), we don’t require the boundedness of the feasible sets, but let the objective functions f (x) satisfy assumption (A4 ). Therefore we can utilize the homotopy continuation method to solve problem (1.1) in unbounded feasible sets. Like for Lemma 2.1 in [2], we get the existence of a homotopy path. Then we should prove the boundedness of the homotopy path. Lemma 2.1. If assumptions (A3 ) and (A4 ) hold, for a given point ξ ∈ Ω , let Ω − (ξ) = {x ∈ Ω : f (x) − f (ξ) ≤ 0}.
Then Ω − (ξ) is bounded. Proof. For any x ∈ Ω − (ξ) ⊂ Ω , by the Taylor expansion, we have f (x) = f (ξ) + ∇ f (ξ)(x − ξ) +
1 2
(x − ξ)T ∇ 2 f (ζ)(x − ξ),
(2.1)
where ζ = ξ + θ(x − ξ) = θx + (1 − θ)ξ, 0 < θ < 1. By the convexity of Ω , we get ζ ∈ Ω . Hence ∀x ∈ Ω − (ξ), by assumption (A4 ), we have f (ξ) ≥ f (x)
1
> f (ξ) + ∇ f (ξ)(x − ξ) + ρkx − ξk2 . 2
(2.2)
Then
∇ f (ξ)(x − ξ) +
1
ρkx − ξk2 < 0,
(2.3)
ρkx − ξk2 < ∇ f (ξ)(ξ − x) ≤ k∇ f (ξ)kkξ − xk.
(2.4)
2
i.e. 1 2
So we have
kx − ξk <
2
ρ
k∇ f (ξ)k.
(2.5)
By the continuity of ∇ f (x) in Ω and the fact that ξ is finite in Ω , we get that Ω − (ξ) is bounded. This completes the proof.
Lemma 2.2. For any given ξ ∈ Ω and x(0) ∈ Ω 0 , the following inequality holds:
kx − ξk2 − kx(0) − ξk2 ≤ 2(x − ξ)T (x − x(0) ),
(2.6)
where x is an arbitrary point in Ω . Proof. It is easy to get the result.
Lemma 2.3. Let H be defined as in (1.3) and assumptions (A1 )–(A4 ) hold. Then for a given w(0) = (x(0) , y(0) ) ∈ Ω 0 × Rm ++ , if 0 is a regular value of H(w, w(0) , µ), then the component x of w is bounded.
M. Su, Q. Hua / Nonlinear Analysis 70 (2009) 3099–3103
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(k) k → ∞ as k → ∞. Proof. If not, then there exists a sequence of points {(x(k) , y(k) , µk )}∞ k=1 such that kx + − + Let Ω (ξ) = {x ∈ Ω : f (x) − f (ξ) > 0}; then Ω = Ω (ξ) ∪ Ω (ξ). Since f (x), g(x) are convex, then for a given point ξ ∈ Ω ,
(ξ − x(k) )T ∇ f (x(k) ) ≤ f (ξ) − f (x(k) ), (k) T
(k)
(2.7)
(k) T
(ξ − x ) ∇ g(x ) ≤ g(ξ) − g(x ) . T
(2.8)
Multiplying the two sides of the first equality of (1.3) by (x(k) − ξ)T , we get
(1 − µk )[(x(k) − ξ)T ∇ f (x(k) ) + (x(k) − ξ)T ∇ g(x(k) )y(k) ] = −µk (x(k) − ξ)T (x(k) − x(0) ). By Lemma 2.2, (2.7)–(2.9), the fact that g(ξ)
T
kx
(k)
− ξ k − kx 2
(0)
(k)
(2.9)
≤ 0 and the second equality of (1.3), we get
− ξ) (x − x(0) ) 2(1 − µk ) = [(ξ − x(k) )T ∇ f (x(k) ) + (ξ − x(k) )T ∇ g(x(k) )y(k) ]
− ξk ≤ 2(x 2
y(k)
T
(k)
µk
≤
2(1 − µk )
≤
2(1 − µk )
=
2(1 − µk )
µk µk µk
[f (ξ) − f (x(k) ) + (g(ξ)T − g(x(k) )T )y(k) ]
(f (ξ) − f (x(k) ) − g(x(k) )T y(k) ) (f (ξ) − f (x(k) )) − 2(1 − µk )g(x(0) )T y(0) .
(2.10)
When x(k) ∈ Ω + (ξ), we have f (ξ) − f (x(k) ) < 0, and consequently
kx(k) − ξk2 ≤ kx(0) − ξk2 − 2(1 − µk )g(x(0) )T y(0) ,
(2.11)
and hence the case where kx(k) k → ∞ in Ω + (ξ) is impossible. Since Ω − (ξ) is bounded, the case where kx(k) k → ∞ in Ω − (ξ) is also impossible. As a result, the component x of w is bounded in Ω for Ω = Ω − (ξ) ∪ Ω + (ξ). Like for Lemma 2.2 and Theorem 2.3 in [2], we have the boundedness of the component y of w in Ω and the convergence of the homotopy method, respectively. In the following, we make the second assumptions:
(B1 ) (B2 ) (B3 ) (B4 )
Ω 0 is nonempty. For any x ∈ ∂Ω , the matrix {∇ gi (x) : i ∈ B(x)} is of full column rank. gi (x), i = 1, . . . , m, are convex. There exists some ξ ∈ Ω such that Ω − (ξ) = {x ∈ Ω : (x − ξ)T ∇ f (x) ≤ 0} is bounded.
Remark 2.2. If Ω is bounded, then assumption (B4 ) holds, for Ω − (ξ) ⊂ Ω . Conversely, the conclusion doesn’t hold. Hence we can solve problem (1.1) in the unbounded sets. In addition, under assumptions (B1 )–(B4 ), we obtain the global convergence results without the convexity of f (x); hence the homotopy continuation method is also applicable to some nonconvex cases. The following Lemma 2.4 and Example 2.1 show that assumptions (B1 )–(B4 ) are weaker than assumptions (A1 )–(A4 ). Lemma 2.4. If Ω satisfies assumptions (A1 )–(A4 ), then it necessarily satisfies assumptions (B1 )–(B4 ). Proof. It is obvious that assumptions (B1 )–(B3 ) are identical to (A1 )–(A3 ); hence we only need to prove that assumption (B4 ) is weaker than assumption (A4 ). Since f (x) is convex, then
(x − ξ)T ∇ f (x) ≥ f (x) − f (ξ).
(2.12)
If x ∈ Ω (ξ), then we have −
f (x) − f (ξ) ≤ 0.
(2.13)
By Lemma 2.1, we get that Ω − (ξ) is bounded. This completes the proof.
Conversely, the conclusion doesn’t hold. This can be illustrated by the following Example 2.1. Example 2.1. f1 (x) = 3(x1 + 3)3 + (x2 − 5)2 ,
and Ω = {(x1 , x2 ) ∈ R : − 2x1 + 2
x22
(2.14)
− 10 ≤ 0, − x2 ≤ 0}.
In this example, it is easy to see that Ω is unbounded and assumptions (B1 )–(B4 ) are satisfied. At the same time, assumptions (A1 )–(A4 ) are not satisfied for this example.
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Remark 2.3. Although assumptions (B1 )–(B4 ) are weaker than (A1 )–(A4 ), in some cases, it is difficult to verify assumptions (B1 )–(B4 ). At the same time, assumptions (A1 )–(A4 ) are easy to verify, as can be seen in the following Example 2.2. Example 2.2. f (x) = 8x21 + 3x1 x2 + 3(x2 − 3)2 ,
and Ω = {(x1 , x2 ) ∈ R2 : − 2x1 + x22 + 10 ≤ 0}. In this example, it is easy to see that Ω is unbounded and assumptions (A1 )–(A4 ) are satisfied. However, assumptions (B1 )–(B4 ) are difficult to verify for this example. Under assumptions (B1 )–(B4 ), we get the global convergence of the homotopy continuation method like in the analysis under assumptions (A1 )–(A4 ), so we omit this here. It should be pointed out that in order to remove the boundedness on Ω , we require that gi (x), i = 1, . . . , m, are convex in assumptions (A1 )–(A4 ) and (B1 )–(B4 ). As for the nonconvex case in [2], i.e. Ω satisfying the so-called normal cone condition, we have the following assumptions:
(C1 ) Ω 0 is nonempty. (C2 ) For any x ∈ ∂Ω , the matrix {∇ gi (x) : i ∈ B(x)} is of full column rank. (C3 ) There exists some ξ ∈ Ω such that Ω − (ξ) = {x ∈ Ω : (x − ξ)T ∇ f (x) ≤ 0}
is bounded and
(x − ξ)T ∇ g(x) < 0. kxk→∞ g(x) (C4 ) For any x ∈ ∂Ω , the normal cone of Ω at x only meets Ω at x, i.e., for any x ∈ ∂Ω , we have lim
x+
X i∈B(x)
yi ∇ gi (x) : yi ≥ 0 for i ∈ B(x) ∩ Ω = {x}.
Note that we remove the convexity of f (x) and g(x) by assumptions (C3 )–(C4 ). Under assumptions (C1 )–(C4 ), we get the global convergence of the homotopy continuation method like in the analysis under assumptions (A1 )–(A4 ) except for the boundedness of the component x of w. We list the results in the following Lemma 2.5. Lemma 2.5. Let H be defined as in (1.3), and let assumptions (C1 )–(C4 ) hold. Then for a given w(0) = (x(0) , y(0) ) ∈ Ω 0 × Rm ++ , if 0 is a regular value of H(w, w(0) , µ), then the component x of w is bounded. (k) k → ∞ as k → ∞. Proof. If not, then there exists a sequence of points {(x(k) , y(k) , µk )}∞ k=1 such that kx Let Ω + (ξ) = Ω \ Ω − (ξ). Since g(x) is convex, then
(ξ − x(k) )T ∇ g(x(k) ) ≤ g(ξ)T − g(x(k) )T .
(2.15)
From the second equality of (1.3), we have y(k) =
µk Y (0) g(x(0) ) . g(x(k) )
(2.16)
Multiplying the two sides of the first equality of (1.3) by (x(k) − ξ)T , we get
(1 − µk )[(x(k) − ξ)T ∇ f (x(k) ) + (x(k) − ξ)T ∇ g(x(k) )y(k) ] = −µk (x(k) − ξ)T (x(k) − x(0) ).
(2.17)
By Lemma 2.2, (2.15)–(2.17), we get
µk (kx(k) − ξk2 − kx(0) − ξk2 ) ≤ 2µk (x(k) − ξ)T (x(k) − x(0) ) = 2(1 − µk )[(ξ − x(k) )T ∇ f (x(k) ) + (ξ − x(k) )T ∇ g(x(k) )y(k) ] " # (0) (0) (k) T (k) (k) T (k) µk Y g (x ) ≤ 2(1 − µk ) (ξ − x ) ∇ f (x ) + (ξ − x ) ∇ g(x ) g(x(k) ) "
= 2(1 − µk ) (ξ − x(k) )T ∇ f (x(k) ) − µk Y (0) g(x(0) )
(x(k) − ξ)T ∇ g(x(k) ) . g(x(k) ) #
(2.18)
By (2.18), we have 2(1 − µk )(x(k) − ξ)T ∇ f (x(k) ) ≤ µk kx(0) − ξk2 − µk kx(k) − ξk2 − µk Y (0) g(x(0) )
(x(k) − ξ)T ∇ g(x(k) ) . g(x(k) )
(2.19)
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When x(k) ∈ Ω + (ξ), if kx(k) k → ∞, by assumption (C3 ), µk ∈ (0, 1] and for the case where ξ is a given point, it is easy to get that there exists some k such that kx(k) − ξk > M, and the right-hand side of (2.19) is strictly smaller than 0 while the left-hand side of (2.19) is strictly larger than 0, a contradiction. Hence the case where kx(k) k → ∞ in Ω + (ξ) is impossible. Since Ω − (ξ) is bounded, the case where kx(k) k → ∞ in Ω − (ξ) is also impossible. Then the component x of w is bounded in Ω for Ω = Ω − (ξ) ∪ Ω + (ξ). This completes the proof. As for how to trace the homotopy path numerically, there have been many predictor–corrector algorithms; see [1,3,4], etc., for references. Hence we omit this in this paper. Acknowledgements The authors express their sincere thanks to Professor Yong Li and Bo Yu for their instruction and many invaluable suggestions. References [1] E.L. Allgower, K. Georg, Numerical Continuation Methods: An Introduction, Springer-Verlag, 1990. [2] G.C. Feng, Z.H. Lin, B. Yu, Existence of interior pathway to the Karush–Kuhn–Tucker point of a nonconvex programming problems, Nonlinear Anal. 32 (6) (1998) 761–768. [3] Z.H. Lin, Z.P. Sheng, L. Yang, Y.Z. Bai, A predictor–corrector algorithm for tracing combined interior homotopy pathway, Math. Numer. Sin. 24 (2002) 405–416. [4] L.T. Watson, S.M. Holzer, M.C. Hansen, Tracking nonlinear equilibrium paths by a homotopy method, Nonlinear Anal. 7 (1983) 1271–1282.
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Solving nonlinear programming problems via a homotopy continuation method under three unbounded conditions Menglong Su a,b,∗ , Qiuling Hua b a Mathematics and Information Science College, Luoyang Normal University, Luoyang 471022, PR China b College of Mathematics, Jilin University, Changchun 130012, PR China
article
a b s t r a c t
info
Article history: Received 14 October 2007 Accepted 15 April 2008
In this paper, we give three unbounded conditions under which we are able to solve the nonlinear programming problems in unbounded sets by a homotopy continuation method. In addition, we also discuss their relations. © 2008 Elsevier Ltd. All rights reserved.
Keywords: Three unbounded conditions Nonlinear programming problems Homotopy continuation method
1. Introduction Consider the following nonlinear programming problems: minn f (x) x ∈R
s.t. gi (x) ≤ 0,
i = 1, . . . , m,
(1.1)
where f : R → R , gi : R → R , i = 1, . . . , m, are three times continuously differentiable. We call a point x∗ a Karush–Kuhn–Tucker point of (1.1) and y∗ are the corresponding Lagrangian multiplier vectors if (x∗ , y∗ ) satisfies n
1
n
1
∇ f (x) + ∇ g(x)y = 0, Yg(x) = 0,
g(x) ≤ 0, y ≥ 0,
(1.2)
where y ∈ Rm , ∇ f (x) = (∂f (x)/∂x)T ∈ Rn , ∇ g(x) = (∇ g1 (x), . . . , ∇ gm (x)) ∈ Rn×m , g(x) = (g1 (x), . . . , gm (x))T ∈ Rm , and Y = diag(y) ∈ Rm×m . In [2], to solve problem (1.2), the authors constructed the following homotopy: ! (1 − µ)(∇ f (x) + ∇ g(x)y) + µ(x − x(0) ) (0) H(w, w , µ) = = 0, (1.3) (0) (0) Yg(x) − µY
g (x
)
(0) = diag(y(0) ), Rm and Rm represent the nonnegative and where w = (x, y) ∈ Rn+m , w(0) = (x(0) , y(0) ) ∈ Ω 0 × Rm ++ and Y + ++ positive orthants of Rm , respectively. But they obtained the global convergence of the homotopy continuation method requiring the boundedness of the feasible set. In this paper, we will attempt to remove the boundedness assumption. To this end, we give three unbounded
∗ Corresponding author at: Mathematics and Information Science College, Luoyang Normal University, Luoyang 471022, PR China. E-mail address: [email protected] (M. Su). 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.04.012
M. Su, Q. Hua / Nonlinear Analysis 70 (2009) 3099–3103
3100
conditions under which we also obtain the global convergence results for the homotopy continuation method. In addition, we discuss the relations of the three unbounded conditions. Throughout this paper, let Ω = {x ∈ Rn : gi (x) ≤ 0, i = 1, . . . , m}, Ω 0 = {x ∈ Rn : gi (x) < 0, i = 1, . . . , m} and ∂Ω = Ω \ Ω 0 be the boundary set of Ω . For any x ∈ ∂Ω , denote the active index set at x by B(x) = {i ∈ (1, . . . , m) : gi (x) = 0}. 2. Main results In this section, we make the first assumptions:
(A1 ) (A2 ) (A3 ) (A4 )
Ω 0 is nonempty. For any x ∈ ∂Ω , the matrix {∇ gi (x) : i ∈ B(x)} is of full column rank. gi (x), i = 1, . . . , m, are convex. There exists some ρ > 0 such that for any x ∈ Ω and d ∈ Rn , we have dT ∇ 2 f (x)d ≥
ρkdk22 .
Remark 2.1. It should be pointed out that under assumptions (A1 )–(A4 ), we don’t require the boundedness of the feasible sets, but let the objective functions f (x) satisfy assumption (A4 ). Therefore we can utilize the homotopy continuation method to solve problem (1.1) in unbounded feasible sets. Like for Lemma 2.1 in [2], we get the existence of a homotopy path. Then we should prove the boundedness of the homotopy path. Lemma 2.1. If assumptions (A3 ) and (A4 ) hold, for a given point ξ ∈ Ω , let Ω − (ξ) = {x ∈ Ω : f (x) − f (ξ) ≤ 0}.
Then Ω − (ξ) is bounded. Proof. For any x ∈ Ω − (ξ) ⊂ Ω , by the Taylor expansion, we have f (x) = f (ξ) + ∇ f (ξ)(x − ξ) +
1 2
(x − ξ)T ∇ 2 f (ζ)(x − ξ),
(2.1)
where ζ = ξ + θ(x − ξ) = θx + (1 − θ)ξ, 0 < θ < 1. By the convexity of Ω , we get ζ ∈ Ω . Hence ∀x ∈ Ω − (ξ), by assumption (A4 ), we have f (ξ) ≥ f (x)
1
> f (ξ) + ∇ f (ξ)(x − ξ) + ρkx − ξk2 . 2
(2.2)
Then
∇ f (ξ)(x − ξ) +
1
ρkx − ξk2 < 0,
(2.3)
ρkx − ξk2 < ∇ f (ξ)(ξ − x) ≤ k∇ f (ξ)kkξ − xk.
(2.4)
2
i.e. 1 2
So we have
kx − ξk <
2
ρ
k∇ f (ξ)k.
(2.5)
By the continuity of ∇ f (x) in Ω and the fact that ξ is finite in Ω , we get that Ω − (ξ) is bounded. This completes the proof.
Lemma 2.2. For any given ξ ∈ Ω and x(0) ∈ Ω 0 , the following inequality holds:
kx − ξk2 − kx(0) − ξk2 ≤ 2(x − ξ)T (x − x(0) ),
(2.6)
where x is an arbitrary point in Ω . Proof. It is easy to get the result.
Lemma 2.3. Let H be defined as in (1.3) and assumptions (A1 )–(A4 ) hold. Then for a given w(0) = (x(0) , y(0) ) ∈ Ω 0 × Rm ++ , if 0 is a regular value of H(w, w(0) , µ), then the component x of w is bounded.
M. Su, Q. Hua / Nonlinear Analysis 70 (2009) 3099–3103
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(k) k → ∞ as k → ∞. Proof. If not, then there exists a sequence of points {(x(k) , y(k) , µk )}∞ k=1 such that kx + − + Let Ω (ξ) = {x ∈ Ω : f (x) − f (ξ) > 0}; then Ω = Ω (ξ) ∪ Ω (ξ). Since f (x), g(x) are convex, then for a given point ξ ∈ Ω ,
(ξ − x(k) )T ∇ f (x(k) ) ≤ f (ξ) − f (x(k) ), (k) T
(k)
(2.7)
(k) T
(ξ − x ) ∇ g(x ) ≤ g(ξ) − g(x ) . T
(2.8)
Multiplying the two sides of the first equality of (1.3) by (x(k) − ξ)T , we get
(1 − µk )[(x(k) − ξ)T ∇ f (x(k) ) + (x(k) − ξ)T ∇ g(x(k) )y(k) ] = −µk (x(k) − ξ)T (x(k) − x(0) ). By Lemma 2.2, (2.7)–(2.9), the fact that g(ξ)
T
kx
(k)
− ξ k − kx 2
(0)
(k)
(2.9)
≤ 0 and the second equality of (1.3), we get
− ξ) (x − x(0) ) 2(1 − µk ) = [(ξ − x(k) )T ∇ f (x(k) ) + (ξ − x(k) )T ∇ g(x(k) )y(k) ]
− ξk ≤ 2(x 2
y(k)
T
(k)
µk
≤
2(1 − µk )
≤
2(1 − µk )
=
2(1 − µk )
µk µk µk
[f (ξ) − f (x(k) ) + (g(ξ)T − g(x(k) )T )y(k) ]
(f (ξ) − f (x(k) ) − g(x(k) )T y(k) ) (f (ξ) − f (x(k) )) − 2(1 − µk )g(x(0) )T y(0) .
(2.10)
When x(k) ∈ Ω + (ξ), we have f (ξ) − f (x(k) ) < 0, and consequently
kx(k) − ξk2 ≤ kx(0) − ξk2 − 2(1 − µk )g(x(0) )T y(0) ,
(2.11)
and hence the case where kx(k) k → ∞ in Ω + (ξ) is impossible. Since Ω − (ξ) is bounded, the case where kx(k) k → ∞ in Ω − (ξ) is also impossible. As a result, the component x of w is bounded in Ω for Ω = Ω − (ξ) ∪ Ω + (ξ). Like for Lemma 2.2 and Theorem 2.3 in [2], we have the boundedness of the component y of w in Ω and the convergence of the homotopy method, respectively. In the following, we make the second assumptions:
(B1 ) (B2 ) (B3 ) (B4 )
Ω 0 is nonempty. For any x ∈ ∂Ω , the matrix {∇ gi (x) : i ∈ B(x)} is of full column rank. gi (x), i = 1, . . . , m, are convex. There exists some ξ ∈ Ω such that Ω − (ξ) = {x ∈ Ω : (x − ξ)T ∇ f (x) ≤ 0} is bounded.
Remark 2.2. If Ω is bounded, then assumption (B4 ) holds, for Ω − (ξ) ⊂ Ω . Conversely, the conclusion doesn’t hold. Hence we can solve problem (1.1) in the unbounded sets. In addition, under assumptions (B1 )–(B4 ), we obtain the global convergence results without the convexity of f (x); hence the homotopy continuation method is also applicable to some nonconvex cases. The following Lemma 2.4 and Example 2.1 show that assumptions (B1 )–(B4 ) are weaker than assumptions (A1 )–(A4 ). Lemma 2.4. If Ω satisfies assumptions (A1 )–(A4 ), then it necessarily satisfies assumptions (B1 )–(B4 ). Proof. It is obvious that assumptions (B1 )–(B3 ) are identical to (A1 )–(A3 ); hence we only need to prove that assumption (B4 ) is weaker than assumption (A4 ). Since f (x) is convex, then
(x − ξ)T ∇ f (x) ≥ f (x) − f (ξ).
(2.12)
If x ∈ Ω (ξ), then we have −
f (x) − f (ξ) ≤ 0.
(2.13)
By Lemma 2.1, we get that Ω − (ξ) is bounded. This completes the proof.
Conversely, the conclusion doesn’t hold. This can be illustrated by the following Example 2.1. Example 2.1. f1 (x) = 3(x1 + 3)3 + (x2 − 5)2 ,
and Ω = {(x1 , x2 ) ∈ R : − 2x1 + 2
x22
(2.14)
− 10 ≤ 0, − x2 ≤ 0}.
In this example, it is easy to see that Ω is unbounded and assumptions (B1 )–(B4 ) are satisfied. At the same time, assumptions (A1 )–(A4 ) are not satisfied for this example.
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Remark 2.3. Although assumptions (B1 )–(B4 ) are weaker than (A1 )–(A4 ), in some cases, it is difficult to verify assumptions (B1 )–(B4 ). At the same time, assumptions (A1 )–(A4 ) are easy to verify, as can be seen in the following Example 2.2. Example 2.2. f (x) = 8x21 + 3x1 x2 + 3(x2 − 3)2 ,
and Ω = {(x1 , x2 ) ∈ R2 : − 2x1 + x22 + 10 ≤ 0}. In this example, it is easy to see that Ω is unbounded and assumptions (A1 )–(A4 ) are satisfied. However, assumptions (B1 )–(B4 ) are difficult to verify for this example. Under assumptions (B1 )–(B4 ), we get the global convergence of the homotopy continuation method like in the analysis under assumptions (A1 )–(A4 ), so we omit this here. It should be pointed out that in order to remove the boundedness on Ω , we require that gi (x), i = 1, . . . , m, are convex in assumptions (A1 )–(A4 ) and (B1 )–(B4 ). As for the nonconvex case in [2], i.e. Ω satisfying the so-called normal cone condition, we have the following assumptions:
(C1 ) Ω 0 is nonempty. (C2 ) For any x ∈ ∂Ω , the matrix {∇ gi (x) : i ∈ B(x)} is of full column rank. (C3 ) There exists some ξ ∈ Ω such that Ω − (ξ) = {x ∈ Ω : (x − ξ)T ∇ f (x) ≤ 0}
is bounded and
(x − ξ)T ∇ g(x) < 0. kxk→∞ g(x) (C4 ) For any x ∈ ∂Ω , the normal cone of Ω at x only meets Ω at x, i.e., for any x ∈ ∂Ω , we have lim
x+
X i∈B(x)
yi ∇ gi (x) : yi ≥ 0 for i ∈ B(x) ∩ Ω = {x}.
Note that we remove the convexity of f (x) and g(x) by assumptions (C3 )–(C4 ). Under assumptions (C1 )–(C4 ), we get the global convergence of the homotopy continuation method like in the analysis under assumptions (A1 )–(A4 ) except for the boundedness of the component x of w. We list the results in the following Lemma 2.5. Lemma 2.5. Let H be defined as in (1.3), and let assumptions (C1 )–(C4 ) hold. Then for a given w(0) = (x(0) , y(0) ) ∈ Ω 0 × Rm ++ , if 0 is a regular value of H(w, w(0) , µ), then the component x of w is bounded. (k) k → ∞ as k → ∞. Proof. If not, then there exists a sequence of points {(x(k) , y(k) , µk )}∞ k=1 such that kx Let Ω + (ξ) = Ω \ Ω − (ξ). Since g(x) is convex, then
(ξ − x(k) )T ∇ g(x(k) ) ≤ g(ξ)T − g(x(k) )T .
(2.15)
From the second equality of (1.3), we have y(k) =
µk Y (0) g(x(0) ) . g(x(k) )
(2.16)
Multiplying the two sides of the first equality of (1.3) by (x(k) − ξ)T , we get
(1 − µk )[(x(k) − ξ)T ∇ f (x(k) ) + (x(k) − ξ)T ∇ g(x(k) )y(k) ] = −µk (x(k) − ξ)T (x(k) − x(0) ).
(2.17)
By Lemma 2.2, (2.15)–(2.17), we get
µk (kx(k) − ξk2 − kx(0) − ξk2 ) ≤ 2µk (x(k) − ξ)T (x(k) − x(0) ) = 2(1 − µk )[(ξ − x(k) )T ∇ f (x(k) ) + (ξ − x(k) )T ∇ g(x(k) )y(k) ] " # (0) (0) (k) T (k) (k) T (k) µk Y g (x ) ≤ 2(1 − µk ) (ξ − x ) ∇ f (x ) + (ξ − x ) ∇ g(x ) g(x(k) ) "
= 2(1 − µk ) (ξ − x(k) )T ∇ f (x(k) ) − µk Y (0) g(x(0) )
(x(k) − ξ)T ∇ g(x(k) ) . g(x(k) ) #
(2.18)
By (2.18), we have 2(1 − µk )(x(k) − ξ)T ∇ f (x(k) ) ≤ µk kx(0) − ξk2 − µk kx(k) − ξk2 − µk Y (0) g(x(0) )
(x(k) − ξ)T ∇ g(x(k) ) . g(x(k) )
(2.19)
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When x(k) ∈ Ω + (ξ), if kx(k) k → ∞, by assumption (C3 ), µk ∈ (0, 1] and for the case where ξ is a given point, it is easy to get that there exists some k such that kx(k) − ξk > M, and the right-hand side of (2.19) is strictly smaller than 0 while the left-hand side of (2.19) is strictly larger than 0, a contradiction. Hence the case where kx(k) k → ∞ in Ω + (ξ) is impossible. Since Ω − (ξ) is bounded, the case where kx(k) k → ∞ in Ω − (ξ) is also impossible. Then the component x of w is bounded in Ω for Ω = Ω − (ξ) ∪ Ω + (ξ). This completes the proof. As for how to trace the homotopy path numerically, there have been many predictor–corrector algorithms; see [1,3,4], etc., for references. Hence we omit this in this paper. Acknowledgements The authors express their sincere thanks to Professor Yong Li and Bo Yu for their instruction and many invaluable suggestions. References [1] E.L. Allgower, K. Georg, Numerical Continuation Methods: An Introduction, Springer-Verlag, 1990. [2] G.C. Feng, Z.H. Lin, B. Yu, Existence of interior pathway to the Karush–Kuhn–Tucker point of a nonconvex programming problems, Nonlinear Anal. 32 (6) (1998) 761–768. [3] Z.H. Lin, Z.P. Sheng, L. Yang, Y.Z. Bai, A predictor–corrector algorithm for tracing combined interior homotopy pathway, Math. Numer. Sin. 24 (2002) 405–416. [4] L.T. Watson, S.M. Holzer, M.C. Hansen, Tracking nonlinear equilibrium paths by a homotopy method, Nonlinear Anal. 7 (1983) 1271–1282.