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Title

Some properties of the solution of the vertical tensor complementarity problem


Abstract

In this paper, we mainly focus on the existence and uniqueness of the vertical tensor complementarity problem. Firstly, combining the generalized-order linear complementarity problem with the tensor complementarity problem, the vertical tensor complementarity problem is introduced. Secondly, we define some sets of special tensors, and illustrate the inclusion relationships. Finally, we show that the solution set of the vertical tensor complementarity problem is bounded under certain conditions, and some sufficient conditions for the existence and uniqueness of the solution of the vertical tensor complementarity problem are obtained from the view of the degree theory and the equal form of the minimum function.

1 Introduction

Over the past few decades, many mathematical researchers have conducted extensive research on the nonlinear complementarity problem. Today, the nonlinear complementarity problem has become a classical problem, and has important applications in operations research and applied science. For more detailed information, one can see [4, 10, 11, 14, 16] and references therein.

In recent years, with the rise of tensor research, a special case of the nonlinear complementarity problem: the tensor complementarity problem, has also received widespread attention and research. The tensor complementarity problem was first proposed in [28] by Song and Qi, i.e., find a vector x ∈ R<sup>n</sup> or show that there is no vector such thatwhere θ denotes the zero vector, A and q ∈ R<sup>n</sup> are a given m order n dimensional tensor and a vector, respectively, and A x<sup>m−1</sup> in [26] is defined aswhich is denoted by TCP( A ,q ). They studied the existence of the solution of the TCP( A ,q ) through the structure of the tensor, and obtained a sufficient and necessary condition for the non-negative symmetric tensor A . Bai et al. [1] studied the global uniqueness and the solvability of the TCP( A ,q ), and obtained two results: (1) if A is a P -tensor, then the TCP( A ,q ) may have more than one solution for some q ∈ R<sup>n</sup> ; (2) if A is a strong P -tensor, then the TCP( A ,q ) has GUS-property for all q ∈ R<sup>n</sup> . The readers can see [3, 13, 17, 22, 29] to know more about the existence of the solution of the TCP( A ,q ). In addition, many researchers are committed to studying the numerical algorithm for solving the TCP( A ,q ), see [5, 8, 9, 15, 17, 19, 30, 32].

Goeleven [12] introduced the generalized-order linear complementarity problem and studied the existence and uniqueness of its solution. The generalized-order linear complementarity problem has important applications in stochastic impulse problems, mixed lubrication problems and singular control problems in bounded intervals, see [23–25, 31].

In this paper, combining the generalized-order linear complementarity problem and the TCP( A ,q ), it is natural to introduce the vertical tensor complementarity problem, i.e., find a vector x ∈ R<sup>n</sup> or show that there is no vector such thatwhere A<sup>1</sup> , A<sup>2</sup> are given m order n dimensional tensors, and q<sub>1</sub> , q<sub>2</sub> ∈ R<sup>n</sup> are given vectors. We denote it by VTCP( ˜ A , ˜ q ) with ˜ A = {A<sup>1</sup> , A<sup>2</sup> } and ˜ q = { q<sub>1</sub> ,q<sub>2</sub> } . We consider the VTCP( ˜ A , ˜ q ) for two reasons: (1) it is equal to the tensor absolute value equation, see [9]; (2) it is the generalization form of the TCP( A ,q ), too. To our knowledge, so far, for the unique solution of the VTCP( ˜ A , ˜ q ), the necessary and sufficient condition is scarce , which is our main motivation. Hence, based on this, our aim is to address this problem, and obtain some useful results on the sufficient condition for the unique solution of the VTCP( ˜ A , ˜ q ).

It is a brief description for the rest of paper. In Section 2, we go over some preliminary knowledge. In Section 3, we introduce some tensor sets with special structures and demonstrate their properties. In Section 4, we show that the solution set of the VTCP( ˜ A , ˜ q ) is bounded under certain conditions, and some sufficient conditions for the existence and uniqueness of the solution of the VTCP( ˜ A , ˜ q ) are obtained on the base of the degree theory and the equal form of minimum function. In Section 5, some conclusions are given to end this paper.


2 Preliminaries

In this section, we review some definitions and notations.

We use A , B , C to represent the tensors, A , B , C to represent the matrices, x , y , z to represent the vectors. We use x<sub>i</sub> to represent the i th element of the vector x , I to represent the unit tensor (including the unit vector and the identity matrix), θ to represent the zero tensor (including the zero vector and the zero matrix). x<sup>[m]</sup> is a vector that x<sub>i</sub><sup>[m]</sup> = ( x<sub>i</sub> )<sup>m</sup> ,i = 1 , ··· ,n . Given a vector x ∈ R<sup>n</sup> , ( x) and (Given x,y ∈ R<sup>n</sup> , x ∧ y (or min { x,y } ) represents the vector, whose i th element is equal to min( x<sub>i</sub> ,y<sub>i</sub> ); x ∨ y (or max { x,y } ) represents the vector, whose i th element is equal to max( x<sub>i</sub> ,y<sub>i</sub> ). x > y , x ≥ y , x < y and x ≤ y mean x<sub>i</sub> > y<sub>i</sub> , x<sub>i</sub> ≥ y<sub>i</sub> , x<sub>i</sub> < y<sub>i</sub> and x<sub>i</sub> ≤ y<sub>i</sub> , respectively, with i = 1 , ··· ,n . R<sub>+</sub><sup>n</sup> := { x ∈ R<sup>n</sup> : x<sub>i</sub> ≥ θ,i = 1 , ··· ,n } and R<sub>++</sub><sup>n</sup> := { x ∈ R<sup>n</sup> : x<sub>i</sub> > θ,i = 1 , ··· ,n } . [ n ] := { 1 , ··· ,n } .

A tensor A = ( a<sub>i<sub>1</sub>i<sub>2</sub>···i<sub>m</sub></sub> ) is a multidimensional array of elements represented as a<sub>i<sub>1</sub>i<sub>2</sub>···i<sub>m</sub></sub> ∈ F , where i<sub>j</sub> = 1 , 2 , ··· ,n<sub>j</sub> , j = 1 , 2 , ··· ,m and F represents a field. In this paper, we consider the problem in the field of real numbers, i.e., F = R . m is called the order of A , and ( n<sub>1</sub> , ··· ,n<sub>m</sub> ) is called the dimension of A . Obviously, when m = 2, the tensor A is a matrix. A is called an m order n dimensional tensor, if n<sub>1</sub> = n<sub>2</sub> = ··· = n<sub>m</sub> . R<sup>[m,n]</sup> represents the set of all m order n dimensional tensors. a<sub>i,···,i</sub> is called the principal diagonal element of A ∈ R<sup>[m,n]</sup> , i = 1 , ··· ,n . A is called a symmetric tensor if a<sub>i<sub>1</sub>i<sub>2</sub>···i<sub>m</sub></sub> is invariant no matter how i<sub>1</sub> ,i<sub>2</sub> , ··· ,i<sub>m</sub> are arranged. Denote the set of all m order n dimensional real symmetric tensors as S<sup>[m,n]</sup> .

The generalized product of tensors [27] is required. Let A and B are n dimensional tensors with order m ( m ≥ 2) and k ( k ≥ 1), respectively. Define the product A·B to be the following tensor C of order ( m − 1)( k − 1) + 1 and dimension n :By Theorem 1.1 and Example 1.1 in [27], the generalized tensor product satisfies the associative law, and A x<sup>m−1</sup> can be written as A · x . By Proposition 1.1 in [27], the generalized tensor product satisfies ( A<sup>1</sup> + A<sup>2</sup> ) · B = A<sup>1</sup> · B + A<sup>2</sup> · B , where A<sup>1</sup> and A<sup>2</sup> have the same order, and also satisfies A · ( B<sup>1</sup> + B<sup>2</sup> ) = A · B<sup>1</sup> + A · B<sup>2</sup> , where A is a matrix.

Based on the generalized product of tensors, the left and right inverses of tensors [2] are defined. Let A ∈ C<sub>t</sub><sup>(n,n)</sup> . If there is a tensor B ∈ C<sub>k</sub><sup>(n,n)</sup> such that A · B = I , where C<sub>k</sub><sup>(m,n)</sup> represents the set of all complex tensors with order k and dimension ( m,n, ··· ,n ), then it is said that A is left inverse of B , and is denoted as B<sup>L<sub>t</sub></sup> ; B is right inverse of A , and is denoted as A<sup>R<sub>k</sub></sup> .

Inspired by the definition of A x<sup>m−1</sup> , we define the following AB as the product of A ∈ R<sup>[m,n]</sup> and B ∈ R<sup>[m−1,n]</sup> :Now, we review the R -tensor [28] and the Z -tensor [6]. ˜ A is called a R -tensor if and only if the following system is inconsistent:A is called a Z -tensor if and only if all its off-diagonal entries are non-positive.



3 Sets of tensors with special structures

In this section, we introduce some sets of tensors with special structures.

Definition 3.1. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> . We say that ˜ A is of1. type V R<sub>0</sub> if A<sup>1</sup> · x ∧ A<sup>2</sup> · x = θ ⇒ x = θ ; 2. type V E if A<sup>1</sup> · x ∧ A<sup>2</sup> · x ≤ θ ⇒ x = θ ; 3. type V P if A<sup>1</sup> · x ∧ A<sup>2</sup> · x ≤ θ ≤ A<sup>1</sup> · x ∨ A<sup>2</sup> · x ⇒ x = θ ; 4. type V P - I if A<sup>1</sup> · x ∧A<sup>2</sup> · x ≤ θ ≤ A<sup>1</sup> · x ∨A<sup>2</sup> · x have no solution in R<sub>+</sub><sup>n</sup> / { θ } ∪ R<sub>−</sub><sup>n</sup> / { θ } ; 5. type V P - II if there is no tensor Z ∈ S<sup>[m−1,n]</sup> , whose principal diagonal elements are not all zeros such that A<sup>1</sup> Z ∧ A<sup>2</sup> Z ≤ θ ≤ A<sup>1</sup> Z ∨ A<sup>2</sup> Z ; 6. type semi-positive if there exists a vector x > θ such that A<sup>1</sup> · x > θ and A<sup>2</sup> · x > θ ; 7. type V Q if the VTCP ( ˜ A , ˜ q ) has a solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .Remark 3.1. For Definition 3.1, the condition of item 1 is equivalent to that the following system is inconsistent:the condition of item 2 is equivalent to that ∀ x 6 = θ , ∃ i ∈ [ n ] satisfies ( A<sup>1</sup> · x )<sub>i</sub> > θ , ( A<sup>2</sup> · x )<sub>i</sub> > θ ; the condition of item 3 is equivalent to that ∀ x 6 = θ , ∃ i ∈ [ n ] satisfies ( A<sup>1</sup> · x )<sub>i</sub> ( A<sup>2</sup> · x )<sub>i</sub> > θ ; the condition of item 4 is equivalent to that ∀ x ∈ R<sub>+</sub><sup>n</sup> / { θ } ∪ R<sub>−</sub><sup>n</sup> / { θ } , ∃ i ∈ [ n ] satisfies ( A<sup>1</sup> · x )<sub>i</sub> ( A<sup>2</sup> · x )<sub>i</sub> > θ ; the condition of item 5 is equivalent to that ∀Z ∈ S<sup>[m−1,n]</sup> with its principal diagonal elements being not all zeros, ∃ i ∈ [ n ] satisfies ( A<sup>1</sup> Z )<sub>i</sub> ( A<sup>2</sup> Z )<sub>i</sub> > θ .

Proposition 3.1. type V E ⊆ type V P ⊆ type V R<sub>0</sub> ; type V P - II ⊆ type V P ⊆ type V P - I . Proof. Assume that ˜ A is of type V E , but not type V P . Then there is a vector x 6 = θ such thatBy the left inequality in (3.2), ˜ A is not type V E . This contradicts the assumption.

Assume that ˜ A is of type V P , but not type V R<sub>0</sub> . Then there is a vector x 6 = θ such that A<sup>1</sup> · x ∧ A<sup>2</sup> · x = θ. Furthermore,which implies that ˜ A is not type V P . This contradicts the assumption.

Thus, based on the above discussion, we can draw a conclusion that type V E ⊆ type V P ⊆ type V R<sub>0</sub> .

Similarly, type V P - II ⊆ type V P ⊆ type V P - I clearly holds.Examples 3.1-3.3 are given to illustrate three aspects: (1) the existence of type V E ; (2) type V P is not necessarily type V E ; (3) type V R<sub>0</sub> is not necessarily type V P .

Example 3.1. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } , whereBy the calculation, we obtainSuppose x 6 = θ , we discuss three cases: − x<sub>1</sub><sup>2</sup>+3 x<sub>1</sub> x<sub>2</sub> = 0 and − x<sub>1</sub><sup>2</sup> +3 x<sub>1</sub> x<sub>2</sub> < 0 . (1) if − x<sub>1</sub><sup>2</sup> + 3 x<sub>1</sub> x<sub>2</sub> > 0 , then 3 x<sub>1</sub> x<sub>2</sub> > 0; (2) if − x<sub>1</sub><sup>2</sup> + 3 x<sub>1</sub> x<sub>2</sub> = 0 , then x<sub>2</sub> 6 = 0 ,x<sub>1</sub><sup>2</sup> − 3 x<sub>1</sub> x<sub>2</sub> + x<sub>2</sub><sup>2</sup> > 0 and 2 x<sub>1</sub><sup>2</sup> − 3 x<sub>1</sub> x<sub>2</sub> + x<sub>2</sub><sup>2</sup> > 0; (3) if − x<sub>1</sub><sup>2</sup> + 3 x<sub>1</sub> x<sub>2</sub> < 0 , then x<sub>1</sub><sup>2</sup> − 3 x<sub>1</sub> x<sub>2</sub> + x<sub>2</sub><sup>2</sup> > 0 and 2 x<sub>1</sub><sup>2</sup> − 3 x<sub>1</sub> x<sub>2</sub> + x<sub>2</sub><sup>2</sup> > 0 .From the above analysis, ˜ A is of type V E .

Example 3.2. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } , whereBy the calculation, we obtainSuppose x 6 = θ , we discuss two cases: x<sub>1</sub> 6 = 0 and x<sub>1</sub> = 0 .(1) if x<sub>1</sub> 6 = 0 , then ( A<sup>1</sup> · x )<sub>1</sub> ( A<sup>2</sup> · x )<sub>1</sub> = x<sub>1</sub><sup>6</sup> + 4 x<sub>1</sub><sup>4</sup> x<sub>2</sub><sup>2</sup> > 0; (2) if x<sub>1</sub> = 0 , then x<sub>2</sub> 6 = 0 , ( A<sup>1</sup> · x )<sub>2</sub> ( A<sup>2</sup> · x )<sub>2</sub> = x<sub>2</sub><sup>6</sup> + 3 x<sub>1</sub><sup>2</sup> x<sub>2</sub><sup>4</sup> + 2 x<sub>1</sub><sup>4</sup> x<sub>2</sub><sup>2</sup> > 0 .From the above analysis, ˜ A is of type V P . Let x = ( − 1 , − 1)<sup>T</sup> . Then ( A<sup>1</sup> x<sup>3</sup> )<sub>1</sub> = − 1 , ( A<sup>2</sup> x<sup>3</sup> )<sub>1</sub> = − 5 , ( A<sup>1</sup> x<sup>3</sup> )<sub>2</sub> = − 2 , ( A<sup>2</sup> x<sup>3</sup> )<sub>2</sub> = − 3 . So, ˜ A is not of type V E .Example 3.3. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } , whereBy the calculation, we obtainSuppose x 6 = θ , we discuss two cases: x<sub>2</sub><sup>2</sup> − x<sub>1</sub> x<sub>2</sub> > 0 and x<sub>2</sub><sup>2</sup> − x<sub>1</sub> x<sub>2</sub> = 0 .(1) if x<sub>2</sub><sup>2</sup> − x<sub>1</sub> x<sub>2</sub> > 0 , then x<sub>2</sub> 6 = 0 , x<sub>2</sub><sup>2</sup> > 0; (2) if x<sub>2</sub><sup>2</sup> − x<sub>1</sub> x<sub>2</sub> = 0 , then x<sub>2</sub> = 0 or x<sub>2</sub> = x<sub>1</sub> 6 = 0 . If x<sub>2</sub> = 0 , then x<sub>1</sub><sup>2</sup> + x<sub>2</sub><sup>2</sup> > 0 and − x<sub>1</sub><sup>2</sup> − 2 x<sub>1</sub> x<sub>2</sub> − x<sub>2</sub><sup>2</sup> < 0 . If x<sub>2</sub> = x<sub>1</sub> 6 = 0 , then x<sub>1</sub><sup>2</sup> + x<sub>2</sub><sup>2</sup> > 0 and − x<sub>1</sub><sup>2</sup> − 2 x<sub>1</sub> x<sub>2</sub> − x<sub>2</sub><sup>2</sup> < 0 .From the above analysis, the corresponding system (3.1) has no solution. So, ˜ A is of type V R<sub>0</sub> . Let x = (1 , 1)<sup>T</sup> . Then ( A<sup>1</sup> · x )<sub>1</sub> ( A<sup>2</sup> · x )<sub>1</sub> = 0 , ( A<sup>1</sup> · x )<sub>2</sub> ( A<sup>2</sup> · x )<sub>2</sub> = − 8 . Thus, ˜ A is not of type V P .

Examples 3.4 and 3.5 are given to illustrate these three aspects: (1) the existence of type V P - II ; (2) type V P is not necessarily type V P - II ; (3) type V P - I is not necessarily type V P .

Example 3.4. Let A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[3,2]</sup> , whereLet Z ∈ S<sup>[2,2]</sup> be an arbitrary symmetric tensor and be of formwhere x<sub>1</sub> and x<sub>3</sub> at least one is not zero. By the calculation, we obtainObviously, at least one of ( x<sub>1</sub> + x<sub>3</sub> )<sup>2</sup> and − x<sub>1</sub> x<sub>3</sub> is greater than zero. Thus, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V P - II .

In Example 3.2, letThen ( A<sup>1</sup> Z )<sub>i</sub> ( A<sup>2</sup> Z )<sub>i</sub> = 0 for all i ∈ { 1 , 2 } . Thus, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } in Example 3.2 is not of type V P - II .Example 3.5. Let A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[3,2]</sup> , whereBy the calculation, we obtainObviously, A<sup>1</sup> · x ∧ A<sup>2</sup> · x ≤ θ ≤ A<sup>1</sup> · x ∨ A<sup>2</sup> · x have no solution in R<sub>+</sub><sup>n</sup> / { θ } ∪ R<sub>−</sub><sup>n</sup> / { θ } . So, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V P - I . Let x = (1 , − 1)<sup>T</sup> . Then ( A<sup>1</sup> · x )<sub>1</sub> ( A<sup>2</sup> · x )<sub>1</sub> = − 1 , ( A<sup>1</sup> · x )<sub>2</sub> ( A<sup>2</sup> · x )<sub>2</sub> = − 2 . Thus, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is not of type V P .

Next, a special tensor set is introduced. Inspired by the P -function [18], the following condition is givenfor all x,y ∈ R<sup>n</sup> and x 6 = y , where G ( x ), F ( x ) are two functions. Now, we define a new tensor set, see Definition 3.2.

Definition 3.2. We say that ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type strong V P if the mappings G ( x ) := A<sup>1</sup> x<sup>m−1</sup> and F ( x ) := A<sup>2</sup> x<sup>m−1</sup> satisfy the condition (3.3) for all x,y ∈ R<sup>n</sup> and x 6 = y .

Proposition 3.2. There is no type strong V P tensor set with odd order.

Proof. Assume that ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type strong V P with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> , and m is an odd number. Obviously,This contradicts the assumption.

Proposition 3.3. type V P - II with even order ⊆ type strong V P ⊆ type V P .

Since there is no type strong V P tensor set with odd order, ˜ A in Example 3.1 is of type V P , but not type strong V P . Examples 3.6 and 3.7 are given to illustrate three aspects: (1) the existence of type strong V P ; (2) type strong V P is not necessarily type V P - II with even order; (3) type V P with even order is not necessarily type strong V P .

Example 3.6. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } , whereA<sup>2</sup> (11:1 0 0 0A<sup>2</sup> (10 0A<sup>2</sup> (2 , 1 , : , :)0 − 1 1 0A<sup>2</sup> (2 , 2 , : , :) = 0 0 0 1 .By the calculation, we obtainSuppose x 6 = y , we discuss two cases: x<sub>1</sub> 6 = y<sub>1</sub> and x<sub>1</sub> = y<sub>1</sub>(1) if x<sub>1</sub> 6 = y<sub>1</sub> , then ( A<sup>1</sup> x<sup>3</sup> − A<sup>1</sup> y<sup>3</sup> )<sub>1</sub> ( A<sup>2</sup> x<sup>3</sup> − A<sup>2</sup> y<sup>3</sup> )<sub>1</sub> = ( x<sub>1</sub><sup>3</sup>y<sub>1</sub> , then x<sub>2</sub> 6 = y<sub>2</sub> andFrom the above analysis, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type strong V P . LetThen ( A<sup>1</sup> Z )<sub>1</sub> ( A<sup>2</sup> Z )<sub>1</sub> = 0 , ( A<sup>1</sup> Z )<sub>2</sub> ( A<sup>2</sup> Z )<sub>2</sub> = 0 . Thus, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is not type V P - II with even order.

Example 3.7. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } , whereBy the calculation, we obtain.

Suppose x 6 = θ , we discuss two cases: x<sub>1</sub> x<sub>2</sub> 6 = 0 and x<sub>1</sub> x<sub>2</sub> = 0 .(1) if x<sub>1</sub> x<sub>2</sub> 6 = 0 , then ( A<sup>1</sup> · x )<sub>1</sub> ( A<sup>2</sup> · x )<sub>1</sub> = x<sub>1</sub><sup>4</sup> x<sub>2</sub><sup>2</sup> > 0 ; (2) if x<sub>1</sub> x<sub>2</sub> = 0 , then ( A<sup>1</sup> · x )<sub>2</sub> ( A<sup>2</sup> · x )<sub>2</sub> = ( x<sub>1</sub><sup>3</sup> + x<sub>2</sub><sup>3</sup> )<sup>2</sup> > 0 .From the above analysis, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V P with even order. Let x = (0 , 1)<sup>T</sup> , y = (1 , 0)<sup>T</sup> . ThenA<sup>1</sup> y<sup>3</sup> 2 A<sup>2</sup> x<sup>3</sup> − A<sup>2</sup> y<sub>2</sub><sup>3</sup> = 0 . Thus, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is not of type strong V P .



4 Solution of the VTCP

In this section, we show that the solution set of the VTCP( ˜ A , ˜ q ) is bounded under certain conditions, and some sufficient conditions for the existence and uniqueness of the solution of the VTCP( ˜ A , ˜ q ) are obtained from the view of the degree theory and the equal form of the minimum function.

To show that the solution set of the VTCP( ˜ A , ˜ q ) is bounded under certain conditions, we give the following result, see Theorem 4.1.

Theorem 4.1. ˜ A is of type V R<sub>0</sub> if and only if the solution set of the VTCP ( ˜ A , ˜ q ) for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> is bounded.

Proof. It is now proved that the following three conditions are equivalent:( i ) ˜ A is of type V R<sub>0</sub> ; ( ii ) ∀ q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> , t > 0, s ∈ R , theois bounded;

( iii ) The solution set of the VTCP( ˜ A , ˜ q ) for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> is bounded.( i ) ⇒ ( ii ): Assume that there exist q<sub>1</sub><sup>∗</sup> ,q<sub>2</sub><sup>∗</sup> ∈ R<sup>n</sup> , t<sup>∗</sup> > 0 and s<sup>∗</sup> ∈ R such that the set τ ( q<sub>1</sub><sup>∗</sup> ,q<sub>2</sub><sup>∗</sup> ,s<sup>∗</sup> ,t<sup>∗</sup> ) is not bounded. Then there is an unbounded sequence x<sup>k</sup> ∈ τ ( q<sub>1</sub><sup>∗</sup> ,q<sub>2</sub><sup>∗</sup> ,s<sup>∗</sup> ,t<sup>∗</sup> ). Since the sequence {<sup>x<sup>k</sup>kx<sup>k</sup>k</sup> } is bounded, there is a convergence subsequence {<sup>x<sup>kj</sup>kx<sup>kj</sup>k</sup> } such that when j → ∞ , x<sup>kj</sup> k<sup>x<sup>kj</sup></sup> k → x<sub>1</sub> 6 = θ and x<sup>k<sub>j</sub></sup> → ∞ . Thus, it follows thatm −≥ θ, A<sup>2</sup>

x<sup>k<sup>j</sup></sup>

x<sup>k<sup>j</sup></sup>x<sup>k<sup>j</sup></sup>



A<sup>1</sup>

x<sup>k<sup>j</sup></sup>

x<sup>k<sup>j</sup></sup>

≤

s<sup>∗</sup>Let j → + ∞ . Then)<sup>m−1T</sup> t<sup>∗</sup> A<sup>2</sup> ( x<sub>1</sub> )<sup>m−1</sup> ≤ θ.

Thus, A<sup>1</sup> ( x<sub>1</sub> )<sup>m−1T</sup> A<sup>2</sup> ( x<sub>1</sub> )<sup>m−1</sup> = θ . This contradicts that ˜ A is of type V R<sub>0</sub> . ( ii ) ⇒ ( iii ): Let s = 0, t = 1. Then the set τ ( q<sub>1</sub> ,q<sub>2</sub> , 0 , 1)

=

n

x : q<sub>1</sub> + A<sup>1</sup> x<sup>m−1</sup> ≥ 0 ,q<sub>2</sub> + A<sup>2</sup> x<sup>m−1</sup> ≥ 0 , q<sub>1</sub> + A<sup>1</sup> x<sup>m−1</sup> T q<sub>2</sub> + A<sup>2</sup> x<sup>m−1</sup> ≤ 0 o =

n

x : q<sub>1</sub> + A<sup>1</sup> x<sup>m−1</sup> ≥ 0 ,q<sub>2</sub> + A<sup>2</sup> x<sup>m−1</sup> ≥ 0 , q<sub>1</sub> + A<sup>1</sup> x<sup>m−1</sup> T q<sub>2</sub> + A<sup>2</sup> x<sup>m−1</sup> = 0 ois bounded, which implies that the solution set of the VTCP( ˜ A , ˜ q ) is bounded.

( iii ) ⇒ ( i ): Assume that ˜ A is not of type V R<sub>0</sub> . There is x 6 = θ such that A<sup>1</sup> · x ∧A<sup>2</sup> · x = θ . Hence, kx ∈ τ ( θ,θ, 0 , 1) for all k ≥ θ . This contradicts the condition ( iii ).

By the above discussion, the conclusion in Theorem 4.1 holds.

Theorem 4.2. ˜ A is of type V E if and only if the VTCP ( ˜ A , ˜ q ) has at most θ solution for all q<sub>1</sub> ≥ θ , q<sub>2</sub> ≥ θ .

Proof. Sufficiency. Assume that ˜ A is not of type V E , i.e., there is a vector x 6 = θ such that A<sup>1</sup> · x ∧A<sup>2</sup> · x ≤ θ . If A<sup>1</sup> · x ∧A<sup>2</sup> · x = θ , then it contradicts that the VTCP( ˜ A , ˜ q ) has at most θ solution for all q<sub>1</sub> ≥ θ , q<sub>2</sub> ≥ θ . If A<sup>1</sup> · x ∧ A<sup>2</sup> · x < θ , then we takeObviously, q<sub>1</sub> ,q<sub>2</sub> ≥ θ , and x is a solution of the VTCP( ˜ A , ˜ q ). This contradicts that the VTCP( ˜ A , ˜ q ) has at most θ solution for all q<sub>1</sub> ≥ θ , q<sub>2</sub> ≥ θ . So, ˜ A is of type V E .

Necessity. Obviously, θ is the solution of the VTCP( ˜ A , ˜ q ) for some q<sub>1</sub> , q<sub>2</sub> ≥ θ . Assume that there exist q<sub>1</sub> , q<sub>2</sub> ≥ θ such that the VTCP( ˜ A , ˜ q ) has a nonzero solution x . Since q<sub>1</sub> ,q<sub>2</sub> ≥ θ and x is a nonzero solution of the VTCP( ˜ A , ˜ q ), we obtainThis contradicts that ˜ A is of type V E tensor. So, the VTCP( ˜ A , ˜ q ) has at most θ solution for all q<sub>1</sub> , q<sub>2</sub> ≥ θ .

Next, from the view of the degree theory, we discuss the existence of the solution of the VTCP( ˜ A , ˜ q ).

Here is a brief review of degree theory. Suppose that Ω is a bounded open set in R<sup>n</sup> . ∂ Ω and Ω represent the boundary of Ω and the set ∂ Ω ∪ Ω, respectively. Let the continuous function f : Ω −→ R<sup>n</sup> be given and p / ∈ f ( ∂ Ω). Then the degree of f = p on Ω can be well defined and denoted as deg ( f, Ω ,p ), which is an integer and implies the number of solutions for f = p , see [10].

Let ˜ A be of type V R<sub>0</sub> ,Then F (<sup>˜A</sup> ) = θ ⇒ x = θ . Let Ω be a bounded open set that contains θ . Then deg F (<sup>˜A</sup> ) , Ω ,θ is defined. Furthermore, deg F (<sub>A</sub><sup>˜</sup> ) , Ω ,θ is independent of the bounded open set Ω. We denote it by VTCP- deg ˜ A .

According to the degree theory, the following results can be obtained.

Theorem 4.3. If ˜ A is of type V R<sub>0</sub> and VTCP- deg ˜ A is nonzero, then ˜ A is of type V Q .Proof. Let q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> andBy Theorem 4.1, the setis bounded. Hence, there is a bounded open set Ω containing the set X and X ∩ ∂ Ω = ∅ . Obviously, F ( x,t ) is a homotopy connecting the mappings F ( x, 1) and F ( x ), where F ( x ) = F ( x, 0). By the homotopy invariance property of the degree [21], we obtainSince VTCP- deg ˜ A is nonzero, together with the property of the degree [21], there is a solution of F ( x, 1) = θ in Ω, which implies that ˜ A is of type V Q .

Theorem 4.4. If ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V R<sub>0</sub> with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> , m > 2 and A<sup>2</sup> is a R -tensor, then ˜ A is of type V Q .

Proof. LetG ( x ) ≤ s o

,

where

F ( x ) = tq<sub>1</sub> + (1 − t ) x + t A<sup>1</sup> x<sup>m−1</sup> and G ( x ) = tq<sub>2</sub> + (1 − t ) d + a A<sup>2</sup> x<sup>m−1</sup> .

Similar to the proof of Theorem 4.1, τ ( q<sub>1</sub> ,q<sub>2</sub> ,s,d,a,t ) is bounded for any q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> , s ∈ R , t ∈ (0 , 1], a > 0 and d > θ . Thus, τ ( q<sub>1</sub> ,q<sub>2</sub> , 0 ,I, 1 ,t ) is bounded for any q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> and t ∈ (0 , 1], which is equivalent to that the set of solutions ofis bounded for any q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> and t ∈ (0 , 1]. Since A<sup>2</sup> is a R -tensor, (4.1) has only one θ solution for any q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> when t = 0. Letwhere q<sub>1</sub> , q<sub>2</sub> ∈ R<sup>n</sup> are any given vectors. Based on the above analysis, the set X = { x : H ( x,t ) = θ,t ∈ [0 , 1] } is bounded in R<sup>n</sup> . So, there is a bounded open set Ω containing the set X and X ∩ ∂ Ω = ∅ . Since A<sup>2</sup> is a R -tensor, H ( x, 0) = θ has only θ solution. Note that I + A<sup>2</sup> x<sup>m−1</sup> > θ in the field around the θ vector, which implies that H ( x, 0) = x in the field around the θ vector. So, degree ( H ( x, 0) , Ω ,θ ) = 1. Obviously, H ( x,t ) is a homotopy connecting the mappingsBy the homotopy invariance property of the degree [21],Thus, by the property of the degree [21], H ( x, 1) = θ has a solution, which implies that the result in Theorem 4.4 holds.Next, some uniqueness conditions of the VTCP( ˜ A , ˜ q ) are obtained.

Theorem 4.5. If ˜ A is of type V P - II with even order, then the VTCP ( ˜ A , ˜ q ) has at most one solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .

Proof. Assume that x and y are the two different solutions of the VTCP( ˜ A , ˜ q ). ThenandFrom (4.2) and (4.3), we obtainandNote that (4.5) is equivalent to thatFrom (4.4) and (4.6), z ∈ S<sup>[m−1,n]</sup> satisfieswhere Z = x<sup>m−1</sup> − y<sup>m−1</sup> . Since the order is even, the principal diagonal elements of Z are not all zeros. This contradicts that ˜ A is of type V P - II . So, the VTCP( ˜ A , ˜ q ) has at most one solution.

Theorem 4.6. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> and m be an even number. If the VTCP ( ˜ A , ˜ q ) for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> has at most one solution, then ˜ A is of type V P - I .

Proof. Assume that ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is not of type V P - I . Then there is a vector x ∈ R<sub>+</sub><sup>n</sup> / { θ } ∪ R<sub>−</sub><sup>n</sup> / { θ } such that A<sup>1</sup> · x ∧ A<sup>2</sup> · x ≤ θ ≤ A<sup>1</sup> · x ∨ A<sup>2</sup> · x . Thus,Let q<sub>1</sub>Then, we need to proveandHere, we only prove (4.9) holds. We consider two cases: x ∈ R<sub>+</sub><sup>n</sup> / { θ } and x ∈ R<sub>−</sub><sup>n</sup> / {}(1) if x ∈ R<sub>+</sub><sup>n</sup> / { θ } , then ( A<sup>1</sup> · x )<sub>+</sub> − ( A<sup>1</sup> · x )<sub>−</sub> = A<sup>1</sup> · x and A<sup>1</sup> · ( x )<sub>+</sub> − A<sup>1</sup> · ( x )<sub>−</sub> = A<sup>1</sup> · ( x )<sub>+</sub> = A<sup>1</sup> · x ;(2) if x ∈ R<sub>−</sub><sup>n</sup> / { θ } , thenandBased on the above analysis, (4.9) holds. By (4.8), (4.9) and (4.10), we obtainandθ, which implies that ( x )<sub>+</sub> and ( x )<sub>−</sub> are two different solutions of the VTCP( ˜ A , ˜ q ). This contradicts that the VTCP( ˜ A , ˜ q ) has at most one solution. So, ˜ A is of type V P - I .

Lemma 4.1. If ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type strong V P , then the VTCP ( ˜ A , ˜ q ) has at most one solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .

Proof. If ˜ A is of type strong V P , thenholds for all x,y ∈ R<sup>n</sup> and x 6 = y . Let x , y be the two different solutions of the VTCP( ˜ A , ˜ q ). Similar to the proof of Theorem 4.5, (4.7) holds. This contradicts that ˜ A is of type strong V P . So, the VTCP( ˜ A , ˜ q ) has at most one solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .

Based on Proposition 3.1, Proposition 3.3, Theorem 4.4 and Lemma 4.1, we can obtain a sufficient condition for the existence and uniqueness of the solution of the VTCP( ˜ A , ˜ q ), see Theorem 4.7.

Theorem 4.7. If ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type strong V P with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> , m > 2 and A<sup>2</sup> is a R -tensor, then the VTCP ( ˜ A , ˜ q ) has a unique solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .

Furthermore, based on Proposition 3.3 and Theorem 4.7, Corollary 4.1 can be obtained.

Corollary 4.1. If ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V P - II with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> , m is an even number and greater than 2, and A<sup>2</sup> is a R tensor, then the VTCP ( ˜ A , ˜ q ) has a unique solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .

In fact, Corollary 4.1 can also be obtained by combining Proposition 3.1, Theorem 4.4 and Theorem 4.5.

Now, we give the following example to illustrate Theorem 4.7, see Example 4.1.

Example 4.1. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } , where1 1 , A<sup>2</sup> (1 , 1 , : , :) = 1 0 0 0 , A<sup>2</sup> (1 , 2 , : , :) = 0 0 0 0 , A<sup>2</sup> (2 , 1 , : , :) = 0 − 1 1 0 , A<sup>2</sup> (2 , 2 , : , :) = 0 0 0 1 .Clearly, from Example 3.6, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type strong V P ,Suppose θ ≤ x 6 = θ and t ≥ 0 , we discuss two cases: x<sub>1</sub> > 0 and x<sub>1</sub> = 0 .(1) if x<sub>1</sub> > 0 , then x<sub>1</sub><sup>3</sup> + t > 0;

(2) if x<sub>1</sub> = 0 , then x<sub>1</sub><sup>3</sup> + t ≥ 0 , but x<sub>2</sub> > 0 ,x<sub>2</sub><sup>3</sup> + t > 0 .Thus, A<sup>2</sup> is a R-tensor. By the above analysis, ˜ A satisfies the conditions of Theorem 4.7 .

Now, we prove that the VTCP ( ˜ A , ˜ q ) has a unique solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> . Let q<sub>1</sub> = ( a<sub>1</sub> ,a<sub>2</sub> )<sup>T</sup> and q<sub>2</sub> = ( b<sub>1</sub> ,b<sub>2</sub> )<sup>T</sup> . We only need to prove that the following system has a unique solutionNote that+=where c = min { a<sub>1</sub> ,b<sub>1</sub> } . So, x<sub>1</sub> = 3 √ − c . Now, we show that(4.11)

has a unique solution, where t = 3 √ − c<sup>2</sup> ≥ 0 . Let f= x<sub>2</sub><sup>3</sup> + b<sub>2</sub> .

Obviously, f<sup>′</sup> ( x<sub>2</sub> ) ≥ g<sup>′</sup> ( x<sub>2</sub> ) ≥ 0 for any x<sub>2</sub> ∈ R . So there is only one x<sub>∗</sub> such that f ( x<sub>∗</sub> ) = g ( x<sub>∗</sub> ) . Clearly, f ( x<sub>2</sub> ) and g ( x<sub>2</sub> ) both have only one zero point, denoted as x<sub>1</sub><sup>∗</sup> and x<sub>2</sub><sup>∗</sup> , respectively. We discusse two cases: f ( x<sub>∗</sub> ) = g ( x<sub>∗</sub> ) ≥ 0 and f ( x<sub>∗</sub> ) = g ( x<sub>∗</sub> ) < 0 .(1) if f ( x<sub>∗</sub> ) = g ( x<sub>∗</sub> ) ≥ 0 , then the unique solution of (4.11) is x<sub>2</sub> = x<sub>1</sub><sup>∗</sup> ; (2) if f ( x<sub>∗</sub> ) = g ( x<sub>∗</sub> ) < 0 , then the unique solution of (4.11) is x<sub>2</sub> = x<sub>2</sub><sup>∗</sup> .Thus, the VTCP ( ˜ A , ˜ q ) has a unique solution x = 3 √ − c,x<sub>1</sub><sup>∗</sup> T or x = 3 √ − c,x<sub>2</sub><sup>∗</sup> T .

Next, we turn to use the equal form of minimum function to obtain the sufficient condition for the existence and uniqueness of solution of the VTCP( ˜ A , ˜ q ). To achieve this goal, the following lemmas are introduced.

Lemma 4.2. If A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> are two Z -tensors, then D<sub>1</sub> ·A<sup>1</sup> + D<sub>2</sub> ·A<sup>2</sup> is a Z -tensor, where D<sub>1</sub> ∈ R<sup>[2,n]</sup> and D<sub>2</sub> ∈ R<sup>[2,n]</sup> are any diagonal matrices with the principal diagonal elements greater than or equal to zero.

Proof. Let D<sub>1</sub> = ( d<sub>i<sub>1</sub>i<sub>2</sub></sub> ) be a diagonal matrix that principal diagonal elements are d<sub>1</sub><sup>(1)</sup> ,d<sub>2</sub><sup>(1)</sup> , ··· , d<sub>n</sub><sup>(1)</sup> and d<sub>i</sub><sup>(1)</sup> ≥ 0 for all i ∈ [ n ]. Then,

where i ∈ [ n ] and α<sub>1</sub> ∈ [ n ]<sup>m−1</sup> . Obviously, D<sub>1</sub> · A<sup>1</sup> is a Z -tensor. Similarly, D<sub>2</sub> · A<sup>2</sup> is also a Z -tensor. Hence, D<sub>1</sub> · A<sup>1</sup> + D<sub>2</sub> · A<sup>2</sup> is a Z -tensor.Lemma 4.3. If ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type semi-positive, then D<sub>1</sub> ·A<sup>1</sup> + D<sub>2</sub> ·A<sup>2</sup> is a semi-positive tensor, where D<sub>1</sub> , D<sub>2</sub> ∈ R<sup>[2,n]</sup> are any diagonal matrices with the principal diagonal elements greater than or equal to zero and D<sub>1</sub> + D<sub>2</sub> 6 = θ .

Proof. Since ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type semi-positive, there exists a vector x > θ such that A<sup>1</sup> x<sup>m−1</sup> > θ and A<sup>2</sup> x<sup>m−1</sup> > θ . Obviously,where D<sub>1</sub> , D<sub>2</sub> ∈ R<sup>[2,n]</sup> are any diagonal matrices with the principal diagonal elements greater than or equal to zero and D<sub>1</sub> + D<sub>2</sub>Lemma 4.4 ([6]) . A Z -tensor is a strong M -tensor if snd only if it is semi-positive.

Lemma 4.5 ([33]) . Let all a<sub>i</sub> , b<sub>i</sub> ∈ R , i = 1 , 2 , ··· ,n . Then there exist λ<sub>i</sub> ∈ [0 , 1] with

P

n i =1 λ<sub>i</sub> = 1 such thatFrom Lemma 4.5, Lemma 4.6 can be easily derived.

Lemma 4.6. Let A<sup>1</sup> , A<sup>2</sup> ∈ T<sup>m,n</sup> and q<sub>1</sub> , q<sub>2</sub> ∈ R<sup>n</sup> . Then there exist D<sub>1</sub> and D<sub>2</sub> such that

min q<sub>1</sub> + A<sup>1</sup> x<sup>m−1</sup>A<sup>2</sup>− minwhere D<sub>1</sub> , D<sub>2</sub> are diagonal matrices that diagonal elements are d<sup>(1)</sup> = d<sub>1</sub><sup>(1)</sup> ,d<sub>2</sub><sup>(1)</sup> , ··· ,d<sub>n</sub><sup>(1)</sup> ∈ [0 , 1]<sup>n</sup> and d<sup>(2)</sup> = d<sub>1</sub><sup>(2)</sup> ,d<sub>2</sub><sup>(2)</sup> , ··· ,d<sub>n</sub><sup>(2)</sup> ∈ [0 , 1]<sup>n</sup> , respectively, and D<sub>1</sub> + D<sub>2</sub> = I .

Theorem 4.8. Let ˜ A = {A<sup>1</sup> , A<sup>2</sup> } be of type semi-positive, where A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> are two Z -tensors. Then the VTCP ( ˜ A , ˜ q ) has a unique positive solution for any q<sub>1</sub> , q<sub>2</sub> < θ .

Proof. By Lemma 4.6, the VTCP( ˜ A , ˜ q ) is equivalent to find a vector x such thatwhere D<sub>i</sub> is a diagonal matrix and the principal diagonal elements d<sup>(i)</sup> := ( d<sub>1</sub><sup>(i)</sup> ,d<sub>2</sub><sup>(i)</sup> , ··· ,d<sub>n</sub><sup>(i)</sup> ) ∈ [0 , 1]<sup>n</sup> with i ∈ { 1 , 2 } and D<sub>1</sub> + D<sub>2</sub> = I . By Lemmas 4.2-4.4, D<sub>1</sub> · A<sup>1</sup> + D<sub>2</sub> · A<sup>2</sup> is a strong M -tensor. Note that − D<sub>1</sub> q<sub>1</sub> − D<sub>2</sub> q<sub>2</sub> > θ . Thus, by Theorem 3.2 in [7],has a unique positive solution, which is equivalent to that the VTCP( ˜ A , ˜ q ) has a unique positive solution.

Now, we give the following example to illustrate Theorem 4.8, see Example 4.2.Example 4.2. Let A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[3,2]</sup> , whereObviously, A<sup>1</sup> and A<sup>2</sup> are two Z -tensors. By the calculation, we obtainLet x = (1 , 2)<sup>T</sup> . Then A<sup>1</sup> x<sup>2</sup> = (1 , 3)<sup>T</sup> and A<sup>2</sup> x<sup>2</sup> = (1 , 1)<sup>T</sup> . Thus, ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type semi-positive. By the above analysis, ˜ A satisfies the conditions of Theorem 4.8 .

Now, we prove that the VTCP ( ˜ A , ˜ q ) has a unique positive solution for any q<sub>1</sub> , q<sub>2</sub> < θ . Let q<sub>1</sub> = ( a<sub>1</sub> ,a<sub>2</sub> )<sup>T</sup> < θ and q<sub>2</sub> = ( b<sub>1</sub> ,b<sub>2</sub> )<sup>T</sup> < θ . We only need to prove that the following system has a unique positive solutionNote thatwhere c = min { a<sub>1</sub> ,b<sub>1</sub> } < 0 . So, the unique positive solution of (4.12) is x<sub>1</sub> = √ − c . Now, we prove thathas a unique positive solution. LetSince f (0) = c + a<sub>2</sub> < 0 and g (0) = c + b<sub>2</sub> < 0 , both f ( x<sub>2</sub> ) and g ( x<sub>2</sub> ) have only two zero points with different signs. It is clear that (4.13) has a unique positive solution x<sub>2</sub> = x<sub>∗</sub> . Thus, the VTCP ( ˜ A , ˜ q ) has a unique positive solution x = √ − c,x<sub>∗</sub> T .

More sufficient conditions for the existence of a unique solution of the tensor equation ( D<sub>1</sub> · A<sup>1</sup> + D<sub>2</sub> · A<sup>2</sup> ) x<sup>m−1</sup> = − D<sub>1</sub> q<sub>1</sub> − D<sub>2</sub> q<sub>2</sub> can be found in [20].

Finally, by the way, we present a sufficient condition for the existence and uniqueness of the solution of the VTCP( ˜ A , ˜ q ) when ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V P - II . For this, Lemma 4.7 is required.

Lemma 4.7. Assume that ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V R<sub>0</sub> , where A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> and m is an even number. If A<sup>1</sup> has an even order right inverse A<sup>R<sub>k</sub>1</sup> ∈ R<sup>[k,n]</sup> such that A<sup>2</sup> · A<sup>R<sub>k</sub>1</sup> is a R -tensor, then ˜ A is of type V Q .

Proof. Let G ( x,t ) = x ∧· x [(−− 1) ] + td and H ( x,t ) = x ∧ A<sup>2</sup> · A<sup>R<sub>k</sub>1</sup>( x − tq<sub>1</sub>( m −k −+ tq<sub>2</sub> ,where t ∈ [0 , 1], d > 0, q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> . Obviously, for a given constant t<sup>∗</sup> ∈ [0 , 1], if vector y is a solution of G ( x,t<sup>∗</sup> ) = θ , then there is a solution x = A<sup>R<sub>k</sub>1</sup> · y [ 1 ( m − 1)( k − 1) ] of A<sup>1</sup> x<sup>m−1</sup> ∧ ( t<sup>∗</sup> d + A<sup>2</sup> x<sup>m−1</sup> ) = θ corresponding to it. If y is a solution of H ( x,t<sup>∗</sup> ), then x = A<sup>R<sub>k</sub>1</sup> · ( y − t<sup>∗</sup> q<sub>1</sub> )[ 1 ( m − 1)( k − 1) ] is a solution ofBy Theorem 4.1, the solution sets of G ( x,t ) = θ and H ( x,t ) = θ are bounded for any t ∈ [0 , 1]. So, there exists a bounded open set Ω that contains the sets

X<sub>1</sub> = { x : G ( x,t ) = θ,t ∈ [0 , 1] } and X<sub>2</sub> = { x : H ( x,t ) = θ,t ∈ [0 , 1] } , and ( X<sub>1</sub> ∪ X<sub>2</sub> ) ∩ ∂ Ω = ∅ . Since A<sup>2</sup> · A<sup>R<sub>k</sub>1</sup> is a R -tensor, there is no solution to the following systemwhich implies that there is no solution to the following systemIn fact, assume that there is an x that satisfies (4.15). Let y = x [ 1 ( m − 1)( k − 1) ] . ThenThus, y is a solution of (4.14), which contradicts that A<sup>2</sup> · A<sup>R<sub>k</sub>1</sup> is a R -tensor. From (4.15),has only the zero solution for any t ∈ [0 , 1] and d > 0. Since d > 0,

A<sup>2</sup> · A

R<sub>k</sub> 1 · x [ 1 ( m − 1)( k − 1) ] + d > θ and G ( x, 1) = x ∧ A<sup>2</sup> · A<sup>R<sub>k</sub>1</sup> · x [ 1 ( m − 1)( k − 1) ] + d = x for x near θ . Thus, degree ( G ( x, 1) , Ω ,θ ) = 1. By the homotopy invariance property of the degree [21], degree ( G ( x, 1) , Ω ,θ ) = degree ( G ( x ) , Ω ,θ ) = 1 ,

where G ( x ) = G ( x, 0). Obviously, H ( x ) = G ( x ), where H ( x ) = H ( x, 0). By the homotopy invariance theorem [21],Thus, ˜ A is of type V Q .

Combining Theorem 4.5 and Lemma 4.7, Proposition 4.1 holds.

Proposition 4.1. Assume that ˜ A = {A<sup>1</sup> , A<sup>2</sup> } is of type V P - II with A<sup>1</sup> , A<sup>2</sup> ∈ R<sup>[m,n]</sup> and m being an even number. If A<sup>1</sup> has an even order right inverse A<sup>R<sub>k</sub>1</sup> ∈ R<sup>[k,n]</sup> such that A<sup>2</sup> · A<sup>R<sub>k</sub>1</sup> is a R -tensor, then the VTCP ( ˜ A , ˜ q ) has a unique solution for all q<sub>1</sub> ,q<sub>2</sub> ∈ R<sup>n</sup> .


5 Conclusions

In this paper, we have introduced the vertical tensor complementarity problem (VTCP), and defined some tensor sets with special structures. We obtained an equivalence condition, i.e., the tenor set ˜ A is of type V R<sub>0</sub> if and only if the solution set of the VTCP( ˜ A , ˜ q ) is bounded. Meanwhile, we obtained some sufficient conditions for the existence and uniqueness of the solution of the VTCP from two aspects: the degree theory and the equal form of minimum function.
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