Classiﬁcation of the nullity for the second order discrete nonlocal problems

. In this paper we investigate the nullity of second order discrete problem with two nonlocal conditions. The classiﬁcation of nullity with respect to rows and columns of discrete problem matrix is presented.

This problem is equivalent to the linear system Au = f . According to S. Roman [?], problem (1)-(2) has a singular matrix A if and only if the condition is satisfied. Here L = (L 1 , L 2 ), u = (u 1 , u 2 ) and functions u 1 , u 2 form any fundamental system of homogeneous equation (1). It is well known that the nullity of singular matrix is nonzero. In articles [?, ?], the nullity and null space of problem (1)-(2) were investigated. There are formulated two classifications of the nullity in [?]. One classification is obtained with respect to rows but another -with respect to columns of matrix A of discrete problem (1)- (2). In this paper we analyze the nullity of problem (1)-(2) with respect to rows and columns together and present its classification.

Classifications of the nullity
Problem (1)- (2) can also be written in the expanded matrix form In [?], classifications of the nullity dim ker A are given as follows.
(2) dim ker A = 1. In this respect, such cases are possible: (a) the row of matrix A that corresponds to the functional L j is a linear combination of rows, that correspond to the operator L, but the row, that corresponds to the functional L 3−j , and rows, that describe the operator L, are linearly independent if and only if (b) the row of matrix A, that corresponds to the functional L 1 (L 2 ), is a linear combination of the row, that corresponds to the functional L 2 (L 1 ), necessarily, and rows, that correspond to the operator L, but the row, that corresponds to the functional L 2 (L 1 ), and rows, that describe the operator L, are linearly independent if and only if (3) dim ker A = 2. Both rows of A, that correspond to L j , j = 1, 2, are linear combinations of rows that describe the operator L (which are linearly independent) if and only if On the other hand [?], discrete problems always have unique solutions. Thus, functions v 1 and v 2 form the particular fundamental system of (1), which always exists.
(2) dim ker A = 1. In this respect, three cases are possible: (a) the next to last column of A is a linear combination of the first n−1 columns of A, but the last column and the first n−1 columns of A are linearly independent if and only if

last (next to last) column of A is a linear combination of the next to last
(last) column, necessarily, and the first n − 1 columns of A, but the next to last (last) column and the first n − 1 columns A are linearly independent if and only if Both the last and next to last columns of A are linear combinations of the first n − 1 columns of A (that are linearly independent) if and only if We can easily observe the following relations.

General classification
By Corollary 1, such a statement follows from Lemma 1 and Lemma 2.
In this respect, both rows that correspond to functionals L j , j = 1, 2, are linear combinations of rows that describe the operator L. Moreover, the last and next to last columns are linear combinations of the first n − 1 columns of A.
Let us investigate problem (1)-(2) with dim ker A = 1. According to Lemma 1, there are three different relationships among rows of A, i.e., two cases are obtained from item (a) with j = 1, 2, and the third case is given by item (b).
We can notice [?, Remark 1], that relation (b), written to the functional L 1 , is always a result of the same relation, written to the functional L 2 , and vice versa. So, we can write relation (b) to the functional L 2 , and then unite it and relation (a) with j = 2 to one relation. Thus, the united relation is as follows.  (1)-(2) where dim ker A = 1.
The row that corresponds to L 1 is a linear combination of rows that describe the operator L, but the rows that describe L 2 and L are linearly independent The row that corresponds to L 2 is a linear combination of rows that describe the operator L and functional L 1 , but the rows that describe L and L 1 are linearly independent The next to last column is is a linear combination of the first n − 1 columns, but the first n − 1 columns and the last column are linearly independent The last column is a linear combination of the first n columns that are linearly independent The row of A, that corresponds to the functional L 2 , is a linear combination of rows, that describe the operator L and functional L 1 . But the rows, that describe the operator L and functional L 1 , are linearly independent.
Moreover, rows of A satisfy this relation if and only if the necessary and sufficient conditions of relation (a) with j = 2 or relation (b) are satisfied: Using Corollary 1, we can simplify these conditions to the conditions Similarly, we can unite relation (b) and relation (c), written to the last column of A, to one relation as follows.
(C2) The last column of A is a linear combination of the first n columns that are linearly independent.
We can similarly obtain the necessary and sufficient conditions of this relation among columns of A. These conditions are given by Thus, for problem (1)-(2) with dim ker A = 1, rows satisfy either relation (a) of Lemma 1 with j = 1 or relation (C1). Similarly, columns satisfy either relation (a) of Lemma 2 or relation (C2). Choosing all the combinations of these relations, we obtain four different relations among rows and columns together. They are given in Table 1. Choosing the relation of rows above and the relation of columns on the left-hand side of the table, we have all the mentioned combinations among rows and columns. There are given the necessary and sufficient conditions for every combination of relations on the corresponding intersections.
Firstly, we investigate case (1). Because D(L)[v] = 0 is valid, we have According to case (1), the inequality L 1 , v 2 = 0 is valid. So, we obtain L 2 , v 1 = 0 from the last equality. But now we have a contrary from the second inequality of (7). Thus, case (1) is impossible. So, cases (2) and (3) remain possible, i.e., the first inequality of (7) is satisfied if and only if either cases (2) or (3) is valid. We can note that this statement is equivalent to the inequality L 1 , v 1 = 0, because other number L 1 , v 2 can obtain any value, i.e., either L 1 , v 2 = 0 or L 1 , v 2 = 0. Finally, we can see that conditions (7) are equivalent to the following conditions Example 1. Let us investigate a differential problem −u ′′ = f (x), x ∈ (0, 1), where f is a real function and γ ∈ R. We introduce the mesh ω h = { x i = ih: i ∈ X n , nh = 1}. Suppose ξ is coincident with the mesh point, i.e., ξ = sh. Let us denote u i = u(x i ) and f i = h 2 f (x i+1 ), i ∈ X n−2 . Then problem (1) can be approximated by a discrete problem According to (3), this problem has a singular matrix A if and only if γξ = 1.
Moreover, we note that functions v 1 = n(1 − x) and v 2 = n(x − 1 + h), x ∈ ω h , are solutions to (4). Thus, the inequality L 1 , v 1 = n = 0 is always satisfied. So, by Table 1, for problem (1) such a corollary follows. Corollary 3. The row of matrix of problem (1), that corresponds to the functional L 2 , is a linear combination of rows, that describe the operator L and functional L 1 , but the rows that describe the operator L and functional L 1 are linearly independent. Moreover, the last column is a linear combination of the first n columns, that are always linearly independent. All these relations are valid if and only if γξ = 1.
We know that homogenous problem (1) (f i = 0, i ∈ X n−2 ) with singular matrix, i.e., γξ = 1, describe the nonzero null space. According to Corollary 3, we can eliminate the equation of (1) that corresponds to the functional L 2 , because it is a linear combination of other (linearly independent) equations. Moreover, we transfer the members with u n to the right-hand side of equality, because they correspond to the last column of discrete problem matrix, which is a linear combination of other columns. Thus, we obtain a linear system A u = g(u n ), where u = (u 0 , u 1 , . . . , u n−1 ) T . Here the matrix A is nonsingular because it is the intersection of linearly independent rows and columns of A. According to linear algebra, the unique solution u = A −1 g(u n ), u n ∈ R, always exists and describes the null space of problem (1).
In general, the obtained classification with respect to rows and columns is very useful for the solution to the null space of problem (1)- (2).