Consider the interpolation problem with a finite dimensional space of univariate func- tions S and a set of knots that is for a given data find a function satisfying the conditions

We say that the set of knots is poised for S if for any data there is a unique function satisfying the conditions (1). A necessary condition of the poisedness is

There are several cases of spaces S of univariate functions for which we have a characterization of all poised sets.

Suppose that is the space of polynomials of degree at most n. We have that Then, according to the Lagrange theorem, all sets of knots are poised.

Now suppose that is the space of spline functions of order n and are the knots of the space (see, e.g., [1] ). Here means that s is piecewise polynomial function of degree at most s vanishes outside of the segment and s belongs to the differentiability class We have that and the set of interpolation knots is poised if and only if the following conditions are satisfied:

This result is due to Schoenberg and Whitney [2] [3] . In the univariate case there are also several characterization results concerning the trigonometric interpolation.

In contrast with this there are no such results in the bivariate case. As an exception one may consider only the Pascal classic theorem, in the interpolation theory inter- pretation (see, e.g., [4] ). To present it let be the space of bivariate polynomials of total degree at most 2. We have that Let also be the space of bivariate linear functions, Then consider any set of 6 nodes in the plane. Construct 3 new nodes as follows:

where is the line passing through the nodes and Then according to the Pascal theorem the 6 nodes are lying in a conic if and only if the 3 nodes are collinear. To arrange the case connected with the parallel pairs of lines one may replace the plane with the projective one. The interpolation version of this result is:

Theorem 1.1 (Pascal) Any 6 nodes in the plane are poised for if and only if the set of respective 3 nodes is poised for

Note that the latter poisedness condition with merely means that are not collinear.

Next we are going to introduce the space of bivariate functions we will consider in this paper. For this we need some preliminaries.

Let us define a strip of triangles. Fix a sequence of points in the plane

for the vertices of triangles. Then consider the n triangles

Note that by a triangle we mean the closed set bounded by the sides of the triangle. The sequence of the triangles

makes a triangulation of the following strip (see Figure 1)

Sometimes it is convenient to call itself a strip, too.

Here we require that the intersection of any neighboring triangles and

is the common side any pair of triangles and

have a single common point which is the vertex and all the other pairs of the triangles are disjoint.

It is easily seen that the sides together with the sides and form the boundary of the strip (see Figure 1). We call these sides boundary sides. The remaining sides of the triangles are called interior sides of the strip. Let us call also the sides and the left and right (boundary) sides of the strip respectively, and denote

For a triangle given in (2) we call the sides and the left and right sides of it, respectively.

Denote by the linear space of continuous piecewise linear functions on More precisely, means that

i)

ii)

Here means the restriction of s on D.

Definition 1.2 A set of nodes is called poised for the space if for any data the interpolation problem

has exactly one solution

Let us denote the interpolation problem (3) by

The aim of this paper is the characterization of all poised sets for the space The following is a necessary condition of the poisedness:

To prove this consider the fundamental functions of the interpolation

defined by the interpolation conditions

where is the symbol of Kronecker.

Obviously the fundamental functions are linearly independent and for the solution of the interpolation problem (3) we have the following formula of Lagrange:

Thus, the fundamental functions form a basis for and we get (4).

Now, let us show that the set of vertices is a poised set for Indeed, having the values of a linear function at the vertices of a triangle we recover it in a unique way on the triangle. On the other hand it is easily seen that the recovered piecewise linear function is continuous on Thus, the dimension of equals to the number of the vertices in the strip

In view of (4) we obtain that any poised set for the space consists of nodes:

Let us call interpolation problem satisfying this condition exact. In the cases and we call the problem underdetermined and over- determined, respectively.

Denote by

the fundamental polynomials with respect to the poised set. They form a basis for the space Thus we have the following representation for any piecewise linear function

In view of this representation the interpolation problem (3) reduces to m linear equations with unknowns, which are the values a piecewise linear function at the vertices. Hence an exact interpolation problem is poised if and only if the following Vandermonde determinant does not vanish:

where

Thus our main problem can be formulated in terms of Vandermonde determinant in the following way:

Characterize all exact sets for which the Vandermonde determinant does not vanish, i.e., (7) holds.

The following two propositions are basic Linear Algebra facts.

Proposition 1.3 Given an exact problem Then each of the following con- ditions is equivalent to the poisedness of:

i) All fundamental functions exist.

ii)

Proposition 1.4 Given a problem Then the following hold:

i) If the problem is underdetermined then there is a function such that

ii) If the problem is overderdetermined then there is a node in for which no fundamental function exists.

For a given strip denote by the following part of it

For a problem denote by the subproblem with the function space and the set of nodes i.e.,

Denote by the linear space of continuous piecewise linear functions on vanishing at the left (boundary) side of the strip, i.e.,

In the similar way one can define the space of functions vanishing at the right side of the strip i.e.,

One may define also the space of functions vanishing at the both left and right sides of the strip i.e.,

By counting the number of vertices of the strip where a function from the space may differ from we get, in the same way as in the proof of (5), that

Denote the interpolation problems with the spaces and a node set by respectively.

As we will see below the poisedness of interpolation problems with boundary con- ditions readily can be reduced to the previous general interpolation problems by just adding two nodes in the left or/and right sides of the strip.

For this purpose it is convenient to use the following notation for the strip

where and are the two vertices of

where and are the two vertices of Denote also

Let us call two interpolation problems equivalent if both they are poised or both they are not poised. By using Proposition 1.3, ii), we readily get

Proposition 2.1 The folllwing pairs of interpolation problems are equivalent:

and

and

and

Note that in the subproblem we have that

where and are the two vertices of The situation is similar with the sets and

Below we bring a main theorem regarding the reduction of an interpolation problem having an exact or overdetermined subproblem.

Theorem 3.1 Suppose that is an exact problem where the strip consists of n triangles and is a subproblem, where.

Then the following hold.

i) If the subproblem is exact and not poised, or overdetermined, then the problem is not poised.

ii) If the subproblem is poised then the problem is poised if and only if the both following two reduced problems

are exact and poised.

Note that in the cases and we have just one reduced problem instead of two.

Proof. Suppose first that the subproblem is exact and not poised, or overdetermined. Then, in view of Propositions 1.3, i), and 1.4, ii), there is a node which does not have a fundamental function. Then, evidently the same node A does not have a fundamental function for the whole node set too. Hence, in view of Proposition 1.3, i), we get that the problem is not poised.

Now consider the case when the subproblem is poised.

Let us assume that the both reduced problems in (8) are poised. Then let us prove that the problem is poised, too. Notice first that the problem is exact. Thus, by following Proposition 1.3, ii), assume that Now, since the subproblem is poised, we conclude that s vanishes on the triangles Therefore s vanishes at the right side of the triangle and at the left side of the triangle Thus we have that and Finally, since the problems in (8) are poised, we obtain that s vanishes on the triangles and Hence s vanishes on all triangles of

Next let us assume that the problem is poised and prove that the both reduced problems in (8) are poised, too. Let us show first that these reduced problems are exact. There are and triangles in the reduced problems (8), respectively. Thus for exactness we need and nodes, respectively, altogether

nodes. It is easily seen that indeed, in two mentioned problems together we have that many nodes. Indeed, this is the number of the nodes in minus the number of the nodes in and plus 4, i.e.,

Latter nodes come from the added nodes in the boundary in the node sets and.

Now, assume by way of contradiction that a subproblem in (8) is not exact. Therefore one of the subproblems, say the first problem in (8), is underdetermined, and another is overdetermined. Then, in view of Proposition 1.4, i), there is a function such that

Thus s here vanishes on and, since of “+2”, also on the right side of the triangle Now let us extend this function s from till the whole strip by defining it to be 0 on the triangles Denote by the extended function. Then it is easily seen that and This, in view of Proposition 1.3, ii), means that the problem is not poised, which contradicts our assumption.

Finally, let us show that the both reduced problems in (8) are poised. Assume by way of contradiction that one of them, say the first, is not poised. Since it is exact, we can use Proposition 1.3, ii), to get that there is a function such that the relation (9) is satisfied. From here we continue in the same way as in the above step, after the relation (9).

Let us denote by “3” the problem with a strip consisting of a triangle and at least three nodes inside. The respective basic problem, denoted briefly by “”, is a “3” problem with exactly three nodes, which are non-collinear.

Then, let us denote by “” the problem with a strip consisting of two triangles and exactly two nodes in each of them. Note that in this case no node can lie in the interior side of the strip. We call a problem “” basic and denote it by “” if in each triangle the line passing through the two nodes there does not intersect the other triangle. Note that in the case of “” problem no node can coincide with a vertex of the strip.

Next consider an interpolation problem denoted by “” where m is the number of 1’s. This is the interpolation problem with a strip consisting of triangles such that each of the first and the last triangles contains exactly two nodes and each of the other m triangles contains exactly a node. Note that in this case no node can be located in an interior side of the strip. We call a problem “” basic and denote it by “” if the line passing through the two nodes in each of the first and the last triangles does not intersect the neighboring triangle.

Note that the “” problem can be considered as a special case of the

“” problem, where

Let us start this section with a simple lemma. Suppose that two nodes of a node set belong to a triangle of Denote by the points of intersection of the line passing through and with the sides of the triangle. Let us call the intersection pair of Denote by the set of the nodes received from by replacing the pair with there. We call the set the line trans- formation of the set with respect to the pair

Two node sets and are called line-equivalent if one of them can be obtained from the other by means of several line transformations.

Lemma 4.1 The exact interpolation problems and are equivalent, if the node sets and are line-equivalent.

Proof. In view of Proposition 1.3, ii), it suffices to verify that

where the node set is the line transformation of the set with respect to an appropriate pair of nodes Now notice that (10) readily follows from the evident fact that

where is the intersection pair of Indeed, each side of this equiva- lence means merely that vanishes on the line passing through and

The proof of the following proposition contains an easy algorithm for verifying the poisedness of basic problems.

Proposition 4.2 The problems “” and “” are always poised.

The problem “” where may be poised or not, depen- ding on the configuration of nodes. Moreover, if the problem is not poised then it becomes poised after changing the location of one of the two nodes in the first or the last triangle, such that the slope of the line passing through the two nodes is changed.

Note that the case here concerns the problem “”

Proof. The case of “” is obvious.

Consider the problem “” (see Figure 2). Suppose that

and, where In view of Lem- ma 4.1 we may replace the node set with the set where and are the intersection pairs of the respective pairs from Since the problem is basic we have that the nodes belong to the four sides of the triangles and which are boundary sides of the strip one node in each side (see Figure 2). Now suppose by way of contradiction that the problem is not poised. This in view of Proposition 1.3, ii), means that there is a function such that while Then it is easily seen that and therefore Assume without loss of generality that Then, since s is linear and we get readily that and Now, in view of the latter condition and we get readily that and Thus s assumes negative values at all the vertices of the triangle and it vanishes at which is a contradiction.

Finally, consider the problem “” where Suppose that and, where

Assume that the problem is not poised. This in view of Proposition 1.3, ii), means that there is a function s such that

It is easily seen that Indeed, implies that s vanishes at the left side of Then, in view of the condition where we conclude that Notice that does not belong to the left (or right) side of the triangle, since the problem is basic. Continuing this way we obtain and thus contradicting our assumption. In the same way we get that

By replacing s, if necessary, with a nonzero constant multiple of it, assume that Now, in view of the condition where the func- tion is determined on Indeed, the nodes are not collinear since the problem is basic. Hence s is determined at the left side of the second triangle Next, we have that where does not belong to the left (or right) side of the triangle. Thus we get that s is determined on Continuing this way we get that s is determined on Now s is determined at the left side of the last triangle By using the condition where we con- clude, in view of (12), that the zero set of s in is a line passing through Thus if the last node lies in this line then we have that s vanishes at too. Hence the condition (11) holds, which, in view of Proposition 1.3, ii), means that the problem is not poised. While the condition implies that s vanishes on which contradicts (12) and whence (11). Thus the problem is poised in this case. This consideration makes clear also the “moreover” part of Proposition.

It is worth mentioning that if the restriction of s on the left side of triangle has a zero (as it will happen in the case of “”) denoted by C then the basic problem is poised, wherever the two last nodes and are situated. Indeed, otherwise we would have that the zero-line of s in the last triangle passes through and C, which contradicts the fact that the problem is basic.

In this section we prove that every exact problem has a subproblem of type “3”, “”, or “”. Next, we show that by applying line-transformations to this subproblem, we can reduce it to a basic subproblem or determine that the given exact problem is not poised. After this, by following the algorithm pointed out in the proof of Proposition 4.2, we may verify whether the basic subproblem is poised or not. If not then, in view of Theorem 3.1, i), we conclude that the given exact problem is not poised either. If the basic subproblem is poised then we may apply the respective reduction, given by Corollaries 5.1, 5.2, or 5.3. Thus we have a complete solution of the poisedness problem.

Theorem 6.1 Assume that is an exact problem. Then it has a subproblem of type “3”, “”, or “”. Moreover, by using line-transformations, we either determine that the problem is not poised, or we reduce the node set to such that the line-equivalent problem has a basic subproblem of type “”, “”, or “”.

Proof. Suppose that a triangle contains more than 3 nodes, or exactly 3 collinear nodes. Then the problem obviously is not poised and also we have a type “3” sub- problem. If a triangle contains exactly 3 non-collinear nodes then we have a type “” basic subproblem.

Now, let us assume that no triangle of contains more than 2 nodes.

Step 1. Let us verify that the problem contains a subproblem of type “”, or “”.

Let us throw away those triangles of which do not contain nodes from Let us throw away also those triangles of which contain just one node from located in an interior side of By saying a node is thrown we mean that it does not belong to the remained triangles.

Notice that only a node located in an interior side of the strip can be thrown. Moreover, this can happen only if the both neighboring triangles were thrown, i.e., if it is the only node in the interior side and also the only node in the both neighboring triangles.

Assume that after this operation the connected blocks of remained triangles in are

where

Now, let us verify that for a block here the number of nodes (from) is greater or equal to the number of triangles plus two.

Assume by way of contradiction that for all blocks in (15) the number of nodes does not exceed the number of triangles plus one. Then we get that the number of nodes in all above connected blocks is less than or equal to the number of triangles in all connected pieces plus, where is the number of blocks in (15).

Next, consider the breaks, i.e., the dropped blocks. For each break we have that the number of nodes, i.e., number of the thrown nodes in the above mentioned operation, is less than or equal to the number of triangles there minus 1. Indeed, a such node (in an interior side of) is thrown if both triangles containing it where thrown. Thus we can assign two thrown triangles to each thrown node. Also note that all the triangles assigned are different. Hence the number of nodes in all above mentioned breaks is less than or equal to the number of the triangles there minus, since there are at least breaks.

Thus we may conclude that the number of nodes in does not exceed the number of triangles there plus This contradicts the assumption that the pro- blem is exact.

Now assume, without loss of generality, that in the first block the number of nodes from is greater or equal to the number of triangles there plus two. Note that if a triangle in this block contains only one node then it is not located in its left or right side, otherwise the triangle would be thrown away before.

Since no triangle contains three nodes, we get that there is a triangle in this block containing two nodes. Consider the first such triangle i.e., a such triangle with the minimal subscript If both the two nodes here are lying in the right side of the triangle then the next triangle contains only these two nodes. Now notice that in the triangles from till the number of nodes equals to i.e., the number of triangles here. This means that there is another triangle in the first block, following containing two nodes, such that at least one node is not lying in the right side of the triangle.

Then consider the first such triangle If one of the two nodes here is in the right side of the triangle then the next triangle also contains two nodes, because otherwise it would be thrown away before. Now notice that in the triangles from till the number of nodes does not exceeds i.e., the number of triangles here plus 1. This means that there is another triangle in this series, denoted by following containing two nodes, which do not lie in its right side. This will be the first triangle of the desired subproblem.

Now notice that in the triangles from till the number of nodes equals to i.e., the number of triangles here plus 1. This means that there is another triangle in the first block, following containing two nodes. Denote by the first such triangle. This will be the last triangle of the desired subproblem. First notice that each triangle between and if there is a such, contains just one node, which, as was mentioned earlier, cannot be located in its left or right side. Therefore no node in belongs to the left side of the triangle, since otherwise would con- tain a node in its right side. Note also that in this case does not coincide with since the latter has no node in its right side.

Thus the subproblem is of type “”, or “” with

Step 2. Next, by using line-transformations, we either reduce this subproblem to a basic subproblem or determine that the problem is not poised.

First suppose that the subproblem is of type “”. If it is not basic then the line through the two nodes in a triangle intersects its interior side. Then by replacing these two nodes with their intersection pair we will have in the other triangle one more node, i.e., three nodes. This case was considered in the beginning of the proof.

Finally, suppose that the subproblem is of type “”, with Now, if this subproblem is not basic then in the same way as in the previous case we may reduce it by line transformation to a problem of the same type, where the number of 1’s equals to Thus we may complete readily the proof by using in- duction on m, where the first step of induction corresponds to the case considered already.

Avdalyan, H. and Hakopian, H. (2016) On Characterization of Poised Nodes for a Space of Bivariate Functions. Advances in Linear Algebra & Matrix Theory, 6, 89-103. http://dx.doi.org/10.4236/alamt.2016.64010