#### Abstract

We study the singularity of multivariate Hermite interpolation of type total degree on nodes with . We first check the number of the interpolation conditions and the dimension of interpolation space. And then the singularity of the interpolation schemes is decided for most cases. Also some regular interpolation schemes are derived, a few of which are proved due to theoretical argument and most of which are verified by numerical method. There are some schemes to be decided and left open.

#### 1. Introduction

Let be the space of all polynomials in variables, and let be the subspace of polynomials of total degree at most . Let be a set of pairwise distinct points in and be a set of nonnegative integers. The Hermite interpolation problem to be considered in this paper is described as follows: Find a (unique) polynomial satisfyingfor given values , where the numbers and are assumed to satisfy

Following [1, 2], such kind of problem is called Hermite interpolation of type total degree. The interpolation problem is called regular if the above equation has a unique solution for each choice of values . Otherwise, the interpolation problem is singular. As shown in [3], the regularity of Hermite interpolation problem implies that it is regular for almost all with . Hence, in this paper, we will call almost -regular if is regular for some . Otherwise, we call -singular. With no confusion, we also call it almost regular or singular for convenience. If is a nontrivial polynomial satisfying (1) with zero interpolation condition, we call a vanishing polynomial with respect to and . Obviously, being singular is equivalent to the existence of a vanishing polynomial of degree no more than .

The research of regularity of multivariate Hermite interpolation is more difficult than Lagrange case, although the latter is also difficult. One of the main reasons is that (2) does not hold in some cases. About the results of multivariate Hermite interpolation, one can refer to [1â€“10] and the references therein. Most recently, authors [5] made further development and gave complete description for the regularity of the interpolation problem on nodes, which is an extension of the results mentioned in [1, 2]. Besides, not any other results appeared for a big number of nodes. This paper is an extension of [5] and we will investigate the singularity of Hermite interpolation for with .

This paper is organized as follows. In Section 2, we consider the singularity of the Hermite interpolation of type total degree and present the main results. In Section 3, we present theoretical proofs for some regular schemes. Finally, in Section 4, we conclude our results.

#### 2. Singularity of Interpolation Schemes

In this section, we will investigate the singularity of Hermite interpolation of type total degree and (2) is always assumed to hold. Hermite interpolation of type total degree is affinely invariant in the sense that the interpolation is singular or regular for one choice of nodes. In what follows, we assume . In this case, since , there must exist a nontrivial quadratic polynomial which vanishes at . Also there exists a nontrivial linear polynomial vanishing at . Given and , if there are vanishing polynomials with respect to and , we will denote by one vanishing polynomial of them.

Here we always assume that no interpolation happens at if the th component of is . Obviously, vanishing polynomials always exist if since the number of the equations is less than the number of the unknowns.

For convenience, we always order with . In [5], authors showed that the inequalitymust hold if is regular, which gives evaluation of in (2). The following theorem implies that inequality (4) is very sharp.

Theorem 1. Given and , if , then is singular.

Proof. We first assume . Then is a vanishing polynomial with respect to and , and Thus, in this case is singular.
If , then . Thus is a vanishing polynomial with respect to and , and Collecting two cases, we complete the proof.

Next, we assume .

Lemma 2. Given and , if , then is singular.

Proof. Let . Then is a vanishing polynomial with respect to and . Moreover This completes the proof.

If , we easily get . The following theorem is due to [5], which will be used in next lemma.

Theorem 3 (see [5]). Assume . Given and , ifthen the Hermite interpolation of type total degree is singular.

This theorem implies that there exists a vanishing polynomial of degree no more than with respect to and if (8) holds.

Lemma 4. Given and , if and , then is singular.

Proof. Let . Then together with all of its partial derivatives of order up to vanishes at the points. For points , since it follows from Theorem 3 that there exists a polynomial with , together with all of its partial derivatives of order up to vanishing at for . Let . Then, and all of its partial derivatives of order up to vanish at for , and Thus, the interpolation is singular.

In what follows, we only need to consider and , which includes

Lemma 5. Given and , suppose that (11) is satisfied; then Hermite interpolation of type total degree is almost regular if and only if and

Proof. Set ; then and . We first check (2). Let Since for and , then We will show that implies . Note that Thus if , then Hencewhere Since for , is monotonically decreasing about . Thus, for due to and .
Since , then for , which means that (2) does not hold for . Thus we only need to consider . In this case Hence, for , (2) does not hold for and holds for only if (14) holds. We will show that it is almost regular in next section. For and , (2) holds only in the case of since (2) holds for . We can show that is singular if . In fact, we can take with as the vanishing polynomial. Obviously . The proof is completed.

Lemma 6. Given and . Suppose that (12) is satisfied and that . Then (i)for , (2) holds only for one form and Hermite interpolation of type total degree is almost regular for ;(ii)for , if , it is singular; if , (2) has three positive integer solutions and corresponding interpolation schemes are almost regular;(iii)for ,(a)if , it is singular;(b)if and , it is also singular;(c)if and , (2) has four positive integer solutions and in these cases are almost regular;(d)if , it is singular;(e)if , (2) holds only for one form and it is almost regular.

Proof. Set and ; then . If , it is easy to check that (2) holds if and only if and . This scheme is almost regular, which will be proved in Theorem 15 of the next section.
For , we first check (2). Let Since for and , thenBy the same argument with Lemma 5, one can show that implies that . The following facts can be checked easily:Since for , (2) does not hold for .
For and , (2) has three positive integer solutions and . These three schemes are almost regular, which can be verified by numerical method; see Remark 7.
Since for , (2) does not hold for and . Similarly, for means that (2) does not hold for and .
For and , (2) has four positive integer solutions which are shown to be almost regular by numerical method presented in Remark 7.
Let us consider the case of .
From the definition of , we obtainThen, (2) does not hold for , .
For , , (2) holds for the form Indeed, this form is the only one since . This scheme is almost regular, which will be proved in the next section.
Finally, we consider the case of . We can show that is singular by taking with for and with for as vanishing polynomial with respect to and . Here we use the fact that if which can be obtained by a simple calculation.
The proof is completed.

Remark 7. Generally speaking, it is difficult to judge the regularity of the interpolation schemes theoretically. For a given , one possible way to decide the regularity is based on numerical method: calculating the vanishing ideal (see [5] for details) or the corresponding Vandermonde matrix, where the points can be selected randomly. However, the former method needs to do symbolic calculation which is little useful for big , and . The latter one needs to judge the singularity of the matrix, which is also difficult if the order is very big. Although so, it is a good way for moderate , , and , which is employed in this paper for some simple cases.

Lemma 8. Given and , suppose that (13) is satisfied and that . Then (i)if , (2) never holds;(ii)if , (2) has finite positive integer solutions listed in Table 2.

Proof. We first check (2). Let Since for and , then In the same way with Lemma 5, one can show that implies that and . By a simple calculation, we have implies that holds for all . Thus we must take to ensure (2) if . Hence, let and we again have implying and . It is easy to get Thus we can obtain the possible pairs satisfying (2); see Figure 1.
By detailed analysis and computation, the solution of (2) can be obtained and is listed in Table 2; see Appendix.

Lemma 9. Let . Given and , suppose that (13) is satisfied and that . Then (i)for , (2) has four positive integer solutions and the corresponding interpolation schemes are singular;(ii)for , (2) has only one positive integer solution and the corresponding interpolation scheme is almost regular;(iii)for and , (2) has finite positive integer solutions and all of them are singular.

Proof. (i) and .
According to the definition of in Lemma 8, we obtain So, (2) never holds if . If , (2) has four positive integer solutions , and . We claim that they are all singular. To show this we can takeas vanishing polynomials with respect to these four solutions, respectively, where (ii) and .
Equation (2) has only one positive integer solution which is almost regular. The regularity can be checked by numerical method mentioned in Remark 7.
(iii) , , and .
We will show that all the schemes in this case are singular. The proof is based on the following three cases.
Caseâ€‰â€‰1. Consider the following: .). If (2) holds, then is singular. In fact we can check that is the desired vanishing polynomial.(2). We consider the case of firstly. In this case, (2) holds only if because and we can check that is singular if . In fact, we can take as the vanishing polynomial of .
We next consider the case of . In this case, must hold; otherwise (2) never holds. If , (2) has two solutions: and . These two schemes are singular, which can be checked by numerical method (see Remark 12). If , then the interpolation scheme is singular, which can be shown to take as the vanishing polynomial. Here the existence of follows from the construction of in (34).
Caseâ€‰â€‰2. .
We first claim that is singular if . Notice that if , then (2) holds only if , which will lead to the singularity of . The vanishing polynomial can be taken as whereIf , then must hold to ensure (2). If or , (2) has four solutions for : , , , and . These four schemes are singular, which can be checked by numerical method (see Remark 12). If , then is singular by taking as the vanishing polynomial, where If , then must hold to ensure (2). In the case of or , (2) has two solutions for : and These two schemes are singular, which can be checked by numerical method (see Remark 12). In the case of , is singular, which can be shown by taking as the vanishing polynomial, where Caseâ€‰â€‰3. Consider the following: .
To ensure (2), must hold. Furthermore, it is easy to check for and , , and that (2) has no solution. If , then is singular and the corresponding vanishing polynomial is taken as where Thus we complete the proof.

Remark 10. The interpolation scheme is first mentioned in [1] and shown to be singular in [11], but one can check that the proof in [11] is not correct. In fact, condition () in [11] holds with equal sign; hence, its poisedness can not be decided by the necessary condition () there. The number of the nodes in this case is 7 not 6.

Remark 11. From the proof in Lemma 9, one can show that the interpolation scheme and is singular by taking whereThis scheme was mentioned in [1, 11] and wrongly claimed to be almost regular in [5].

Remark 12. The singularity of interpolation schemes , , , , , , , , and can be verified by numerical method. Here we notice that not all of them need to prove independently because we have the following observation: Nine nodes can be selected as

Lemma 13. Given and . Suppose that (13) is satisfied and that . Then (2) has finite positive integer solutions for any and all except one scheme are almost regular.

Proof. By setting , , and in (2), we obtain where . This equation has finite positive integer solutions for any , which is listed in Table 1 for .
All the schemes except are almost regular. The singularity of , , and can be proved by numerical method. The 7 points can be selected as , , , , , , and . Then the Vandermonde matrix is singular by symbolical computation, implying is singular.

Remark 14. In [2], Lorentz presented a conjecture (Conjecture ) which gives a necessary and sufficient condition about the singularity of multivariate Hermite interpolation. Although the scheme is singular, it can not be checked by the conjecture from [2]. Hence Conjectureâ€‰â€‰8 in [2] is not correct.

#### 3. The Proof of Regularity of Some Interpolation Schemes

In this section, we will prove that is almost -regular and and are both almost 3-regular. To this end, we need to choose carefully and then prove the regularity of .

Theorem 15. Given and , then is almost -regular.

Proof. Let and define , for . Finally, let and . We will show the regularity of . To this end, we assume that is a polynomial of degree 2 and satisfies all the homogenous interpolation conditions. It only needs to show .
Due to for , is of form Thus we have which lead to for . Then should be a zero polynomial, which completes the proof.

To prove the regularity of and , we need the following lemma.

Lemma 16 (see [1, 4]). Let ; then interpolating the value of a function and all of its partial derivatives of order up to at each of the three vertices of a triangle as well as the value of the function and all of its derivatives of order up to at a fourth point lying anywhere in the interior of the triangle by polynomials from , is regular.

In fact the fourth point can lie anywhere except on the three edges of the triangle.

Theorem 17. Let , , , , , , , , and ; then is regular.

Proof. Suppose that is a polynomial of degree no more than such that That is, satisfies the homogeneous interpolation conditions. To prove this theorem, we only need to show .
Consider the value of on the plane . It follows from (55) that According to Lemma 16, vanishes on the plane , which implies that can be divided by . Similarly, can be divided by and , respectively. Hence can be written as with .
If , by taking into (55) we obtain Thus and hence , which will prove the theorem for .
If , taking into (55) will giveThen consider the plane . It follows from Lemma 16 and the equations that can be divided by . Similarly, can be also divided by and , respectively. Thus can be written as with .
By collecting the above results, we have with . If , then satisfies , which implies that and the theorem will be proved for . Otherwise, for , satisfiesby taking into (55).
Clearly, (60) is the same interpolation problem as (55), but with a smaller . Thus we can end the proof by repeating the above process or by induction.

By similar proof, we can get the following theorem and the proof is omitted.

Theorem 18. Let , , , , , , , , and ; then is regular.

Theorems 17 and 18 imply that and are both almost 3-regular.

#### 4. Conclusion

In this paper, we consider the singular problem of multivariate Hermite interpolation of total degree. We make a detailed investigation for Hermite interpolation problem of type total degree on nodes in .

Given with and , the following results are derived in this paper:(1)If , then is singular or (2) does not hold (see Lemma 2).(2)Suppose , then(a) is singular if ; see Lemma 4;(b) for