We study the singularity of multivariate Hermite interpolation of type total degree on m nodes with 3+d<m≤d(d+3)/2. 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.
National Natural Science Foundation of China1130105361432003111710521127106061272371Fundamental Research Funds for the Central Universities1. Introduction
Let Πd be the space of all polynomials in d variables, and let Πnd be the subspace of polynomials of total degree at most n. Let X={X1,X2,…,Xm} be a set of pairwise distinct points in Rd and p={p1,p2,…,pm} be a set of m nonnegative integers. The Hermite interpolation problem to be considered in this paper is described as follows: Find a (unique) polynomial f∈Πnd satisfying(1)∂α1+α2+⋯+αd∂x1α1⋯∂xdαdfXq=cq,α,1≤q≤m,0≤α≤pq,for given values cq,α, where the numbers pq and n are assumed to satisfy(2)n+dd=∑q=1mpq+dd.
Following [1, 2], such kind of problem is called Hermite interpolation of type total degree. The interpolation problem (p,X) is called regular if the above equation has a unique solution for each choice of values {cq,α,1≤q≤m,0≤|α|≤pq}. Otherwise, the interpolation problem is singular. As shown in [3], the regularity of Hermite interpolation problem (p,X) implies that it is regular for almost all X⊂Rd with |X|=m. Hence, in this paper, we will call p almost d-regular if (p,X) is regular for some X⊂Rd. Otherwise, we call pd-singular. With no confusion, we also call it almost regular or singular for convenience. If f is a nontrivial polynomial satisfying (1) with zero interpolation condition, we call f a vanishing polynomial with respect to X and p. Obviously, (p,X) being singular is equivalent to the existence of a vanishing polynomial of degree no more than n.
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 [1–10] and the references therein. Most recently, authors [5] made further development and gave complete description for the regularity of the interpolation problem on m=d+k(k≤3) 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 m=d+k≤d(d+3)/2 with d≥3,k≥4.
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 m=d+k≤d(d+3)/2. In this case, since m<d+22, there must exist a nontrivial quadratic polynomial Q which vanishes at X1,X2,…,Xm. Also there exists a nontrivial linear polynomial L vanishing at Xk+1,Xk+2,…,Xk+d. Given X~={Xi1,Xi2,…,Xiτ}⊂X and p~={pi1,pi2,…,piτ}, if there are vanishing polynomials with respect to p~ and X~, we will denote by ftdX~p~ one vanishing polynomial of them.
Here we always assume that no interpolation happens at Xir if the rth component pir of p~ is -1. Obviously, vanishing polynomials always exist if (3)∑j=1τpij+dd<t+dt,since the number of the equations is less than the number of the unknowns.
For convenience, we always order 0≤p1≤p2≤⋯≤pm with pm≥1. In [5], authors showed that the inequality(4)n≥pm+pm-1+1must hold if (p,X) is regular, which gives evaluation of n in (2). The following theorem implies that inequality (4) is very sharp.
Theorem 1.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, if n>pm+pm-1+1, then p is singular.
Proof.
We first assume pk+1≤pm. Then f=Qpk+1Lpm-pk is a vanishing polynomial with respect to X and p, and (5)degf=2pk+1+pm-pk≤pm+pk+2≤pm+pm-1+2≤n.Thus, in this case p is singular.
If pk=pk+1=⋯=pm, then n>pm+pm-1+1=2pm+1. Thus Qpm+1 is a vanishing polynomial with respect to X and p, and (6)degQpm+1=2pm+2≤n.Collecting two cases, we complete the proof.
Next, we assume n=pm+pm-1+1.
Lemma 2.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, if pk+1≤pm-1, then p is singular.
Proof.
Let f=Qpk+1·Lpm-pk. Then f is a vanishing polynomial with respect to X and p. Moreover (7)degf=2pk+1+pm-pk=pm+pk+2≤pm+pm-1+1=n.This completes the proof.
If pk+1>pm-1, we easily get pk=pk+1=⋯=pm-1. The following theorem is due to [5], which will be used in next lemma.
Theorem 3 (see [5]).
Assume m≥2. Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, if(8)p1+p2+⋯+pm+m≤ndthen the Hermite interpolation of type total degree is singular.
This theorem implies that there exists a vanishing polynomial of degree no more than n with respect to p and X if (8) holds.
Lemma 4.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, if pk=pk+1=⋯=pm-1 and pk-1+2≤pm, then p is singular.
Proof.
Let P1=Qpk-1+1. Then P1 together with all of its partial derivatives of order up to pk-1 vanishes at the m points. For d+1 points Xk,Xk+1,…,Xk+d, since (9)pk-pk-1-1+pk+1-pk-1-1+⋯+pm-pk-1-1+d+1=dpm-1+pm-d+1pk-1=dpm-1+pm+d-1pk-1-2dpk-1≤dpm-1+pm+d-1pm-2-2dpk-1≤dpm-1+pm-2pk-1-1,it follows from Theorem 3 that there exists a polynomial P2 with deg(P2)≤pm-1+pm-2pk-1-1, together with all of its partial derivatives of order up to pi-pk-1-1 vanishing at Xi for i=k,k+1,…,m. Let f=P1P2. Then, f and all of its partial derivatives of order up to pi vanish at Xi for i=1,2,…,m, and (10)degf=2pk-1+1+pm-1+pm-2pk-1-1≤pm+pm-1+1≤n.Thus, the interpolation is singular.
In what follows, we only need to consider pk=pk+1=⋯=pm-1 and pk-1+1≥pm, which includes(11)pk-1+1=pk=⋯=pm,n=2pm+1,(12)pk-1=⋯=pm-1=pm-1,n=2pm,(13)pk-1=pk=⋯=pm,n=2pm+1.
Lemma 5.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, suppose that (11) is satisfied; then Hermite interpolation of type total degree is almost regular if and only if d=3,k=5 and(14)p1,p2,…,p8=p-1,p-1,p-1,p-1,p,p,p,p.
Proof.
Set pk=⋯=pm=p(p≥1); then pk-1=p-1 and n=2p+1. We first check (2). Let (15)Gkd,p=∑j=1mpi+dd-n+dd=1d!∑j=1k-2∏i=1dpj+i+∏i=1dp+i-1+d+1∏i=1dp+i-∏i=1d2p+1+i.Since pj≤pk-1 for j=1,2,…,k-2 and d+k≤d(d+3)/2, then (16)Gkd,p·d!≤k-1∏i=1dp+i-1+d+1∏i=1dp+i-∏i=1d2p+1+i≤12d+2d-1∏i=1dp+i-1+d+1∏i=1dp+i-∏i=1d2p+1+i≔Fd,p.We will show that F(d,p)<0 implies F(d+1,p)<0. Note that (17)∏i=1dp+i-1=pp+d∏i=1dp+i.Thus if F(d,p)<0, then (18)∏i=1dp+i<p+d∏i=1d2p+1+i1/2d+2d-1p+d+1d+p.Hence(19)Fd+1,p=12d+3d∏i=1d+1p+i-1+d+2∏i=1d+1p+i-∏i=1d+12p+1+i=12d+3d∏i=0dp+i+d+2∏i=1d+1p+i-∏i=1d+12p+1+i=12d+3dp+d+2p+d+1∏i=1dp+i-∏i=1d+12p+1+i<1/2d+3dp+d+2p+d+1p+d∏i=1d2p+1+i1/2d+2d-1p+d+1d+p-∏i=1d+12p+1+i=∏i=1d2p+1+i1/2d+2d-1p+d+1d+p·Hd,p,where (20)Hd,p=12d+3dp+d+2p+d+1·p+d-2p+d+2·12d+2d-1p+d+1d+p=-p2d2p+2d2+dp-4p-4.
Since ∂H/∂d=-p/2(2dp+4d+p)<0 for d≥2, H(d,p) is monotonically decreasing about d. Thus, H(d,p)<0 for d≥2 due to H(2,p)=-p(p+2)<0 and F(d+1,p)<0.
Since F(4,p)=-2p(p+1)2(p+2)<0, then Gk(d,p)≤F(d,p)<0 for d≥4, which means that (2) does not hold for d≥4. Thus we only need to consider d=3. In this case (21)6Gk3,p≤k-1∏i=13p+i-1+4∏i=13p+i-∏i=132p+1+i=kp3+3kp2-5p3+2kp-15p2-10p=-p3-3p2-2p<0,k=4;0,k=5;p3+3p2+2p,k=6.Hence, for d=3, (2) does not hold for k=4 and holds for k=5 only if (14) holds. We will show that it is almost regular in next section. For d=3 and k=6, (2) holds only in the case of p2≤p-2 since (2) holds for {p-1,p-1,p-1,p-1,p,p,p,p}. We can show that p is singular if p2≤p-2. In fact, we can take f=Qp-1·f33Xp~ with p~={-1,-1,0,0,0,1,1,1,1} as the vanishing polynomial. Obviously deg(f)=2(p-1)+3=2p+1. The proof is completed.
Lemma 6.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}. Suppose that (12) is satisfied and that pk-1=⋯=pm-1=p≥0. Then
for p=0, (2) holds only for one form {0,0,…,0︸(1/2)d(d+1),1} and Hermite interpolation of type total degree is almost regular for d≥2;
for p=1, if d≥7, it is singular; if 3≤d≤6, (2) has three positive integer solutions and corresponding interpolation schemes are almost regular;
for p≥2,
if d≥5, it is singular;
if d=4 and p>6, it is also singular;
if d=4 and 2≤p≤6, (2) has four positive integer solutions and in these cases p are almost regular;
if d=3,k=4,6, it is singular;
if d=3,k=5, (2) holds only for one form (p-1,p,p,p,p,p,p,p+1) and it is almost regular.
Proof.
Set pk-1=⋯=pm-1=p(p≥0) and pm=p+1; then n=2p+2. If p=0, it is easy to check that (2) holds if and only if m=d(d+1)/2+1 and pm=1,pi=0,i=1,2,…,d(d+1)/2. This scheme is almost regular, which will be proved in Theorem 15 of the next section.
For p≥1, we first check (2). Let (22)Gkd,p=∑j=1mpi+dd-n+dd=1d!∑j=1k-2∏i=1dpj+i+d+1∏i=1dp+i+∏i=1dp+1+i-∏i=1d2p+2+i.Since pj≤pk-1 for j=1,2,…,k-2 and d+k≤d(d+3)/2, then(23)Gkd,p·d!≤dd+32-1∏i=1dp+i+∏i=1dp+1+i-∏i=1d2p+2+i≔Fd,p.
By the same argument with Lemma 5, one can show that F(d,p)<0 implies that F(d+1,p)<0. The following facts can be checked easily:(24)Fd,p=-3p+4p+3p+2p+131p3+257p2+440p-420,d=7-p-137p+150p+4p+3p+2p+1,d=6-p+1p+2p+312p2+23p-80,d=5-2p-6p+3p+2p+1,d=4.
Since F(7,p)<0 for p≥1, (2) does not hold for d≥7.
For p=1 and 3≤d≤6, (2) has three positive integer solutions {0,1,…,1︸6,2}(d=3),{1,…,1︸11,2}(d=4) and {1,…,1︸26,2}(d=6). These three schemes are almost regular, which can be verified by numerical method; see Remark 7.
Since F(5,p)<0 for p≥2, (2) does not hold for d≥5 and p≥2. Similarly, F(4,p)<0 for p>6 means that (2) does not hold for d=4 and p>6.
For d=4 and 2≤p≤6, (2) has four positive integer solutions (25)1,1,2,…,2︸11,3,1,3,…,3︸12,4,3,4,…,4︸12,5,6,…,6︸13,7,which are shown to be almost regular by numerical method presented in Remark 7.
Let us consider the case of d=3.
From the definition of Gk(d,p), we obtain(26)3!·G43,p=∑j=12∏i=13pj+i+4∏i=13p+i+∏i=13p+i+1-∏i=132p+2+i≤6∏i=13p+i+∏i=13p+1+i-∏i=132p+2+i=-pp+1p+2<0.Then, (2) does not hold for d=3, k=4.
For d=3, k=5, (2) holds for the form (27)p-1,p,p,p,p,p,p,p+1.Indeed, this form is the only one since p1≤p2=p3=p. This scheme is almost regular, which will be proved in the next section.
Finally, we consider the case of d=3,k=6. We can show that p is singular by taking Qp·f23Xp1 with p1={-1,-1,-1,0,0,0,0,0,1} for p3<p and Qp-1·f43Xp2 with p2={-1,-1,1,1,1,1,1,1,2} for p3=p as vanishing polynomial with respect to X and p. Here we use the fact that p2<p-1 if p3=p 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 p, 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 d,m, and p. 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 d, m, and p, which is employed in this paper for some simple cases.
Lemma 8.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, suppose that (13) is satisfied and that pk-1=⋯=pm=p≥2. Then
if d≥21, (2) never holds;
if 4≤d≤20, (2) has finite positive integer solutions listed in Table 2.
Proof.
We first check (2). Let (28)Gkd,p=∑j=1mpi+dd-n+dd=1d!∑j=1k-2∏i=1dpj+i+d+2∏i=1dp+i-∏i=1d2p+1+i.Since pj≤pk-1 for j=1,2,…,k-2 and d+k≤d(d+3)/2, then (29)Gkd,p·d!≤dd+32∏i=1dp+i-∏i=1d2p+1+i≔Fd,p.In the same way with Lemma 5, one can show that F(d,p)<0 implies that F(d+1,p)<0 and F(d,p+1)<0. By a simple calculation, we have (30)F21,2<0,F10,3<0,F8,4<0,F7,5<0,F6,6<0,F5,8<0,F4,20<0,F20,2=0,F19,2>0,F9,3>0,F7,4>0,F6,5>0,F5,6>0,F4,8>0,F4,19>0.F(d,p)>0 implies that F(d~,p)>0 holds for all d~<d. Thus we must take p1≤p-1 to ensure (2) if F(d,p)>0. Hence, let (31)F~d,p=∏i=1dp-1+i+dd+3-22∏i=1dp+i-∏i=1d2p+1+iand we again have F~(d,p)<0 implying F~(d+1,p)<0 and F~(d,p+1)<0. It is easy to get (32)F~20,2<0,F~10,3<0,F~7,4<0,F~6,5<0,F~4,18<0.Thus we can obtain the possible pairs (d,p) 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.
The possible pairs (d,p) satisfying (2) in the case of d>3 and p≥2.
Lemma 9.
Let d=3. Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}, suppose that (13) is satisfied and that pk-1=⋯=pm=p≥2. Then
for k=4, (2) has four positive integer solutions and the corresponding interpolation schemes are singular;
for k=5, (2) has only one positive integer solution and the corresponding interpolation scheme is almost regular;
for k=6 and p<10, (2) has finite positive integer solutions and all of them are singular.
Proof.
(i) d=3 and k=4.
According to the definition of Gk(d,p) in Lemma 8, we obtain (33)G43,p≤-16p+1p+2p-9.So, (2) never holds if p>9. If p≤9, (2) has four positive integer solutions {2,2,3,3,3,3,3},{2,4,4,4,4,4,4},{5,6,6,6,6,6,6}, and {9,9,9,9,9,9,9}. We claim that they are all singular. To show this we can take(34)f1=f43Xp1·f33Xp2,f2=f33Xp2·f33Xp3·f33Xp4,f3=f43Xp1·f43Xp5·f43Xp7·L,f4=f43Xp1·f43Xp5·f43Xp6·f43Xp7·f33Xp2as vanishing polynomials with respect to these four solutions, respectively, where (35)p1=1,1,2,1,1,1,1,p2=0,0,0,1,1,1,1,p3=0,1,1,1,1,0,0,p4=0,1,1,0,0,1,1,p5=1,2,1,1,1,1,1,p6=1,1,1,2,1,1,1,p7=2,1,1,1,1,1,1.
(ii) d=3 and k=5.
Equation (2) has only one positive integer solution {0,0,1,2,2,2,2,2} which is almost regular. The regularity can be checked by numerical method mentioned in Remark 7.
(iii) d=3, k=6, and p5=⋯=p9=p<10.
We will show that all the schemes in this case are singular. The proof is based on the following three cases.
Case 1. Consider the following: p4<p.
p4≤p-2. If (2) holds, then p is singular. In fact we can check that (36)Qp-2·f53Xp8,p8=0,0,0,0,2,2,2,2,2
is the desired vanishing polynomial.
p4=p-1. We consider the case of p3<p4 firstly. In this case, (2) holds only if p1≤p3-1=p-3 because (37)p+33·5+p-1+33+p-2+33·3-2p+1+33=16p+1p2-4p+6,
and we can check that p is singular if p1≤p-4. In fact, we can take (38)Qp-3·f73Xp9,p9=0,1,1,2,3,3,3,3,3as the vanishing polynomial of p.
We next consider the case of p3=p4. In this case, p2≤p3-2 must hold; otherwise (2) never holds. If p2=p3-2, (2) has two solutions: 0,1,3,3,4,4,…,4 and 1,2,4,4,5,…,5. These two schemes are singular, which can be checked by numerical method (see Remark 12). If p2<p3-2, then the interpolation scheme p is singular, which can be shown to take (39)Qp-3·f73Xp10,p10=-1,-1,2,2,3,3,3,3,3as the vanishing polynomial. Here the existence of f73Xp10 follows from the construction of f1 in (34).
Case 2. p3<p4=⋯=p9=p.
We first claim that p is singular if p3≤p-3. Notice that if p3=p-3, then (2) holds only if p2<p3, which will lead to the singularity of p. The vanishing polynomial can be taken as (40)Qp-3·f33Xp11·f23Xp12·f23Xp13,where(41)p11=-1,-1,0,0,0,1,1,1,1,p12=-1,-1,-1,1,0,0,0,0,0,p13=-1,-1,-1,0,1,0,0,0,0.
If p3=p-2, then p2<p3 must hold to ensure (2). If p3=p2+1 or p2+2, (2) has four solutions for p<10: {0,3,4,6,…,6}, {2,5,6,8,…,8}, {1,1,3,5,…,5}, and {3,3,5,7,…,7}. These four schemes are singular, which can be checked by numerical method (see Remark 12). If p3≥p2+3, then p is singular by taking (42)Qp-4·f33Xp11·f33Xp14·f33Xp15as the vanishing polynomial, where (43)p14=-1,-1,0,1,1,1,1,0,0,p15=-1,-1,0,1,1,0,0,1,1.
If p3=p-1, then p2<p3-2 must hold to ensure (2). In the case of p3=p2+3 or p3=p2+4, (2) has two solutions for p<10: {2,3,7,8,…,8} and {3,4,8,9,…,9}. These two schemes are singular, which can be checked by numerical method (see Remark 12). In the case of p3≥p2+5, p is singular, which can be shown by taking (44)Qp-5·f43Xp16·f43Xp17·f33Xp18as the vanishing polynomial, where (45)p16=-1,-1,1,1,1,1,1,1,2,p17=-1,-1,1,1,1,1,1,2,1,p18=-1,-1,0,1,1,1,1,0,0.
Case 3. Consider the following: p3=⋯=p9.
To ensure (2), p2<p-4 must hold. Furthermore, it is easy to check for p≤10 and p2=p-5, p-6, and p-7 that (2) has no solution. If p2≤p-8, then p is singular and the corresponding vanishing polynomial is taken as (46)Qp-7·f43Xp16·f43Xp17·f33Xp18·f43Xp19,where (47)p19=-1,-1,2,1,1,1,1,1,1.Thus we complete the proof.
Remark 10.
The interpolation scheme 9,9,9,9,9,9,9 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 (4.5) in [11] holds with equal sign; hence, its poisedness can not be decided by the necessary condition (4.5) 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 d=3 and p=3,3,3,3,3,3 is singular by taking (48)f23Xp20·f23Xp21·f23Xp22·L,where(49)p20=1,0,0,0,0,0,p21=0,1,0,0,0,0,p22=0,0,1,0,0,0.This scheme was mentioned in [1, 11] and wrongly claimed to be almost regular in [5].
Remark 12.
The singularity of interpolation schemes 0,1,3,3,4,4,…,4, 1,2,4,4,5,…,5, {0,3,4,6,…,6}, {2,5,6,8,…,8}, {1,1,3,5,…,5}, {3,3,5,7,…,7}, {1,6,9,10,…,10}, {2,3,7,8,…,8}, and {3,4,8,9,…,9} can be verified by numerical method. Here we notice that not all of them need to prove independently because we have the following observation: (50)0,1,3,3,4,4,…,4is singular⟹1,2,4,4,5,…,5is singular;0,3,4,6,…,6is singular⟹2,5,6,8,…,8is singular;1,1,3,5,…,5is singular⟹3,3,5,7,…,7is singular;2,3,7,8,…,8is singular⟹3,4,8,9,…,9is singular.Nine nodes can be selected as(51)Xi=xi,yi,ziT,i=1,2,…,5,X6=1,0,0T,X7=0,1,0T,X8=0,0,1T,X9=0,0,0T.
Lemma 13.
Given X={X1,X2,…,Xm} and p={p1,p2,…,pm}. Suppose that (13) is satisfied and that pk-1=⋯=pm=p=1. Then (2) has finite positive integer solutions for any d≥3 and all except one scheme are almost regular.
Proof.
By setting p1=p2=⋯=pk0=0, pk0+1=⋯=pk0+k1=1, and n=3 in (2), we obtain (52)3+dd=k00+dd+k11+dd,where d+2≤k1≤d+k,k0+k1=m. This equation has finite positive integer solutions for any d≥3, which is listed in Table 1 for 3≤d≤10.
All the schemes except d=4,k0=0,k1=7 are almost regular. The singularity of d=4, k0=0, and k1=7 can be proved by numerical method. The 7 points can be selected as X1=(0,0,0,0)T, X2=(1,0,0,0)T, X3=(0,1,0,0)T, X4=(0,0,1,0)T, X5=(0,0,0,1)T, X6=(x1(1),x2(1),x3(1),x4(1))T, and X7=(x1(2),x2(2),x3(2),x4(2))T. Then the Vandermonde matrix is singular by symbolical computation, implying (1,1,1,1,1,1,1) is singular.
The solution of (2) in the case of pk-1=⋯=pm=1.
d
k0
k1
d
k0
k1
d
k0
k1
3
0
5
4
5
6
4
0
7
5
8
8
5
2
9
6
14
10
6
7
11
6
0
12
7
16
13
7
8
14
7
0
15
8
21
16
8
12
17
8
3
18
9
30
19
9
20
20
9
10
21
9
0
22
10
33
23
10
22
24
10
11
25
10
0
26
All the solutions of (2) in Lemma 8.
d
k
p
n
Singular/regular
4
6
(0,1,2,2,…,2)
5
Almost regular
(2,3,3,…,3)
7
Almost regular
4
7
(1,4,4,…,4)
9
4
8
(0,1,1,1,1,2,2,…,2)
5
Almost regular
(1,1,1,3,3,…,3)
7
Almost regular
(1,2,2,2,3,3,…,3)
7
Almost regular
(2,3,3,4,4,…,4)
9
(3,4,5,5,5,…,5)
11
(4,6,6,…,6)
13
4
9
(1,1,1,4,4,…,4)
9
(2,3,3,3,3,4,4…,4)
9
(3,3,3,5,5…,5)
11
(3,3,6,6…,6)
13
(6,7,8,8,…,8)
17
(8,8,9,9,9,…,9)
19
4
10
(0,0,0,0,0,0,2,2,…,2)
5
Almost regular
(0,1,1,1,1,1,1,1,2,2…,2)
5
Almost regular
(1,1,1,1,2,2,3,3,…,3)
7
Almost regular
(2,2,…,2,3,3,3,3,3,3)
7
Almost regular
(1,1,2,2,2,2,2,3,3…,3)
7
Almost regular
(1,1,1,3,3,4,4,…,4)
9
(2,3,3,3,3,3,3,4,4,…,4)
9
(1,2,2,2,3,4,4,…,4)
9
(1,2,2,4,5,5,…,5)
11
(4,4,4,4,6,6,…,6)
13
(4,5,5,5,5,5,6,6,…,6)
13
(0,3,6,7,7,…,7)
15
(6,7,7,7,7,8,8,…,8)
17
(7,7,7,9,9,9,…,9)
19
(8,9,9,9,10,10,10,…,10)
21
(10,11,11,12,12,12,…,12)
25
(12,13,14,14,…,14)
29
(14,16,16,16,…,16)
33
5
7
(2,2,…,2)
5
Almost regular
5
12
(0,0,0,1,1,1,2,2,…,2)
5
Almost regular
(1,1,1,1,1,1,1,2,2,…,2)
5
Almost regular
(0,0,1,3,3,…,3)
7
(0,2,2,2,3,3,…,3)
7
(3,3,4,4,…,4)
9
5
14
(5,6,6,…,6)
13
5
15
(4,4,6,6,…,6)
13
6
12
(1,1,2,2,…,2)
5
Almost regular
6
15
(1,1,1,1,1,1,2,2,…,2)
5
Almost regular
6
17
(0,1,2,3,3,…,3)
7
6
18
(0,0,0,0,0,0,0,1,2,2,2,…,2)
5
Almost regular
(1,1,1,1,1,1,1,1,1,1,2,2,…,2)
5
Almost regular
6
19
(0,1,2,2,2,2,3,3,…,3)
7
6
20
(0,1,1,1,1,1,3,3,…,3)
7
(1,3,3,4,4,…,4)
9
6
21
(0,0,0,0,0,0,0,1,1,1,1,1,2,2,…,2,4)
9
(0,1,2,2,2,2,2,2,2,3,3,…,3)
7
(1,…,1︸14,2,…,2︸13)
5
Almost regular
7
15
(2,2,…,2)
5
7
22
(0,0,0,0,1,1,1,1,2,2,…,2)
5
(1,1,1,1,1,1,1,1,1,2,2,…,2)
5
7
23
(2,2,3,3,…,3)
7
8
23
(1,1,1,2,2,…,2)
5
8
27
(1,1,1,1,1,1,1,1,2,2,…,2)
5
8
31
(0,0,0,0,0,0,0,0,0,1,1,2,2,…,2)
5
(1,…,1︸13,2,…,2︸26)
5
(3,3,…,3)
7
8
35
(0,…,0︸9,1,…,1︸7,2,…,2︸27)
5
(1,…,1︸18,2,…,2︸25)
5
9
31
(0,0,1,1,2,2,…,2)
5
9
40
(0,…,0︸7,1,…,1︸7,2,…,2︸35)
5
(0,0,1,…,1︸13,2,…,2︸34)
5
9
43
(3,3,…,3)
7
10
38
(1,1,1,2,…,2︸45)
5
10
43
(1,…,1︸45,2,…,2︸44)
5
10
48
(1,…,1︸45,2,…,2︸44)
5
(1,…,1︸15,2,…,2︸43)
5
10
53
(0,…,0︸11,1,…,1︸8,2,…,2︸44)
5
(1,…,1︸21,2,…,2︸42)
5
11
45
(2,…,2︸56)
5
11
56
(0,0,0,0,0,0,1,1,1,1,1,1,2,…,2︸55)
5
(1,…,1︸11,2,…,2︸54)
5
12
56
(2,…,2︸68)
5
13
79
(0,0,0,0,0,0,0,1,1,1,1,2,…,2︸81)
5
(1,…,1︸12,2,…,2︸80)
5
14
92
(0,0,0,1,1,1,1,1,1,1,2,…,2︸96)
5
14
99
(0,0,0,1,…,1︸15,2,…,2︸98)
5
15
99
(2,…,2︸114)
5
15
114
(0,…,0︸8,1,…,1︸8,2,…,2︸113)
5
(1,…,1︸17,2,…,2︸112)
5
16
117
(2,2,…,2)
5
16
125
(1,…,1︸9,2,…,2︸132)
5
16
133
(1,…,1︸18,2,…,2︸131)
5
17
137
(2,…,2︸154)
5
Remark 14.
In [2], Lorentz presented a conjecture (Conjecture 8) which gives a necessary and sufficient condition about the singularity of multivariate Hermite interpolation. Although the scheme d=4,k0=0,k1=7,n=3 is singular, it can not be checked by the conjecture from [2]. Hence Conjecture 8 in [2] is not correct.
3. The Proof of Regularity of Some Interpolation Schemes
In this section, we will prove that {0,0,…,0︸d(d+1)/2,1} is almost d-regular and {p-1,p,p,p,p,p,p,p+1} and {p-1,p-1,p-1,p-1,p,p,p,p} are both almost 3-regular. To this end, we need to choose X carefully and then prove the regularity of p.
Theorem 15.
Given X={X1,X2,…,Xm} and p={0,0,…,0︸d(d+1)/2,1}, then p is almost d-regular.
Proof.
Let ei=(0,0,…,0,1i,0,…,0)T∈Rd and define Xi,i=ei, Xi,j=ei+ej for 1≤i<j≤d. Finally, let Xm=(0,0,…,0)T and X=Xi,j,1≤i≤j≤d,Xm. We will show the regularity of p. To this end, we assume that f(X) is a polynomial of degree 2 and satisfies all the homogenous interpolation conditions. It only needs to show f≡0.
Due to Dαf(Xm)=0 for |α|≤1, f(X) is of form (53)fX=XTAX,A=aijd×d,aij=aji.Thus we have (54)fXi,i=eiTAei=aii=0i=1,2,…,d,fXi,j=ei+ejTAei+ej=aii+2aij+ajj=0,which lead to aij=0 for 1≤i,j≤d. Then f should be a zero polynomial, which completes the proof.
To prove the regularity of {p-1,p,p,p,p,p,p,p+1} and {p-1,p-1,p-1,p-1,p,p,p,p}, we need the following lemma.
Lemma 16 (see [1, 4]).
Let d=2; then interpolating the value of a function and all of its partial derivatives of order up to p 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 p+1/p-1 at a fourth point lying anywhere in the interior of the triangle by polynomials from Π2p+22/Π2p+12, is regular.
In fact the fourth point can lie anywhere except on the three edges of the triangle.
Theorem 17.
Let X={X1=1,1,1T, X2=1,0,0T, X3=0,1,0T, X4=0,0,1T, X5=1,1,0T, X6=(1,0,1)T, X7=(0,1,1)T, X8=(0,0,0)T}, and p={p-1,p,p,p,p,p,p,p+1}(p≥1); then (p,X) is regular.
Proof.
Suppose that f is a polynomial of degree no more than 2p+2 such that (55)DαfX1=0,α≤p-1,DγfX8=0,γ≤p+1,DβfXi=0,β≤p,i=2,3,…,7.That is, f satisfies the homogeneous interpolation conditions. To prove this theorem, we only need to show f≡0.
Consider the value of f on the plane x3=0. It follows from (55) that (56)DγfX8=0,γ≤p+1,DβfXi=0,β≤p,i=2,3,5.According to Lemma 16, f vanishes on the plane x3=0, which implies that f can be divided by x3. Similarly, f can be divided by x1 and x2, respectively. Hence f can be written as f=x1x2x3f1 with degf1=2p-1.
If p=1, by taking f=x1x2x3f1 into (55) we obtain (57)f1Xi=0,i=1,5,6,7.Thus f1=0 and hence f=0, which will prove the theorem for p=1.
If p>1, taking f=x1x2x3f1 into (55) will give(58)Dαf1X1=0,α≤p-1,Dγf1X8=0,γ≤p-2,Dβf1Xi=0,β≤p-2,i=2,3,4,Dμf1Xi=0,μ≤p-1,i=5,6,7.
Then consider the plane x3-1=0. It follows from Lemma 16 and the equations (59)Dγf1X4=0,γ≤p-2,Dβf1Xi=0,β≤p-1,i=1,6,7,that f1 can be divided by x3-1. Similarly, f1 can be also divided by x1-1 and x2-1, respectively. Thus f1 can be written as f1=(x1-1)(x2-1)(x3-1)f2 with degf2=2p-4.
By collecting the above results, we have f=x1x2x3(x1-1)(x2-1)(x3-1)f2 with degf2=2p-4. If p=2, then f2 satisfies f2(X8)=0, which implies that f2=0 and the theorem will be proved for p=2. Otherwise, for p>2, f2 satisfies(60)Dαf2X1=0,α≤p-4,Dγf2X8=0,γ≤p-2,Dβf2Xi=0,β≤p-3,i=2,3,…,7,by taking f into (55).
Clearly, (60) is the same interpolation problem as (55), but with a smaller p. 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 X={X1=(1,1,1)T, X2=(1,0,0)T, X3=(0,1,0)T, X4=(0,0,1)T, X5=(1,1,0)T, X6=(1,0,1)T, X7=(0,1,1)T, X8=(0,0,0)T}, and p={p-1,p-1,p-1,p-1,p,p,p,p}(p≥1); then (p,X) is regular.
Theorems 17 and 18 imply that {p-1,p,p,p,p,p,p,p+1} and {p-1,p-1,p-1,p-1,p,p,p,p} 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 m=d+k≤1/2d(d+3) nodes in Rd.
Given p=p1,p2,…,pm with m=d+k≤1/2d(d+3) and p1≤p2≤⋯≤pm, the following results are derived in this paper:
If pk+1≤pm-1, then p is singular or (2) does not hold (see Lemma 2).
Suppose pk=pk+1=⋯=pm-1, then
p is singular if pk-1+2≤pm; see Lemma 4;
for pk-1+1=pk=⋯=pm,n=2pm+1, p is almost regular if and only if d=3,k=5, and p=p-1,p-1,p-1,p-1,p,p,p,p; see Lemma 5; the regularity is given in Theorem 18;
if pk=pk-1=⋯=pm-1=pm-1=p-1, the following 9 interpolation schemes are almost regular (see Lemma 6):(61)0,0,…,0︸1/2dd+1,1,0,1,…,1︸6,2,d=3,1,…,1︸11,2d=4,1,…,1︸26,2d=6,1,1,2,…,2︸11,3d=4,1,3,…,3︸12,4d=4,3,4,…,4︸12,5d=4,6,…,6︸13,7d=4,p-1,p,p,p,p,p,p,p+1d=3;
other interpolation schemes are singular;
pk=pk-1=⋯=pm-1=pm=p.
If p≥2, then (2) never holds for d≥21 and has finite solutions for 4≤d≤20 listed in Table 2.
If p≥2,d=3, and k=4,5, only one interpolation scheme 0,0,1,2,2,2,2,2 is almost regular and other schemes are singular. If p≥2,d=3, and k=6, (2) has finite solutions for p<10 and all of them are singular; see Lemma 9.
If p=1, then (2) has finite solutions listed in Table 1. All except one scheme are almost regular. See Lemma 13 for this case.
Given p, how to decide the regularity theoretically remains difficult and will be our future research project.
AppendixSolutions of (2) in Lemma 8
Table 2 presents all the solutions of (2) in Lemma 8. A small quotient of them is decided by numerical method, but most of them are left open.
Competing Interests
The authors declare that they have no competing interests.
Acknowledgments
This project is supported by NNSFC (nos. 11301053, 61432003, 11171052, 11271060, and 61272371) and “the Fundamental Research Funds for the Central Universities.”
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