Birkhoff interpolation

In mathematics, Birkhoff interpolation is an extension of polynomial interpolation. It refers to the problem of finding a polynomial [math]\displaystyle{ P(x) }[/math] of degree [math]\displaystyle{ d }[/math] such that only certain derivatives have specified values at specified points:

[math]\displaystyle{ P^{(n_i)}(x_i) = y_i \qquad\mbox{for } i=1,\ldots,d, }[/math]

where the data points [math]\displaystyle{ (x_i,y_i) }[/math] and the nonnegative integers [math]\displaystyle{ n_i }[/math] are given. It differs from Hermite interpolation in that it is possible to specify derivatives of [math]\displaystyle{ P(x) }[/math] at some points without specifying the lower derivatives or the polynomial itself. The name refers to George David Birkhoff, who first studied the problem in 1906.

Existence and uniqueness of solutions

In contrast to Lagrange interpolation and Hermite interpolation, a Birkhoff interpolation problem does not always have a unique solution. For instance, there is no quadratic polynomial [math]\displaystyle{ P(x) }[/math] such that [math]\displaystyle{ P(-1)=P(1)=0 }[/math] and [math]\displaystyle{ P^{(1)}(0)=1 }[/math]. On the other hand, the Birkhoff interpolation problem where the values of [math]\displaystyle{ P^{(1)}(-1), P(0) }[/math] and [math]\displaystyle{ P^{(1)}(1) }[/math] are given always has a unique solution.

An important problem in the theory of Birkhoff interpolation is to classify those problems that have a unique solution. Schoenberg formulates the problem as follows. Let [math]\displaystyle{ d }[/math] denote the number of conditions (as above) and let [math]\displaystyle{ k }[/math] be the number of interpolation points. Given a [math]\displaystyle{ d\times k }[/math] matrix [math]\displaystyle{ E }[/math], all of whose entries are either [math]\displaystyle{ 0 }[/math] or [math]\displaystyle{ 1 }[/math], such that exactly [math]\displaystyle{ d }[/math] entries are [math]\displaystyle{ 1 }[/math], then the corresponding problem is to determine [math]\displaystyle{ P(x) }[/math] such that

[math]\displaystyle{ P^{(j)}(x_i) = y_{i,j} \qquad\forall (i,j) / e_{ij} = 1 }[/math]

The matrix [math]\displaystyle{ E }[/math] is called the incidence matrix. For example, the incidence matrices for the interpolation problems mentioned in the previous paragraph are:

[math]\displaystyle{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \qquad\mathrm{and}\qquad \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}. }[/math]

Now the question is: Does a Birkhoff interpolation problem with a given incidence matrix [math]\displaystyle{ E }[/math] have a unique solution for any choice of the interpolation points?

The case with [math]\displaystyle{ k=2 }[/math] interpolation points was tackled by George Pólya in 1931. Let [math]\displaystyle{ S_m }[/math] denote the sum of the entries in the first [math]\displaystyle{ m }[/math] columns of the incidence matrix:

[math]\displaystyle{ S_m = \sum_{i=1}^k \sum_{j=1}^m e_{ij}. }[/math]

Then the Birkhoff interpolation problem with [math]\displaystyle{ k=2 }[/math] has a unique solution if and only if [math]\displaystyle{ S_m\geqslant m \quad\forall m }[/math]. Schoenberg showed that this is a necessary condition for all values of [math]\displaystyle{ k }[/math].

Some examples

Consider a differentiable function [math]\displaystyle{ f(x) }[/math] on [math]\displaystyle{ [a,b] }[/math], such that [math]\displaystyle{ f(a)=f(b) }[/math]. Let us see that there is no Birkhoff interpolation quadratic polynomial such that [math]\displaystyle{ P^{(1)}(c)=f^{(1)}(c) }[/math] where [math]\displaystyle{ c=\frac{a+b}{2} }[/math]: Since [math]\displaystyle{ f(a)=f(b) }[/math], one may write the polynomial as [math]\displaystyle{ P(x)=A(x-c)^2+B }[/math] (by completing the square) where [math]\displaystyle{ A,B }[/math] are merely the interpolation coefficients. The derivative of the interpolation polynomial is given by [math]\displaystyle{ P^{(1)}(x)=2A(x-c)^2 }[/math]. This implies [math]\displaystyle{ P^{(1)}(c)=0 }[/math], however this is absurd, since [math]\displaystyle{ f^{(1)}(c) }[/math] is not necessarily [math]\displaystyle{ 0 }[/math]. The incidence matrix is given by:

[math]\displaystyle{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}_{3\times3} }[/math]

Consider a differentiable function [math]\displaystyle{ f(x) }[/math] on [math]\displaystyle{ [a,b] }[/math], and denote [math]\displaystyle{ x_0=a,x_2=b }[/math] with [math]\displaystyle{ x_1\in[a,b] }[/math]. Let us see that there is indeed Birkhoff interpolation quadratic polynomial such that [math]\displaystyle{ P(x_1)=f(x_1) }[/math] and [math]\displaystyle{ P^{(1)}(x_0)=f^{(1)}(x_0),P^{(1)}(x_2)=f^{(1)}(x_2) }[/math]. Construct the interpolating polynomial of [math]\displaystyle{ f^{(1)}(x) }[/math] at the nodes [math]\displaystyle{ x_0,x_2 }[/math], such that [math]\displaystyle{ \displaystyle P_1(x) = \frac{f^{(1)}(x_2)-f^{(1)}(x_0)}{x_2-x_0}(x-x_0)+f^{(1)}(x_0) }[/math]. Thus the polynomial : [math]\displaystyle{ \displaystyle P_2(x) = f(x_1) + \int_{x_1}^x\!P_1(t)\;\mathrm{d}t }[/math] is the Birkhoff interpolating polynomial. The incidence matrix is given by:

[math]\displaystyle{ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}_{3\times3} }[/math]

Given a natural number [math]\displaystyle{ N }[/math], and a differentiable function [math]\displaystyle{ f(x) }[/math] on [math]\displaystyle{ [a,b] }[/math], is there a polynomial such that: [math]\displaystyle{ P(x_0)=f(x_0) }[/math] and [math]\displaystyle{ P^{(1)}(x_i)=f^{(1)}(x_i) }[/math] for [math]\displaystyle{ i=1,\cdots,N }[/math] with [math]\displaystyle{ x_0,x_1,\cdots,x_N\in[a,b] }[/math]? Construct the Lagrange/Newton polynomial (same interpolating polynomial, different form to calculate and express them) [math]\displaystyle{ P_{N-1}(x) }[/math] that satisfies [math]\displaystyle{ P_{N-1}(x_i)=f^{(1)}(x_i) }[/math] for [math]\displaystyle{ i=1,\cdots,N }[/math], then the polynomial [math]\displaystyle{ \displaystyle P_N(x) = f(x_0) + \int_{x_0}^x\! P_{N-1}(t)\;\mathrm{d}t }[/math] is the Birkhoff interpolating polynomial satisfying the above conditions. The incidence matrix is given by:

[math]\displaystyle{ \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 0 & \cdots & 0 \\ \end{pmatrix}_{N\times N} }[/math]

Given a natural number [math]\displaystyle{ N }[/math], and a [math]\displaystyle{ 2N }[/math] differentiable function [math]\displaystyle{ f(x) }[/math] on [math]\displaystyle{ [a,b] }[/math], is there a polynomial such that: [math]\displaystyle{ P^{(k)}(a)=f^{(k)}(a) }[/math] and [math]\displaystyle{ P^{(k)}(b)=f^{(k)}(b) }[/math] for [math]\displaystyle{ k=0,2,\cdots,2N }[/math]? Construct [math]\displaystyle{ P_1(x) }[/math] as the interpolating polynomial of [math]\displaystyle{ f(x) }[/math] at [math]\displaystyle{ x=a }[/math] and [math]\displaystyle{ x=b }[/math], such that [math]\displaystyle{ P_1(x)=\frac{f^{(2N)}(b)-f^{(2N)}(a)}{b-a}(x-a) +f^{(2N)}(a) }[/math]. Define then the iterates [math]\displaystyle{ \displaystyle P_{k+2}(x)=\frac{f^{(2N-2k)}(b)-f^{(2N-2k)}(a)}{b-a}(x-a) +f^{(2N-2k)}(a) + \int_a^x\!\int_a^t\! P_k(s)\;\mathrm{d}s\;\mathrm{d}t }[/math] . Then [math]\displaystyle{ P_{2N+1}(x) }[/math] is the Birkhoff interpolating polynomial. The incidence matrix is given by:

[math]\displaystyle{ \begin{pmatrix} 1 & 0 & 1 & 0 \cdots \\ 1 & 0 & 1 & 0 \cdots \\ \end{pmatrix}_{2\times N} }[/math]