Multidimensional Interpolation

4.3.4 Multidimensional Interpolation

Polynomial interpolation for functions of several independent variables are generally more difficult than the one-dimensional case. There is in general a lack of uniqueness. In par- ticular, it may not suffice to require that the interpolation points are distinct; see Prob- lem 4.3.9 (b).

As a simple example, consider the problem of interpolating a function given at three distinct points p i = (x i ,y i ) , i = 1 : 3, by a linear function in two variables,

p(x, y) =c 1 +c 2 x +c 3 y.

This leads to the linear system V c = f , where

V = 1x 2 y 2 ,

f = f 2 . 1x 3 y 3 c 3 f 3

0. But 1 2 det(V ) is just the area of the triangle with vertices (x i ,y i ) , i = 1 : 3. If this area is zero, then the three points lie on a line and the problem has either infinitely many solutions or no solution.

Much of the theory for univariate interpolation can be generalized to multidimensional interpolation problems provided that the function is specified on a Cartesian (tensor) product grid. For simplicity, we first concentrate on functions f (x, y) of two variables, but the extension to more dimensions is not difficult. Assume that we are given function values

(4.3.39) where x i , i = 1 : n, and y j , j = 1 : m, are distinct points. We seek a polynomial p(x, y)

f ij = f (x i ,y j ), i = 1 : n, j = 1 : m,

of degree at most n − 1 in x and at most m − 1 in y that interpolates these values. Such a polynomial has the form

n −1 m −1

p(x, y) =

where the mn coefficients c ij are to be determined. Since the number of coefficients equal the number of interpolatory conditions we can expect the polynomial p(x, y) to be uniquely determined. To show this it suffices to show that any polynomial q(x, y) of degree n − 1 in x and m − 1 in y that vanishes at the mn distinct points (x i ,y j ) , i = 1 : n, and y j ,

j = 1 : m, must vanish identically. If we want to compute p(x, y) for given values of x and y, then we can proceed as follows. For each j = 1 : m, use univariate interpolation to determine the values p(x, y j ) , where p(x, y) is the interpolation polynomial in (4.3.40). Next, the m values p(x, y j ) ,

396Chapter 4. Interpolation and Approximation j = 1 : m, are interpolated using univariate interpolation, which determines p(x, y).

Note that since the points x i and y j are distinct all univariate interpolation polynomials are uniquely determined. It is clear that we will obtain the same result, whether we interpolate first for x and then for y or vice versa. Clearly this approach can also be used in more than two dimensions.

In many cases we are not satisfied with obtaining p(x, y) for specific values of x and y, but want to determine p(x, y) as a polynomial in x and y. We can then use the above procedure algebraically to derive a Newton formula for a tensor product interpolation problem in two variables. We set [x i ;y j ]f = f (x i ,y j ) , and define bivariate divided

differences [x 1 ,...,x ν ;y 1 ,...,y µ ]f , by recurrences, separately for each variable. We start by forming, for each y j , j = 1 : m, divided differences with respect to the x variable:

[x 1 ;y 1 ]f [x 1 ,x 2 ;y 1 ]f · · · [x 1 ,x 2 ,...,x n ;y 1 ]f , [x 1 ;y 2 ]f [x 1 ,x 2 ;y 2 ]f · · · [x 1 ,x 2 ,...,x n ;y 2 ]f ,

[x 1 ;y m ]f [x 1 ,x 2 ;y m ]f · · · [x 1 ,x 2 ,...,x n ;y m ]f .

These are used to form the Newton polynomials

i & −1

p(x, y j ) =

[x 1 ,...,x i ;y j ]f

which give, for any x, the values of the interpolation polynomial p(x, y) for y 1 ,...,y m . Next we form in each column above the divided differences with respect to y:

[x 1 ;y 1 ]f

· · · [x 1 ,...,x n ;y 1 ]f , [x 1 ;y 1 ,y 2 ]f

[x 1 ,x 2 ;y 1 ]f

· · · [x 1 ,...,x n ;y 1 ,y 2 ]f , .. .

[x 1 ;y 1 ,...,y m ]f [x 1 ,x 2 ;y 1 ,...,y m ]f · · · [x 1 ,...,x n ;y 1 ,...,y m ]f . If these nm divided differences are used for Newton interpolation in the y variable, we

obtain Newton’s interpolation formula in two variables,

i j & −1 & −1 p(x, y) =

nm

[x 1 ,...,x i ;y 1 ,...,y j ]f

(x −x ν )

(y −y µ ), (4.3.41)

where empty products have the value 1. Note that it is indifferent in which order the divided differences are formed. We could equally well have started to form divided differences with respect to y.

Remainder formulas can be derived from the corresponding one-dimensional error formulas; see Isaacson and Keller [208, Sec. 6.6]. For f sufficiently smooth there exist

4.3. Generalizations and Applications 397 values ξ, ξ ′ , η, η ′ such that

∂ n f (ξ, y) 2 n (x

f (x, η) µ =1 (y −y µ ) R(x, y) =

Lagrange’s interpolation formula can also be generalized for the tensor product case. We have

ν ) & (y µ ) p(x, y) =

Clearly p(x, y) assumes the values f (x i ,y j ) for i = 1 : n, j = 1 : m. As the polynomial is of degree n − 1 in x and m − 1 in y, it must equal the unique interpolating polynomial.

Therefore, the remainder must also be the same as for the Newton formula. The Lagrange formula (4.3.44) is easily extended to three and more variables.

The interpolation problem we have treated so far specifies the maximum degree of p(x, y) in x and y separately. Instead we could specify the total degree to be at most n − 1. Then the interpolation polynomial must have the form

n −1 n −i−1

p(x, y) =

There are 1 2 n(n + 1) coefficients to determine in (4.3.45). We shall show that with the “triangular” array of interpolation points (x i ,y j ) , i+j = 1 : n, the interpolation polynomial is uniquely determined.

The Newton formula (4.3.41) can be generalized to the case when for each i, i = 1 : n, the interpolation given points are (x i ,y j ) ,j=1:m i with 1 ≤ m i ≤ m, with a slightly more

complicated remainder formula. A particularly interesting case is when m i = n − i + 1, i.e., the interpolation points form a triangle. This gives rise to the interpolating polynomial

j & −1 p(x, y) =

i & −1

[x 1 ,...,x i ;y 1 ,...,y j ]f

(x −x ν )

(y −y µ ), (4.3.46)

2≤i+j≤n+1

with remainder formula

∂ −i n f (ξ i ,η i ) −1 ν (x −x ν ) µ =1 (y −y µ ) R(x, y) =

(n − i)! This formula is due to Biermann [26].

Interpolation formulas for equidistant points x i =x 0 + ih and y j =y 0 + jk can readily be obtained from Newton’s formula (4.3.41). Using the points (x i ,y j ) , i = 0 : n,

i = 0 : m, we get

p(x j 0 + ph, y + qk) = 4 x 4 y f (x 0 ,y 0 ). (4.3.48)

i =0 j =0

398 Chapter 4. Interpolation and Approximation

Example 4.3.8.

Formulas for equidistant points can also be obtained by using the operator formulation of Taylor’s expansion:

h 2 D 2 2 2 3 + 3 x + 2hkD x D y +k D y f 0,0 + O(h +k ). An interpolation formula, exact for all functions p(x, y) in (4.3.40) with m = n = 3, can

be obtained by replacing in Taylor’s formula the derivatives by difference approximations valid for quadratic polynomials,

f (x 0 + ph, y 0 + qh) ≈ f 0,0 + p(f 1,0 −f −1,0 ) + q(f 2 2 0,1 −f 0,−1 )

1 2 + p (f 1,0 − 2f 0,0 +f −1,0 )

+ pq(f 1,1 4 −f 1,−1 −f −1,1 +f −1,−1 )

+ q 2 (f 0,1 − 2f 0,0 +f

This formula uses function values in nine points. (The proof of the expression for approxi- mating the mixed derivative D x D y f 0,0 is left as an exercise; see Problem 4.3.11.)

An important case, to be treated in Sec. 5.4.4, is interpolation formulas in two or more dimensions with function values specified on the vertices and sides of a simplex. These play a fundamental role in the finite element method.