Problem 3.

Interpolation and numerical integration in 2D.

a) Let Pn,n = {Σ-o-j¹|aj R 0≤ij≤n} denote the set of 2-variate

polynomials in z and y that are of degree ≤n in z and of degree ≤n in y.

For a given 2D square [0, 1]2 with a square mesh (r.. y) = (ih, jh) for 0 ≤i, j≤n

and h = 1/n, and f = C([0, 1]2), consider the interpolation problem: find p € Pn,n

such that

(6)

P(Z₁yj) = f(x,y) for all 0≤i,j≤n.

We recall from the lectures that this problem has a solution

12 12

p(z,y)=f(x,ye)Lx (T) Le(y),

k-0 1-0

where Le (2) is a univariate polynomial in z of degree n and Le(y) is a univariate

polynomial of degree n in y.

Describe L₁ (2) and (y) and show that p(x, y) in (7) with the functions L and

Le that you define indeed is a solution of (6).

b) Leaning on the fact that any univariate polynomial p of degree ≤n that has n+1

or more zeros is equal to the 0 function, meaning p = 0, show that the solution to

the 2 dimensional interpolation problem (6) is unique.

Hint: Let q € Pn, be another solution and consider the polynomial r = p-q-

Then r € Pn,n, so it can be written

12 n

r(z,y) = Σcjz¹y

1-0 j-0/nfor some c₁ € R. For any fixed ye, for 0 ≤ ≤n, the function r(x, y) is a

univariate polynomial in z of degree ≤n, namely,

n

12

r(x,y) = Σ(Σ)n - Σαμπ

= (v².

j-0

1-0

-(Me)

How many zeros does r(x, y) have, what does this imply about the coefficients

a(ye) for all 0 ≤ i ≤n, and how can this be used to show that c = 0 for all

0 ≤i, j≤n?

c) The composite trapezoidal rule in 2D on the square mesh presented above can be

described as follows:

12

[..

f(x, y)dx dy ≈ [Ë"*" ["“_ p., (x,y) dx dy =: T(n),

1-1-1-1

where pij € P₁,1 denotes the unique polynomial that goes through f in the four

points (-1,3-1), (Fi, Yj-1), (Fi-1, y₁) and (zi, yj).

Determine pij(x,y) and verify that the 2D square-mesh trapezoidal rule is given

by

12

T(n)=ΐΣΣ(f(-1,31

si−1, Yj−1) + ƒ(zi, Yj−1) + f(x−1, 9j) +;

i-1 j-1

+ f(24.3))).

d) For na = 2¹+*, 8 = 1,2,..., 7, compute T(n.) to estimate the integral

I 1 = exp(-(x - sin(y²))³3) dr dy.

Since we do not know the alue of I, we approximate the error E(8) = |I-T(n,),

using the pseudo-reference solution I := T(210). Estimate numerically the order of

convergence r in

E(n)= en+O(n-(r+1))

for example, through estimating the slope of the curve (log(n,), log(E(n.)) in a

plot, or by studying the ratio

log(E(n¸)) – log(E(n,-1))

log(n.) -log(n-1)

which typically becomes more accurate for larger values of s. (See this link for

more on using loglog plots for convergence estimates.)