Of course, we discussed this topic in Math So this partially a review. Apolynomialofdegreed in two variable x and y is defined by. α ij x i y j.

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1 Math 5620 Spring 2019 Notes of 4/5/19 Polynomials in Two Variables Of course, we discussed this topic in Math So this partially a review. Apolynomialofdegreed in two variable x and y is defined by p d (x, y) = α ij x i y j. i+j d d I PCt 4 doo t d E t do Y By contrast, the tensor product of two polynomials in x and y, ofdegreesm and n, respectively, is given by p mn (x, y) = m n α ij x i y j. i=0 O j=1 Pi Lx 4 Loot L E to Y t dat Y Setting one variable equal to a constant gives apolynomialofdegreem or n in the other variable. The coordinate directions matter! If p mn is restricted to a line y = mx + b then the resulting polynomial is of degree mn in x. m th Math 5620 Spring 2019 Notes of 4/5/19 page 1 i

2 Polynomials of degree d have specific names for small values of d, thesameasinthecase of polynomials in one variable: d Name # coefficients 0 constant 1 linear 2 quadratic 3 cubic 4 quartic 5 quintic n I too 3 Loo 1L ot tho Y I htt ht 2 Math 5620 Spring 2019 Notes of 4/5/19 page 2

3 Tensor Product Polynomials Tensor product polynomials with m = n have corresponding names preceded with bi, as in bilinear, biquadratic, bicubic, etc. Examples: EYE in O i 9 2 How many coefficients does a tensor product polynomial have if m = n = d? Colt 2 Pmn mu Utc coeffs polyn of degree 4 in k variables DIY Math 5620 Spring 2019 Notes of 4/5/19 page 3

4 We already considered continuous piecewise linear functions on triangles and piecewise bilinear functions on rectangles. Let s look close at joining the pieces (or patches) continuously along common edges. Joining polynomials along lines and curves a i 99 bilinear Math 5620 Spring 2019 Notes of 4/5/19 page 4

5 linear interpolation d z Cti Yi Zi i l 6 Pki c4 Zi Plt 4 doo 140 4, I Y X X Y 4,2 x y xixsys Ys Xo X I t Y R Ey yz 4 4 IVI A t Bt ta t DR t Exy Fy2 O

6 difitx.si Mi Fei 01 it Otaki cxiy T Him fi o E i II

7 C 0 Quadratic Elements on Triangles A oases µ Math 5620 Spring 2019 Notes of 4/5/19 page 5

8 C 0 Cubic Elements on Triangles. f of A Math 5620 Spring 2019 Notes of 4/5/19 page 6

9 C 1 Elements Some problems require higher smoothness than continuity. For example, the deformation of a loaded and clamped elastic plate is described by the Variational Principle 1 ( I(u) = u 2 2 xx +2u 2 xy + u 2 yy 2fu ) dxdy. R Its Euler Equation is u xxxx +2u xxyy + u yyyy = f(x, y) The approximating space requires differentiable basis functions. The underlying principle is that the order of the Euler equation is twice that of the highest derivative in the variational principle, and the required smoothness of the approximating function is one less than the degree of highest derivative in the variational principle. Of course, we might want a higher degree of differentiability in our approximating function for other reasons as well, for example if we want to display the approximating function. Reflections on a surface have one degree less of differentiability than the differentiability of the surface. So we might even want a twice differentiable function, for example, if it describes the shape of a car or plane. Math 5620 Spring 2019 Notes of 4/5/19 page 7

10 Bicubic C 1 elements on rectangles We ll first redo bilinear C 0 in a way that carries over to bicubic C 1. For simplicity, we focus on the unit square. We can transform from the unit square to any rectangle by a linear change in x and y. recall the dichotomy of interpolation versus computation of the data. They use the same spaces. In one case the data, i.e., the coefficients, are known, in the other they are determined as the solution of a large linear system. Define the partial interpolation operators P x f(x, y) =(1 x)f(0,y)+xf(1,y) P y f(x, y) =(1 y)f(x, 0) + yf(x, 1) Then the bilinear element is given by Qf = P x P y f. Co Oeo yo I R f cry Py c y fcx.ytyfct.tl l X fl y floaty food 1 t H 4 fit o t y fa l Math 5620 Spring 2019 Notes of 4/5/19 page 8

11 l t l 4 f Coco t l x y f oil t t l Y f fo t Xy f Iii

12 C 1 piecewise bicubic Hermite element to obtain a C 1 approximating function we start with cubic Hermite interpolation in one variable. Suppose s L h 0,h 1, h 0, and h 1 are cubic polynomials with the properties h i (j) =δ ij h i (j) =0 h i (j) =0 h i (j) =δ ij Exercise: compute the h i and h i. Then the interpolation operators in x and y are given by H x f(x, y) =f(0,y)h 0 (x)+f(1,y)h 1 (x)+f x (0,y) h 0 (x)+f x (1,y) h 1 (x) H y f(x, y) =f(x, 0)h 0 (y)+f(x, 1)h 1 (y)+f x (x, 0) h 0 (y)+f x (x, 1) h 1 (y) Then the resulting bicubic operator is Hf(x, y) =H x H y f(x, y) We get (exercise): Math 5620 Spring 2019 Notes of 4/5/19 page 9

13 Hf(x, y) = f(0, 0)h 0 (x)h 0 (y)+f(0, 1)h 0 (x)h 1 (y) + f(1, 0)h 1 (x)h 0 (y)+f(1, 1)h 1 (x)h 1 (y) + f x (0, 0) h 0 (x)h 0 (y)+f x (0, 1) h 0 (x)h 1 (y) + f x (1, 0) h 1 (x)h 0 (y)+f x (1, 1) h 1 (x)h 1 (y) + f y (0, 0)h 0 (x) h 0 (y)+f y (0, 1)h 0 (x) h 1 (y) + f y (1, 0)h 1 (x) h 0 (y)+f y (1, 1)h 1 (x) h 1 (y) + f xy (0, 0) h 0 (x) h 0 (y)+f xy (0, 1) h 0 (x) h 1 (y) + f xy (1, 0) h 1 (x) h 0 (y)+f xy (1, 1) h 1 (x) h 1 (y) I Math 5620 Spring 2019 Notes of 4/5/19 page 10

14 Smoothness Between Elements Math 5620 Spring 2019 Notes of 4/5/19 page 11