Approximation on Finite Elements Bruce A Finlayson Rehnberg

Approximation on Finite Elements Bruce A. Finlayson Rehnberg Professor of Chemical Engineering

The function x^2 exp(y-0. 5) looks like this when plotted:

Approximation on finite elements • Break the region into small blocks, and color each block according to an average value in the block. • The approximation depends on the number of blocks.

Here is what we expect in a contour plot of the function:

This is for N x N blocks, N=4

N =8

N = 16

N = 32

N = 64

N = 128

This is mesh refinement. • Notice how the picture got better and better the more squares we took. • We approximated the function on each block - a finite element approximation. • We get a better approximation when we use small finite elements. • As the number of blocks increases, the picture approaches that of a continuous function.

To Review: N = 4, 8, 16, and 32:

Let functions in the block be bilinear functions of u and v. • • • N 1 = (1 - u) (1 - v) N 2 = u (1 - v) N 3 = u v N 4 = (1 - u) v For example: N 3(1, 1) = 1; N 3(0, 1) = N 3(1, 0) = N 3(0, 0)=0

N = 4, bilinear interpolation

N = 8, bilinear interpolation

N = 16, bilinear interpolation

Compare constant interpolation on finite elements with bilinear interpolation on finite elements. Constant interpolation with 32 x 32 = 1024 blocks. Bilinear interpolation with 4 x 4 = 16 blocks.

Instead of matching the function at the block-corners, find the best interpolant minimizing the mean square difference between the approximation and the exact function. Still use finite elements, but bilinear approximations.

What do you do if you don’t know the function? Suppose you want to minimize the difference between the approximation and exact function and their derivatives.

One can still find the best finite element approximation that minimizes this integral. It won’t fit the function exactly anywhere, nor the first derivative, but it will minimize the integral.

Calculus of Variations The function that satisfies this differential equation: minimizes this integral (this must be proved for each equation): The same approach can be taken: to satisfy the differential equation, one approximates the integral on the finite element blocks and finds the minimum.

We choose finite element functions which satisfy the boundary conditions, and then find the values of the parameters that make the integral a minimum.

The solution with linear elements on 312 triangles (177 nodes) is:

The solution with linear elements on 1248 triangles (665 nodes) is:

Finite Element Variational Method • Divide the domain into small regions. • Write a low degree polynomial on each small region: constant, bilinear, biquadratic. These are the basis functions. • Write the solution as a series of basis functions. • Determine the coefficients by minimizing an integral. (The trick is to know what integral to use. )

Galerkin Finite Element Method • If a variational principle exists, the Galerkin method is the same as the variational method. • The same finite elements can be used. • Now the residual is made orthogonal to each basis function; this applies when there is no integral to be minimized or made stationary.

Conclusion - Three Basic Ideas • Write the solution in a series of functions, each of which is defined over small elements, using low-order polynomials. • Minimize some integral to solve a differential equation (or use Galerkin or MWR). • Increase the number of basis functions in order to show convergence.