Download - Approximation on Finite Elements Bruce A. Finlayson Rehnberg Professor of Chemical Engineering
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Approximation on Finite Elements
Bruce A. Finlayson
Rehnberg Professor of
Chemical Engineering
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The function x^2 exp(y-0.5)looks like this when plotted:
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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.
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Here is what we expect in a contour plot of the function:
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This is for N x N blocks, N=4
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N =8
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N = 16
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N = 32
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N = 64
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N = 128
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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.
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To Review: N = 4, 8, 16, and 32:
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Let functions in the block be bilinear functions of u and v.
• N1 = (1 - u) (1 - v)
• N2 = u (1 - v)
• N3 = u v
• N4 = (1 - u) v
• For example:
• N3(1,1) = 1; N3(0,1) = N3(1,0) = N3(0,0)=0
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N = 4, bilinear interpolation
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N = 8, bilinear interpolation
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N = 16, bilinear interpolation
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Compare constant interpolation on finite elements with bilinear interpolation on finite elements.
Constant interpolation with 32x32 = 1024 blocks.
Bilinear interpolation with 4x4 = 16 blocks.
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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.
∫ ( y − y2ex ) dx dy
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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.
∫ ( y − y2ex ) dx dy + α ∫ ⎝
⎛dydx −
dyex
dx ⎠⎞2
dx dy
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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.
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Calculus of VariationsThe 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.
∂2 z∂x2 +
∂2 z∂x2 = (2 + x2 ) e(y − 0.5)
12∫ ⎣
⎡⎢ ⎝⎛∂z∂x ⎠
⎞2
+ ⎝⎛∂z∂y ⎠
⎞2
⎦⎤⎥ dx dy + ∫ z [ (2 + x2) e(y − 0.5)] dx dy
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We choose finite element functions which satisfy the boundary conditions,
and then find the values of the parameters that make the integral a minimum.
z(0,y) = 0; z(x,0) = x2 e−0.5
z(1,y) = e(y−0.5); z(x,1) = x2 e0.5
z = ∑zi Ni (x,y)
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The solution with linear elements on 312 triangles (177 nodes) is:
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The solution with linear elements on 1248 triangles (665 nodes) is:
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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.)
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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.
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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.