Lecture 4 FEM in 1D: FEM and solution of algebraic systems ... · Lecture 4 FEM in 1D: FEM and solution of algebraic systems (Lecture notes taken by Peter Gerakios and Steven Burgess)

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Lecture 4

FEM in 1D: FEM and solution of algebraic systems

(Lecture notes taken by Peter Gerakios and Steven Burgess)

• Bilinear and linear description of FEM: a(u,φ) = l(φ).

General description of weak equation is as follows:

, ,

, , , , 0 .

An energy equation J(u) = ½ a(u,u) – l(u) can be shown to be admissible.

Example 1. In the string equation:

,

2

12 ,

Example 2. Another example is the fin problem:

0

,

It can be shown that the energy equation J(u) = ½ a(u,u) – l(u) is also admissible.

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Example 3. Another example can be found in two dimensions:

u u f |

We can use Green’s theorem to get: Let | 0.

,

Note: J(u) is admissible if

12 ,

where both a(.,.) and l(.) are real-valued functions. Use the linear and symmetric properties

12 , ,

12 ,

0 , , 0 0 0 ,

, 0

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• Element matrices from a(u,φ) = l(φ).

When we restrict u to some specific element, .

The functions (Nie) are the linear shape functions

,

The summation can be pulled out due to linearity

,

This can be written as

, , , ,

And this gives the element matrix k:

, ,, , and

, ,

OR, alternatively using the variational approach:

where the k coefficients are identical to the a(Ni,Nj)’s.

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W

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• General Outline of FEM.

Goal is to begin with a differential equation and rewrite in the weak or energy form. There are six major components/steps to this procedure.

I. Input Data a. i = nod(e , ie), where i is the system node #, ie is the element node #, and e is the

element matrix # b. the components of the points in the domain c. other parameters in the model

II. Find element matrices a. , , b. The dimension of matrix k depends on the dimensions of the model as well as

the shape functions.

III. Assemble matrix Ann x nn and right hand side dnn x 1 where nn is the total number of nodes a. This can be done via the assembly by elements or nodes

IV. Adjust for given boundary conditions a. ,

# .

V. “Solve” Au = d

VI. Analyze Output -- Graphically

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• Gauss elimination.

Solve

First step is to find U and then use backward substitution to solve

However, the operation count is of the order

Example:

The first step would be to obtain a sequence of row operations to obtain U.

∏ ∏

Consider the following augmented system:

3 1 01 3 1

0 1 3|123

3 1 0

083 1

0 1 3

|1733

3 1 0

083 1

0 0218

|

173

318

Now utilize backward substitution to solve for the three unknowns.

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• Tridiagonal systems.

0

0

0 00

0

000 0

And this gives the equation L(Ux) = d. Let Ux = y and solve Ly = d by forward substitution. Now solve Ux = y by backward substitution. This is possible if the alpha’s and gamma’s are known.

Tridiagonal algorithm:

, ,

2,

1, 1

Note: For a point-wise tridiagonal system, the operation count is of order 5n.

• SOR iterative method.

Let Au = d be written in component form

.

The SOR method should be used for sparse matrices (for each i most aij are zero) and for diagonally dominate matrices (for each i aii ≥∑ ). Even in these cases the method may not converge to the solution!

Let uim+1/2 be the solution of

∑ / ∑ . (1)

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Next, let ω be greater than or equal to 1.0 and less than 2.0. Good choices of this parameter can accelerate convergence (it takes less steps to reach a convergence test).

1 / . (2)

SOR Algorithm.

Choose an initial vector u0

For m = 1, maxm

For i = 1, n

Solve (1) for uim+1/2

Solve (2) for uim+1

End

Test for convergence

End.

Example. Consider the algebraic system n = 3 unknowns

3 1 01 3 1

0 1 3

123

.

Use the initial choice u0 = [ 1 1 1]T and do one iteration.

3u11 = 1 + 1 u2

0 = 1 + 1, which implies u11 = 2/3

3u21 = 2 + 1 u1

1 + 1 u30 = 2 + 2/3 +1, which implies u2

1=11/9

3u31 = 3 + 1 u2

1 = 3 + 11/9, which implies u31=38/27.

Next repeat this until difference in successive iterations is “suitably” small.