2. Non-Linear Equations
a. Matrix Algorithms; Matrix Structures
Simultaneous non-linear equations are solved using the abbreviated Taylor Series. This requires an iterative solution, beginning with approximations for the unknowns, and refining them until some correction criteria is met.
Equations D-1 and D-2 are used to set up and solve non-linear equations.
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[C] x [U] = [K] |
Equation D-1 |
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[U] = [CTC]-1 x [CTK] |
Equation D-2 |
Matrix elements are modified from their linear counterparts to incorporate the Taylor Series:
- Each [C] matrix element is a partial derivative of a function with respect to each variable evaluated at an approximation.
- Each [K] matrix element is the difference between the function's constant and that computed from approximations.
- The [U] matrix has corrections which are used to update the approximations.
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For these three functions in three unknowns: |
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The matrices are: |
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After [U] is determined, its elements are used to update the approximations. The [C] and [K] matrices are recomputed and the process repeated until the corrections reach an acceptable level.
b. Example
Determine x and y values which satisfy these equations.
Start with initial approximations of x = 4 and y = 3. Continue until the corrections do not exceed 0.1.
This are the same equations used in the Chapter C's Taylor Series two variable example. Let’s see if we get the same answers.
(1) Matrix contents
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(2) Partial derivatives with respect to each unknown
(3) First Iteration
Build [C] from partial derivative coefficients at the initial approximations
Compute expression constants at the initial approximations
Compute [K] elements
Transpose [C], then multiply [CT] by C.
Determine the inverse of [CTC].
A 2 x 2 matrix can be inverted using the Determinant Method.
Multiply [CT] and [K].
Determine [U] by multiplying [CTC]-1 and [CTK].
The corrections exceed 0.1 so continue with a second iteration.
(4) Second Iteration
Update initial approximations
x = 4+1.167 = 5.167
y = 3-0.833 = 2.167
Recompute [C] and [K] matrices
Compute [U] matrix
The corrections still exceed 0.1 so continue with a third iteration.
(5) Third Iteration
Updated approximations
x = 5.167-0.164 = 5.003
y = 2.167-0.161 = 2.006
Recompute [C] and [K] matrices
Compute [U] matrix
The corrections are less than 0.1 so the iterations can terminate.
(5) Final Values
x = 5.167-0.003 = 5.000
y = 2.006-0.006 = 2.000
These are the same as the example in Chapter C.
Ta-daa!