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Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141
Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone Text: Linear Algebra With Applications, Second Edition Otto Bretscher
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Friday, May 2 Differential Equations
No Hand-In-Homework Main Idea: The Jordan canonical form is behind the solution of differential equations. Key Words: Taylor Polynomial. Goal: Learn to understand where those factors of xn come from that you have to stick into the solution of differential equations.
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Previous Assignment 1. Write the matrix of differentiation with respect to the basis. B1 = e3x, B2 = e 3x (1+x), B3 = e 3x (1+x+x2/2), B4 = e 3x(1+x+x2/2+x3/3!), B5 = e 3x (1+x+x2/2+x3/3!+x4/4!)
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B1 B B3 B B5 B1 | 3 | | | | | B2 | 1 | 3 | | | | B3 | | 1 | 3 | | | B4 | | | 1 | 3 | | B5 | | | | 1 | 3 |
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So after taking the transpose we have
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2. Give all the possible Jordan Canonical form which have
p(x) = (x-1)4 (x-2)2 for their characteristic polynomial.
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3. Find a matrix P such that P -1 A P is in Jordan Canonical form where
| | A = | | | |
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| -2-x | Det [ A - x I] = | x | | x | | 2-x x | | | Det [ A - x I] = (2-x) | x |
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| | Det [ A - x I] = (2-x) | x | | x | Det [ A - x I] = (2-x) ( -4 x + x2 + 4 ) Det [ A - x I] = (2-x) ( x - 2 ) 2 The eigen values are 2,2,2.
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x = 2 | | a | A - 2I = | | b | | | c | | | -b | ~ | | a-2b | | | c+2b | p q | /2 | -b/2 | ~ | | a-2b | | | c+2b | | x | | -1 | |-3/2 | | y | = p | 1 | + q | 0 | | z | | 0 | | 1 | | x | | -1 | | -3 | | y | = p | 1 | + q | 0 | | z | | 0 | | 2 |
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| x | | -p - 3 q | | y | = | p | | z | | 2q | To extend we have to have -p-3q -2p = 0 and 2q+2p = 0 So p+q = 0 | 2 | | -1/2 | | -1 | use V1 = | 1 | V2 = | 0 | V3 = | 1 | |-2 | | 0 | | 0 | | / | P = | | | | | | Use P = | | | |
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P-1 A P = | | -1 | | | | | | | | | | | | | | | |
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| | | | (1/4) | | | | | | | | | | | | (1/4) | | = | | | | | |
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New Material: Given a system of differential equations x' = 4 x - 2 y y' = 3 x - 1 y |x|' = | 4 -2 | |x| |y| | 3 -1 | |y|
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Let |x| = | 2 1 || u| |y| | 3 1 || v| Substitute this in to the above equation giving | 2 1 | |u|' = | 4 -2 | | 2 1 | | u | | 3 1 | |v| | 3 -1 | | 3 1 | | v | |u|' = | 2 1 | -1 | 4 -2 | | 2 1 | | u | |v| | 3 1 | | 3 -1 | | 3 1 | | v | |u|' = -| 1-1 | | 2 2 | | u | |v| |-3 2 | | 3 2 | | v | |u|' = -|-1 0 | | u | |v| | 0 -2 | | v | |u|' = | 1 0 | | u | |v| | 0 2 | | v |
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This is the system u' = u and v' = 2 v.
u = C1 et v = C2 e2t | x | = | 2 1 | | C1 e t | | y | | 3 1 | | C2 e2t |
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Theory: To solve X' = A X. Substitute X = P Y (PY)' = A(PY) PY' = A P Y y' = P -1 A P Y Choose P such that P -1 A P = diagonal [ d1, d2,... dn]
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| y1| | C1 ed1| | y2| | C2 ed2| | . | | | | yn| | Cn edn| | C1 ed1 | | C2 ed2 | and X = P | | | | | Cn edn |
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If A cannot be diagonalized, put A in Jordan canonical form
P-1 A P | a | | | . a 1 | | | _______ a _|_______ | | |a 1 | | | |____a_|_______________| | | b | | | . b | | | b 1 | | | b | | | |
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The solution is constructed from the solution of a single Jordan block
The solution is constructed from the solution of a single Jordan block. For example. | a 1 0 | | 0 a 1 | | 0 0 a | | 1+x+x2/2| This has solution e at | x | | |
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The other two solutions are the same thing but the top entries being zero.
| | | 0 | e at | 1+x | and e at | 0 | | | | 1 | We can write this in matrix form as |1+x+x2/ | | C1 e at | | 1+x x 0 | | C2 e at | | | | C3 e at |
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The solution for Y is the sum of the solution for the partial blocks.
|P2 o o o o o o o o ||C1eat | |P1 P1 o o o o o o o || C2eat | |P0 P0 P0 o o o o o o || C3eat | | o o o P1 o o o o o || C4eat | | o o o P0 P0 o o o o || C5eat | | o o o o o P3 o o o || C6ebt | | o o o o o P2 P2 o o || C7ebt | | o o o o o P1 P1 P1 o || C8ebt | | o o o o o P0 P0 P0 P0 || C9ebt| And finally X = PY.
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FINAL EXAM Wednesday, May 7, 2003 2:15 to 4:15 1324 Howe Hall
(RIGHT HERE IN THIS ROOM)
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