Instructor: Irvin Roy Hentzel Office 432 Carver Phone

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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 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, Second Edition Otto Bretscher

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.

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!)

B1 B2 B3 B4 B5 ------------------------------------ B1 | 3 | | | | | B2 | 1 | 3 | | | | B3 | | 1 | 3 | | | B4 | | | 1 | 3 | | B5 | | | | 1 | 3 |

So after taking the transpose we have | 3 1 0 0 0 | | 0 3 1 0 0 | | 0 0 3 1 0 | | 0 0 0 3 1 | | 0 0 0 0 3 |

2. Give all the possible Jordan Canonical form which have p(x) = (x-1)4 (x-2)2 for their characteristic polynomial.

| 1 1 0 0 0 0 | | 1 1 0 0 0 0 | | 1 1 0 0 0 0 | | 0 1 1 0 0 0 | | 0 1 1 0 0 0 | | 0 1 0 0 0 0 | | 0 0 1 1 0 0 | | 0 0 1 0 0 0 | | 0 0 1 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 0 2 0 | | 0 0 0 0 2 0 | | 0 0 0 0 2 0 | | 0 0 0 0 0 2 | | 0 0 0 0 0 2 | | 0 0 0 0 0 2 | | 1 1 0 0 0 0 | | 1 0 0 0 0 0 | | 0 1 0 0 0 0 | | 0 1 0 0 0 0 | | 0 0 1 0 0 0 | | 0 0 1 0 0 0 | | 0 0 0 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 0 2 0 | | 0 0 0 0 2 0 | | 0 0 0 0 0 2 | | 0 0 0 0 0 2 |

| 1 1 0 0 0 0 | | 1 1 0 0 0 0 | | 1 1 0 0 0 0 | | 0 1 1 0 0 0 | | 0 1 1 0 0 0 | | 0 1 0 0 0 0 | | 0 0 1 1 0 0 | | 0 0 1 0 0 0 | | 0 0 1 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 0 2 1 | | 0 0 0 0 2 1 | | 0 0 0 0 2 1 | | 0 0 0 0 0 2 | | 0 0 0 0 0 2 | | 0 0 0 0 0 2 | | 1 1 0 0 0 0 | | 1 0 0 0 0 0 | | 0 1 0 0 0 0 | | 0 1 0 0 0 0 | | 0 0 1 0 0 0 | | 0 0 1 0 0 0 | | 0 0 0 1 0 0 | | 0 0 0 1 0 0 | | 0 0 0 0 2 1 | | 0 0 0 0 2 1 | | 0 0 0 0 0 2 | | 0 0 0 0 0 2 |

3. Find a matrix P such that P -1 A P is in Jordan Canonical form where | -2 -4 -6 | A = | -2 0 -3 | | 4 4 8 |

| -2-x -4 -6 | Det [ A - x I] = | -2 -x -3 | | 4 4 8-x | | 2-x 0 2-x | | 1 0 1 | Det [ A - x I] = (2-x) | -2 -x -3 |

| 1 0 0 | Det [ A - x I] = (2-x) | -2 -x -1 | | 4 4 4-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.

x = 2 | -4 -4 -6 | a | A - 2I = | -2 -2 -3 | b | | 4 4 6 | c | | 2 2 3 | -b | ~ | 0 0 0 | a-2b | | 0 0 0 | c+2b | p q | 1 1 3/2 | -b/2 | ~ | 0 0 0 | a-2b | | 0 0 0 | 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 |

| 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 | | 2 -1/2 -1 | P = | 1 0 1 | | -2 0 0 | | 4 -1 -2 | Use P = | 2 0 2 | | -4 0 0 |

P-1 A P = | 4 -1 -2 | -1 | -2 -4 -6 | | 4 -1 -2 | | 2 0 2 | | -2 0 -3 | | 2 0 2 | | -4 0 0 | | 4 4 8 | | -4 0 0 |

| 0 0 -1 | | 8 2 -4 | (1/4) | -4 -4 -6 | | 4 2 4 | | 0 2 1 | | -8 -4 0 | | 8 4 0 | | 2 1 0 | (1/4) | 0 8 0 | = | 0 2 0 | | 0 0 8 | | 0 0 2 |

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|

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 |

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 |

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]

| y1| | C1 ed1| | y2| | C2 ed2| | . | | . | | yn| | Cn edn| | C1 ed1 | | C2 ed2 | and X = P | . | | . | | Cn edn |

If A cannot be diagonalized, put A in Jordan canonical form P-1 A P | a 1 . |. . . . . . | | . a 1 |. . . . . . | | _______ a _|_______ . . . . | | . . . |a 1 | . . . . | | . . . |____a_|_______________| | . . . | b 1 . . | | . . . . . | . b 1 . | | . . . . . | . . b 1 | | . . . . . | . . . b | | . . . . . | . . . . |

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 | 1+x | | 1 |

The other two solutions are the same thing but the top entries being zero. | 0 | | 0 | e at | 1+x | and e at | 0 | | 1 | | 1 | We can write this in matrix form as |1+x+x2/2 0 0 | | C1 e at | | 1+x 1+x 0 | | C2 e at | | 1 1 1 | | C3 e at |

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.

FINAL EXAM Wednesday, May 7, 2003 2:15 to 4:15 1324 Howe Hall (RIGHT HERE IN THIS ROOM)