# 2250-1 7:30am Lecture Record Week 13 S2012

Last Modified: April 10, 2012, 13:04 MDT.    Today: August 16, 2018, 00:44 MDT.

### Week 13, April 9 to 13: Sections 6.1,6.2,7.2, 7.3

#### Thu, April 12: Exam 3, part I: problems 1,2

```Topics from chapter 10, EPbvp7.6. Problems 1,2 are on Laplace theory,
but there is some contact with chapter 5, sections 5.1 to 5.4. See the online
sample exam for details. Problems 3,4,5 will be on April 19.

Sample Exam 3 for S2012
PDF: sample exam 3, all five problems. (182.4 K, pdf, 14 Nov 2010)```

## Mon and Tue, Apr 9-10: Sections 6.1, 6.2, 7.1

```REVIEW from Last Week
Conversion Methods to Create a First Order System
The position-velocity substitution.
How to convert second order systems.
How to convert nth order scalar differential equations.
Non-homogeneous terms and the vector matrix system
u' = Au + F(t)
Non-linear systems and the vector-matrix system
u' = F(t,u)
Example: The system u'=Au, A=matrix([[2,1],[0,3]]);

Systems of two differential equations
The Laplace resolvent method for systems.
Solving the resolvent equation for L(x), L(y).
Cramer's Rule
Matrix inversion
Elimination
Example: Solving a 2x2 dynamical system
Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]).
Dynamical system scalar form is
x' = 2x + y,
y' = 3y,
x(0)=1, y(0)=2.
The equations for L(x), L(y)
(s-2)L(x)  +  (-1)L(y)=1,
(0)L(x)  + (s-3)L(y)=2
REMARK: Laplace resolvent method shortcut.
How to solve the [resolvent] equations for L(x), L(y).
Cramer's Rule
Matrix inversion
Elimination
delta=(s-2)(s-3), delta1=s-1, delta2=2(s-2)
L(x) = -1/(s-2)+2/(s-3), L(y)=2/(s-3)
Backward table and Lerch's theorem
Answers: x(t) = - e^{2t} + 2 e^{3t},
y(t) = 2 e^{3t}.
Edwards-Penney Shortcut Method in Example 5, 7.1. Uses Chapter 1+5 methods.
This is the Cayley-Hamilton-Ziebur method. See below.
Solve w'+p(t)w=0 as w = constant / integrating factor.
Then  y' -2y=0 ==> y(t) = 2 exp(3t)
Stuff y(t) into the first DE to get the linear DE
x' - 2x = 2 exp(3t)
Superposition: x(t)=x_h(t)+x_p(t),
x_h(t)=c exp(2t),
x_p(t) = d1 exp(t) = 2 exp(3t) by undetermined coeff.
Then x(t)= - exp(2t) + 2 exp(3t).

Cayley-Hamilton Theorem
A matrix satisfies its own characteristic equation.
ILLUSTRATION: det(A-r I)=0 for the previous example
is (2-r)(3-r)=0 or r^2 -5r + 6=0. Then C-H says
A^2 - 5A + 6I = 0.
Cayley-Hamilton-Ziebur Method
ZIEBUR'S LEMMA.
The components of u in u'=Au are linear combinations of
the atoms created by Euler's theorem applied to the
roots of the characteristic equation det(A-rI)=0.
THEOREM. Solve u'=Au without complex numbers or eigenanalysis.
The solution of u'=Au is a linear combination of atoms
times certain constant vectors [not arbitrary vectors].
u(t)=(atom_1)vec(c_1)+ ... + (atom_n)vec(c_n)

PROBLEM: Solve by Cayley-Hamilton-Ziebur the 2x2 dynamical system
x' = 2x + y,
y' = 3y,
x(0)=1, y(0)=2.
The characteristic equation is (2-lambda)(3-lambda)=0
with roots lambda = 2,3
Euler's theorem implies the atoms are exp(2t), exp(3t).
Ziebur's Theorem says that
u(t) = exp(2t) vec(u_1) + exp(3t) vec(u_2)
where vectors u_1, u_2 are to be determined from the matrix
A = matrix([[2,1],[0,3]]) and initial conditions x(0)=1, y(0)=2.

ZIEBUR ALGORITHM.
To solve for u_1, u_2 in the example, differentiate the
equation u(t) = exp(2t) u_1 + exp(3t) u_2 and set t=0
in both relations. Then u'=Au implies
u_0 =    u_1  +   u_2,
Au_0 = 2 u_1 + 3 u_2.
These equations can be solved by elimination.
u_1 = (3 u_0 -Au_0), u_2 = (Au_0 - 2 u_0)
= vector([-1,0])     = vector([2,2])
Vectors u_1, u_2 are recognized as eigenvectors of A for
lambda=2 and lambda=3, respectively, after studying chapter 6.

ZIEBUR SHORTCUT [Edwards-Penney textbook method, Example 5 in 7.1]
x(t) = k1 exp(2t) + k2 exp(3t).
Use the first DE to solve for y(t):
y(t) = x'(t) - 2x(t)
=  2 k1 exp(2t) + 3 k2 exp(3t)
- 2 k1 exp(2t) - 2 k2 exp(3t))
=   k2 exp(3t)
For example, x(0)=1, y(0)=2 implies k1 and k2 are
defined by
k1 + k2 = 1,
k2 = 2,
which implies k1 = -1, k2 = 2, agreeing with a previous
solution formula.
```
```EIGENANALYSIS WARNING
Reading Edwards-Penney Chapter 6 may deliver the wrong ideas
about how to solve for eigenpairs. The examples emphasize a
clever shortcut, which does not apply in general to solve for
eigenpairs.

HISTORY. Chapter 6 originally appeared in the 2280 book
as a summary, which assumed a linear algebra course. The
chapter was copied without changes into the Edwards-Penney
Differential Equations and Linear Algebra textbook, which you
currently own. The text contains only shortcuts. There is
no discussion of a general method for finding eigenpairs.
You will have to fill in the details by yourself. The online
lecture notes and slides were created to fill in the gap.

Lecture: Fourier's Model. Intro to Eigenanalysis, Ch6.
Examples and motivation.
Ellipse, rotations, eigenpairs.
General solution of a differential equation u'=Au and eigenpairs.
Fourier's model.
History.
J.B.Fourier's 1822 treatise on the theory of heat.
The rod example.
Physical Rod: a welding rod of unit length, insulated on the
lateral surface and ice packed on the ends.
Define f(x)=thermometer reading at loc=x along the rod at t=0.
Define u(x,t)=thermometer reading at loc=x and time=t>0.
Problem: Find u(x,t).
Fourier's solution assume that
f(x) = 17 sin (pi x) + 29 sin(5 pi x)
= 17 v1 + 29 v2
Packages v1, v2 are vectors in a vector space V of functions on [0,1].
Fourier computes u(x,t) by re-scaling v1, v2 with numbers Lambda_1,
Lambda_2 that depend on t. This idea is called Fourier's Model.

u(x,t) = 17 ( exp(-pi^2 t) sin(pi x)) + 29 ( exp(-25 pi^2 t) sin (5 pi x))
= 17 (Lambda_1 v1) + 29 (Lambda_2 v2)

Eigenanalysis of u'=Au is the identical idea.
u(0) = c1 v1 + c2 v2  implies
u(t) = c1 exp(lambda_1 t) v1 + c2 exp(lambda_2 t) v2
Fourier's re-scaling idea from 1822, applied to u'=Au,
replaces v1 and v2 in the expression
c1 v1 + c2 v2
by their re-scaled versions to obtain the answer
c1 (Lambda1 v1) + c2 (Lambda2 v2)
where
Lambda1 = exp(lambda_1 t), Lambda2 = exp(lambda_2 t).
```
```Main Theorem on Fourier's Model

THEOREM. Fourier's model
A(c1 v1 + c2 v2) = c1 (lambda1 v1) + c2 (lambda2 v2)
with v1, v2 a basis of R^2 holds [for all constants c1, c2]
if and only if
the vector-matrix system
A(v1) = lambda1 v1,
A(v2) = lambda2 v2,
has a solution with vectors v1, v2 independent
if and only if
the diagonal matrix D=diag(lambda1,lambda2) and
the augmented matrix P=aug(v1,v2) satisfy
1. det(P) not zero [then v1, v2 are independent]
2. AP=PD

THEOREM. The eigenvalues of A are found from the determinant
equation
det(A -lambda I)=0,
which is called the characteristic equation.
THEOREM. The eigenvectors of A are found from the frame
sequence which starts with B=A-lambda I [lambda a root of
the characteristic equation], ending with last frame rref(B).

The eigenvectors for lambda are the partial derivatives of
the general solution obtained by the Last Frame Algorithm,
with respect to the invented symbols t1, t2, t3, ...
```
```Algebraic Eigenanalysis Section 6.2.
Calculation of eigenpairs to produce Fourier's model.
Connection between Fourier's model and a diagonalizable matrix.
How to find the variables lambda and v in Fourier's model using
determinants and frame sequences.
Solved in class: examples similar to the problems in 6.1 and 6.2.
Web slides and problem notes exist for the 6.1 and 6.2 problems.
Examples where A has an eigenvalue of multiplicity greater than one.
```

## Wed, Apr 11: Eigenanalysis. First Order Systems. Sections 6.1, 6.2, 7.1, 7.3

``` Projection: glass-breaking video. Wine glass experiment. Tacoma narrows.
Video: Wine glass breakage (QuickTime MOV) (96.8 K, mov, 31 Mar 2008)       Video: Wine glass experiment (12mb mpg, 2min) (12493.8 K, mpg, 01 Apr 2008)       Video: Tacoma Narrows Bridge Nov 7, 1940 (18mb mpg, 4min) (18185.8 K, mpg, 01 Apr 2008)       Video: Resonance #17, Wine Glass and Tacoma Narrows (29min Annenburg CPB)
```
```Eigenanalysis Examples
Solving DE System u' = Au by Eigenanalysis
Example: Solving a 2x2 dynamical system
Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,3],[0,4]]).
Dynamical system scalar form is
x' = 2x + 1y,
y' = 3y,
x(0)=1, y(0)=2.
Find the eigenpairs (2, v1), (3,v2) where v1=vector([1,0])
and v2=vector([1,1]).
THEOREM. The solution of u' = Au in the 2x2 case is
u(t) = c1 exp(lambda1 t) v1 + c2 exp(lambda2 t) v2
APPLICATION:
u(t) = c1 exp(2t) v1 + c2 exp(4t) v2
[ 1 ]            [ 1 ]
u(t) = c1 e^{2t} [   ] + c2 e^4t} [   ]
[ 0 ]            [ 1 ]
which means
x(t) = c1 exp(2t) + 3 c2 exp(4t),
y(t) = 2 c2 exp(4t).
Diagonalization Theory
In the case of a 2x2 matrix A,
FOURIER'S MODEL is
A(c1 v1 + c2 v2) = c1(lambda1 v1) + c2(lambda2 v2)
where v1,v2 are a basis for the plane
equivalent to DIAGONALIZATION
AP=PD, where D=diag(lamba1,lambda2), P=augment(v1,v2),
where det(P) is not zero
equivalent to EIGENPAIR EQUATIONS
A(v1)=lambda1 v1, A(v2)=lambda2 v2,
where vectors v1,v2 are independent
Examples
Given the eigenpairs of A, find A via AP=PD.
Given P, D, then find A.
Given A, then find P, D.
Cayley-Hamilton topics, Section 6.3.
Computing powers of matrices.
Stochastic matrices.
Example of 1984 telecom companies ATT, MCI, SPRINT with discrete
dynamical system u(n+1)=A u(n). Matrix A is stochastic.
EXAMPLE:
[ 6  1  5 ]               [ a(t) ]
10 A = [ 2  7  1 ]        u(t) = [ m(t) ]
[ 2  2  4 ]               [ s(t) ]

Meaning: 60% stay with ATT and 20% switch to MCI, 20% switch to SPRINT.
70% stay with MCI and 20% switch to SPRINT, 10% switch to ATT.
40% stay with SPRINT and 50% switch to ATT, 10% switch to MCI.
Powers of A and the meaning of A^n x_0 for the telecom example.