Week 13: 23 Nov, 24 Nov, 25 Nov,

# 2250 Lecture Record Week 13 F2009

Last Modified: November 28, 2009, 06:09 MST.    Today: July 17, 2018, 17:24 MDT.

## 23 Nov: Algebraic Eigenanalysis Sections 6.1,6.2

```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.
Slides and problem notes exist for 6.1 and 6.2 problems. See the web site.

Lawrence Page's pagerank algorithm, google web page rankings.
Relationship to the power method, stochastic matrices and eigenanalysis.
```

## 24 Nov: Algebraic Eigenanalysis. Systems of Differential Equations. Sections 6.2,7.1,7.2,7.3

```Algebraic Eigenanalysis Section 6.2.
Calculation of eigenpairs to produce Fourier's model.
Connection between Fourier's model and a diagonalizable matrix.
Further calculation examples.
How to deal with examples where A has an eigenvalue of multiplicity
greater than one.

Cayley-Hamilton topics.
Computing powers of matrices.
Stochastic matrices.
Example of 1984 telecom companies MCI, SPRINT, ATT with discrete
dynamical system u(n+1)=A u(n). Matrix A is stochastic.

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 + 3y,
y' = 4y,
x(0)=1, y(0)=2.
Find the eigenpairs (2, v1), (4,v2) where v1=vector(1,0])
and v2=vector([3,2]).
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 ]            [ 3 ]
u(t) = c1 e^{2t} [   ] + c2 e^4t} [   ]
[ 0 ]            [ 2 ]
which means
x(t) = c1 exp(2t) + 3 c2 exp(4t),
y(t) = 2 c2 exp(4t).
```

## 25 Nov: Systems of Differential Equations. Ziebur's Lemma. Sections 7.1,7.2,7.3

```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([[2,3],[0,4]]).
Dynamical system scalar form is
x' = 2x + 3y,
y' = 4y,
x(0)=1, y(0)=2.
Laplace resolvent method
The shortcut equations
Solving for L(x), L(y)
Backward table and Lerch's theorem
Chapter 1+5 Method
Solve w'+p(t)w=0 as w = constant / integrating factor.
Then  y(t) = 2 exp(4t)
Stuff y(t) into the first DE to get the linear DE
x' - 2x = 6 exp(4t)
Superposition: x(t)=x_h(t)+x_p(t),
x_h(t)=c exp(2t),
x_p(t) = d1 exp(4t) = 3 exp(4t) by undetermined coeff.
Then x(t)= -2 exp(2t) + 3 exp(4t).
Cayley-Hamilton 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 Ziebur's Lemma the 2x2 dynamical system above.
The characteristic equation is (2-lambda)(4-lambda)=0
with roots lambda = 2,4
Euler's theorem implies the atoms are exp(2t), exp(4t).
Ziebur's Lemma says that
u(t) = exp(2t) u_1 + exp(4t) u_2
where vectors u_1, U_2 are to be determined from A and
the initial conditions x(0)=1, y(0)=2.

ZIEBUR ALGORITHM.
To solve for u_1, u_2 in the example, differentiate the
equation and set t=0 in both relations. Then u'=Au
implies
u_0 =   u_1 +   u_2,
Au_0 = 2 u_1 + 4 u_2.
These equations can be solved by elimination.
u_1 = -(Au_0 - 4 u_0)/2, u_2 = (Au_0 - 2 u_0)/2
= vector([-2,0])          = vector([3,2])
These are recognized as eigenvectors of A for lambda=2
and lambda=4, respectively.

ZIEBUR SHORTCUT [a textbook method]
x(t) = k1 exp(2t) + k2 exp(4t).
Use the first DE to solve for y(t):
y(t) = (1/3)(x'(t) - 2x(t))
=  (1/3)(2 k1 exp(2t) + 4 k2 exp(4t) -
2 k1 exp(2t) - 2 k2 exp(4t))
=  (2/3) k2 exp(4t)
For example, x(0)=1, y(0)=2 implies k1 and k2 are
defined by
k1 + k2 = 1,
(2/3) k2 = 2,
which implies k1 = -2, k2 = 3, agreeing with a previous
solution formula.

Survey of Methods for solving a 2x2 dynamical system
1. Cayley-Hamilton method for u'=Au
Solution: u(t)=(atom_1)vec(c_1)+ ... + (atom_n)vec(c_n)
Atoms: They are constructed by Euler's theorem from roots of det(A-rI)=0
Vectors: Symbols vec(c_1), ..., vec(c_n) are not arbitrary. They are
determined from A and u(0). Algorithm outlined above for 2x2.
2. Laplace resolvent L(u)=(s I - A)^(-1) u(0)
3. Eigenanalysis  u(t) = exp(lambda_1 t) v1 + exp(lambda_2 t) v2
4. Putzer's method for the 2x2 matrix exponential.
Solution of u'=Au is: u(t) = exp(A t)u(0)
THEOREM: exp(A t) = r1(t) I + r2(t) (A-lambda_1 I),
Lambda Symbols: lambda_1 and lambda_2 are the roots of det(A-lambda I)=0.
The DE System:
r1'(t) = lambda_1 r1(t),         r1(0)=0,
r2'(t) = lambda_2 r2(t) + r1(t), r2(0)=0

Topics from linear systems:
Brine tank models.
Recirculating brine tanks.
Pond pollution.
Home heating.
Earthquakes.
Railway cars.
All are 2x2 or 3x3 or nxn system applications that can be solved by Laplace methods.
```
```Engineering models
The job-site cable hoist example [delayed]
Sliding plates example  [delayed]
Home heating example  [more coming]
```