# 2250-1 7:30am Lecture Record Week 12 S2012

Last Modified: April 04, 2012, 05:01 MDT.    Today: July 17, 2018, 17:24 MDT.

## Apr 2 and 3: Sections 5.5, 5.6

```Exam 3 Review
Patrick will do exam 3 review this week on Thursday, JTB 140 7:30 to 9:25am.

Exam 3 part I: April 12
The problems are 1,2 on the sample exam below, which matches 3,4 on the S2010 exam key.
See the Fall 2010 exam 3 solution key for another sample exam, in which the problem
numbering is the same as your exam.

Exam 3 part II: April 19
The Thursday exam is at the usual 7:30am time in JTB 140.
The problems are 3,4,5 on the sample exam below, which matches 1,2,5 on the S2010 exam key.
```

## Mon, Apr 2: Variation of parameters. Sections 5.5, 10.4

``` More Laplace Examples
Continuing 10.3, 10.5 examples from last lecture.
Review of the Laplace resolvent method for 2x2 systems
Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au.
x' = 2x + y, y' = 3y ==> u'=Au
Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au.
u'=Au where A=<2,0|1,3> and u(0)=<1,2>
The general vector-matrix DE Model u'=Au
BASIC RESULT: Laplace of u(t) = Resolvent matrix times vector u(0)
DEFINITION: Resolvent matrix = inverse(sI - A)
Solve the systems by ch1 methods for x(t), y(t):
x' = 2x, x(0)=100,
y' = 3y, y(0)=50.
Answer: x = 100 exp(2t), y = 50 exp(3t)
x' = 2x+y, x(0)=1,
y' = 3y, y(0)=2.
Answer: y(t) = 2 exp(3t) and x(t) is the solution of the linear
integrating factor problem x'(t)=2x(t)+2 exp(3t).
Review of problem 10.2-16, a 3x3 system that can be solved with
the resolvent equation shortcut.
RULE: Use Cramer's rule or matrix inversion to solve the resolvent
equation for the vector of components L(x), L(y), L(z). Any
linear algebra problem Bu=c where B contains symbols should
be solved this way, unless B is triangular.

Variation of parameters
The second order formula.
Application to y''=1+x
How to calculate y_p(x) from the five parameters
y1(x)
y2(x)
W(x) = y1(x)y2'(x)-y1'(x)y2(x)
A(x) = coefficient in the DE of y''
f(x) = input or forcing term, the RHS of the DE
See (33) in section 5.5.
```
```    Undetermined Coefficients
Which equations can be solved
Intro to the basic trial solution method
Laplace solution of y'' + y = 1+x
[use x''(t)+x(t) = 1+t, x(0)=x'(0)=0]
How to find the atoms in y_p(x).
How to find the atoms in y_h(x)
THEOREM. Solution y_h(x) is a linear combination of atoms.
THEOREM. Solution y_p(x) is a linear combination of atoms.
THEOREM. (superposition)  y = y_h + y_p
EXAMPLE. How to find a shortest expression for y_p(x) using
Laplace's method.
Details for x''(t)+x(t) = 1+t, to
obtain the trial solution x(t)=A+Bt
BASIC METHOD. Given a trial solution with undetermined coefficients,
find a system of equations for d1, d2, ... and solve it.
Report y_p as the trial solution with substituted
METHODS to FIND the SOLUTION.
Laplace solution of y''(t) + y(t) = f(t) when
f(t)=linear combination of atoms
We get L(y) = polynomial fraction. By partial fractions and
the backward table, L(y)=L(g(t)), where g(t)=linear combination
of atoms! This proves the general result:

THEORY. y = y_h + y_p, and each is a linear combination of atoms.

How to find the homogeneous solution y_h(x) from the characteristic equation.
How to determine the form of the shortest trial solution for y_p(x)
METHOD 1. Laplace theory. It works, but it is slow. Central to this
method is dividing L(f(t)) by the characteristic polynomial,
then writing this polynomial fraction as L(g(t)). The homogeneous
solution atoms are shrunk from g(t) to obtain the trial solution.
METHOD 2. A rule for finding y_p(x) from f(x) and the DE.
We get rid of Laplace theory's t-variable and use x.
Finding a trial solution with fewest symbols.
Rule I. Assume the right side f(x) of the differential equation is a
linear combination of atoms. Make a list of all distinct atoms that appear in
the derivatives f(x), f'(x), f''(x), ... . Multiply these k atoms by
undetermined coefficients d_1, ... , d_k, then add to define a trial solution y.

This rule FAILS if one or more of the k atoms is a solution of
the homogeneous differential equation.

Rule II. If Rule I FAILS, then break the k atoms into groups
with the same base atom. Cycle through the groups, replacing atoms
as follows. If the first atom in the group is a solution of the homogeneous
differential equation, then multiply all atoms in the group by factor x. Repeat
until the first atom is not a solution of the homogeneous differential equation.
Multiply the constructed k atoms by symbols d_1, ... , d_k and add to define trial solution y.

Explanation: The relation between the Rule I + II trial solution and
the book's table that uses the mystery factor x^s.
EXAMPLES.
y'' = x
y'' + y = x exp(x)
y'' - y = x exp(x)
y'' + y = cos(x)
y''' + y'' = 3x + 4 exp(-x)

THEOREM. Suppose a list of k atoms is generated from the
atoms in f(x), using Rule I. Then the shortest trial
solution has exactly k atoms.

EXAMPLE. How to find a shortest trial solution using
Rules I and II.

Details for x''(t)+x(t) = t^2 + cos(t), obtaining
the shortest trial solution
x(t)=d1+d2 t+d3 t^2+d4 t cos(t) + d5 t sin(t).
How to use dsolve() in maple to check the answer.
EXAMPLE. Suppose the DE has order n=4 and the homogeneous
equation has solution atoms cos(t), t cos(t), sin(t),
t sin(t). Assume f(t) = t^2 + cos(t). What is the
shortest trial solution?
EXAMPLE. Suppose the DE has order n=2 and the homogeneous
equation has solution atoms cos(t), sin(t). Assume
f(t) = t^2 + t cos(t).
What is the shortest trial solution?
EXAMPLE. Suppose the DE has order n=4 and the homogeneous
equation has solution atoms 1, t, cos(t), sin(t).
Assume
f(t) = t^2 + t cos(t).
What is the shortest trial solution?

```

## Wed Apr 4: Resonance, Section 5.6

```Undetermined coefficients
Examples continued from the previous lecture.
EXAMPLES.
y'' = x
y'' + y = x exp(x)
y'' - y = x exp(x)
y'' + y = cos(x)
y''' + y'' = 3x + 4 exp(-x)

Shortest trial solution.
Two Rules to find the shortest trial solution.
1. Compute the atoms in f(x). The number k of atoms found
is the number needed in the shortest trial solution.
2. Correct groups with the same base atom, by
multiplication by x until the group contains no atom
which is a solution of the homogeneous problem
[eliminate homogeneous DE conflicts].
The x^s mystery factor in the book's table. The number s is the
multiplicity of the root in the homogenous DE characteristic
equation, which constructed the base atom of the group.
Reference:
Edwards-Penney, Differential Equations and Boundary Value
Problems, 4th edition, section 3.7 [math 2280 textbook].
Extra pages supplied by Pearson with bookstore copies of
the 2250 textbook. Also available as a xerox copy in case
your book came from elsewhere. Check-out the 2280 book in
the math center or the Math Library. All editions of the
book have identical 3.7 and 7.6 sections.
```
```Transform Terminology
Convolution theorem and x'' + 4x = cos(t), x(0)=x'(0)=0.
Input
Output
Transfer Function
Circuits EPbvp3.7:
Electrical resonance.
Derivation from mechanical problems 5.6.
THEOREM: omega = 1/sqrt(LC).
Impedance, reactance.
amplitude
Transfer function.
Input and output equation.

Wine Glass Experiment
The lab table setup
Speaker.
Amplifier with volume knob.
Wine glass.
x(t)=deflection from equilibrium of the radial component of the
glass rim, represented in polar coordinates, orthogonal to
the speaker front.
mx'' + cx' + kx = F_0 cos(omega t)  The model of the wine glass
m,c,k are properties of the glass sample itself
omega = frequency generator knob adjustment
Theory of Practical Resonance
The equation is
mx''+cx'+kx=F_0 cos(omega t)
THEOREM. The limit of x_h(t) is zero at t=infinity
THEOREM. x_p(t) = C(omega) cos(omega t - phi)
C(omega) = F_0/Z, Z^2 = A^2+B^2, A and B are the
undetermined coefficient answers for trial solution
x(t) = A cos(omega t) + B sin(omega t).
THEOREM. The output x(t) = x_h(t) + x_p(t) is graphically
just x_p(t) = C(omega) cos(omega t - phi) for large t.
Therefore, x_p(t) is the OBSERVABLE output.
THEOREM. The amplitude C(omega) is maximized over all possible
input frequencies omega>0 by the single choice
omega = sqrt(k/m - c^2/(2m^2)).
DEFINITION. The practical resonance frequency is the number omega
defined by the above square root expression.
```
``` 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)
Slides: Basic undetermined coefficients, draft 5 (153.6 K, pdf, 03 Mar 2012)
Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)

## Fri Apr 6: Sections 7.1, 7.2

```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.
Systems of two differential equations
Solving a system from Chapter 1 methods
The Laplace resolvent method for systems.
Cramer's Rule,
Matrix inversion methods.
EXAMPLE: Solving a 2x2 dynamical system using Laplace's resolvent method.
Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]).
EXAMPLE: Problem 10.2-16,
This problem is a 3x3 system for x(t), y(t), z(t) solved
by Laplace theory methods. The resolvent formula
(sI - A) L(u(t)) = u(0)
with u(t) the fixed 3-vector with components x(t), y(t), z(t),
amounts to a shortcut to obtain the equations for L(x(t)),
L(y(t)), L(z(t)). After the shortcut is applied, in which
Cramer's Rule is the method of choice, to find the formulas,
there is no further shortcut: we have to find x(t), for example,
by partial fractions and the backward table, followed by Lerch's
theorem.
```

## Mon Apr 9: Sections 6.1, 7.1

```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.
Chapter 1: Linear integrating factor method.
Laplace resolvent method
The shortcut equations
Solving for L(x), L(y)
Backward table and Lerch's theorem
Answers: x(t) = - e^{2t} + 2 e^{3t},
y(t) = 2 e^{3t}.
Chapter 1+5 Shortcut Method [appears in Edwards-Penney]
Solve w'+p(t)w=0 as w = constant / integrating factor.
Then  y(t) = 2 exp(3t)
Stuff y(t) into the first DE to get the linear DE
x' - 2x = 6 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).
```
Variation of Parameters and Undetermined Coefficients references
Slides: Basic undetermined coefficients, draft 5 (153.6 K, pdf, 03 Mar 2012)
Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)

Extra Credit Maple Project: Tacoma narrows. Explore an alternative explanation for what caused the bridge to fail, based on the hanging cables.