# 2250-4 12:25pm Lecture Record Week 10 F2010

Last Modified: November 09, 2010, 21:20 MST.    Today: July 17, 2018, 17:24 MDT.

## 1,2 and 4 Nov: Laura and Michal

Exam 2, problems 4,5 at 12:50pm in 335 JWB on Monday, 1 Nov or Wednesday 3 Nov. Also possible is 7:25am in WEB 103 on Thursday, 4 Nov. Study the S2010 exam solution key for similar problems.

## 2 Nov: Tables, Rules of Laplace's Calculus 10.2, 10.3.

```  Partly solved 5.4-34.
The DE is 3.125 x'' + cx' + kx=0. The characteristic equation
is 3.125r^2 + cr + kr=0 which factors into 3.125(r-a-ib)(r-a+ib)=0
having complex roots a+ib, a-ib. Problems 32, 33 find the numbers
a, b from the given information. This is an inverse problem, one
in which experimental data is used to discover the differential
equation model. The book uses its own notation for the symbols
a,b: a ==> -p and b ==> omega1. Because the two roots a+ib, a-ib
determine the quadratic equation, then c and k are known in terms
of symbols a,b. See also the web site FAQ for more details.
Partly solved 5.4-20
The problem breaks into two distinct initial value problems:
(1)   2x'' + 16x' + 40x=0, x(0)=5, x'(0)=4
Characteristic equation  2(r^2+8r+20)=0. Roots r=-4+2i,r=-4-2i.
Solution Atoms=e^{-4t}cos 2t, e^{-4t}sin 2t. Underdamped.

(2)   2x'' + 0x' + 40x=0, x(0)=5, x'(0)=4
Characteristic equation 2(r^2+0+20)=0. Roots r=sqrt(20)i,r=-sqrt(20)i.
Solution Atoms=cos( sqrt(20)t), sin( sqrt(20)t).
Each system has general solution a linear combination of the solution atoms.
Evaluate the constants in the linear combination, in each of the two
cases, using the initial conditions x(0)=5, x'(0)=4. There are two linear algebra
problems to solve.
Answers: (1)  Coefficients 5, 2  for 2x'' + 16x' + 40x=0
Amplitude sqrt(5^2 + 12^2) = 13
(2)  Coefficients 5, 2/sqrt(5) for 2x'' + 0x' + 40x=0
Amplitude sqrt(5^2 + 4/5) = sqrt(129/5)
Write each solution in phase-amplitude form, a trig problem. See section
5.4 for specific instructions. The book's answers:
(1) tan(alpha) = 5/12   (2) tan(alpha) = 5 sqrt(5)/2
Laplace Theory
A brief Laplace table.
Forward table.
Backward table.
Extensions of the Table.
Laplace rules.
Linearity.
The s-differentiation theorem (d/ds)L(f(t))=L((-t)f(t)).
Shift theorem.
Parts theorem.
Finding Laplace integrals using Laplace calculus.
Solving differential equations by Laplace's method.
Basic Theorems of Laplace Theory
Lerch's theorem
Linearity.
The s-differentiation theorem (d/ds)L(f(t))=L((-t)f(t)).
Shift theorem L(exp(at)f(t)) = L(f(t))|s->(s-a)
Parts theorem L(y')=sL(y)-y(0)
```

## 2 Nov: Applications of Laplace's method from 10.3, 10.4, 10.5.

``` Solving y' = -1, y(0)=2
Solving y''+y=0, y(0)=0, y'(0)=1
Solving y''+y=1, y(0)=y'(0)=0
Solving y''+y=cos(t), y(0)=y'(0)=0
Computing Laplace integrals
Solving an equation L(y(t))=expression in s for y(t)
Dealing with complex roots and quadratic factors
Partial fraction methods
Using the s-differentiation theorem
Using the shifting theorem
Harmonic oscillator y''+a^2 y=0
```

## 4 Nov: Mechanical oscillators. Resonance. Beats. Sections 10.4, 10.5

```Basic Theorems of Laplace Theory
Periodic function theorem
Proof
Some engineering functions
unit step
ramp
sawtooth wave
other periodic waves next Monday
Convolution theorem application
L(cos t)L(sin t) = L(0.5 t sin(t))
Applications
How to solve differential equations
LRC Circuit
Second shifting rule
Specialized models.
Pure Resonance x''+x=cos(t)
Solution explosion, unbounded solution x=(1/2) t sin t.
Resonance examples: Soldiers marching in cadence, Tacoma narrows bridge,
Wine Glass Experiment. Theodore Von Karman and vortex shedding.
Cable model of the Tacoma bridge, year 2000. Resonance explanations.
Beats x''+x=cos(2t)
Graphics for beats [x=sin(10 t)sin(t/2)], slowly-oscillating envelope,
rapidly oscillating harmonic with time-varying amplitude.
```

## 4 Nov: Piecewise Functions. Shifting. Section 10.5 and EPbvp supplement 7.6.

``` Piecewise Functions
Unit Step
Pulse
Ramp
Second shifting Theorems
e^{-as}L(f(t))=L(f(t-a)step(t-a))
L(g(t)step(t-a))=e^{-as}L(g(t+a))
Piecewise defined periodic waves
Square wave
Triangular wave
Sawtooth
Rectified sine
Half-wave rectified sine
Parabolic wave
Periodic function theorem
Laplace of the square wave, tanh function.
Laplace Resolvent Method.
--> This method is considered a shortcut for systems by Laplace's method.
--> It is also a convenient way to solve systems with maple.
Consider problem 10.2-16
x'=x+z, y'=x+y, z'=-2x-z, x(0)=1, y(0)=0, z(0)=0
Write this as a matrix differential equation
u'=Bu, u(0)=u0
Then
u:=vector([x,y,z]);
B:=matrix([[1,0,1],[1,1,0],[-2,0,-1]]); u0:=vector([1,0,0]);
If we think of the matrix differential equation as a scalar equation, then
its Laplace model is
-u(0) + s L(u(t)) = BL(u(t))
or equivalently
sL(u(t)) - B L(u(t)) = u0
Write s = sI where I is the 3x3 identity matrix. Then the Laplace model is
(aI - B) L(u(t)) = u0
which is called the Resolvent Equation.
The RESOLVENT is the inverse of the matrix multiplier on the left:
Resolvent == inverse(sI - B)
It is so-named because the vector of Laplace answers is
vector([L(x),L(y),L(z)]) = L(u(t)) = inverse(sI - B) u0
Briefly,
Laplace of u(t) = RESOLVENT x u(0)
Intro to the Laplace resolvent method for 2x2 systems
Remarks on problem 10.2-16, a 3x3 system that can be solved with
the resolvent equation shortcut.
Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au.
Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au.
The general vector-matrix DE Model u'=Au
Laplace of u(t) = Resolvent x u(0)
Resolvent = 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).
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.
```