## 2280 12:55pm Lectures Week 9 S2016

Last Modified: March 12, 2016, 11:40 MST.    Today: October 22, 2018, 12:07 MDT.
``` Edwards-Penney, sections 7.1 to 7.6 and 4.1 to 4.3
The textbook topics, definitions and theoremsEdwards-Penney 7.1 to 7.5 (21.0 K, txt, 22 Feb 2015)Edwards-Penney 4.1 to 4.3 (10.7 K, txt, 05 Jan 2015)```

#### Monday: Topics from last week

```Maple Example to find roots of the characteristic equation
Consider the recirculating brine tank example:
20 x' = -6x + y,
20 y' = 6x - 3y
The maple code to solve the char eq:
A:=(1/20)*Matrix([[-6,1],[6,-3]]);
linalg[charpoly](A,r);
solve(%,r);
```
```Applications of Laplace's method from 7.3, 7.4, 7.5
Convolution theorem
DEF. Convolution of f and g = f*g(t) = integral of f(x)g(t-x) from x=0 to x=t
THEOREM. L(f(t))L(g(t))=L(convolution of f and g)
Application:   L(cos t)L(sin t) = L(0.5 t sin(t))
Second shifting Theorems
e^{-as}L(f(t))=L(f(t-a)u(t-a)) Backward table
L(g(t)u(t-a))=e^{-as}L(g(t+a)) Forward table
```

``` EXAMPLES: Second Shift Theorem.
Forward table
L(sin(t)u(t-Pi)) = e^{-Pi s} L(sin(t)|t->t+Pi)
= e^{-Pi s} L(sin(t+Pi))
= e^{-Pi s} L(sin(t)cos(Pi)+sin(Pi)cos(t))
= e^{-Pi s} L(-sin(t))
= e^{-Pi s} ( -1/(s^2+1))
Backward table
L(f(t)) = e^{-2s}/s^2
= e^{-2s} L(t)
= L(t u(t)|t->t-2)
= L((t-2)u(t-2))
Therefore f(t) = (t-2)u(t-2) = ramp at t=2.

Laplace Resolvent Method.Slides: Laplace resolvent method (81.0 K, pdf, 14 Mar 2016)
The general vector-matrix DE Model u'=Au
Laplace of u(t) = Resolvent times u(0)
Resolvent = inverse(sI - A)
Chapter 1 methods for solving 2x2 systems
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).
Laplace Resolvent Method
Consider problem 7.2-16
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
(sI - B) L(u(t)) = u0
which is called the Resolvent Equation.

For the example, the resolvent equation is

Matrix([[s-1,0,-1],[-1,s-1,0],[2,0,s+1]]).Vector([L(x),L(y),L(z)]) = Vector([1,0,0]);

DEF. 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
= L(u(t)) = inverse(sI - B) times vector u0
Briefly,
Laplace of VECTOR u(t) = RESOLVENT MATRIX times VECTOR u(0)
ADVICE: 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.

```

#### Monday: Resolvent. Dirac Impulse. Delta and hammer hits.

``` Hammer hits and the Delta function
Definition of delta(t)
delta(t) = idealized injection of energy into a system at
time t=0 of impulse=1.
A hammer hit model in mechanics:
Camshaft impulse in a car engine
How delta functions appear in circuit calculations
Start with Q''+Q=E(t) where E is a switch. Then differentiate to get
I''+I=E'(t). Term E'(t) is a Dirac impulse.
Paul Dirac (1905-1985) and impulses
Laurent Schwartz (1915-2002) and distribution theory
Riemann Stieltjes integration theory: making sense of the Dirac delta.
Def: RS-integral equals the limit of RS-sums as N-->infinity and mesh-->zero.
RS-sum = sum of terms  f(x_i)(alpha(x_i)-alpha(x_{i-1})) where alpha(x) is
the monotonic RS integrator.
Why int( f(t) delta(t-a), t=-infinity .. infinity) = f(a)
The symbol delta(t-a) makes sense only under an integral sign.

Engineering models
Short duration impulses: Injection of energy into a mechanical or electrical model.
Definition: The impulse of force f(t) on interval [a,b] equals the
integral of f(t) over [a,b]
An example for f(t) with impulse 5 is defined by
f(t) = (5/(2h))pulse(t,-h,h)
EXAMPLE. The Laplace integral of f(t) and its limit as h --> 0.
EXAMPLE. The delta function model x''(t) + 4x(t) = 5 delta(t-t0),
x(0)=0, x'(0)=0. The model is a mass on a spring with no
damping. It is at rest until time t=t0, when a short duration
impulse of 5 is applied. This starts the mass oscillation.
EXAMPLE. The delta impulse model from EPbvp 7.6,
x''(t) + 4x(t) = 8 delta(t-2 pi), x(0)=3, x'(0)=0.
The model is a mass on a spring with no damping. The mass is moved
to position x=3 and released (no velocity). The mass oscillates until
time t=2Pi, when a short duration impulse of 8 is applied. This
alters the mass oscillation, producing a piecewise output x(t).
# How to solve it with dsolve in maple.
de:=diff(x(t),t,t)+4*x(t)=f(t);f:=t->8*Dirac(t-2*Pi);
ic:=x(0)=3,D(x)(0)=0; dsolve({de,ic},x(t));
convert(%,piecewise,t);
Details of the Laplace calculus in maple: inttrans package.
with(inttrans): f:=x->cos(omega*t)+8*Dirac(t-2*Pi);
G:=laplace(f(t),t,s); invlaplace(G,s,t);
de:=diff(x(t),t,t)+4*x(t)=f(t);
laplace(de,t,s);
subs(ic,%);
solve(%,laplace(x(t),t,s));
CALCULATION. Phase amplitude conversion [see EP 5.4]
x(t) = 3 cos(2t) until hammer hit at t=2Pi. It has amplitude 3.
x(t) = 3 cos(2t)+4 sin(2t) after the hit. It has amplitude 5.
= 5 cos(2t - arctan(4/3))

An RLC circuit model
Q'' + 110 Q' + 1000 Q = E(t)
Differentiate to get [see EPbvp 3.7]
I'' + 100 I' + 1000 I = E'(t)
When E(t) is a switch, then E'(t) is a Dirac delta.
```

#### Tuesday-Wednesday: Problem session. Sections 7.1 to 7.5.

``` Forward and Backward Table Applications
Review of previously solved problems.
Problem 7.1-18. Trig identity 2 sin (theta) cos(theta) = sin(2 theta)
used for L(sin(3t)cos(3t)).
Problem 7.1-28. Splitting a fraction into backward table entries.
Partial Fractions and Backward Table Applications
Problem 7.2-24. L(f)=1/(s(s+1)(s+2)) solved by the three methods for
partial fractions: sampling, atom method, Heaviside cover-up.
Problem 7.2-9. Solve x''+3x'+2x=t, x(0)=0, x'(0)=2. Get the resolvent
equation
(s^2+3s+2)L(x)=2+L(t)
L(x)=(1+2s^2)/(s^2(s+2)(s+1))
L(x)=A/s + B/s^2 + C/(s+2) + D(s+1)
L(x)=L(A+Bt+C e^{-2t} +D e^{-t})
Solve for A,B,C,D by the sampling method (partial fraction method).
Shifting Theorem and u-substitution Applications
Problem 7.3-8. L(f)=(s-1)/(s+1)^3
See #18 details for a similar problem.
Problem 7.3-18. L(f)=s^3/(s-4)^4.
L(f) = (u+4)^3/u^4  where u=s-4
L(f) = (u^3+12u^2+48u+64)/u^4
L(f) = (1/s + 12/s^2 + 48/s^3 + 64/s^4) where s --> (s-4)
L(f)=L(e^{4t}(1+12t+48t^2/2+64t^3/6)) by shifting thm
Problem 7.3-8. L(f)=(s+2)/(s^2+4s+5)
L(f) = (s+2)/((s+2)^2 + 1)
L(f) = u/(u^2 + 1)  where u=s+2
L(f) = s/(s^2 + 1) where s --> s+2
L(f) = L(e^{-2t} cos(t))  by shifting thm
S-differentiation theorem
Problem 7.4-21. Similar to Problem 7.4-22.
Clear fractions, multiply by (-1), then:
(-t)f(t) = -exp(3t)+1
L((-t)f(t)) = -1/(s-3) + 1/s
(d/ds)F(s) = -1/(s-3) + 1/s
F(s) = ln(|s|/|s-3|)+c
To show c=0, use this theorem:
THEOREM. The Laplace integral has limit zero at s=infinity.
Convolution theorem
THEOREM. L(f(t)) L(g(t)) = L(convolution of f and g)
Example. L(cos t)L(sin t) = L(0.5 t sin t)
Example: 7.4-36. x''+4x=f(t), x(0)=x'(0)=0 has solution
x(t)=0.5 int(sin(2u)f(t-u),u=0..t)

Periodic function theorem application
Problem 7.5-28.
Find L(f(t)) where f(t) = t on 0 <= t < a and f(t)=0 on a <= t < 2a,
with f(t) 2a-periodic [f(t+2a)=f(t)].
Details
According to the periodic function theorem, the answer is
found from maple integration:
L(f) = int(t*exp(-s*t),t=0..a)/(1-exp(-2*a*s));

Piecewise Functions
Unit Step: u(t)=1 for t>=0, u(t)=0 for t<0.
Pulse: pulse(t,a,b)=u(t-a)-u(t-b)
Ramp: ramp(t-a)=(t-a)u(t-a)
Problem Session: Periodic function theorem
Laplace of the square wave. Problem 7.5-25. Done earlier.
Laplace of the sawtooth wave. Problem 7.5-26. Done earlier.
Method: (d/dt) sawtooth = square wave
The use the parts theorem.
Or, use the Integral theorem.
Laplace of the staircase function. Problem 7.5-27. Done earlier.
This is floor(t/a). The Laplace answer is
L(floor(t/a))=(1/s)/(exp(as)-1))
This answer can be verified by maple code
inttrans[laplace](floor(t/a),t,s);
Laplace of the sawtooth wave, revisited.
Identity: floor(t) = staircase with jump 1.
Identity: t - floor(t) = saw(t) = sawtooth wave
General:  t - a*floor(t/a) = a*saw(t/a) = sawtooth wave of period a.
Problem 7.5-28. Details revisited.
f(t)=t on 0 <= t <= a,
f(t)=0 on a <= t <= 2a
According to the periodic function theorem, the answer is
found from maple integration:
int(t*exp(-s*t),t=0..a)/(1-exp(-2*a*s));
A better way to solve the problem is to write a formula for
f'(t) and use the s-differentiation rule. We  get for a=1
f'(t) = (1/2)(1+sqw(t))
and then
sL(f(t)) = (1/(2s))(1+tanh(s/2))
L(f(t)) = (1/(2s^2))(1+tanh(s/2))
ALTERNATIVE
Use the Laplace integral theorem, which says the answer is (1/s)
times the Laplace answer for the 2a-periodic function g(t)=1 on
[0,a], g(t)=0 on [a,2a]. We check that g(t)=(1/2)(1+sqw(t/a)).
```
```  Problem Session: Second Shifting Theorem Applications
Second shifting Theorems
e^{-as}L(f(t))=L(f(t-a)u(t-a))  Requires a>=0.
L(g(t)u(t-a))=e^{-as}L(g(t+a))  Requires a>=0.
Problem 7.5-3. L(f)=e^{-s}/(s+2)
Problem 7.5-4. L(f) = (e^{-s} - e^{2-2s})/(s-1)
Problem 7.5-22. f(t)=t^3 pulse(t,1,2)
Problem 7.5-4.
F_1(s) = exp(-s)/(s-1) = exp(-as)L(exp(t)) with a=1
= L(exp(t-1)u(t-1)) by the second shifting theorem
F_2(s) = exp(2-2s)/(s-1) = exp(-2s)/s with shift s --> s-1
= L(1 u(t-2)) [2nd shifting theorem] shift s --> s-1
= L( exp(t) 1 u(t-2)) by the first shifting theorem
F=F_1 - F_2 = L(exp(t-1)u(t-1)-exp(t)u(t-2))
f(t) =  exp(t-1)u(t-1)-exp(t)u(t-2)
```

```   Problem 7.5-22.
f(t)=t^3 pulse(t,1,2)
= t^3 u(t-1) - t^3 u(t-2)
L(t^3 u(t-1)) = exp(-s)L((t+1)^3) 2nd shifting theorem
L(t^3 u(t-2)) = exp(-2s)L((t+2)^3) 2nd shifting theorem
Details to be finished in class. Pascal's triangle and (a+b)^3.
Function notation and dummy variables.

Dirac Applications
x''+x=5 Delta(t-1), x(0)=0,x'(0)=1
THEOREM.
The Laplace integral has limit zero at t=infinity, provided
f(t) is of exponential order. The Laplace of the Dirac impulse
violates this theorem's hypothesis, because L(delta(t))=1.

```
```  Cayley-Hamilton-Ziebur Method: Ch4, sections 4.1, 4.2

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
(2I-A)(3I-A)=0 or 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)Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (137.4 K, pdf, 14 Mar 2016)
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(v_1) + exp(3t) vec(v_2)
where vectors v_1, uv_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 v_1, v_2 in the example, differentiate the
equation u(t) = exp(2t) v_1 + exp(3t) v_2 and set t=0
in both relations. Then u'=Au implies
u_0 =    v_1  +   v_2,
Au_0 = 2 v_1 + 3 v_2.
These equations can be solved by elimination.
v_1 = (3 u_0 -Au_0), v_2 = (Au_0 - 2 u_0)
= vector([-1,0])     = vector([2,2])
Vectors v_1, v_2 are recognized as eigenvectors of A for
lambda=2 and lambda=3, respectively.

ZIEBUR SHORTCUT [Edwards-Penney textbook method, Example 5 in 4.1]
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.
Cayley-Hamilton topics, Section 6.3.
Cayley-Hamilton Theorem
A matrix satisfies its own characteristic equation.

Proof of the Ziebur Lemma for 2x2 matrices.
```