## 2280 12:55pm Lectures Week 9 S2018

Last Modified: March 07, 2018, 12:17 MST.    Today: December 14, 2018, 01:23 MST.
``` 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-Tuesday: Piecewise Functions, Convolution, Resolvent

```DEF: Piecewise Continuous Function
Existence of the Laplace integral.
One-sided and two-sided Laplace integral
Freeway example, suspension collides with a ramp.

One-sided and two-sided Laplace integral
Freeway example, suspension collides with a ramp.

DEF. Gamma function
Gamma(t) = integral x=0 to x=infinity  x^{t-1} e^{-x} dx
Gamma(n)=(n-1)!, generalizes the factorial function
DEF. Mellin transform
{Mf}(s)= phi(s)=integral x=0 to x=infinity  x^{s-1} f(x) dx
DEF. Two-sided Laplace transform
{Bf}(s) = {Mf(-ln(x))}(s) = integral x=0 to x=infinity x^{s-1}f(-ln x) dx

DEF. Unit step u(t-a)=1 for t>=a, else zero
DEF. Ramp t->(t-a)u(t-a)

Backward table problems: examples
Forward table problems: examples
Computing Laplace integrals L(f(t)) with rules
Solving an equation L(y(t))=expression in s for y(t)
Partial fraction methods
Trig identities and their use in Laplace calculations
Hyperbolic functions and Laplace calculations
Why the forward and backward tables don't have cosh, sinh entries
```
```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).

The three-tank problem has a special form called
a cascade because the system can be solved by Ch1
methods: Linear Integrating Factor method and shortcuts.

x' = b0 - b1 x,  Tank 1 amount x(t)
y' = b1 x - b2 y, Tank 2 amount y(t),
z' = b2 y - b3 z, Tank 3 amount z(t).

In system form: u'=Au, A=3x3 band matrix. All such
systems can be solved by the same method. Laplace
also works, not much worse for details.

Coupled Mechanical Oscialltions
EXAMPLE. Forced system of undamped oscillators
m1 x'' = - k1 x + k2 (y-x),
m2 y'' = -k2(y-x) + f(t)
INSTANCE m1=2, m2=1, k1=4, k2=2, f(t)=40sin(3t)
2 x'' = - 6x + 2y,
y'' = 2x - 2y + 40 sin(3t)
Can be solved by Laplace's method, but tedious.
This can be converted to a 4x4 system solving for
four variables x,x',y,y'. The first order system
can be solved by a variety of methods, Laplace being
just one method.

Recirculating Brine Tank: 2 Lakes Pollution Example
Consider the recirculating brine tank example:
20 x' = -6x + y,
20 y' = 6x - 3y
The is u'=Au in vector-matrix form.
We have to solve for roots of the characteristic equation

Maple Example
The maple code to solve the char eq:
A:=(1/20)*Matrix([[-6,1],[6,-3]]);
chareq:=linalg[charpoly](A,r);
solve(chareq,r);

We will also solve the system by Laplace methods, getting the answers
X:=c1*exp(-(9+alpha)*t/40)+c2*exp((-9+gamma)*t/40);
Y:=c3*exp((-9+gamma)*t/40)+c4*exp(-(9+gamma)*t/40);
[gamma]=[sqrt(33)];
The method: Resolvement method below.

Maple code
eq1:=20*diff(x(t),t)=-6*x(t) + y(t);
eq2:=20*diff(y(t),t)=6*x(t) - 3*y(t);
dsolve([eq1,eq2],[x(t),y(t)]);

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. The determinants
must be evaluated by the cofactor method, due to symbols.

```
``` 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)
L(u(t-a)) = (1/s) exp(-as) [for a >= 0 only]
Integral Theorem
L(int(g(x),x=0..t)) = s L(g(t))
Applications to computing ramp(t-a)
L(ramp(t-a)) = (1/s^2) exp(-as) [for a >= 0 only]
Piecewise defined periodic waves
Square wave: f(t)=1 on [0,1), f(t)=-1 on [1,2), 2-periodic
Triangular wave: f(t)=|t| on [-1,1], 2-periodic
Sawtooth wave: f(t)=t on [0,1], 1-periodic
Rectified sine: f(t)=|sin(kt)|
Half-wave rectified sine: f(t)=sin(kt) when positive, else zero.
Parabolic wave
Periodic function theorem
Proof details
Laplace of the square wave. Problem 7.5-25.

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) dx 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))
This is a BEATS example from 3.6, solved by Laplace theory.
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.
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.
==> This method is a shortcut for solving systems by Laplace's method.
==> It is also a convenient way to solve systems with maple.Slides: Laplace resolvent method (81.0 K, pdf, 14 Mar 2016)```

#### Friday: Dirac Impulse. Energy injection like hammer hits.

``` Hammer hits and the Delta impulsen
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 impulses 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 impulse 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 impulse.
```