# 2250-1 7:30am Lecture Record Week 11 S2012

Last Modified: April 10, 2012, 12:34 MDT.    Today: September 24, 2018, 02:14 MDT.

## Thu, Apr 1:Patrick

Exam 2, problems 4,5.

## Mon, Mar 26: Piecewise Functions. Section 10.5 and EPbvp supplement 7.6.

```Laplace Resolvent Method.
--> This method is considered a shortcut for solving 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
(sI - 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.
```

## Mar 27,28: Piecewise Functions. Shifting. Section 10.5 and EPbvp supplement 7.6. Delta function and hammer hits.

```Piecewise Functions
Unit Step: step(t)=1 for t>=0, step(t)=0 for t<0.
Pulse: pulse(t,a,b)=step(t-a)-step(t-b)
Ramp: ramp(t-a)=(t-a)step(t-a)
L(step(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]
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.
Periodic function theorem
Proof
Application to the square wave.
Convolution theorem
Application:   L(cos t)L(sin t) = L(0.5 t sin(t))
```
``` 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.
Hammer hit models
How delta functions enter into circuit calculations
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 [a.b] equals the integral of f(t) over [a,b]
An example when f(t) has impulse 5, defined by
f(t) = (5/(2h))pulse(t,-h,h)
Laplace integral of f(t) and its limit as h --> 0.  Answer is the Dirac delta.
The delta function model
x''(t) + 4x(t) = 5 delta(t-t0),
x(0)=0, x'(0)=0
The delta function model from EPbvp 7.6
x''(t) + 4x(t) = 8 delta(t-2 pi),
x(0)=3, x'(0)=0
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));
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))  Previously discussed how to convert in class. See 5.4.
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 step, then E'(t) is a Dirac delta.
Resonance examples
x'' + x = cos(t)
Pure resonance, unbounded solution x(t) = 0.5 t sin(t)
mx'' + cx' + kx = F_0 cos(omega t)
Practical resonance, all solutions bounded, but x(t)
can have extremely large amplitude when omega is tuned
to the frequency omega = sqrt(k/m - c^2/(2m^2))
LQ'' + RQ' + (1/C)Q = E_0 sin(omega t)
Practical resonance, all solutions bounded, but the current
I(t)=dQ/dt can have large amplitude when omega is tuned
to the resonant frequency omega = 1/sqrt(LC).
later lecture, with slides and video:
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.
```

## Mar 27,28: Problem session. Sections 10.4, 10.5.

``` Forward and Backward Table Applications
Review of previously solved problems.
Problem 10.1-18. Trig identity 2 sin (theta) cos(theta) = sin(2 theta)
used for L(sin(3t)cos(3t)).
Problem 10.1-28. Splitting a fraction into backward table entries.
Partial Fractions and Backward Table Applications
Problem 10.2-24. L(f)=1/(s(s+1)(s+2)) solved by the three methods for
partial fractions: sampling, atom method, Heaviside cover-up.
Problem 10.2-9. Solve x''+3x'+2x=t, x(0)=0, x'(0)=2. Get 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.
Shifting Theorem and u-substitution Applications
Problem 10.3-8. L(f)=(s-1)/(s+1)^3
See #18 details for a similar problem.
Problem 10.3-18. 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 10.3-8. 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 10.4-21. Similar to Problem 10.4-22.
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: 10.4-36. 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
Laplace of the square wave. Problem 10.5-25.
Problem 10.5-28.
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: step(t)=1 for t>=0, step(t)=0 for t<0.
Pulse: pulse(t,a,b)=step(t-a)-step(t-b)
Ramp: ramp(t-a)=(t-a)step(t-a)
Periodic function theorem
Laplace of the square wave. Problem 10.5-25. Done earlier.
Laplace of the sawtooth wave. Problem 10.5-26.
Method: (d/dt) sawtooth = square wave
The use the parts theorem.
Laplace of the staircase function. Problem 10.5-27.
This is floor(t/a). The Laplace answer is
L(floor(t/a))=1/(s(exp(as)-1))
This answer can be verified in maple by the code
with(inttrans):
laplace(floor(t/a),t,s);
Laplace of the sawtooth wave, revisited.
Identity: sqw(t) = floor(t) = staircase with jump 1.
Identity: t - floor(t) = saw(t) = sawtooth wave
General:  t - a*floor(t/a) = saw(t,a) = sawtooth wave of period a.
Problem 10.5-28. Details
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))
```

## Fri, Mar 30: Laplace theory, Dirac, Variation of parameters, Undetermined Coefficients. Sections 10.5,7.1,5.5

```  Second Shifting Theorem Applications
Problem 10.5-3. L(f)=e^{-s}/(s+2)
Problem 10.5-4. L(f) = (e^{-s} - e^{2-2s})/(s-1)
Problem 10.5-22. f(t)=t^3 pulse(t,1,2)
Second shifting Theorems
e^{-as}L(f(t))=L(f(t-a)step(t-a))  Requires a>=0.
L(g(t)step(t-a))=e^{-as}L(g(t+a))  Requires a>=0.
Problem 10.5-4.
F_1(s) = exp(-s)/(s-1) = exp(-as)L(exp(t)) with a=1
= L(exp(t-1)step(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 step(t-2)) [2nd shifting theorem] shift s --> s-1
= L( exp(t) 1 step(t-2)) by the first shifting theorem
F=F_1 - F_2 = L(exp(t-1)step(t-1)-exp(t)step(t-2))
f(t) =  exp(t-1)step(t-1)-exp(t)step(t-2)
Problem 10.5-22.
f(t)=t^3 pulse(t,1,2)
= t^3 step(t-1) - t^3 step(t-2)
L(t^3 step(t-1)) = exp(-s)L((t+1)^3) 2nd shifting theorem
L(t^3 step(t-2)) = exp(-2s)L((t+2)^3) 2nd shifting theorem
Details were finished in class. Pascal's triangle and (a+b)^3.
Function notation and dummy variables.

Piecewise Applications
Staircase or floor function
Sawtooth wave
Square wave
Dirac Applications
x''+x=5 Delta(t-1), x(0)=0,x'(0)=1
THEOREM. The Laplace integral has limit zero at t=infinity.
Piecewise Functions
Unit Step: step(t)=1 for t>=0, step(t)=0 for t<0.
Pulse: pulse(t,a,b)=step(t-a)-step(t-b)
Ramp: ramp(t-a)=(t-a)step(t-a)
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.
```
```    Undetermined Coefficients
Which equations can be solved
Intro to the basic trial solution method
Laplace solution of x'' + 9x = 30 sin(2t)
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
```

Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (164.5 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.
Variation of Parameters and Undetermined Coefficients references
Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)