## 2250 8:05am Lectures Week 12 S2015

Last Modified: April 06, 2015, 05:22 MDT.    Today: December 16, 2018, 17:31 MST.

### Week 12: Sections 10.4, 10.5, EPbvp7.6, 6.1, 6.2

``` Edwards-Penney, sections 10.4, 10.5, EPbvp7.6, 6.1, 6.2
The textbook topics, definitions and theoremsEdwards-Penney 10.1, 10.2, 10.3, 10.4, 10.5 (20.5 K, txt, 18 Dec 2013)Edwards-Penney 6.1, 6.2 (7.6 K, txt, 18 Dec 2013)```

#### Mon-Wed: Continue,. Sections 10.4 and EPbvp supplement 7.6.

``` DEF. Gamma function
Gamma(t) = integral x=0 to x=infinity  x^{t-1} e^{-x}
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)
DEF. Two-sided Laplace transform
{Bf}(s) = {Mf(-ln(x))}(s) = integral x=0 to x=infinity x^{s-1}f(-ln x)
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

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 10.5-25.

Applications of Laplace's method from 10.3, 10.4, 10.5
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.
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 (88.1 K, pdf, 03 Mar 2012)
Intro to the Laplace resolvent shortcut for 2x2 systems
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 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).
```

#### Wednesday: Resolvent. Dirac Impulse. Section 10.5 and EPbvp supplement 7.6. Dirac Delta and hammer hits.

```Laplace Resolvent Method
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.
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.

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 Dirac delta appears 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 Delta.
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 Dirac 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.
```

#### Wed: Problem session. Sections 10.1 to 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 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 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 application
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: 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 10.5-25. Done earlier.
Laplace of the sawtooth wave. Problem 10.5-26. Done earlier.
Method: (d/dt) sawtooth = square wave
The use the parts theorem.
Or, use the Integral theorem.
Laplace of the staircase function. Problem 10.5-27. 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 10.5-28. 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 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)
Problem 10.5-4.
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 10.5-22.
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 were 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 delta function
violates this theorem's hypothesis, because L(delta(t))=1.
```

#### Friday: Sections 6.1, 7.1

```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);
EIGENANALYSIS WARNING
Reading Edwards-Penney Chapter 6 may deliver the wrong ideas
about how to solve for eigenpairs. The examples emphasize a
clever shortcut, which does not apply in general to solve for
eigenpairs.

HISTORY. Chapter 6 originally appeared in the 2280 book
as a summary, which assumed a linear algebra course. The
chapter was copied without changes into the Edwards-Penney
Differential Equations and Linear Algebra textbook, which you
currently own. The text contains only shortcuts. There is
no discussion of a general method for finding eigenpairs.
You will have to fill in the details by yourself. The online
lecture notes and slides were created to fill in the gap.

Lecture: Fourier's Model. Intro to Eigenanalysis, Ch6.
Examples and motivation.
Ellipse, rotations, eigenpairs.
General solution of a differential equation u'=Au and eigenpairs.
Fourier's model.
History.
J.B.Fourier's 1822 treatise on the theory of heat.
The rod example.
Physical Rod: a welding rod of unit length, insulated on the
lateral surface and ice packed on the ends.
Define f(x)=thermometer reading at loc=x along the rod at t=0.
Define u(x,t)=thermometer reading at loc=x and time=t>0.
Problem: Find u(x,t).
Fourier's solution. Let's assume that
f(x) = 17 sin (pi x) + 29 sin(5 pi x)
= 17 v1 + 29 v2
Packages v1, v2 are vectors in a vector space V of functions on [0,1].
Fourier computes u(x,t) by re-scaling v1, v2 with numbers Lambda_1,
Lambda_2 that depend on t. This idea is called Fourier's Model.

u(x,t) = 17 ( exp(-pi^2 t) sin(pi x)) + 29 ( exp(-25 pi^2 t) sin (5 pi x))
= 17 (Lambda_1 v1) + 29 (Lambda_2 v2)

Eigenanalysis of u'=Au is the identical idea.
u(0) = c1 v1 + c2 v2  implies
u(t) = c1 exp(lambda_1 t) v1 + c2 exp(lambda_2 t) v2
Fourier's re-scaling idea from 1822, applied to u'=Au,
replaces v1 and v2 in the expression
c1 v1 + c2 v2
by their re-scaled versions to obtain the answer
c1 (Lambda1 v1) + c2 (Lambda2 v2)
where
Lambda1 = exp(lambda_1 t), Lambda2 = exp(lambda_2 t).

Main Theorem on Fourier's Model

THEOREM. Fourier's model
A(c1 v1 + c2 v2) = c1 (lambda1 v1) + c2 (lambda2 v2)
with v1, v2 a basis of R^2 holds [for all constants c1, c2]
if and only if
the vector-matrix system
A(v1) = lambda1 v1,
A(v2) = lambda2 v2,
has a solution with vectors v1, v2 independent
if and only if
the diagonal matrix D=diag(lambda1,lambda2) and
the augmented matrix P=aug(v1,v2) satisfy
1. det(P) not zero [then v1, v2 are independent]
2. AP=PD

THEOREM. The eigenvalues of A are found from the determinant
equation
det(A -lambda I)=0,
which is called the characteristic equation.

THEOREM. The eigenvectors of A are found from the frame
sequence which starts with B=A-lambda I [lambda a root of
the characteristic equation], ending with last frame rref(B).

The eigenvectors for lambda are the partial derivatives of
the general solution obtained by the Last Frame Algorithm,
with respect to the invented symbols t1, t2, t3, ...

Algebraic Eigenanalysis Section 6.2.
Calculation of eigenpairs to produce Fourier's model.
Connection between Fourier's model and a diagonalizable matrix.
How to find the variables lambda and v in Fourier's model using
determinants and frame sequences.
Solved in class: examples similar to the problems in 6.1 and 6.2.
Web slides and problem notes exist for the 6.1 and 6.2 problems.
Examples where A has an eigenvalue of multiplicity greater than one.
```
```Laplace theory references
Slides: Laplace and Newton calculus. Photos. (200.2 K, pdf, 03 Mar 2012) Slides: Intro to Laplace theory. Calculus assumed. (144.8 K, pdf, 25 Mar 2015) Slides: Laplace rules (160.3 K, pdf, 03 Mar 2012) Slides: Laplace table proofs (169.6 K, pdf, 03 Mar 2012) Slides: Laplace examples (133.2 K, pdf, 27 Mar 2015) Slides: Piecewise functions and Laplace theory (108.5 K, pdf, 03 Mar 2013) MAPLE: Optional Maple Lab 7. Laplace applications (0.0 K, pdf, 31 Dec 1969) Manuscript: DE systems, examples, theory (730.9 K, pdf, 10 Apr 2014) Slides: Laplace resolvent method (88.1 K, pdf, 03 Mar 2012) Slides: Laplace second order systems (288.1 K, pdf, 03 Mar 2012) Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 09 Apr 2014) Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008) Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008) Manuscript: Heaviside's method 2008 (352.3 K, pdf, 07 Jan 2014) Manuscript: Laplace theory 2008 (500.9 K, pdf, 16 Mar 2014) Transparencies: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010) Text: Laplace theory problem notes for Chapter 10 (17.7 K, txt, 17 Mar 2014) Text: Final exam study guide (8.3 K, txt, 06 Jan 2015)
Systems of Differential Equations references
Manuscript: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014) Slides: Laplace resolvent method (88.1 K, pdf, 03 Mar 2012) Slides: Laplace second order systems (288.1 K, pdf, 03 Mar 2012) Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 09 Apr 2014) Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008) Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)```
```References for Eigenanalysis and Systems of Differential Equations.
Sildes: Algebraic eigenanalysis (187.6 K, pdf, 03 Mar 2012) Slides: What's eigenanalysis 2008 (174.2 K, pdf, 03 Mar 2012) Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (137.7 K, pdf, 28 Mar 2015) Slides: Laplace resolvent method (88.1 K, pdf, 03 Mar 2012) Slides: Laplace second order systems (288.1 K, pdf, 03 Mar 2012) Manuscript: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014) Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 09 Apr 2014) Text: Lawrence Page's pagerank algorithm (0.7 K, txt, 06 Oct 2008) Text: History of telecom companies (1.9 K, txt, 03 Apr 2013) Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008) Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)
```