## 2280 8:05am Lectures Week 4 S2015

Last Modified: February 09, 2015, 12:40 MST.    Today: January 16, 2019, 07:01 MST.
Topics
The textbook definitions and theorems
Edwards-Penney 2.4, 2.5, 2.6 (11.1 K, txt, 18 Dec 2013)
Edwards-Penney 3.1, 3.2, 3.3, 3.4 (16.2 K, txt, 04 Jan 2015)
PDF: Week 4 Examples (0.0 K, pdf, 31 Dec 1969)

### Monday: Numerical Methods: Sections 2.4, 2.5, 2.6. Maple examples.

WORKED EXAMPLE
y'=1-x-y, y(0)=3, solution y=2-x+exp(-x).
We will make a dot table by hand and also by machine.
The handwritten work for the Euler Method and the Improved Euler
Method [Heun's Method] are available:
Jpeg: Handwritten example y'=1-x-y, y(0)=3, Euler and Heun (427.8 K, jpg, 16 Dec 2012) The basic reference is
Slides: Numerical methods (149.7 K, pdf, 26 Jan 2014) EULER METHOD Let f(x,y)=1-x-y, the right side of the differential equation. Use step size h=0.2 from x=0 to x=0.4. The dot table has 3 rows. Table row 1: x0=0, y0=3 Taken from initial condition y(0)=3 Table row 2: x1=0.2, y1=2.6 Compute from x1=x0+h, y1=y0+hf(x0,y0)=3+0.2(1-0-3) Table row 3: x2=0.4, y2=2.24 Compute from x2=x1+h, y2=y1+hf(x1,y1)=2.6+0.2(1-0.2-2.6) MAPLE EULER: [0, 3], [.2, 2.6], [.4, 2.24] HEUN METHOD Let f(x,y)=1-x-y, the right side of the differential equation. Use step size h=0.2 from x=0 to x=0.4. The dot table has 3 rows. Table row 1: x0=0, y0=3 Taken from initial condition y(0)=3 Table row 2: x1=0.2, y1=2.62 Compute from x1=x0+h, tmp=y0+hf(x0,y0)=2.6, y1=y0+h(f(x0,y0)+f(x1,tmp))/2=2.62 Table row 3: x2=0.4, y2=2.2724 Compute from x2=x1+h, tmp=y1+hf(x1,y1)=2.62+0.2*(1-0.2-2.62) y2=y1+h(f(x1,y1)+f(x2,tmp))/2=2.2724 MAPLE HEUN: [0, 3], [.2, 2.62], [.4, 2.2724] RK4 METHOD Let f(x,y)=1-x-y, the right side of the differential equation. Use step size h=0.2 from x=0 to x=0.4. The dot table has 3 rows. The only Honorable way to solve RK4 problems is with a calculator or computer. A handwritten solution proceeds like this Table row 1: x0=0, y0=3 Taken from initial condition y(0)=3 FIRST DOT = [0,3] Table row 2: x1=0.2, y1=2.618733333 Compute from x1=x0+h, 5 lines of RK4 code: k1=h*f(x0,y0)=0.2*f(0,3)=0.2*(1-0-3)=-0.4 k2=h*f(x0+h/2,y0+k1/2)=0.2*f(0.1,3-0.4)=-0.3800000000 k3=h*f(x0+h/2,y0+k2/2)=0.2*f(0.1,3-0.38)=-0.3820000000 k4=h*f(x0+h,y0+k3)=0.2*f(0.2,3-0.382)=-0.3636000000 y1=y0+(1/6)*(k1+2*k2+2*k3+k4)=2.618733333 0.2 2.618733333 MAPLE RK4: [0, 3], [.2, 2.618733333] Table row 3: x2=0.4, y2=2.270324271 Compute from x2=x1+h, 5 lines of RK4 code MAPLE RK4: [0, 3], [.2, 2.618733333], [.4, 2.270324271] EXACT SOLUTION We solve the linear differential equation y'=1-x-y by the integrating factor method to obtain y=2-x+exp(-x). # MAPLE evaluation of y=2-x+exp(-x), the exact solution. F:=x->2-x+exp(-x);[0.0,F(0.0)],[0.2,F(0.2)],[0.4,F(0.4)]; # Answer [[0., 3.], [0.2, 2.618730753], [0.4, 2.270320046]] COMPARISON GRAPHIC The three results for Euler, Heun, RK4 are compared to the exact solution y=2-x+exp(-x) in the
GRAPHIC: y'=1-x-y Compare Euler-Heun-RK4 (17.8 K, jpg, 10 Dec 2012) The comparison graphic was created with this
MAPLE TEXT: y'=1-x-y by Euler-Heun-RK4 (1.1 K, txt, 10 Dec 2012)
Third lecture on numerical methods. Solved Problems.
Theory for RK4
Historical events: Heun, Runge and Kutta
How Simpson's Rule provides RK4, using Predictors and Correctors.
Why we don't read the proof of RK4.
Numerical project example y'=-2xy, y(0)=2.
Handwritten example for y'=1-x-y, y(0)=3
References for numerical methods:
Slides: Numerical methods (149.7 K, pdf, 26 Jan 2014)
Jpeg: Handwritten example y'=1-x-y, y(0)=3, Euler and Heun (427.8 K, jpg, 16 Dec 2012) Additional References for numerical methods:
Manuscript: Numerical methods y'=F(x) and y'=f(x,y) (659.1 K, pdf, 29 Nov 2014)
Manuscript: Numerical methods RECT, TRAP, SIMP for y'=F(x) (305.2 K, pdf, 23 Jan 2014)
Manuscript: Numerical methods to compute Pi, ln2 and e (205.9 K, pdf, 26 Jan 2014)
Manuscript: Jules Verne problem, skydiving, lunar lander (278.2 K, pdf, 01 Jan 2015)
Slides: Jules Verne problem and escape velocity (124.2 K, pdf, 01 Jan 2015)
Matlab: Numerical DE in matlab m-file from Professor W. Nesse (3.1 K, m, 28 Jan 2014)
Maple: Ejected baggage nonlinear drag example, numerical DE from Professor Gustafson (0.8 K, mpl, 28 Jan 2014)
How to use maple at home (7.3 K, txt, 05 Dec 2012)

### Tuesday-Wednesday: Ch 3

EULER SOLUTION ATOMS and INDEPENDENCE.
Def. atom=x^n(base atom)
base atom = 1, exp(ax), cos(bx), sin(bx), exp(ax) cos(bx),
exp(ax) sin(bx) for nonzero real a and positive real b
THEOREM. Euler solution atoms are independent.
EXAMPLE. Show 1, x^2, x^9 are independent by 3 methods.
EXAMPLE. Independence of 1, x^2, x^(3/2) by the Wronskian test.
EXAMPLE.
Equation y''+10y'=0 has general solution y=c1 + c2 exp(-10x)
The Euler solution atoms for this example are 1 and exp(-10x).
Differential equations like this one have general solution a
linear combination of atoms.

TOOLKIT for SOLVING LINEAR CONSTANT DIFFERENTIAL EQUATIONS
Picard: Order n of a DE = dimension of the solution space.
General solution = linear combination n independent atoms.
Euler's theorem(s), an algorithm for solution atoms.
Summary for Higher Order Differential Equations

Slides: Atoms, Euler's theorem, 7 examples (130.5 K, pdf, 25 Feb 2013)
Slides: Base atom, atom, basis for linear DE (117.6 K, pdf, 03 Mar 2012) EXAMPLE. The equation y'' +10y'=0. How to solve y'' + 10y' = 0, chapter 1 methods. Idea: Let v=x'(t) to get a first order DE in v and a quadrature equation x'(t)=v(t). Solve the first order DE by the linear integrating factor method. Then insert the answer into x'(t)=v(t) and continue to solve for x(t) by quadrature. Vector space of functions: solution space of a differential equation. A basis for the solution space of y'' + 10y'=0 is {1,exp(-10x)} PICARDS THEOREM The theorem of Picard and Lindelof for y'=f(x,y) has an extension for systems of equations, which applies to scalar higher order linear differential equations. THEOREM [Picard] A homogeneous nth order linear differential equation with continuous coefficients has a general solution written as a linear combination of n independent solutions. This means that the solution space of the differential equation has dimension n. PICARD FAILURE. Although Picard's structure theorem does not provide an algorithm for construction of independent solutions, the theorems of Euler do that. Combined, there is an easy path to finding a basis for the solution space of an nth order linear differential equation. EULER'S THEOREM for CONSTRUCTING a SOLUTION BASIS It says y=exp(rx) is a solution of ay'' + by' + cy = 0 <==> r is a root of the characteristic equation ar^2+br+c=0. REAL EXPONENTIALS: If the root r is real, then the exponential is a real solution. THEOREM. A real root r=a (positive, negative or zero) produces one Euler solution atom exp(ax). COMPLEX EXPONENTIALS: If a nonreal root r=a+ib occurs, a complex number, then there is a conjugate root a-ib. The pair of roots produce two real solutions from EULER'S FORMULA (a trig topic): exp(i theta) = cos(theta) + i sin(theta) Details to obtain the two solutions will be delayed. The answer is THEOREM. A conjugate root pair a+ib,a-ib produces two independent Euler solution atoms exp(ax) cos(bx), exp(ax) sin(bx). HIGHER MULTIPLICITY For roots of the characteristic equation of multiplicity greater than one, there is a correction to the answer obtained in the two theorems above: Multiply the answers from the theorems by powers of x until the number of Euler solution atoms produced equals the multiplicity. EXAMPLE: If r=3,3,3,3,3 (multiplicity 5), then multiply exp(3x) by 1, x, x^2, x^3, x^4 to obtain 5 Euler solution atoms. EXAMPLE: If r=5+3i,5+3i (multiplicity 2), then there are roots r=5-3i,5-3i, making 4 roots. Multiply the two Euler atoms exp(5x)cos(3x), exp(4x)sin(3x) by 1, x to obtain 4 Euler solution atoms. SHORTCUT: The characteristic equation can be synthetically formed from the differential equation ay''+by'+cy=0 by the formal replacement y ==> 1, y' ==> r, y'' ==> r^2. EULER'S SOLUTION ATOMS Leonhard Euler described a complete solution to finding n independent solutions in the special case when the coefficients are constant. The Euler solutions are called atoms in these lectures. THE TERM ATOM. The term atom abbreviates Euler solution atom of a linear differential equation. The main theorem says that the answer to a homogeneous constant coefficient linear differential equation of higher order is a linear combinations of atoms. EULER SOLUTION ATOMS for LINEAR DIFFERENTIAL EQUATIONS DEF. Base atoms are 1, exp(a x), cos(b x), sin(b x), exp(ax)cos(bx), exp(ax)sin(bx). DEF: atom = x^n (base atom) for n=0,1,2,... THEOREM. Euler solution atoms are independent. THEOREM. Solutions of constant-coefficient homogeneous differential equations are linear combinations of a complete set of Euler solution atoms. EXAMPLE. The equation y''+10y'=0 has characteristic equation r^2+10r=0 with roots r=0, r=-10. Then Euler's theorem says exp(0x) and exp(-10x) are solutions. By vector space dimension theory, the Euler solution atoms 1, exp(-10x) are a basis for the solution space S of the differential equation. Then the general solution is y = a linear combination of the Euler solution atoms y = c1 (1) + c2 (exp(-10x)).

### Wednesday-Friday: Ch 3

SOLUTION ATOMS and INDEPENDENCE.
Def. atom=x^n(base atom), n=0,1,2,3,...
where for a nonzero real and b>0,
base atom = 1, cos(bx), sin(bx),
exp(ax), exp(ax) cos(bx), exp(ax) sin(bx)
"atom" abbreviates
"Euler solution atom of a linear differential equation"
THEOREM. Atoms are independent.
EXAMPLE. Show 1, x^2, x^9 are independent [atom theorem]
EXAMPLE. Show 1, x^2, x^(3/2) are independent [Wronskian test]
PARTIAL FRACTION THEORY. MAPLE ASSIST. MUPAD ASSIST.
top:=x-1; bottom:=(x+1)^2*(x^2+1)^2;
convert(top/bottom,parfrac,x);  # Maple
EXAMPLE. Solve y''+10y'=0 for general solution y=c1 + c2 exp(-10x)
TOOLKIT for SOLVING LINEAR CONSTANT DIFFERENTIAL EQUATIONS
Picard: Order n of a DE = dimension of the solution space.
General solution = linear combination n independent atoms.
Euler's theorem(s), an algorithm for finding solution atoms.
Summary for Higher Order Differential Equations

Slides: Atoms, Euler's theorem, 7 examples (130.5 K, pdf, 25 Feb 2013)
Slides: Base atom, atom, basis for linear DE (117.6 K, pdf, 03 Mar 2012) Second order equations. Homogeneous equation. Harmonic oscillator example y'' + y=0. Picard-Lindelof theorem. Dimension of the solution space. Structure of solutions. Non-homogeneous equation. Forcing term. Nth order equations. Solution space theorem for linear differential equations. Superposition. Independence and Wronskians. Independence of atoms. Main theorem on constant-coefficient equations THEOREM. Solutions are linear combinations of atoms. Euler's substitution y=exp(rx). Shortcut to finding the characteristic equation. Euler's basic theorem: y=exp(rx) is a solution <==> r is a root of the characteristic equation. Euler's multiplicity theorem: y=x^n exp(rx) is a solution <==> r is a root of multiplicity n+1 of the characteristic equation. How to solve any constant-coefficient nth order homogeneous differential equation. 1. Find the n roots of the characteristic equation. 2. Apply Euler's theorems to find n distinct solution atoms. 2a. Find the base atom for each distinct real root. Multiply each base atom by powers 1,x,x^2, ... until the number of atoms created equals the root multiplicity. 2b. Find the pair of base atoms for each conjugate pair of complex roots. Multiply each base atom by powers 1,x,x^2, ... until the number of atoms created equals the root multiplicity. 3. Report the general solution as a linear combination of the n atoms. Derivations: Spring-Mass System and RLC Circuit See the textbook section 3.7 Constant coefficient equations with complex roots. Applying Euler's theorems to solve a DE. Examples of order 2,3,4. Exercises 3.1, 3.2, 3.3. 3.1-34: y'' + 2y' - 15y = 0 3.1-36: 2y'' + 3y' = 0 3.1-38: 4y'' + 8y' + 3y = 0 3.1-40: 9y'' -12y' + 4y = 0 3.1-42: 35y'' - y' - 12y = 0 3.1-46: Find char equation for y = c1 exp(10x) + c2 exp(100x) 3.1-48: Find char equation for y = l.c. of atoms exp(r1 x), exp(r2 x) where r1=1+sqrt(2) and r2=1-sqrt(2). 3.2-18: Solve for c1,c2,c3 given initial conditions and general solution. y(0)=1, y'(0)=0, y''(0)=0 y = c1 exp(x) + c2 exp(x) cos x + c3 exp(x) sin x. 3.2-22: Solve for c1 and c2 given initial conditions y(0)=0, y'(0)=10 and y = y_p + y_h = -3 + c1 exp(2x) + c2 exp(-2x). 3.3-8: y'' - 6y' + 13y = 0 (r-3)^2 +4 = 0 3.3-10: 5y'''' + 3y''' = 0 r^3(5r+3) = 0 3.3-16: y'''' + 18y'' + 81 y = 0 (r^2+9)(r^2+9) = 0 Check all answers with Maple/MuPad, using this example: de:=diff(y(x),x,x,x,x)+18*diff(y(x),x,x)+81*y(x) = 0; # Maple/MuPad dsolve(de,y(x)); # maple only ode::solve(de,y(x)); # MuPad only 3.3-32: Theory of equations and Euler's method. Char equation is r^4 + r^3 - 3r^2 -5r -2 = 0. Use the rational root theorem and long division to find the factorization (r+1)^3(r-2)=0. Check the root answer in Maple/MuPad, using the code solve(r^4 + r^3 - 3*r^2 -5*r -2 = 0,r); The answer is a linear combination of 4 atoms, obtained from the roots -1,-1,-1,2.
Topics
html: Problem notes S2015 (2.5 K, html, 23 Dec 2014)

#### Monday: Damped and Undamped Motion. Section 3.4

Lecture: Applications. Damped and undamped motion.
Last time: Theory of equations and 5.3-32.
Spring-mass equation,
Spring-mass DE derivation
Spring-mass equation mx''+cx'+kx=0 and its physical parameters.
Spring-mass system,
my'' + cy' + ky = 0
harmonic oscillation,
y'' + omega^2 y = 0,
Forced systems.
Forcing terms in mechanical systems. Speed bumps.
Harmonic oscillations: sine and cosine terms of frequency omega.
Damped and undamped equations. Phase-amplitude form.
Slides:
Shock-less auto.
Rolling wheel on a spring.
Swinging rod.
Mechanical watch.
Bike trailer.
Physical pendulum.
Slides: Unforced vibrations 2008 (647.6 K, pdf, 27 Feb 2014)
Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)
Slides: Forced damped vibrations (264.0 K, pdf, 08 Mar 2014)
Slides: Forced vibrations and resonance (253.0 K, pdf, 08 Mar 2014)
Slides: Resonance and undetermined coefficients, cafe door, pet door, phase-amplitude (178.0 K, pdf, 08 Mar 2014)
Slides: Electrical circuits (112.9 K, pdf, 08 Mar 2014)