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# 2280 12:55pm Lectures Week 3 S2015

Last Modified: January 25, 2015, 13:06 MST.    Today: March 19, 2019, 15:32 MDT.
```Topics
Sections 2.3, 2.4, 2.5, 2.6
The textbook topics, definitions, examples and theoremsEdwards-Penney 2.1, 2.2, 2.3 (15.5 K, txt, 17 Dec 2014)Edwards-Penney 2.4, 2.5, 2.6 (11.1 K, txt, 18 Dec 2013)PDF: Week 3 Examples (104.2 K, pdf, 04 Feb 2015)```

### Monday: Newton Kinematic Models. Projectiles. Jules Verne. Section 2.3.

```Continue 2.2 + Drill and Review
Phase diagram for y'=y^2(y^2-4)
Phase line diagram
Threaded curves
Labels: stable, unstable, funnel, spout, node

Phase line diagrams.
Phase diagram.
Section 2.3: Newton's force and friction models
Isaac Newton ascent and descent kinematic models.
Free fall with no air resistance F=0.
Linear air resistance models F=kx'.
Non-linear air resistance models F=k|x'|^2.
The tennis ball problem.
Does it take longer to rise or longer to fall?Text: Bolt shot Example 2.3-3 (1.0 K, txt, 16 Dec 2012)Slides: Newton kinematics with air resistance. Projectiles. (138.9 K, pdf, 11 Jan 2015)
A rocket from the earth to the moonSlides: Jules Verne Problem (124.2 K, pdf, 01 Jan 2015)
Reading assignment:
Proofs of 2.3 theorems in the textbook and derivation of details for
the rise and fall equations with air resistance.Problem notes for Chapter 2 (10.8 K, txt, 22 Dec 2014)

Mon-Tue: Jules Verne Problem. Sections 2.4, 2.5, 2.6. Algorithms for y'=F(x)

Numerical Solution of y'=f(x,y)
Two problems will be studied.
First problem
y' = -2xy, y(0)=2
Symbolic solution y = 2 exp(-x^2)
This problem appears in the Week 3 homework.
Second problem
y' = (1/2)(y-1)^2, y(0)=2
Symbolic solution y = (x-4)/(x-2)
Not assigned, only a lecture example.
Numerical Solution of y'=F(x)
Example: y'=2x+1, y(0)=1
Symbolic solution y=x^2 + x + 1.
Dot table. Connect the dots graphic.
The exact answers for y(x)=x^2+x+1 are
(x,y) = [0., 1.], [.1, 1.11], [.2, 1.24], [.3, 1.39],
[.4, 1.56], [.5, 1.75], [.6, 1.96], [.7, 2.19], [.8,
2.44], [.9, 2.71], [1.0, 3.00]
Maple support for making a connect-the-dots graphic.
Example: L:=[[0., 1.], [2,3], [3,-1], [4,4]]; plot(L);JPG Image: connect-the-dots graphic (11.2 K, jpg, 11 Sep 2010)
Example: Find y(2) when y'=x exp(x^3), y(0)=1.
No symbolic solution!
How to draw a graphic with no solution formula?
Make the dot table by approximation of the integral of F(x).
REFERENCE:Slides: Numerical methods (149.7 K, pdf, 26 Jan 2014)
RECTANGULAR RULE
int(F(x),x=a..b) = F(a)(b-a) approximately
for small intervals [a,b]
Geometry and the Rectangular Rule
Example: y'=2x+1, y(0)=1
Rectangular rule applied to y(1)=1+int(F(x),x=0..1)
for y'=F(x), in the case F(x)=2x+1.
Dot table steps for h=0.1, using the rule 10 times.
Answers: (x,y) = [0, 1], [.1, 1.1], [.2, 1.22], [.3, 1.36],
[.4, 1.52], [.5, 1.70], [.6, 1.90], [.7, 2.12], [.8, 2.36],
[.9, 2.62], [1.0, 2.90]
The correct answer y(2)=3.00 was approximated as 2.90.
Where did [.2,1.22] come from?
y(.2) = y(.1)+int(F(x),x=0.1 .. 0.2) [exactly]
= y(.1)+(0.2-0.1)F(0.1) [approximately, RECT RULE]
= y(.1)+0.1(2x+1) where x=0.1
= 1.1 + 0.1(1.2)  [approx, from data [.1,1.1]]
= 1.22
Rect, Trap, Simp rules from calculus
RECT
Replace int(F(x),x=a..b) by rectangle area (b-a)F(a)
TRAP
Replace int(F(x),x=a..b) by trapezoid area (b-a)(F(a)+F(b))/2
SIMP
Replace int(F(x),x=a..b) by quadratic area
(b-a)(F(a)+4F(a/2+b/2)+F(b))/6

The Euler, Heun, RK4 rules from this course: how they relate to
calculus rules RECT, TRAP, SIMP
Numerical Integration   Numerical Solutions of DE
RECT                    Euler
TRAP                    Heun [modified Euler]
SIMP                    Runge-Kutta 4 [RK4]

Example: y'=3x^2-1, y(0)=2 with solution y=x^3-x+2.
Example: y'=2x+1, y(0)=1 with solution y=x^2+x+1.
Dot tables,  connect the dots graphic.
How to draw a graphic without knowing the solution equation for y.

What to do when int(F(x),x) has no formula?
Key example y'=x exp(x^3), y(0)=2.
Challenge: Can you integrate F(x) = x exp(x^3)?
Making the dot table by approximation of the integral of F(x).
Accuracy: Rect, Trap, Simp rules have 1,2,4 digits resp.
Maple code for the RECT rule
Applied to the quadrature problem y'=2x+1, y(0)=1.
# Quadrature Problem y'=F(x), y(x0)=y0.
# Group 1, initialize.
F:=x->2*x+1:
x0:=0:y0:=1:h:=0.1:Dots:=[x0,y0]:n:=10:

# Group 2, repeat n times. RECT rule.
for i from 1 to n do
Y:=y0+h*F(x0);
x0:=x0+h:y0:=Y:Dots:=Dots,[x0,y0];
od:

# Group 3, display dots and plot.
Dots;
plot([Dots]);
Example 1, for your study:
Problem:  y'=x+1, y(0)=1
It has a dot table with x=0, 0.25, 0.5, 0.75, 1 and
y= 1, 1.25, 1.5625, 1.9375, 2.375.
The exact solution y = 0.5(1+(x+1)^2) has values
y=1, 1.28125, 1.625, 2.03125, 2.5000.
Determine how the dot table was constructed and identify
which rule, either Rect, Trap, or Simp, was applied.
Example 2, for your study:
Problem:  y'=x exp(x^3), y(0)=2
Find the value of y(2)=2+int(x*exp(x^3),x=0..1) to 4 digits.
Elementary integration won't find the integral, it has to be done
numerically. Choose a method and obtain 2.781197xxxx.
MAPLE ANSWER CHECK
F:=x->x*exp(x^3);
int(F(x),x=0..1);  # Re-prints the problem. No answer.
evalf(%);          # ANS=0.7811970311 by numerical integration.

Wed-Fri: Sections 2.5, 2.5, 2.6. Algorithms for y'=f(x,y)

Second lecture on numerical methods
Study problems like y'=-2xy, which have the form y'=f(x,y).
New algorithms are needed. Rect, Trap and Simp won't work,
because of the variable y on the right.
Euler, Heun, RK4 algorithms
Computer implementation in maple
Geometric and algebraic ideas in the derivations.
Numerical Integration   Numerical Solutions of y'=f(x,y)
RECT                    Euler
TRAP                    Heun [modified Euler]
SIMP                    Runge-Kutta 4 [RK4]
Reference for the ideas is thisSlides: Numerical methods (149.7 K, pdf, 26 Jan 2014)
Numerical Solution of y'=f(x,y)
Two problems will be studied.
First problem
y' = -2xy, y(0)=2
Symbolic solution y = 2 exp(-x^2)
Second problem
y' = (1/2)(y-1)^2, y(0)=2
Symbolic solution y = (x-4)/(x-2)
MAPLE TUTOR for NUMERICAL METHODS
# y'=-2xy, y(0)=2, by Euler, Heun, RK4
with(Student[NumericalAnalysis]):
InitialValueProblemTutor(diff(y(x),x)=-2*x*y(x),y(0)=2,x=0.5);
# The tutor compares exact and numerical solutions.
Examples
Web references contain two kinds of examples.
The first three are quadrature problems dy/dx=F(x).
y'=3x^2-1, y(0)=2, solution y=x^3-x+2
y'=exp(x^2), y(0)=2, solution y=2+int(exp(t^2),t=0..x).
y'=2x+1, y(0)=3 with solution y=x^2+x+3.
The fourth is of the form dy/dx=f(x,y), which requires a
non-quadrature algorithm like Euler, Heun, RK4.
y'=1-x-y, y(0)=3, solution y=2-x+exp(-x).
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 isSlides: 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 is not available (and won't be).
Table row 1: x0=0, y0=3
Taken from initial condition y(0)=3
Table row 2: x1=0.2, y1=2.618733333
Compute from x1=x0+h, 5 lines of RK4 code
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 by the integrating factor
method to obtain y=2-x+exp(-x).
# MAPLE evaluation of y=2-x+exp(-x)
F:=x->2-x+exp(-x);[[j*0.2,F(j*0.2)] \$j=0..2];
# 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 theGRAPHIC: y'=1-x-y Compare Euler-Heun-RK4 (17.8 K, jpg, 10 Dec 2012)
The comparison graphic was created with thisMAPLE TEXT: y'=1-x-y by Euler-Heun-RK4 (1.1 K, txt, 10 Dec 2012)

Friday: Sections 2.4, 2.5, 2.6. Maple examples.

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.
References for numerical methods:Slides: Numerical methods (149.7 K, pdf, 26 Jan 2014)How to use maple at home (7.3 K, txt, 05 Dec 2012)Jpeg: Handwritten example y'=1-x-y, y(0)=3, Euler and Heun (427.8 K, jpg, 16 Dec 2012)

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