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2250 8:05am Lectures Week 3 S2015

Last Modified: February 01, 2015, 08:54 MST.    Today: November 22, 2017, 17:07 MST.
Topics
  Sections 2.2, 2.3
  The textbook topics, definitions, examples and theorems
Edwards-Penney 2.1, 2.2, 2.3 (15.5 K, txt, 17 Dec 2014) Sections 2.4, 2.5, 2.6 The textbook topics, definitions, examples and theorems
Edwards-Penney 2.4, 2.5, 2.6 (10.9 K, txt, 26 Jan 2015)
Examples: Week 3 (342.7 K, png, 04 Feb 2015)

Week 3: Sections 2.3, 2.4, 2.5, 2.6

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

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.
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 moon
Slides: Jules Verne Problem (124.2 K, pdf, 01 Jan 2015)
Manuscript: Earth to the moon, Skydiving, Lunar lander (278.2 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 2.3 (8.9 K, txt, 28 Jan 2015) Midterm 1 sample exam is yet to be written. Samples exist for 5+ years, 3 exams per semester. Found at the course web site:
HTML: 2250 midterm exam samples S2015 (22.2 K, html, 23 Feb 2015)

Mon-Wed: 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.

Thursday: Ziwen Zhu

Lab 3.

Friday: 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 are the
Slides: 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. ==> Left off here on Friday, will complete the lecture on Monday 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 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 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 y'=1-x-y 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 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)

Monday: 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)