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2280 12:55pm Lectures Week 3 S2017

Last Modified: October 17, 2016, 03:23 MDT.    Today: November 20, 2017, 15:53 MST.
Topics
  Sections 2.3, 2.4, 2.5, 2.6
  The textbook topics, definitions, examples and theorems
Edwards-Penney 2.1, 2.2, 2.3 (15.8 K, txt, 17 Dec 2014)
Edwards-Penney 2.4, 2.5, 2.6 (11.3 K, txt, 18 Dec 2013)
PDF: Week 3 Examples (98.3 K, pdf, 17 Oct 2016)

Week 3: Sections 2.3, 2.4, 2.5, 2.6

Mon-Tue: Newton Kinematic Models. Projectiles. Jules Verne. Section 2.3.

Continue 2.1, 2.2 + skipped topics
Lecture on 2.2:
  Theory of autonomous DE y'=f(y)
     Picard's theorem and non-crossing of solutions.
     Direction fields and translation of solutions
  Constructing Euler's threaded solution diagrams
     No direction field is needed to draw solution curves
 
       We throw out the threaded solution rule used in chapter 1,
       replace it by two rules from calculus and a theorem:
          1. If y'(x)>0, then y(x) increases.
          2. If y'(x)<0, then y(x) decreases.
          THEOREM. For y'=f(y), a threaded solution starting with
              y'(0)>0 must satisfy y'(x)>0 for x>0. A similar result
              holds for y'(0)<0.
     Definition: phase line diagram, phase diagram,
       Calculus tools: f'(x) pos/neg ==> increasing/decreasing
       DE tool: solutions don't cross
       Maple tools for production work.
  Stability theory of autonomous DE y'=f(y)
    Stability of equilibrium solutions.
    Stable and unstable classification of equilibrium solutions.
    funnel, spout, node,
  How to construct Phase line diagrams
  How to make a phase diagram graphic
    Inventing a graph window
    Invention of the grid points
    Using the phase line diagram to make the graphic
        calculus tools
        DE tools

  Phase diagram for y'=y^2(y^2-4)
     DEF: Phase line diagram
     DEF: Phase Diagram
     Threaded curves
     Labels: stable, unstable, funnel, spout, node

Partial Fraction Theory

  Separation of variables and partial fractions (Delayed from week 2)
      Exercise solution problem 2.1-8
      The equation y'=7y(y-13), y(0)=17
      F(x) = 7, G(y) = y(y-13)
      Separated form y'/G(y) = F(x)
      Answer check using the Verhulst solution
          P(t) = aP_0/(bP_0 + (a-b P_0)exp(-at))
      Separation of variables details.
      Partial fraction details for 1/((u(u-13)) = A/u + B/(u-13)

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

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. (170.8 K, pdf, 27 Jan 2016) A rocket from the earth to the moon
Slides: 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) 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 (138.0 K, pdf, 15 Feb 2017) 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 this
Slides: Numerical methods (138.0 K, pdf, 15 Feb 2017) 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). ==> Left off here on Friday, will continue on Monday 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:
Slides: Numerical methods (138.0 K, pdf, 15 Feb 2017) 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 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)