ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 1 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engr/Math/Physics 25 Chp9: ODE’s Numerical Solns

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 2 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Learning Goals  List Characteristics of Linear, MultiOrder, NonHomgeneous Ordinary Differential Equations (ODEs)  Understand the “Finite-Difference” concept that is Basis for All Numerical ODE Solvers  Use MATLAB to determine Numerical Solutions to Ordinary Differential Equations (ODEs)

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 3 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Differential Equations  Ordinary Diff Eqn  Partial Diff Eqn  PDE’s Not Covered in ENGR25 Discussed in More Detail in ENGR45  Examining the ODE, Note that it is: LINEAR → y, dy/dt, d 2 y/dt 2 all raised to Power of 1 2 nd ORDER → Highest Derivative is 2 NONhomogenous → RHS  0; –i.e., y(t) has a FORCING Fcn f(t) has CONSTANT CoEfficients

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 4 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Solving 1 st Order ODEs - 1  Given the Simple ODE with No Zero Order (i.e., “y”) term An INITIAL Condition  Can Solve by SEPARATING the VARIABLES  Now use the IC in the Limits of Integ.  AND Integrating Both Sides Note the use of DUMMY VARIABLES of INTEGRATION α and β

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 5 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Solving 1 st Order ODEs - 2  Integrating  Separating The Variables sometimes works for 1 st Order Eqns  The Function on the RHS of the 1 st Order ODE is the FORCING Function Function only of t Can be a CONSTANT

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 6 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Solving 1 st Order ODEs - 3  Consider the 1 st Order ODE with a “Zero” Order Term and a Forcing Fcn  This is the GENERAL Eqn  By Theorems of Linear ODEs Let  y p (t)  ANY Solution to the General ODE Called the “Particular” Solution  y c (t)  The Solution to the General Eqn with f(t) = 0 The “Complementary Solution” or the “Natural” (UnForced) Response  i.e., y c is the Soln to the “Homogenous” Eqn

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 7 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 1 st Order Response Eqns  Given y p and y c then the TOTAL Solution to the ODE  Consider the Case Where the Forcing Function is a Constant f(t) = A  Now Solve the ODE in Two Parts for y p & y c  For the Particular Soln, Notice that a CONSTANT Fits the Eqn:

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 8 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 1 st Order Response Eqns cont  Sub Into the General (Particular) Eqn y p and dy p /dt  Next, Divide the Homogeneous Eqn by  ·y c to yield (on whtbd)  Next Separate the Variables & Integrate  Recognize LHS as a Natural Log; so  Next Take “e” to The Power of the LHS & RHS

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 9 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 1 st Order Response Eqns cont  Then   is called the TIME CONSTANT  Thus the Solution for a Constant Forcing Fcn  For This Solution Examine Extreme Cases t = 0 t →   The Latter Case is Called the Steady-State Response All Time-Dependent Behavior has dissipated

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 10 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Higher Order, Linear ODE’s  The GENERAL Higher Order ODE Where the derivative CoEfficients, the g i (t), may be constants, including Zero  IF an analytical Solution Exists Then use the same “linear” methodology as for the First Order Eqn

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 11 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Higher Order, Linear ODE’s  Where as Before y c (t) is the solution to Complementary Eqn y p (t) is ANY single solution to the FULL, Orginal Eqn that includes the Force-Fcn  e.g.:

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 12 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods For More Info On Higher Order  Hi-Order ODEs usually do NOT have Analytical solns, except in special cases Consider a 2 nd order, Linear, NonHomogenous, Constant CoEfficient ODE of the form –ODE’s with these SPECIFIC Characteristics can ALWAYS be Solved Analytically  See APPENDIX for more details  These Methods used in ENGR43

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 13 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Numerical ODE Solutions  Today we’ll do some MTH25  We’ll “look under the hood” of NUMERICAL Solutions to ODE’s  The BASIC Game- Plan for even the most Sophisticated Solvers: Given a STARTING POINT, y(0) Use ODE to find the slope dy/dt at t=0 ESTIMATE y 1 as

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 14 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Numerical Solution - 1  Notation  Exact Numerical Method (impossible to achieve) by Forward Steps  Now Consider slope y n+1 tntn ynyn t n+1 t tt

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 15 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Numerical Solution - 2  The diagram at Left shows that the relationship between y n, y n+1 and the CHORD slope y n+1 tntn ynyn t n+1 t tt  The problem with this formula is we canNOT calculate the CHORD slope exactly We Know Only Δt & y n, but NOT the NEXT Step y n+1 The Analyst Chooses Δt Chord Slope Tangent Slope

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 16 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Numerical Solution -3  However, we can calculate the TANGENT slope at any point FROM the differential equation itself  The Basic Concept for all numerical methods for solving ODE’s is to use the TANGENT slope, available from the R.H.S. of the ODE, to approximate the chord slope  Recognize dy/dt as the Tangent Slope

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 17 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Euler Method – 1 st Order  Solve 1 st Order ODE with I.C.  ReArranging  Use: [Chord Slope]  [Tangent Slope at start of time step]  Then Start the “Forward March” with Initial Conditions

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 18 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 19 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Euler Example  Consider 1 st Order ODE with I.C.  Use The Euler Forward-Step Reln  See Next Slide for the 1 st Nine Steps For Δt = 0.1  But from ODE  So In This Example:

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 20 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Euler Exmpl Calc ntntn ynyn f n = – y n +1 y n+1 = y n +  t∙f n Plot Slope

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 21 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Euler vs Analytical  The Analytical Solution

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 22 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Analytical Soln  Let u = −y+1  Then  Sub for y & dy in ODE  Separate Variables  Integrate Both Sides  Recognize LHS as Natural Log  Raise “e” to the power of both sides

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 23 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Analytical Soln  And  Thus Soln u(t)  Sub u = 1−y  Now use IC  The Analytical Soln

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 24 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Predictor-Corrector - 1  Again Solve 1 st Order ODE with I.C.  Mathematically  This Time Let: Chord slope  average of tangent slopes at start and END of time step  BUT, we do NOT know y n+1 and it appears on the BOTH sides of the Eqn... Avg of the Tangent Slopes at (t n,y n ) & (t n+1,y n+1 )

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 25 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Predictor-Corrector - 2  Use Two Steps to estimate y n+1  First → PREDICT* Use standard Euler Method to Predict  Then Correct by using y* in the Avg Calc  Then Start the “Forward March” with the Initial Conditions

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 26 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Predictor-Corrector Example  Solve ODE with IC  The Corrector step  The next Step Eqn for dy/dt = f(t,y)= –y+1  Numerical Results on Next Slide

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 27 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Predictor-Corrector Example n Slope

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 28 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Predictor-Corrector  Greatly Improved Accuracy

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 29 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods ODE Example:  Euler Solution with ∆t = 0.25, y(t=0) = 37  The Solution Table

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 30 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Compare Euler vs. ODE45 Euler SolutionODE45 Solution Euler is Much LESS accurate

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 31 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Compare Again with ∆t = Euler SolutionODE45 Solution Smaller ∆T greatly improves Result

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 32 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods MatLAB Code for Euler % Bruce Mayer, PE % ENGR25 * 04Jan11 % file = Euler_ODE_Numerical_Example_1201.m % y0= 37; delt = 0.25; t= [0:delt:10]; n = length(t); yp(1) = y0; % vector/array indices MUST start at 1 tp(1) = 0; for k = 1:(n-1) % fence-post adjustment to start at 0 dydt = 3.9*cos(4.2*yp(k))^2-log(5.1*tp(k)+6); dydtp(k) = dydt % keep track of tangent slope tp(k+1) = tp(k) + delt; dely = delt*dydt delyp(k) = dely yp(k+1) = yp(k) + dely; end plot(tp,yp, 'LineWidth', 3), grid, xlabel('t'),ylabel('y(t) by Euler'),... title('Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)')

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 33 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods MatLAB Command Window for ODE45 >> dydtfcn 3.9*(cos(4.2*yf))^2-log(5.1*tf+6); >> [T,Y] = ode45(dydtfcn,[0 10],[37]); >> plot(T,Y, 'LineWidth', 3), grid, xlabel('T by ODE45'), ylabel('Y by ODE45')

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 34 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods All Done for Today Carl Runge Carl David Tolmé Runge Born: 1856 in Bremen, Germany Died: 1927 in Göttingen, Germany

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 35 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engr/Math/Physics 25 Appendix

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 36 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order Linear Equation  Need Solutions to the 2 nd Order ODE  As Before The Solution Should Take This form  If the Forcing Fcn is a Constant, A, Then Discern a Particular Soln  Verify y p  Where y p  Particular Solution y c  Complementary Solution  For Any const Forcing Fcn, f(t) = A 

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 37 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods The Complementary Solution  The Complementary Solution Satisfies the HOMOGENOUS Eqn  Look for Solution of this type  Need y c So That the “0 th ”, 1 st & 2 nd Derivatives Have the SAME FORM so they will CANCEL (i.e., Divide-Out) in the Homogeneous Eqn  Sub Assumed Solution (y = Ge st ) into the Homogenous Eqn  Canceling Ge st  The Above is Called the Characteristic Equation

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 38 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Complementary Solution cont  A value for “s” That SATISFIES the CHARACTERISTIC Eqn ensures that Ge st is a SOLUTION to the Homogeneous Eqn  Recall the Homogeneous Eqn  If Ge st is indeed a Solution Then Need The Characteristic Eqn  Solve For s by Quadratic Eqn  In terms of the Discriminant γ

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 39 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Complementary Solution cont.2  Given the “Roots” of the Homogeneous Eqn  In the Unstable case the response will grow exponentially toward ∞ This is not terribly interesting  If the Solution is Stable, need to Consider three Sets of values for s based on the sign of γ 1. γ > 0 → s 1, s 2 REAL and UNequal roots 2. γ = 0 → s 1 = s 2 = s; ONE REAL root 3. γ < 0 → Two roots as COMPLEX CONJUGATES  Can Generate STABLE and UNstable Responses Stable UNstable

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 40 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Complementary Soln Cases 1&2  For the Linear, 2 nd Order, Constant Coeff, Homogenous Eqn  By the Methods of MTH4 & ENGR43 Find Solutions to the ODE by discriminant case: 1.Real & Unequal Roots (Stable for Neg Roots) 2.Single Real Root (Stable for Neg Root)

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 41 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Complementary Soln Case - 3  Using the Euler Identity: And Collecting Terms find 3.Complex Conjugate Roots of the form: s = a ± jω (Stable for Neg a) a, ω, B 1, B 2 all Constants (a & ω are KNOWN)

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 42 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order Solution  For the Linear, 2 nd Order, Constant Coeff, Homogenous Eqn  Can Find Solution based Upon the nature of the Roots of the Characteristic Eqn  To Find the Values of the Constants Need TWO Initial Conditions (ICs) The ZERO Order IC The 1 st Order IC

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 43 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Properly Apply Initial conditions  The IC’s Apply ONLY to the TOTAL Solution  Many times It’s EASY to forget to add the PARTICULAR solution BEFORE applying the IC’s Do NOT neglect y p (t) prior to IC’s

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 44 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE Example - 1  The Homogeneous Equation  And the IC’s  Then the Soln Form Given Real & UnEqual Roots  The Characteristic Eqn and Roots

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 45 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE Example - 2  From the Zero Order IC  To Use the 1 st Order IC need to take Derivative  Now Have 2 Eqns for A 1 & A 2  Then at t = 0  Solve w/ MATLAB BackDivision

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 46 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2nd Order ODE Example - 3  MATLAB session  Or  The Response Curve >> C = [1,1; -3,-6]; >> b = [0.5; -8.5]; >> A = C\b A = >> A_6 = 6*A A_6 = Be sure to check for correct IC’s  Starting-Value & Slope

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 47 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SuperSUMMARY-1  See Appendix for FULL Summary  Find ANY Particular Solution to the ODE, y p (often a CONSTANT)  Homogenize ODE → set RHS = 0  Assume y c = Ge st ; Sub into ODE  Find Characteristic Eqn for y c ; a 2 nd order Polynomial

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 48 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SuperSUMMARY-2  Find Roots to Char Eqn Using Quadratic Formula (or MATLAB)  Examine Nature of Roots to Reveal form of the Eqn for the Complementary Solution: Real & Unequal Roots → y c = Decaying Constants Real & Equal Roots → y c = Decaying Line Complex Roots → y c = Decaying Sinusoid

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 49 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SuperSUMMARY-3  Then the TOTAL Solution: y = y c + y p  All TOTAL Solutions for y(t) include 2 Unknown Constants  Use the Two INITIAL Conditions to generate two Eqns for the 2 unknowns  Solve the Total Solution for the 2 Unknowns to Complete the Solution Process

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 50 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SUMMARY-1  If NonHomogeneous Then find ANY Particular Solution  Next HOMOGENIZE the ODE  The Soln to the Homog. Eqn Produces the Complementary Solution, y c  Assume y c take this form

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 51 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SUMMARY-2  Subbing yc = Ae s t into the Homog. Eqn yields the Characteristic Eqn  Find the TWO roots that satisfy the Char Eqn by Quadratic Formula  Check FORM of Roots  If s 1 & s 2 → REAL & UNequal Decaying Contant(s)

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 52 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SUMMARY-3  If s 1 & s 2 → REAL & Equal, then s 1 = s 2 =s Decaying Line  If s 1 & s 2 → Complex Conjugates then Decaying Sinusoid  Add Particlular & Complementary Solutions to yield the Complete Solution

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 53 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods 2 nd Order ODE SUMMARY-4  To Find Constant Sets: (G 1, G 2 ), (m, b), (B 1, B 2 ) Take for COMPLETE solution  Find Number-Values for the constants to complete the solution process Yields 2 eqns in 2 for the 2 Unknown Constants

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 54 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Finite Difference Methods - 1  Another way of thinking about numerical methods is in terms of finite differences.  Use the Approximation  And From the Differential Eqn  From these two equations obtain:  Recognize as the Euler Method

ENGR-25_Lec-23_ODEs_Euler_Numerical.pptx 55 Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods Finite Difference Methods - 2  Could make More Accurate by Approximating dy/dt at the Half-Step as the average of the end pts  Recognize as the Predictor-Corrector Method  Then Again Use the ODE to Obtain