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Chp9: ODE’sNumerical Solns

Bruce Mayer, PE

Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

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)

Ordinary Diff EqnDifferential Equations

- 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, d2y/dt2 all raised to Power of 1
- 2nd ORDER → Highest Derivative is 2
- NONhomogenous→ RHS 0;
- i.e., y(t) has a FORCING Fcn f(t)
- has CONSTANT CoEfficients

Given the Simple ODE with

No Zero Order (i.e., “y”) term

An INITIAL Condition

Solving 1st Order ODEs - 1- AND Integrating Both Sides

- Now use the IC in the Limits of Integ.

- Can Solve by SEPARATING the VARIABLES

- Note the use of DUMMY VARIABLES of INTEGRATIONα and β

IntegratingSolving 1st Order ODEs - 2

- Separating The Variables sometimes works for 1st Order Eqns
- The Function on the RHS of the 1st Order ODE is the FORCING Function
- Function only of t
- Can be a CONSTANT

Consider the 1st Order ODE with a “Zero” Order Term and a Forcing FcnSolving 1st Order ODEs - 3

- yp(t) ANY Solution to the General ODE
- Called the “Particular” Solution
- yc(t) The Solution to the General Eqn with f(t) = 0
- The “Complementary Solution” or the “Natural” (UnForced) Response
- i.e., yc is the Soln to the “Homogenous” Eqn

- This is the GENERAL Eqn
- By Theorems of Linear ODEs Let

Given yp and yc then the TOTAL Solution to the ODE1st Order Response Eqns

- Consider the Case Where the Forcing Function is a Constant
- f(t) = A
- Now Solve the ODE in Two Parts for yp & yc

- For the Particular Soln, Notice that a CONSTANT Fits the Eqn:

Sub Into the General (Particular) Eqn yp and dyp/dt 1st Order Response Eqns cont

- Next Separate the Variables & Integrate

- Recognize LHS as a Natural Log; so

- Next, Divide the Homogeneous Eqn by ·yc to yield (on whtbd)

- Next Take “e” to The Power of the LHS & RHS

Then 1st Order Response Eqns cont

- For This Solution Examine Extreme Cases
- t = 0
- t →

- is called the TIME CONSTANT
- Thus the Solution for a Constant Forcing Fcn

- The Latter Case is Called the Steady-State Response
- All Time-Dependent Behavior has dissipated

Higher Order, Linear ODE’s

- The GENERAL Higher Order ODE
- Where the derivative CoEfficients, the gi(t), may be constants, including Zero
- IF an analytical Solution Exists Then use the same “linear” methodology as for the First Order Eqn

Higher Order, Linear ODE’s

- Where as Before
- yc(t) is the solution to Complementary Eqn
- yp(t) is ANY single solution to the FULL, OrginalEqn that includes the Force-Fcn
- e.g.:

For More Info On Higher Order

- Hi-Order ODEs usually do NOT have Analytical solns, except in special cases
- Consider a 2nd 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

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 slopedy/dtat t=0
- ESTIMATE y1 as

NotationNumerical Solution - 1

- Exact Numerical Method (impossible to achieve) by Forward Steps

yn+1

yn

- Now Consider slope

t

tn

tn+1

Dt

Numerical Solution - 2

- The diagram at Left shows that the relationship between yn, yn+1 and the CHORD slope

yn+1

Tangent Slope

yn

Chord

Slope

- The problem with this formula is we canNOT calculate the CHORD slope exactly
- We Know Only Δt & yn, but NOT the NEXT Step yn+1

t

tn

tn+1

Dt

The AnalystChooses Δt

However, we can calculate the TANGENT slope at any point FROM the differential equation itselfNumerical Solution -3

- 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

Solve 1st Order ODE with I.C.Euler Method – 1st Order

- ReArranging

- Use: [Chord Slope] [Tangent Slope at start of time step]

- Then Start the “Forward March” with Initial Conditions

Consider 1st Order ODE with I.C.Euler Example

- But from ODE

- So In This Example:

- Use The Euler Forward-Step Reln

- See Next Slide for the 1st Nine Steps For Δt = 0.1

Euler vs Analytical

- The Analytical Solution

Let u = −y+1

Then

Analytical Soln- Integrate Both Sides

- Recognize LHS as Natural Log

- Sub for y & dy in ODE

- Raise “e” to the power of both sides

- Separate Variables

Again Solve 1st Order ODE with I.C.Predictor-Corrector - 1

- Mathematically

Avg of the Tangent Slopes at (tn,yn) & (tn+1,yn+1)

- This Time Let: Chord slope average of tangent slopes at start and END of time step

- BUT, we do NOT know yn+1 and it appears on the BOTH sides of the Eqn...

Use Two Steps to estimate yn+1

First → PREDICT*

Use standard Euler Method to Predict

Predictor-Corrector - 2- Then Correct by using y* in the AvgCalc

- Then Start the “Forward March” with the Initial Conditions

Solve ODE with ICPredictor-Corrector Example

- The Corrector step

- The next Step Eqn for dy/dt = f(t,y)= –y+1

- Numerical Results on Next Slide

Predictor-Corrector

- Greatly Improved Accuracy

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)')

MatLAB Command Window forODE45

>> dydtfcn = @(tf,yf) 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')

All Done for Today

CarlRunge

Carl David Tolmé Runge

Born: 1856 in Bremen, Germany

Died: 1927 in Göttingen, Germany

Appendix

Time For

Live Demo

Bruce Mayer, PE

Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

Need Solutions to the 2nd Order ODE2nd Order Linear Equation

- If the Forcing Fcn is a Constant, A, Then Discern a Particular Soln

- As Before The Solution Should Take This form

- Verify yp

- Where
- yp Particular Solution
- yc Complementary Solution

- For Any const Forcing Fcn, f(t) = A

The Complementary Solution Satisfies the HOMOGENOUS EqnThe Complementary Solution

- Look for Solution of this type

- Sub Assumed Solution (y = Gest) into the Homogenous Eqn

- Need yc So That the “0th”, 1st & 2nd Derivatives Have the SAME FORM so they will CANCEL (i.e., Divide-Out) in the Homogeneous Eqn

- Canceling Gest

- The Above is Called the Characteristic Equation

A value for “s” That SATISFIES the CHARACTERISTIC Eqn ensures that Gest is a SOLUTION to the Homogeneous Eqn

Recall the Homogeneous Eqn

Complementary Solution cont- The Characteristic Eqn
- Solve For s by Quadratic Eqn

- In terms of the Discriminant γ

- If Gest is indeed a Solution Then Need

Given the “Roots” of the Homogeneous EqnComplementary Solution cont.2

- 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 → s1, s2 REAL and UNequal roots
- 2. γ = 0 → s1 = s2 = s; ONE REAL root
- 3. γ < 0 → Two roots as COMPLEX CONJUGATES

- Can Generate STABLE and UNstable Responses
- Stable

- UNstable

For the Linear, 2nd Order, Constant Coeff, Homogenous EqnComplementary Soln Cases 1&2

- By the Methods of MTH4 & ENGR43 Find Solutions to the ODE by discriminant case:

Real & Unequal Roots (Stable for Neg Roots)

Single Real Root (Stable for Neg Root)

Complementary Soln Case - 3

Complex Conjugate Roots of the form: s = a ± jω (Stable for Neg a)

- Using the Euler Identity:And Collecting Terms find

- a, ω, B1, B2 all Constants(a & ω are KNOWN)

For the Linear, 2nd Order, Constant Coeff, Homogenous Eqn2nd Order Solution

- To Find the Values of the Constants Need TWO Initial Conditions (ICs)
- The ZERO Order IC

- Can Find Solution based Upon the nature of the Roots of the Characteristic Eqn

- The 1st Order IC

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 yp(t) prior to IC’s

The Homogeneous Equation2nd Order ODE Example - 1

- The Characteristic Eqn and Roots

- And the IC’s

- Then the Soln Form Given Real & UnEqual Roots

From the Zero Order IC2nd Order ODE Example - 2

- Then at t = 0

- Now Have 2 Eqns for A1 & A2

- To Use the 1st Order IC need to take Derivative

- Solve w/ MATLABBackDivision

MATLAB session2nd Order ODE Example - 3

- The Response Curve

>> C = [1,1; -3,-6];

>> b = [0.5; -8.5];

>> A = C\b

A =

-1.8333

2.3333

>> A_6 = 6*A

A_6 =

-11.0000

14.0000

Be sure to check for correct IC’s Starting-Value & Slope

- Or

2nd Order ODE SuperSUMMARY-1

- See Appendix for FULL Summary
- Find ANY Particular Solution to the ODE, yp (often a CONSTANT)
- Homogenize ODE → set RHS = 0
- Assume yc = Gest; Sub into ODE
- Find Characteristic Eqn for yc; a 2nd order Polynomial

2nd 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 → yc = Decaying Constants
- Real & Equal Roots → yc = Decaying Line
- Complex Roots → yc = Decaying Sinusoid

2nd Order ODE SuperSUMMARY-3

- Then the TOTAL Solution: y = yc + yp
- 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 TotalSolution for the 2 Unknowns to Complete the Solution Process

If NonHomogeneous Then find ANY Particular Solution2nd Order ODE SUMMARY-1

- The Soln to the Homog. Eqn Produces the Complementary Solution, yc
- Assume yc take this form

- Next HOMOGENIZE the ODE

Subbing yc = Aest into the Homog. Eqn yields the Characteristic Eqn2nd Order ODE SUMMARY-2

- Check FORM of Roots
- If s1 & s2→ REAL & UNequal

- Find the TWO roots that satisfy the Char Eqn by Quadratic Formula

- Decaying Contant(s)

If s1 & s2→ REAL & Equal, then s1 = s2 =s2nd Order ODE SUMMARY-3

- Decaying Sinusoid
- Add Particlular & Complementary Solutions to yield the Complete Solution

- Decaying Line
- If s1 & s2→ Complex Conjugates then

To Find Constant Sets: (G1, G2), (m, b), (B1, B2) Take for COMPLETE solution2nd Order ODE SUMMARY-4

- Find Number-Values for the constants to complete the solution process

- Yields 2 eqns in 2 for the 2 Unknown Constants

Another way of thinking about numerical methods is in terms of finite differences.

Use the Approximation

Finite Difference Methods - 1- And From the Differential Eqn

- From these two equations obtain:

- Recognize as the Euler Method

Could make More Accurate by Approximating dy/dt at the Half-Step as the average of the end ptsFinite Difference Methods - 2

- Then Again Use the ODE to Obtain

- Recognize as the Predictor-Corrector Method

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