ME451 Kinematics and Dynamics

of Machine Systems

Numerical Integration.Stiffness: Implicit Methods.

October 30, 2013

Radu SerbanUniversity of Wisconsin-Madison

2

Before we get started… Last Time:

Recovering constraint reaction forces Started discussion on numerical integration

Today Stiff differential equations Implicit numerical integration formulas

Assignments: Homework 9 – 6.3.3, 6.4.1 – due November 4 (12:00pm) Matlab 7 – due tonight, October 30, Learn@UW (11:59pm) Project 1 – due Wednesday, November 6, Learn@UW (11:59pm)

Your simEngine2D can now perform complete Kinematic analysis! Test the provided visualization function (available on the SBEL course webpage)

Miscellaneous Draft proposals for the Final Project due on Friday, November 1 Midterm 2 – Wednesday, November 6, 12:00pm in ME 1143 Review session – Monday, November 4, 6:30pm in ME 1143

Numerical Integration

4

Basic Concept

IVP

In general, all we can hope for is approximating the solution at a sequence of discrete points in time Uniform grid (constant step integration) Adaptive grid (variable step integration)

The numerical solution is then the sequence of approximations

Basic idea: somehow turn the differential problem into an algebraic problem (approximate the derivatives)

5

Simplest method: Forward Euler

Starting from the IVP

Use the simplest approximation to the derivative

Rewrite the above as

and use ODE to obtain

Forward Euler Methodwith constant step-size

6

FE: Geometrical Interpretation

IVP

Forward Euler integration formula

7

FE: Example

Solve the IVP

using Forward Euler (FE) with a step-size

Compare to the exact solution

8

Forward Euler: Effect of Step-Size

0 5 10 15 20 25 30 35 40 45 50-0.06

-0.04

-0.02

0

0.02

0.04

0.06

h = 0.001h = 0.01h = 0.1

% IVP (RHS + IC)f = @(t,y) -0.1*y + sin(t);y0 = 0;tend = 50; % Analytical solutiony_an = @(t) (0.1*sin(t) - cos(t) + exp(-0.1*t)) / (1+0.1^2); % Loop over the various step-size values and plot errorscolors = [[0, 0.4, 0]; [1, 0.5, 0]; [0.6, 0, 0]];Figure, hold on, box onh = [0.001 0.01 0.1];for ih = 1:length(h) tspan = 0:h(ih):tend; y = zeros(size(tspan)); err = zeros(size(tspan)); y(1) = y0; err(1) = 0; for i = 2:length(tspan) y(i) = y(i-1) + h(ih) * f(tspan(i-1), y(i-1)); err(i) = y(i) - y_an(tspan(i)); end plot(tspan, err, 'color', colors(ih,:));endlegend('h = 0.001', 'h = 0.01', 'h = 0.1');

FE errors for different values of the step-size

9

0 5 10 15 20 25 30 35 40 45 50-6

-4

-2

0

2

4

6

h = 0.1h = 1h = 5

FE: Effect of Step-Size

% IVP (RHS + IC)f = @(t,y) -0.1*y + sin(t);y0 = 0;tend = 50; % Loop over the various step-size values and plot errorscolors = [[0, 0.4, 0]; [1, 0.5, 0]; [0.6, 0, 0]];Figure, hold on, box onh = [0.1 1.0 5.0];for ih = 1:length(h) tspan = 0:h(ih):tend; y = zeros(size(tspan)); y(1) = y0; for i = 2:length(tspan) y(i) = y(i-1) + h(ih) * f(tspan(i-1), y(i-1)); end plot(tspan,y, 'color', colors(ih,:))endlegend('h = 0.1', 'h = 1', 'h = 5');

FE accuracy at different values of the step-size

Stiff Differential Equations

11

A Simple IVP Example

Consider the IVP

whose analytical solution is

5 integration steps with Forward Euler formula, , , Compare the global error (difference between analytical and numerical solution)

Error for

0 0.01873075307798 0.03032004603564 0.03681163609403 0.03972896411722 0.04019944117144

Error for

0 0.36787944117144 0.13533528323661 0.04978706836786 0.01831563888873 0.00673794699909

Error for

0 2.04978706836786 -3.99752124782333 8.00012340980409 -15.99999385578765 32.00000030590232

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A Simple IVP Example

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-10000

-5000

0

5000

Forward EulerAnalytical Solution

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Stiff Differential Equations

Problems for which explicit integration methods (such as Forward Euler) do not work well Other explicit formulas: Runge-Kutta (RK4), DOPRI5, Adams-

Bashforth, etc.

Stiff problems require a different class of integration methods: implicit formulas The simplest implicit integration formula: Backward Euler (BE)

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BE: Geometrical Interpretation

IVP

Forward Euler integration formula

Backward Euler integration formula

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A Simple IVP Example

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-10000

-5000

0

5000

Forward Euler Analytical SolutionBackward Euler

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

16

Forward Euler vs. Backward Euler

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IVP

Also known as Gear method (DIFSUB – 1971)

Family of implicit linear multi-step formulas

BDF of 1st order:

BDF of 2nd order:

BDF of 3rd order:

BDF of 4th order:

BDF of 5th order:

Backward-Difference Formulas (BDF)

C.W. (Bill) Gearb. 1935

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Curtiss & Hirschfelder Example

Forward Euler

Backward Euler

C.F. Curtiss and J.O. Hirschfelder – “Integration of Stiff Equations”Proc. Nat. Acad. Sci, USA (1952)

Stability and Accuracy of Numerical Integrators

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Two Key Properties of a Numerical Integrator

Two properties are relevant when considering a numerical integrator for finding an approximation of the solution of an IVP

Stability

Accuracy

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Stability of a Numerical Integrator

The problem: How big can the integration step-size be without the numerical solution

blowing up?

Tough question, answered in a Numerical Analysis class

Different integration formulas, have different stability regions

We would like to use an integration formula with large stability region: Example: Backward Euler, BDF methods, Newmark, etc.

Why not always use these methods with large stability region? There is no free lunch: these methods are implicit methods that require the

solution of an algebraic problem at each step

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Accuracy of a Numerical Integrator

The problem: How accurate is the formula that we are using? If we decrease , how will the accuracy of the numerical solution improve? Tough question, answered in a Numerical Analysis class

Examples: Forward and Backward Euler: accuracy RK45: accuracy

Why not always use methods with high accuracy order? There is no free lunch: these methods usually have very small stability regions Therefore, they are limited to using very small values of

Implicit Integration of Stiff ODEs

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Implicit Integration and Nonlinear Systems[Examples]

Backward Euler (BE) formula:

Apply one step of BE for the following IVPs ()

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Implicit Integration and Nonlinear Systems[Example 1]

Backward Euler (BE) formula:

Apply one step of BE for the following IVP ()

This example shows that discretization with an implicit integration formula requires the solution of an algebraic equation

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Implicit Integration and Nonlinear Systems[Example 2]

Backward Euler (BE) formula:

Apply one step of BE for the following IVP ()

This example shows that the solution of the resulting algebraic equation raises additional problems (selecting one of multiple possible solutions)

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Implicit Integration and Nonlinear Systems[Example 3]

Backward Euler (BE) formula:

Apply one step of BE for the following IVP ()

This example shows that, in general, the resulting nonlinear algebraic problem can only be solved numerically.

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Implicit Integration and Nonlinear Systems[Examples] Backward Euler (BE) formula:

Apply one step of BE for the following IVPs ()

Upon discretization with an implicit integration formula (such as BE), the approximation to the solution of the DE at the new time is obtained by solving an algebraic equation

In general, the resulting (system of) nonlinear equation(s) can only be solved numerically (using for example the Newton-Raphson method)

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Implicit Integration: Conclusions

Stiff problems require the use of implicit integration methods

Because they have very good stability, implicit integration methods allow for step-sizes that could be orders of magnitude larger than those needed if using explicit integration

However, for most real-life IVPs, discretization using an implicit integration formula leads to another nasty problem:

To find the solution at the new time, we must solve a nonlinear algebraic problem

This brings back into the picture the Newton-Raphson method (and its variants) We have to deal with providing a good starting point (initial guess), computing the

matrix of partial derivatives, etc.

Solution of IVPs in Matlab

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General Call to a Matlab ODE Solver

IVP:

Example of using Matlab’s ode45:

Use odeset to create an ODE options structure to change default values for various solver parameters

[t,y] = ode45(odefun,tspan,y0,options)

function dydt = odefun(t,y)

[t0 tend]

y0

options = odeset(…)

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Example of IVP Solution with Matlab

>> [t,y]=ode45(‘rhs’,[0 20],[2;0]);>> plot(t, y);

0 2 4 6 8 10 12 14 16 18 20-3

-2

-1

0

1

2

3

function yd = rhs(t, y)yd = [y(2); (1-y(1)^2)*y(2)-y(1)];

Convert to 1st order ODE

Convert to vector form

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ODE Solvers in MATLAB

Solver Problem Type

Order of Accuracy When to use

ode45 Nonstiff Medium Most of the time. This should be the first solver tried

ode23 Nonstiff Low For problems with crude error tolerances or for solving moderately stiff problems.

ode113 Nonstiff Low to high For problems with stringent error tolerances or for solving computationally intensive problems

ode15s Stiff Low to medium If ods45 is slow because the problem is stiff

ode23s Stiff Low If using crude error tolerances to solve stiff systems and the mass matrix is constant

ode23t Moderately stiff Low For moderately stiff problems is you need a solution

without numerical damping

ode23tb Stiff Low If using crude error tolerances to solve stiff systems

Use ode45 for non-stiff problems and ode15s for stiff problems

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A Stiff Problem

% specify RHS functionrhs = @(t,y) [-1,-1;1,-5000]*y; % specify accuracy and turn on stats option options = odeset('RelTol',1e-6,'Stats','on'); % call the ode45 solverfprintf('\n---- ode45 solver statistics ----\n')tic, [~,~] = ode45(rhs, [0, 5] , [1; 1], options); toc % call the ode15s solverfprintf('\n---- ode15s solver statistics ----\n')tic, [~,~] = ode15s(rhs, [0, 5] , [1; 1], options); toc

---- ode45 solver statistics ----7557 successful steps504 failed attempts48367 function evaluationsElapsed time is 0.564820 seconds.

---- ode15s solver statistics ----139 successful steps3 failed attempts288 function evaluations1 partial derivatives27 LU decompositions284 solutions of linear systemsElapsed time is 0.188120 seconds.

On a stiff problem, an explicit solver is forced to take small steps to ensure stability an implicit solver can take much larger steps, which reduces overall time to

solution; but it must solve algebraic equations at each time step