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551614:Advanced Mathematics for Mechatronics
Numerical solution for ODEsSchool of Mechanical Engineering
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Prescribed text :
Numerical Method for Engineering, Seventh Edition, Steven C.Chapra, Raymond P. Canale.,McGraw Hill 2014
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Recommended reading :
Numerical Methods for Engineers and Scientists, Amos
Gilat and Vish Subramaniam, Wiley,2008
Numerical Methods Using MATLAB, John.H.Methews.,
Kurtis D.Fink.,Prentice Hall , Fourth Edition,2004
ระเบียบวธีิเชิงตวัเลขในงานวศิวกรรม, ปราโมทย์ เดชะ
อาํไพ., สาํนกัพิมพแ์ห่งจุฬาลงกรณ์มหาวิทยาลยั พ.ศ.2556
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Course Description
Truncation errors and the Taylor seriesNumerical Solutions for ODEsFinite difference methodOptimization
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Errors in Numerical Solutions
Truncation ErrorsRound-Off ErrorsTotal Error• Local Error• Global Error
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Truncation Errors :The Taylor series
( ) ( ) ( ) ( ) +
∆++
∆+
∆+=+
iii xn
nn
xxii dx
fdnx
dxfdx
dxdfxxfxf
!!2!1 2
22
1
( ) ( )ii xfxf ≅+1
Zero-order approximation
First-order approximation
( ) ( )ix
ii dxdfxxfxf
!11∆
+=+
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The Taylor series
( ) 2.125.05.015.01.0 234 +−−−−= xxxxxfapproximation
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The Remainder for the Taylor Series Expansion
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Truncation Errors: Example3 5 7 9
sin( )3! 5! 7! 9!x x x xx x= − + − + +
sin 0.52359886 6π π = =
sin 0.56π =
Taylor’s series expansion:
The exact value:
The Zero-order approximation;
The truncation error: 0.5 0.5235988 0.0235988TRE = − = −
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Truncation Errors: Example3 5 7 9
sin( )3! 5! 7! 9!x x x xx x= − + − + +
6sin 0.49967426 6 3!π π π = − =
sin 0.56π =
Taylor’s series expansion:
The exact value:
The first-order approximation;
The truncation error: 0.5 0.4996742 0.0003258TRE = − =
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Round-Off Errors
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Total Error
TrueError TrueSolution NumericalSolution= −
The true error:
The true relative error:
TrueRelativeError TrueSolution NumericalSolutionTrueSolution
−=
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Total Error
The Global Error is the total discrepancy due to past as well as present stepsThe Local Error refers to the error incurred over a single step. It is calculated with a Taylor series expansion.
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The Global ErrorThe relative global error:
%TheGlobalError 100TrueSolution NumericalSolutionTrueSolution
−= ×
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The Total Error: Example
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The Total Error: Example
3.21875 5.25%TheGlobalError 100 63.13.21875
−= × =
The percent relative global error (Step 1);
The percent relative global error (Step 2);
3 5.875%TheGlobalError 100 95.83
−= × =
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The Total Error: Example
3 22 12 20 8.5y x x x= − + − +
( ) ( ) ( )2 3 42 3 4
2 3 42! 3! 4!i i i
localx x x
x x xd f d f d fEdx dx dx
∆ ∆ ∆= + +
Exact estimates of the errors in Euler’s method
2
4
4
6 24 2012 24
12
y x xy x
d fdx
= − + −= − +
= −
Problem statement:
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The Total Error: Example
22
2
33
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6(0) 24(0) 20 (0.5) 2.52
12(0) 24 (0.5) 0.53 212 (0.5) 0.03125
4 3 2
l
l
l
E
E
E
− + −= = −
− += =
×−
= = −× ×
2 3 4 2.03125localtotal l l lE E E E= + + = −
The percent relative local error (Step 1);
%TheLocalError= 100 63.1localtotalE
TrueSolution× =
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The Total Error: Example
22
2
33
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6(0.5) 24(0.5) 20 (0.5) 1.18752
12(0.5) 24 (0.5) 0.3753 2
12 (0.5) 0.031254 3 2
l
l
l
E
E
E
− + −= = −
− += =
×−
= = −× ×
2 3 4 0.84375localtotal l l lE E E E= + + = −
The percent relative local error (Step 2);
0.84375%TheLocalError= 100 28.13
−× =
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Numerical Solutions for ODEs
Introduction to Ordinary Differential Equations (ODE)
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Study Objectives
• Solve Ordinary differential equation (ODE) problems.
• Appreciate the importance of numerical method in solving ODE.
• Assess the reliability of the different techniques.
• Select the appropriate method for any particular problem.
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Computer Objectives
• Develop programs to solve ODE.• Use software packages to find the solution of
ODE
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Learning Objectives of Lesson 1
Recall basic definitions of ODE,• order, • linearity• initial conditions, • solution,
Classify ODE based on( order, linearity, conditions)
Classify the solution methods
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Partial Derivatives
u is a function of more than one
independent variable
Ordinary Derivatives
v is a function of one independent variable
Derivatives
Derivatives
dtdv
yu∂∂
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Partial Differential Equations
involve one or more partial derivatives of unknown functions
Ordinary Differential Equations
involve one or more
Ordinary derivatives of
unknown functions
Differential EquationsDifferentialEquations
162
2
=+ tvdt
vd 02
2
2
2
=∂∂
−∂∂
xu
yu
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Ordinary Differential Equations
)cos()(2)(5)(
)()(:
2
2
ttxdt
tdxdt
txd
etvdt
tdvExamples
t
=+−
=−
Ordinary Differential Equations (ODE) involve one or more ordinary derivatives of unknown functions with respect to one independent variable
x(t): unknown function
t: independent variable
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Order of a differential equation
1)(2)()(
)cos()(2)(5)(
)()(:
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2
2
2
2
=+−
=+−
=−
txdt
tdxdt
txd
ttxdt
tdxdt
txd
etxdt
tdxExamples
t
The order of an ordinary differential equations is the order of the highest order derivative
Second order ODE
First order ODE
Second order ODE
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A solution to a differential equation is a function that satisfies the equation.
Solution of a differential equation
0)()(:
=+ txdt
tdxExample
0)()(
)(:Proof
)(
=+−=+
−=
=
−−
−
−
tt
t
t
eetxdt
tdx
edt
tdx
etxSolution
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Linear ODE
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2
:
( )( )
( ) ( )( ) 1
t
Examples
dx tx t e
dt
d x t dx tx t
dt dt
− =
− + =
An ODE is linear ifThe unknown function and its derivatives appear to power one
No product of the unknown function and/or its derivatives
Linear ODE
Non-linear ODE
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Nonlinear ODE
1)()()(
2)()(5)(
1))(cos()(:ODEnonlinear of Examples
2
2
2
2
=+−
=−
=−
txdt
tdxdt
txd
txdt
tdxdt
txd
txdt
tdx
An ODE is linear ifThe unknown function and its derivatives appear to power one
No product of the unknown function and/or its derivatives
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Uniqueness of a solution
byay
xydt
xyd
==
=+
)0()0(
0)(4)(2
2
In order to uniquely specify a solution to an n th order differential equation we need n conditions
Second order ODE
Two conditions are needed to uniquely specify the solution
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Auxiliary conditions
Boundary Conditions
• The conditions are not at one point of the independent variable
Initial Conditions
all conditions are at one point of the independent variable
auxiliary conditions
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Boundary-Value and Initial value Problems
Boundary-Value Problems
• The auxiliary conditions are not at one point of the independent variable
• More difficult to solve than initial value problem
5.1)2(,1)0(2 2
===++ −
xxexxx t
Initial-Value Problems
The auxiliary conditions are at one point of the independent variable
5.2)0(,1)0(2 2
===++ −
xxexxx t
same different
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Classification of ODEODE can be classified in different ways
Order• First order ODE• Second order ODE• Nth order ODE
Linearity• Linear ODE• Nonlinear ODE
Auxiliary conditions• Initial value problems• Boundary value problems
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Analytical Solutions
Analytical Solutions to ODE are available for linear ODE and special classes of nonlinear differential equations.
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Numerical Solutions
Numerical method are used to obtain a graph or a table of the unknown functionMost of the Numerical methods used to solve ODE are based directly (or indirectly) on truncated Taylor series expansion
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Classification of the MethodsNumerical Methods
for solving ODE
Multiple-Step Methods
Estimates of the solution at a particular step are based on information on more than one step
One-Step Methods
Estimates of the solution at a particular step are entirely based on information on the previous step
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More Lessons in this unit
Taylor series methodsMidpoint and Heun’s methodRunge-Kutta methodsMultiple step MethodsSolving systems of ODEBoundary value Problems
39
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The sequence of events in the application of
ODEs for engineering problem solving
Differential equations
Analytical methods• Exact solution
Numerical methods Approximate solution
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5.82012215.81045.0
23
234
+−+−=
++−+−=
xxxyxxxxy
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5.82012215.81045.0
23
234
+−+−=
++−+−=
xxxyxxxxy
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