numerical methodslecture 9: numerical solution of ode’s math259 numerical analysis. 20. the final...

Numerical Methods

Mustafa YILMAZSchool of Mechanical Engineering

University of Marmara

Lecture 9 – Numerical Solution of ODE’s

MATH259 Numerical AnalysisLecture Numerical Solution of ODE s9: ’ 2

Example: the falling parachutist

v: dependent variable (function)t: independent variable

Basics

dtdx

=y

vmcg

dtdv

−=

0kxdtdxc

dtxdm 2

2=++

Differential equation: an equation composed of an unknown function and its derivatives

Ordinary differential equation: if there is only one independent variable

Partial differential equation: if there are two or more independent variables

Order of ODE: the order of the highest derivative in the equation

Example: second order ODE

Reduction of order: higher-order ODE can be reduced to a system of 1st-order ODE

0kxdtdxc

dtxdm 2

2=++

=++

=

0kxcydtdym

dtdxy


Why Study Differential Equations?

0sinθlg

dtθd2

2=+

θ of change of rate : dtdθ

Many physical phenomena are best formulated mathematically in terms of their rate of change (which is derivative)!

Example: motion of a swinging pendulum

( )θ of change of rate of change of rate dtdθ of change of rate :

dtθd2

2


ODE and Engineering Practice

Fundamental lawsEmpirical observations ODE SolutionsAnalytical/numerical

methods

Sequence of the application of ODEs for engineering problems

Independent variable: spatial and temporal


Ordinary Differential Equations

0,

0

2/1

2

2

2

=++=

=++

−=

Kxcydtdymy

dtdx

Kxdtdxc

dtxdm

vm

Kgdtdv

ODEs order 1st of Set

Order

damper mass Spring

is derivative highest since Order

brick falling Exampleρ


Noncomputer Methods for Solving ODEs

vmcg

dtdv

−= dtvmcgv ∫

−=

0=+ sinθlg

dtθd2

2

One particular useful analytical integration technique: linearization

an(x)y(n) + an-1(x)y(n-1) + … +a1(x)y’+ a0(x)y = f(x)

Differential equation Integration SolutionconversionAnalytical integration

techniques

This can be solved analytically!

0θlg

dtd

2

2=+

θ

( ) ( )( )tc/me1c

gmtv −−=

Sinθ ≈ θ if θ is small

(non-linear) (linear)


Solution by Integration15810450 234 ++−+−= x.xxx.y

differentiation

For an nth-order ODE, n conditions are required to obtain a unique solution

x.xxxdxdy 5820122 23 +−+−=

( )dx.xxxy ∫ +−+−= 5820122 23

Cx.xxx.y ++−+−= 5810450 234

Multiple solutions

integration

Initial-value problem

All n conditions are specified at a same value of x n conditions occur at different x

Boundary-value problem


1st Order ODEs

• We are going to look at first order (and first degree) ODEs

• These take the general form:

• Some examples:

),( yxfdxdy

=

xedxdy

xydxdy

ydxdy

yxdxdy

x 4

7

2

)2sin(3

+=

+=

=

+=


1st Order ODEs

• The equation is not enough to give us a specific curve– whether we use numerical or analytical methods

• e.g. has the solution

• e.g. has the solution

and we need to be able to also work out the value of C or of A to get the full solution

),( yxfdxdy

=

xdxdy

= Cxy +=2

2

ydxdy 2=

xAey 2=


So how do we use NM?

• We see the ‘true’ solution is obtained by integrating

– which provides a constant of integration necessary for a full solution

• We need boundary conditions to be able to fully solve the

problem

• There are two broad types (for general DEs)

• Initial value problems

• Boundary value problems


Boundary Conditions

– Initial value problems

• We have one known condition for the problem (for

example, at x=0, y=2)

• This is what we are going to look at

– Boundary value problems

• We know values at two points, for example a metal bar

with two different temperatures at the two ends


Consider

L=0.1HE=5V

R=10 Ω

1st order ODE

Solution is:

Steady State Soln. :

ERtidt

tdiL =+ )()(

)1()( / LRtss eIti−−=

REI ss =

)(ti


• Notice that can be rearranged to be:

• which we can see is of the same form as our general

case

• with

ERtidt

tdiL =+ )()(

LiRE

dtdi −

=

),( yxfdxdy

=

LiREtifyxf

txiy−

=↔

↔↔

),(),(

We don’t have to use x and y as our symbols!


Graphically ( a reminder !)

i

Iss

t0

)(ti


• Notice that we’ve chosen an example that has an

analytical solution.

• This means that if we practice finding a numerical

solution to the problem we have an accurate solution to

check it against.

– Obviously if we already have the analytical solution we

wouldn’t normally bother to do it numerically!


First Step

• Returning to our problem:

• Closing the switch gives our initial condition of t=0, i=0 so

– di/dt at t=0 is E/L , the initial gradient

• So from this we could estimate a value of i for a small

increment in t (∆t)

LRtiE

dttdi )()( −

=


First Step

i

ti0 = 0

i1

∆t

Gradient = E/L

i1 = (E/L) . ∆t

0(t0) (t1)

Solving the general case

this next point would be y1=y0+f(x0,y0)∆x

),( yxfdxdy

=


More Steps

• Since we now know i1 we can find the gradient at t=∆t :

– di/dt = (E-i1R) / L

• Hence we can predict the next value (i2) using the same

process i2=i1+ ((E-i1R) / L) ∆t and, for all steps, the formula is

in+1=in+ ((E-inR) / L) ∆t

• The theory behind this approach comes from Taylor’s

Theorem (we use Taylor Series taken to the first order) but the

method just described is known as Euler’s Method


Euler’s Method

• The formula (the recurrence relation) for solving the general case

of is yn+1=yn+f(xn,yn)∆x

• We obtain a series of discrete points (x0,y0), (x1,y1),…, (xn,yn),

(xn+1,yn+1) which can be linked together to give an approximation to

the desired curve

• Notice that in general to calculate the next point we need the x and

y values of the previous point. This means we must also keep track

of the x values via xn+1=xn+ ∆x

),( yxfdxdy

=


The final result

• So, for our electrical example, since we know i0=0 we can solve & plot!

• We do this numerically by– Asking the user the limits ( a & b )

• a=t0=0 in our specific example

• b is the stopping point

– Requesting the interval size (called ∆x in our general case but sometimes h is also used)

• ∆x =∆t in our specific example

– Start at the first point (a), calculate the value at a+ ∆x, and continue until b is reached

– We print out the answer


The Unified Form of Solving ODE

( )dxyx,fdy =

( )hh,y,xyy iii1i φ+=+

Depending on function φ, there are different solutions.

The Runge-Kutta methods are the most popular.

00ii xx at y condition initial and xxh =−= +1

( )yx,fdxdy

=


The Euler’s Method

( )hy,xfyy iiii +=+1

( )iixx

y,xfdxdy

i

===

φ

Error estimate:( )

nn

ni2i

iii Rhn!yh

2!yhyyy +++

′′+′+=+ 1

( ) ( ) ( )( )( )

nnii

1-n3ii2ii

iiii Rhn!

y,xfh

3!y,xfh

2!y,xfhy,xfyy ++

′′+

′++=+ 1

Et = Et,2 + Et,3 + … + Et,n


Example of Euler’s Method

8.520x12x2xdxdy 23 +−+−=


Tank mixing problem

tccV

Vcc

ccV

Vdtdc

iinii

in

∆

−+=

−=

+ )(

)(

1tank

tank


Mixing tank

∆t Error Etat t=600

300 1.4150 0.61100 0.3950 0.1930 0.1115 0.05510 0.0365 0.0183 0.011


Quantify the error

( ) ( )

order? is error So

on centred for series Taylor Write

...6

)(2

)()()()( 321

1

+∆′′′

+∆′′

+∆′+=+

+

ttcttcttctctc

c c

iiiii

ii


Not quite – accumulation of errors

( )

( ) Ntc

ttcEEEEE

N

i

Ntotal

2

2

321

2

2)(

...

∆′′

≅

∆′′

=

++++=

∑

error daccumulate total so error, this has step each But

order 1st waserror the that beforesaw We


Euler Error

Local errorlk=yk-y*k-1(tk)

Global errorek=yk-y(tk)

ty

y’=λy

Chart1

0000

0.10.10.10.1

0.20.20.20.2

0.30.30.30.3

0.40.40.40.4

0.50.50.50.5

0.60.60.60.6

0.70.70.70.7

0.80.80.80.8

0.90.90.90.9

1111

1.11.11.11.1

1.21.21.21.2

1.31.31.31.3

1.41.41.41.4

y'

y'

y'

y'

t

y

0.5

0.6

0.7

0.8

0.552585459

0.6631025508

0.7736196427

0.8841367345

0.6107013791

0.7328416549

0.8549819307

0.9771222065

0.6749294038

0.8099152845

0.9449011653

1.0798870461

0.7459123488

0.8950948186

1.0442772883

1.1934597581

0.8243606354

0.9892327624

1.1541048895

1.3189770166

0.9110594002

1.0932712802

1.2754831603

1.4576950403

1.0068763537

1.2082516245

1.4096268952

1.611002166

1.1127704642

1.3353245571

1.5578786499

1.7804327428

1.2298015556

1.4757618667

1.7217221778

1.9676824889

1.3591409142

1.6309690971

1.9027972799

2.1746254628

1.502083012

1.8024996144

2.1029162168

2.4033328192

1.6600584614

1.9920701536

2.3240818459

2.6560935382

1.8346483338

2.2015780006

2.5685076673

2.9354373341

2.0275999834

2.4331199801

2.8386399768

3.2441599735

Sheet1

y' =y

dty0y1y2y3

0.10.50.60.70.8

ty'y'y'y'

00.50.60.70.8

0.10.5525854590.66310255080.77361964270.8841367345

0.20.61070137910.73284165490.85498193070.9771222065

0.30.67492940380.80991528450.94490116531.0798870461

0.40.74591234880.89509481861.04427728831.1934597581

0.50.82436063540.98923276241.15410488951.3189770166

0.60.91105940021.09327128021.27548316031.4576950403

0.71.00687635371.20825162451.40962689521.611002166

0.81.11277046421.33532455711.55787864991.7804327428

0.91.22980155561.47576186671.72172217781.9676824889

11.35914091421.63096909711.90279727992.1746254628

1.11.5020830121.80249961442.10291621682.4033328192

1.21.66005846141.99207015362.32408184592.6560935382

1.31.83464833382.20157800062.56850766732.9354373341

1.42.02759998342.43311998012.83863997683.2441599735

Sheet1

y'

y'

y'

y'

t

y

Sheet2

Sheet3


Euler Stability

kk

k

kkk

t

hyy

hyyhyy

eytyyy

)1(

)1(

)(

0

1

0

λ

λλ

λ λ

+=

+=+=

==′

+

that So

Euler


Effect of Step Size on Euler’s Method

8.520x12x2xdxdy 23 +−+−=

Comparing h = 0.5 and h=0.25 Error of y(5) as a function of number of steps

h = 0.5


However

• Whilst the methods are OK, the error may be large

• Error in every step

– because only using first order of Taylor Series

• Error accumulates because use approximate (i.e. errored) values

to generate following step


A better solution…

• Is to use higher orders of the series

– We can use 2nd or higher orders of the Taylor series (error is

then related to ∆t2)

• A more popular approach is to use Runge-Kutta

methods


Systems of ODEs

( )

( )

( )n2nn

n2

n2

y,...,y,yx,fdx

dy

y,...,y,yx,fdxdy

y,...,y,yx,fdxdy

1

122

111

=

=

=

sinxedxdyc

dxydb

dx3yda dx2

23+=++ −

Example:

( )122

1

1

1 cy-by-sinxeadx

dy

ydxdy

ydxdy

2dx +=

=

=

−

Engineering applications can involve thousands of simultaneous ODEs!

conversion

(n equations require n initial conditions)


Euler’s Method for System of ODEs


Example of System of ODEs

( )

−=

=

34

4

y16.1sindt

dy

ydt

dy3

−=

=

12

2

16.1ydt

dy

ydtdy1

y1 = θ, y2 = dθ/dt

0116 =+ sinθdtθd2

2.

(linear approximation when θ is small)

y3 = θ, y4 = dθ/dt

(exact conversion)

small initial θ large initial θ


Sets of ODEs

So ODE problems can be reduced to a set of N first-order ODEs

Not completely specified –needs boundary conditions• initial value problems

• boundary value problems

Niyyyxfdx

xdyNi

i ,...,1),...,,()( 21 == equations for


General Procedure

• Re-write the dy and dx terms as Δy and Δx and multiply by Δx

• Literally doing this is Euler’s method

Niyyyxfdx

xdyNi

i ,...,1),...,,()( 21 == equations for

xyxfyy

yxfxxy

yxfdx

xdy

iiii ∆+=

=∆

∆

=

+ ),(

),()(

),()(

1


Improvements to Euler’s Method

Euler

Heun’s method (predictor-corrector)

Procedurecalc yi+1 with Euler (predictor)calc slope at yi+1calc average slopeuse this slope to calc new yi+1 (corrector)

xyxfyy

yxfdx

xdy

iiii ∆+=

=

+ ),(

),()(

1


Home Work of Chapter 25From the text book:

25.1, 25.2, 25.4, 25.5, 25.7a, 25.10

Additional:

Numerical MethodsBasicsSlayt Numarası 3Slayt Numarası 4Ordinary Differential EquationsSlayt Numarası 6Slayt Numarası 71st Order ODEs1st Order ODEsSo how do we use NM?Boundary ConditionsConsiderSlayt Numarası 13Graphically ( a reminder !)Slayt Numarası 15First StepFirst StepMore StepsEuler’s MethodThe final resultThe Unified Form of Solving ODEThe Euler’s MethodExample of Euler’s MethodTank mixing problemMixing tankQuantify the errorNot quite – accumulation of errorsEuler ErrorEuler StabilityEffect of Step Size on Euler’s MethodHoweverA better solution…Systems of ODEsEuler’s Method for System of ODEsExample of System of ODEsSets of ODEsGeneral ProcedureImprovements to Euler’s MethodHome Work of Chapter 25

numerical methodslecture 9: numerical solution of ode’s math259 numerical analysis. 20. the final...

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