Numerical Solutions of Ordinary Differential Equations 1Dr.G.Suresh Kumar@KL University 7/3/2015

Numerical Solutions of Numerical Solutions of Ordinary Differential Ordinary Differential

EquationsEquations

Numerical Solutions of Numerical Solutions of Ordinary Differential Ordinary Differential

EquationsEquations

1Dr.G.Suresh Kumar@KL University

04/19/23

Ordinary Differential Ordinary Differential EquationsEquations

Equations which are composed of an unknown function and its derivatives are called differential equations.

Differential equations play a fundamental role in engineering because many physical phenomena are best formulated mathematically in terms of their rate of change.

vm

cg

dt

dv

v - dependent variable

t - independent variable

04/19/23 2Dr.G.Suresh Kumar@KL University

Types of Differential Types of Differential EquationsEquations

• When a function involves one independent

variable, the equation is called an ordinary

differential equation (ODE).

•A partial differential equation (PDE)

involves two or more independent

variables.


Model Ordinary Differential Model Ordinary Differential EquationsEquations


Why Numerical solutions of Why Numerical solutions of ODE ?ODE ?

Many differential equations cannot be solved analytically, in which case we have to satisfy ourselves with an approximation to the solution.

04/19/23 Dr.G.Suresh Kumar@KL University 5

Initial value problemInitial value problem

Initial value problem associated with first order differential equations is


.y)y(x

y)f(x,dx

dy

00

Methods of solving IVPMethods of solving IVPTaylor series Picard successive approximationEulerModified EulerRunge-Kutta (RK)


Taylor Series MethodTaylor Series MethodThe Taylor series method is a

straight forward adaptation of classic calculus to develop the solution as an infinite series.

The method is not strictly a numerical method but it is used in conjunction with numerical schemes.


Taylor Series MethodTaylor Series Method

The Taylor series expansion of y(x) about

x = x0 is


...xy 4!

x-xxy

3!

x-x

xy 2!

x-xxy x-xxyxy

0(IV)

40

0

30

0

20

000

Problem#1Problem#1

The differential equation y'(x) = x + y satisfying y(0)=1. Determine the Taylor series solution. Also compare with the exact solution.

{Analytical solution: y(x)=2ex - x -1(Cauchy linear differential equation).}


SolutionSolution

First, we find derivatives of y' = x + y,


xyxy

xyxy

xy1xy

xyxxy

(IV)

2.0y0y

20y0y

2110y10y

1100yx0y

(IV)

SolutionSolutionThe Taylor series expansion of y(x)

about x = x0 is


...xy 4!

x-xxy

3!

x-x

xy 2!

x-xxy x-xxyxy

0(IV)

40

0

30

0

20

000

... 12

x

3

x xx 1

... 2 4!

x2

3!

x2

2!

x1x 1xy

432

432

Comparison of solutionsComparison of solutions

The exact solution is

Therefore, Taylor series solution and exact solution are equal.


...24

x

3

xxx1

1)(x...4!

x

3!

x

2!

xx12y(x)

432

432

Picard’s MethodPicard’s Method


00 y)y(x

y)f(x,dx

dy

The initial value Problem

is equivalent to the following integral equation

.y(t))dtf(t,yy(x)x

x

0

0

Picard’s MethodPicard’s Method


The Picard’s nth approximate solution is

... 3, 2, 1, n ,(t))dtyf(t,y(x)yx

x1n0n

0

.n as y(x)(x)yn

Problem#1Problem#1Using Picard’s process of successive

approximation, obtain a solution up to the fifth approximation of the equation


y xdx

dy

such that y = 1 when x = 0. Check your answer by finding the exact solution.

720

x

60

x

12

x

3

xxx1(x)y

65432

5 Ans.

Exact solution y(x) = 2ex – (x+1).

Application ProblemApplication Problem

For the free-falling bungee jumper with linear drag, compute the velocity at 10 seconds of a free-falling parachutist of mass 80kg and drag coefficient 10 kg/s at 10 seconds, with initial velocity 20 m/s, using

(i) Taylor series method(ii) Picard’s method.


SolutionSolutionLet v(t) be the velocity and ‘a’ be the

acceleration of the parachutist at any time t.

Then the total force acting on the parachutist is equal to the sum of the

downward force FD due to gravity andupward force FU due to air resistance.

i.e. F = FD + FU

FD = mg

FU α v (linear drag) FU = - Cd v(t).


SolutionSolutionBy Newton second law of motion F =

ma.ma = mg – Cd v

Therefore,


vm

Cg

dt

dv d

Given that m = 80kg, Cd = 10 kg/s,

and initial velocity v(0)=20 m/s.Now solve the above IVP using

Taylor and Picard’s method.

Euler’s MethodEuler’s Method

,...2,1),(

)(

:MethodEuler

,...2,1)(:Determine

)( :condition initial with the

),()(':ODEorder first Given the

:Problem

1

00

0

00

iforyxfhyy

xyy

iforihxyy

xyy

yxfxy

iiii

i


04/19/23 21

Interpretation of Euler Interpretation of Euler MethodMethod

x0 x1 x2 x

y0

y1

y2

Dr.G.Suresh Kumar@KL University

04/19/23 22


hx0 x1 x2 x

y0

y1=y0+hf(x0,y0)

Slope=f(x0,y0)

hf(x0,y0)

y1


04/19/23 23


hx0 x1 x2 x

y0

y1=y0+hf(x0,y0)

Slope=f(x0,y0)

hf(x0,y0)y1

h

y2=y1+hf(x1,y1)Slope=f(x1,y1)

y2

hf(x1,y1)




Problem#1Problem#1Using Euler’s method, find an

approximate value of y corresponding to x=1, given that yʹ = x + y such that y=1 when x=0.

Ans: Taking h = 0.1, theny(0.1) = 1.1, y(0.2) = 1.22, y(0.3) = 1.36y(0.4) = 1.53, y(0.5) = 1.72, y(0.6) =

1.94y(0.7) = 2.19, y(0.8) = 2.48, y(0.9) =

2.81y(1.0) = 3.18.


Problem#2Problem#2

A cup of a coffee originally has temperature of 68oC. The temperature of the ambient is 21oC and the thermal constant is 0.017oC/min. Determine the temperature of the coffee from t = 0 to 10 minutes insteps of 2 minutes.

Sol. The differential equation is Tʹ(t) = -k(T - Ta) = -0.017(T - 21) T(0) = 68


Improvements of Euler’s methodImprovements of Euler’s methodA fundamental source of error in Euler’s

method is that the derivative at the beginning of the interval is assumed to apply across the entire interval.

Two simple modifications are available to circumvent this shortcoming:

Heun’s (Modified Euler) MethodMidpoint (Improved Polygon) Method


Modified Euler’s MethodModified Euler’s MethodTo improve the estimate of the slope, determine two

derivatives for the interval:(1) At the initial point(2) At the end pointThe two derivatives are then averaged to obtain an improved estimate of the slope for the entire interval.

hyxfyxf

yy

hyxfyy

iiiiii

iiii

2

),(),(:

),( :0

111

01

Corrector

Predictor


29

To improve the estimate of the slope, determine two derivatives for the interval:

At the initial point At the end point

The two derivatives are then averaged to obtain an improved estimate of the slope for the entire interval.

hyxfyxf

yy

hyxfyy

iiiiii

iiii

2

),(),(:

),( :0

111

01

Corrector

Predictor

Euler Modified methodEuler Modified method

30

hyxfyxf

yy

hyxfyy

iiiiii

iiii

2

),(),(:

),( :

0111

1

01

Corrector

Predictor

Heun’s method (improved)Heun’s method (improved)

,...2,1 2

),(),(:

),( :1

11

01

1

jh

yxfyxfyy

hyxfyyj

iiiii

j

iiii

iCorrector

Predictor

Note that the corrector can be iterated to improve the accuracy of yi+1 .

Original Huen’s:

Problem#1Problem#1Using Modified Euler’s method, find an

approximate value of y when x= 0.3, given that yʹ = x + y and y=1 when x=0.

Sol. Compare the IVP with the general IVP

yʹ = f(x, y), y(x0) = x0, we have

f(x, y) = x + y, x0 = 0 and y0 = 1.

Taking the step size h = 0.1, then x0 = 0, x1 = 0.1, x2 = 0.2, x3 = 0.3. Now we find the corresponding value of y, using modified Euler’s method.

31

SolutionSolutionStep-1. To find y1 = y(x1) = y(0.1)

Predictor formula (Euler’ formula)y1

(0) = y0 + h f(x0, y0)

= 1 + 0.1 f(0, 1) = 1+0.1(0+1) =1.1Corrector formula y1

(k) = y0 + (h/2) [f(x0, y0) + f(x1, y1(k-1))]

y1(1) = y0 + (h/2) [f(x0, y0) + f(x1, y1

(0))]

= 1 + (0.1/2) [f(0,1) + f(0.1, 1.1)] = 1 + 0.05 [0+1 + 0.1+1.1] =1.11

32

SolutionSolutiony1

(2) = y0 + (h/2) [f(x0, y0) + f(x1, y1(1))]

= 1 + (0.1/2) [f(0,1) + f(0.1, 1.11)] = 1 + 0.05 [0+1 + 0.1+1.11] =1.1105.y1

(3) = y0 + (h/2) [f(x0, y0) + f(x1, y1(2))]

= 1 + (0.1/2) [f(0,1) + f(0.1, 1.1105)] = 1 + 0.05 [0+1 + 0.1+1.1105] =1.1105.Therefore, y1

= 1.1105.

33

SolutionSolutionStep-2. To find y2 = y(x2) = y(0.2)

Predictor formula (Euler’ formula)y2

(0) = y1 + h f(x1, y1)

= 1.1105 + 0.1 f(0.1, 1.1105) = 1.1105+0.1(0.1 + 1.1105) =1.2316.Corrector formula y2

(k) = y1 + (h/2) [f(x1, y1) + f(x2, y2(k-1))]

y2(1) = y1 + (h/2) [f(x1, y1) + f(x2, y2

(0))]

= 1.2316 + (0.1/2) [f(0.1,1.1105) + f(0.2, 1.2316)] = 1.2316 + 0.05 [0.1+1.1105 + 0.2+1.2316] =1.2426.

34

SolutionSolutiony2

(2) = y1 + (h/2) [f(x1, y1) + f(x2, y2(1))]

= 1.1105 + (0.1/2) [f(0.1,1.1105) + f(0.2, 1.2426)]

= 1.1105 + 0.05 [0.1+1.1105 + 0.2+1.2426] =1.2432. y2

(3) = y1 + (h/2) [f(x1, y1) + f(x2, y2(2))]

= 1.1105 + (0.1/2) [f(0.1,1.1105) + f(0.2, 1.2432)]

= 1.1105 + 0.05 [0.1+1.1105 + 0.2+1.2432] =1.2432.Therefore, y2 = y(0.2)= 1.2432.

Similarly, y3 = y(0.3) = 1.4004.35

Application ProblemApplication ProblemA storage tank (Fig.) contains a liquid

at depth y where y = 0 when the tank is half full. Liquid is withdrawn at a constant flow rate Q=500m3/h to meet demands. The contents are resupplied at a sinusoidal rate 3Q sin2(t). The surface area of the tank is 1200 m2. Determine the depth of the tank from t= 0 to t=6 in steps of 2 hours, using modified Euler’s method.


Problem#1Problem#1


3Q sin2(t) Q =500 m3/h

SolutionSolutionLet y(t) be the depth of the tank at any time t

and A be the surface area of the tank.

The depth of the tank is changed when volume of the liquid in the tank changed. Therefore,


rate flowout – rate inflow in volume

change of Rate

QQ (t)sin3Ay(t)dt

d 2

SolutionSolutionSince A is constant, it implies that

Given that Q=500m3/h, A =1200 m2 and y(0) = 0. Substituting these value, we have


A(t)sin

A

3

dt

dy 2 QQ

0y(0) 12

5(t)sin

4

5

dt

dy 2

Now solving the above initial value problem, using modified Euler’s method.

Runge - Kutta Runge - Kutta MethodMethod


Carl RungeCarl Runge(1856-1927)(1856-1927)

Martin Wilhelm Kutta

(1867-1944)


IntroductionIntroductionRunge-Kutta methods are very popular because of

their good efficiency; and are used in most computer programs for differential equations.

These methods agree with Taylor’s series solution up to the term in hr, where r differs method to method and is called the order of the method.

Euler’s and modified Euler’s are called the first and second order Runge-Kutta methods.

The fourth order R-K method is most commonly used and is often referred to as R-K method only.


Working RuleWorking RuleTo find increment ‘k’ of y corresponding

to an increment h of x by R-K method from the initial value problem

y´ = f(x, y) with y(x0) = y0 is as follows

Calculate k1 = h f(x0, y0)

k2 = h f(x0+0.5h, y0+0.5k1)

k3 = h f(x0+0.5h, y0+0.5k2)

k4 = h f(x0+h, y0+k3)


Working RuleWorking RuleFinally compute k = 1/6 (k1 + 2k2 + 2k3 + k4)

is the weighted mean of k1, k2, k3 and k4.

Therefore the required approximate value is

y1 = y(x1) = y(x0 + h) = y0 + k.



Graphical RepresentationGraphical Representation

xi xi + h/2 xi + h

k1

k2

k3

k4

4321 226

1kkkkk

k

04/19/23 45

Problem#1Problem#1Using Runge-Kutta method of

fourth order, solve yʹ = (y2 – x2)/(y2 + x2) with y(0)=1 at x = 0.2, 0.4.

Sol : - Compare the given problem with the general first order initial value problem yʹ=f(x, y) satisfying y(x0)=y0, we have

f(x, y) = (y2 – x2)/(y2 + x2) x0 = 0, y0 =1.

Here we have to find value of y at x=0.2, 0.4.04/19/23 46Dr.G.Suresh Kumar@KL University

SolutionSolutionTaking step size h = 0.2.Step(1) To find y(0.2) k1 = h f(x0, y0)

= 0.2 f(0, 1) = 0.2 (12-02)/(12+02) = 0.2 k2 = h f(x0+0.5h, y0+0.5k1)

= 0.2 f(0+0.5(0.2), 1+ 0.5(0.2)) = 0.2 f(0.1, 1.1) = 0.19672


SolutionSolution

k3 = h f(x0+0.5h, y0+0.5k2)

= 0.2 f(0+0.5(0.2), 1+0.5(0.19672))

= 0.2 f(0.1, 1.09836) = 0.1967 k4 = h f(x0+h, y0+k3)

= 0.2 f(0+0.2, 1+0.1967) = 0.2 f(0.2, 1.1967) = 0.1891


SolutionSolutionk = 1/6 (k1 + 2k2 + 2k3 + k4)

= 1/6 (0.2 +2(0.19672)+2(0.1967)+0.1891)

= 0.19599.Hence y1 = y0 + k

y(0.2) = 1 + 0.19599 = 1.19599.Step(2) To find y(0.4) k1 = h f(x1, y1)

= 0.2 f(0.2, 1.19599) = 0.1891


solutionsolutionk2 = h f(x1+0.5h, y1+0.5k1)

= 0.2 f(0.2+0.5(0.2), 1.19599+ 0.5(0.1891)) = 0.2 f(0.3, 1.29054) = 0.1795k3 = h f(x1+0.5h, y1+0.5k2)

= 0.2 f(0.2+0.5(0.2), 1.19599+0.5(0.1795)) = 0.2 f(0.3, 1.28574) = 0.1793


SolutionSolutionk4 = h f(x1+h, y1+k3)

= 0.2 f(0.2+0.2, 1.19599+0.1793) = 0.2 f(0.4, 1.37529) = 0.1688k = 1/6 (k1 + 2k2 + 2k3 + k4)

= 1/6 (0.1891 +2(0.1795)+2(0.1793)+0.1688)

= 0.1792.Hence y2 = y1 + k

y(0.4) = 1.19599 + 0.1792 = 1.37519.


Problem#2Problem#2Apply Runge-Kutta method to find

an approximate value of y for x=0.2 in steps of 0.1, if yʹ = x + y2 given that y=1 where

x = 0.Sol : - Compare the given problem with the

general first order initial value problem yʹ=f(x, y) satisfying y(x0)=y0, we have

f(x, y) = x + y2 , x0 = 0, y0 =1.

Here we have to find value of y at x = 0.2.04/19/23 52Dr.G.Suresh Kumar@KL University

SolutionSolutionTaking step size h = 0.1.Step(1) To find y(0.1) k1 = h f(x0, y0)

= 0.1 f(0, 1) = 0.1 (0 + 12) = 0.1 k2 = h f(x0+0.5h, y0+0.5k1)

= 0.1 f(0+0.5(0.1), 1+ 0.5(0.1)) = 0.2 f(0.05, 1.05) = 0.1152


SolutionSolution

k3 = h f(x0+0.5h, y0+0.5k2)

= 0.1 f(0+0.5(0.1), 1+0.5(0.1152))

= 0.1 f(0.05, 1.0576) = 0.11685 k4 = h f(x0+h, y0+k3)

= 0.1 f(0+0.1, 1+0.11685) = 0.1 f(0.1, 1.11685) = 0.13473


SolutionSolutionk = 1/6 (k1 + 2k2 + 2k3 + k4)

= 1/6 (0.1 +2(0.1152)+2(0.11685)+0.13473)

= 0.11647 ≈ 0.1165.Hence y1 = y0 + k

y(0.1) = 1 + 0.1165 = 1.1165.Step(2) To find y(0.2) k1 = h f(x1, y1)

= 0.1 f(0.1, 1.1165) = 0.1347


solutionsolutionk2 = h f(x1+0.5h, y1+0.5k1)

= 0.1 f(0.1+0.5(0.1), 1.1165+ 0.5(0.1347)) = 0.2 f(0.15, 1.1838) = 0.1551k3 = h f(x1+0.5h, y1+0.5k2)

= 0.1 f(0.1+0.5(0.1), 1.1165+0.5(0.1551)) = 0.1 f(0.15, 1.194) = 0.1575


SolutionSolutionk4 = h f(x1+h, y1+k3)

= 0.1 f(0.1+0.1, 1.1165+0.1576) = 0.1 f(0.2, 1.2741) = 0.1823k = 1/6 (k1 + 2k2 + 2k3 + k4)

= 1/6 (0.1347 +2(0.1551)+2(0.1576)+0.1823)

= 0.1571.Hence y2 = y1 + k

y(0.2) = 1.1165 + 0.1571 = 1.2736.04/19/23 Dr.G.Suresh Kumar@KL University 57

Documents

Numerical Solutions of Ordinary Differential Equations 1Dr.G.Suresh Kumar@KL University 7/3/2015