DSAS 4 Runge-Kutta Formula For Differential...

NCTU Department of Electrical and Computer Engineering Senior Course <Dynamic System Analysis and Simulation>

By Prof. Yon-Ping Chen

4-1

4. Runge-Kutta Formula For Differential Equations

A. Euler Formula

B. Runge-Kutta Formula

C. An Example for Fourth-Order Runge-Kutta Formula

To solve the differential equations numerically, the most useful formula is called

Runge-Kutta formula which has been widely used in numerical analysis.

A. Euler Formula

For a dynamic system without input, it is generally expressed as the following

first-order differential equation:

( ) ( )( )txtftx ,=& , x(t0)=α (4-1)

where t=t0 is the initial time and x(t0)=α is the initial condition. The problem to solve

x(t) in (4-1) for t>t0 is called the initial value problem, or IVP in brief. For example,

the following equation

( ) ( ) ttxtx +−=& , x(0)=1 (4-2)

is an IVP and its solution can be obtained in closed form as below:

( ) 12 −+= − tetx t , t≥0 (4-3)

However, if (4-1) is given more complicated, such as

( ) ( ) ( )tsinttxtx ⋅+−= 2& , x(0)=1 (4-4)

then it is often difficult or impossible to find x(t) in closed form since most of the

differential equations do not have closed form solution. To determimne the solution,

numerical methods are required.

The simplest numerical method is called the Euler formula, which was propsed

by Euler in 1768. With the use of fixed grid size ∆t=h, the grid points along t are

orderly denoted as t0, t1, t2,… , where

hitt ⋅+= 0i , i=1,2,3,…… (4-5)

and the value x(t) at ti is defined as

( )ii txx = , i=0,1,2,3,…… (4-6)



4-2

where x0=x(t0)=α is given. According to the definition of derivative, we have

( ) ( ) ( )t

txttxlimtxt ∆

∆∆

−+=→0

& , (4-7)

which implies the derivative of x(t) at t=ti can be approximately expressed as

( ) ( ) ( ) ( ) ( )h

xx

h

txtx

t

txttxtx iiiiii

i

−=

−=

−+≈ ++ 11

∆∆

& (4-8)

Substituting (4-8) into (4-1) at t=ti leads to

( )iiii x,tf

h

xx≈

−+1 , x0=α (4-9)

and then

( )ii xtfhxx ,ii ⋅+=+1 , x0=α (4-10)

which is the well kown Euler formula. Clearly, in order to achieve a solution more

precisely, the grid size h should be as small as possible. However, reducing the grid

size h is inefficient since the cost of calculation time may increase tremendously.

Now, let’s introduce Taylor’s expansion to explain the error caused by Euler

formula (4-10). According to Taylor’s expansion, a function y(t) continuous at t=ti can

be expressed as

( ) ( ) ( ) ( ) LL +−++−+−+= ninii ttattattaatx 2

210 (4-11)

whose higher order derivatives are

( ) ( ) ( ) ( ) LL& +−++−+−+= −12321 32 n

inii ttnattattaatx (4-12)

( ) ( ) ( )

( ) ( ) LL

L&&

+−−++

+−⋅+−⋅+=−2

2432

1

34232n

in

ii

ttann

ttattaatx (4-13)

( ) ( ) ( )

( )( ) ( ) LL

L&&&

+−−−++

+−⋅⋅+−⋅⋅+⋅=−3

2543

21

34523423n

in

ii

ttannn

ttattaatx (4-14)

Viewing from the above equations, we have ( )itxa =0 and ( ) ( )ik

k txk

a!

1= for

k=1,2,…,∞, and thus, (4-11) can be rewritten as

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) LL

&&& +−++−+−+= n

ii

n

ii

iii ttn

txtt

txtttxtxtx

!!2

2 (4-15)

Let t=ti+1, we have



4-3

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) LL

L&&

&

+−++

+−+−+=

+

+++

nii

in

iii

iiiii

ttn

tx

tttx

tttxtxtx

1

2111

2

!

! (4-16)

i.e.,

( ) ( ) ( ) ( )

( ) ( )2

12

1

2

hOhxtfx

hn

xtfh

xtfhxtfxx

iii

niin

iiiiii

++=

+++′

++=−

+

,!

,

!

,, LL

(4-17)

where

( ) ( ) ( ) ( )LL +++

′=

−nii

nii h

n

xtfh

xtfhO

!

,

!

, 122

2 (4-18)

Comparing (4-17) with (4-10), we know that the term ( )2hO is the error of Euler

formula, which is proportional to h2.

B. Runge-Kutta Formula

In order to reduce the error to h3, further modify Euler form (4-10) as the

following form:

( ) ( )hxhtfxtfh

xx δγββ +++=−+

iiiiii , , 10

1 , (4-19)

where β0, β1, γ and δ are variables to be determined and a new term ( )hxhtf δγ ++ ii ,

is included. Since ( )( )txtf , depends on x(t) and t, its Taylor’s expansion at x(t)=xi

and t=ti is given as the following form

( )( ) ( ) ( )( ) ( ) ( )( )( )

( )( ) ( ) ( ) ( )( )iittxitttixx

iitxittixit

xtxttattaxtxa

ttxtxattaxtxattaaty,tf

−−+−+−+

−−+−+−+−+=232

20

( ) ( )( ) ( )( ) L+−+−−+ 32 ixxxiitxx xtxaxtxtta (4-20)

where all the coefficients a0, at, ax, att, atx … are constant. Its partial derivatives are

( )( ) ( )( )

( ) ( )( ) ( )( ) ( )( ) ( )( ) L+−+−−+

−+−+−+=∂

∂=

2

2

2

32

itxxiittx

itttitxittt

t

xtxaxtxtta

ttaxtxattaa

t

tx,tftx,tf

(4-21)



4-4

( )( ) ( )( )( )

( ) ( )( ) ( )( ) ( )( ) ( )( ) L+−+−−+

−+−+−+=

∂∂=

2

2

32

2

ixxxiitxx

ittxixxitxx

x

xtxaxtxtta

ttaxtxattaa

tx

txtftxtf

,,

(4-22)

( )( ) ( )( ) ( )( )

( ) ( )( ) L+−+−⋅+=∂

∂=

∂∂=

ittxittttt

ttt

xtxattaat

txtf

t

txtftytf

2232

2

2 ,,,

(4-23)

( )( ) ( )( )

( )( )( )

( ) ( )( ) L+−+−+=∂

∂=

∂∂∂=

itxxittxtx

xtx

xtxattaa

t

txtf

ttx

txtftytf

22

2 ,,,

(4-24)

( )( ) ( )( )

( )( )( )( )

( )( ) ( )( ) L+−⋅+−+=

∂∂

=∂

∂=

ixxxitxxxx

xxx

xtxattaa

tx

txtf

tx

txtftxtf

2322

2

2 ,,,

(4-25)

Clearly, all the coefficients can be derived from the above partial derivatives and

expressed as ( )ii xtfa ,=0 , ( )iitt xtfa ,= , ( )iixx xtfa ,= , ( )iitttt xtfa ,!2

1= ,

( )iitxtx xtfa ,= , ( )iixxxx xtfa ,!2

1= , …. Substituting them into (4-20) yields

( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )2

2

1iiittiiixiiitii ttx,tf

!xtxx,tfttx,tfx,tftx,tf −+−+−+=

( )( ) ( )( ) ( ) ( )( )

( )( ) ( ) ( )( ) L+−+−+

−+−−+

33

2

3

1

3

1

2

1

iiixxxiiittt

iiixxiiiitx

xtxx,tf!

ttx,tf!

xtxx,tf!

xtxttx,tf (4-26)

Let t=ti+γh and x(t)=xi+δh, then

( )( ) ( ) ( ) ( )

( ) ( ) ( )3222

22

,2

1 ,

,2

1 , , ,

hOxtfhxtfh

xtfhxtfhxtfhxtf

hxhtf

xxtx

ttxt

+++

+++=

++

iiii

iiiiiiii

ii ,

δγδ

γδγ

δγ

(4-27)

Clearly, (4-19) can be rewritten as

( ) ( )[ ] ( ) ( )[ ]

( ) ( ) ( )[ ]( )4

211

21

3

11

2

101

,3,6,33

,2,22

,

hO

xtfxtfxtf!

h

xtfxtf!

hxtfhxx

iixxiitxiitt

iixiitiiii

+

+++

++++=+

δβγδβγβ

δβγβββ

(4-28)



4-5

If we want to take the precision to h2, from (4-17) and (4-28) we have

( ) ( ) ( )( ) ( )[ ]

( ) ( )[ ] ( )311

2

10

321

,2,22

,

2

hOxtfxtf!

h

xtfhx

hOh!

x,tfhx,tfxx

iixiit

iii

iiiiii

+++

++≈

+′

++≈+

δβγβ

ββ (4-29)

Since ( ) ( )iii xtftx ,=& , the term ( )ii x,tf ′ can be changed into the following form:

( ) ( ) ( ) ( )

( ) ( ) ( )iiiixiit

iiixiitii

xtfxtfxtf

txxtfxtfxtf

,,,

,,,

+=+=′ &

(4-30)

As a result, from (4-29) and (4-30) we obtain

( ) ( ) ( ) ( )( )

( ) ( )[ ] ( ) ( )[ ]iixiitii

iiiixiitii

xtfxtf!

hxtfh

xtfxtfxtf!

hhx,tf

,2,22

,

,,,2

11

2

10

2

δβγβββ +++=

++ (4-31)

which leads to

110 =+ ββ (4-32)

( ) ( ) ( ) ( ) ( )iiiiiiiiii xtfxtfxtfxtfxtf xtxt , , , ,2,2 11 +=+ δβγβ (4-33)

i.e., 12 1 =γβ (4-34)

( )ii xtf ,2 1 =δβ (4-35)

Since there are four variables β0, β1, γ and δ in three equations (4-32), (4-34) and

(4-35), the choice of these variables is not unique. The most common one is

( )ii xtf , ,1 ,2

110 ==== δγββ (4-36)

i.e., from (4-28) we have the approximate ssolution as below:

( ) ( )( )ii xthfxhtfh

xtfh

xx ,2

,21 ++++=+ iiiiii , , (4-37)

which is called the second-order Runge-Kutta formula. For the convenience of

programming, the formula is often rearranged as below:

( )101 2

1kkxx ++=+ ii (4-38)

where

( )( )

++⋅=⋅=

01

0

,

kx,htfhk

xtfhk

ii

ii (4-39)



4-6

In case that a higher precision is required, we oftem employ the higher order

Runge-Kutta formula. For example, if a precision of h4 is needed, then the

fourth-order Runge-Kutta formula must be used, which is given as

( )32101 226

1kkkkxx ++++=+ ii (4-40)

where

( )

( )

+⋅=

++⋅=

++⋅=⋅=

231

2

010

,22

22 ,

kx,tfhk

kx,

htfhk

kx,

htfhkx,tfhk

iiii

iiii

(4-41)

This formula has been widely applied to a lot of applications in engineering due to its

simplicity and acceptable accuracy.

C. An Example for Fourth-Order Runge-Kutta Formula

Next, let’s use an example to show the programming of the fourth-order

Runge-Kutta formula in MATLAB. Consider the following equation:

( ) ( )( ) ( ) 1+−== txttxftx ,& , x(0)=0.5 (4-42)

and find the solution x(t) for t≥0. The closed form solution can be obtained as

( ) 150 +−= −tetx . (4-43)

Now let’s apply the fourth-order Runge-Kutta formula to solve (4-42) numerically. If

the step size is set as h=0.01, then the precision will be in the order of h4=10−8.

According to (4-40), we have

( )32101 226

1kkkkxx ++++=+ ii (4-44)

where

( ) ( ) ( )

( ) ( )( ) ( )11

12

1222

12

1222

11,

2223

1112

0001

0

−+−=++−=+⋅=

−+−=

+

+−=

++⋅=

−+−=

+

+−=

++⋅=

−−=+−=⋅=

kxhkxh,kx,tfhk

kxh

kxh

kx,

htfhk

kxh

kxh

kx,

htfhk

xhxhxtfhk

iiii

iiii

iiii

iiii



4-7

The programming in MATLAB is given as below:

=================================================== >> % Fourth order Runge-Kutta method >> x0=0.5; % initial condition x(0)=0.5 >> h=0.01; % step size 0.01sec >> k0=-h*(x0-1); k1=-h*(x0+k0/2-1); >> k2=-h*(x0+k1/2-1); k3=-h*(x0+k2-1); >> x(1)=x0+(k0+2*k1+2*k2+k3)/6; >> e(1)=x(1)-(-0.5*exp(-h)+1); % numerical error at t=h >> t(1)=h; >> for n=1:599 % total simulation time 6 sec >> k0=-h*(x(n)-1); k1=-h*(x(n)+k0/2-1); >> k2=-h*(x(n)+k1/2-1); k3=-h*(x(n)+k2-1); >> x(n+1)=x(n)+(k0+2*k1+2*k2+k3)/6; >> t(n+1)=(n+1)*h; % t-axis >> e(n+1)=x(n+1)-(-0.5*exp(-t(n+1))+1); % numerical error >> end >> plot(t,x); xlabel(‘t’); ylabel(‘x(t)’)

0 1 2 3 4 5 60.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

t

x(t)

>> plot(t,e); xlabel(‘t’); ylabel(‘e(4-t)’)

0 1 2 3 4 5 6-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0x 10

-11

t

e(t)

===========================================================



4-8

From the above numerical results of e(t), it is truth that the precision is less than

10−10 which is indeed around the order of h4 (10−8). In MATLAB, some instructions are

provided to solve the differential equations, such as ode23 and ode45. In fact, these

instructions are also implemented by the concept of Runge-Kutta method. Now, let’s

use the instruction ode23 to solve the system (4-42) and use the instruction ode45 to

solve the system described below:

( ) ( )( ) ( ) ( )tsinsintxtxtftx 1050., +−==& , x(0)=0.5 (4-45)

whose solution can not be expressed by closed form. The programming in MATLAB

is given as below:

========================================= Create m-file: first1.m function dx=first1(t,x) dx=-x+1; Create m-file: first2.m function dx=first2(t,x) dx=-x+0.5*sin(sin(10*t));

>> % key in the following instructions >> [t,x]=ode23(@first1,[0:0.01:6],0.5) >> plot(t,x); xlabel(‘t’); ylabel(‘x(t)’)

0 1 2 3 4 5 60.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

t

x(t)

>> % key in the following instructions >> [t,x]=ode45(@first2,[0:0.01:6],0.5) >> plot(t,x); xlabel(‘t’); ylabel(‘x(t)’)



4-9

0 1 2 3 4 5 6-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

t

x(t)

=====================================================

Problems

P.4-1 Derive the following third-order Runge-Kutta formula:

( )2101 46

1kkkxx ii +++=+

where

( )

( )102

01

0

222

,

kkx,htfhk

kx,

htfhk

xtfhk

ii

ii

ii

+−+⋅=

++⋅=

⋅=

P.4-2 Solve the following system:

( ) ( )( ) ( ) ( )tsinetxtxtftx t 52 10., −+−==& , x(0)=1

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