1MOHAN LAL SUKHADIA UNIVERSITY

TOPIC :RUNGE-KUTTA METHOD

SUBMITTED BY :SHUBHAM TOMAR

SUBMITTED TO:Mrs. DIPTI MA’AM

2CONTENTS :

• Introduction

• Example of Second-order Runge-kutta method

• Fourth order Runge-kutta method

• Example of fourth order Runge-kutta method

• Illustration of Heun’s Method

• Illustration of Runge-Kutta second order

• Illustration of Runge Kutta fourth order

• Introduction• Runge-kutta method are popular because of efficiency.

• It is single step method as Euler’s method.

• Developed by two German mathematicians Runge and kutta .

• Also called R-K method.

• Runge-kutta method distinguished by their order

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 Illustrating Heun’s Method• Also Known as Runge Kutta

Method• Given dy/dx=f(x,y)and

y(x1 )= y1

• Draw straight line from (x1,y1) with a slope s1=f (x1,y1).

• Draw straight line from (x2,y2)

• Let it cut vertical line through x1+h at slope s2=(x2,y2’)

• Now connect (x1,y1) and (x2,y2) which is slope (s1+s2)/2

• We got y2=y1+(s1+s2)

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Illustrating Runge-Kutta 2th order method• Also Known as Polygon Method• Draw straight line from (x1,y1)

find the value of y where this line cuts vertical line erected at x1+h/2 call it yh

• Calculate f(x1+h/2, yh) which is slope of solution curve at this point

• Go back to (x1,y1) and draw a straight line with slope s2

• We have yi+1=yi+hf(xi+h/2,yi+sih/2 )

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 Illustrating Runge-Kutta 4th order method• Give the point A=(xi,yi) on the

solution curve.• Draw straight line cut the

vertical line erected at xi+h/2 at B this slope s2= f(x1+h/2, yi+sih/2.

• Draw another straight line starting at A with slope s3

• Go back to (x1,y1) and draw a straight line with slope s2

• Let this straight line cut the vertical line erected at xi+h at D

• Find the slope s4 of the solution curve at D

• We haveyi+1=yi+h/6(s1+2s2,2s3+s4)

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Eg: find the numerical solution of the initial value problem described as

y+x/y-x y(0)=1 At x=0.4 and taking h=0.2

Solution:f(x,y)=y+x/y-x h=0.2 x0=0 and y0=1

We have to calculate first : y1=y0+(k1+k2)

K1 =f(x0,y0) =(1+0)/(1-0) =1

K2 =f(x0+h,y0+hk1) =f(0+.2,1+0.2*1) =1.24

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Y1 =1+0.2(1+1.4)=24

X1 =0+0.2=0.2 h=0.2

now calculate y2 =y1+(k1+k2)k1=f(x1,y1) k2=f(x1+h,y1+hk1) =(y+x)/(y-x) =f(0.2+0.2, 1.24+0.2*1.3846) =(1.24+0.2)/1.24-0.2 =1.7162 =1.3846

y2=1.24+(1.3846+1.7162) =1.5500

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 Runge-Kutta 4th order methodyn+1=yn+(k1+2k2+2k3+k4)

k1=f(xn, yn)

k2=f(xn+h/2, yn+h)

k3=f(xn+h/2, yn+h)

k4=f(xn+h, yn+hk3)

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Eg: find the numerical solution of the initial value problem described as y+x/y-x y(0)=1 At x=0.4 and taking h=0.2

Solution:

k1= f(x0,y0) =(1+0)/(1-0) =1

k2=k2=f(xn+h/2, yn+h) =f(0+.2/2, 1+*1) = (0.1, 1+0.1) = 1.2

k3= f(x0+h/2, y0+h) =f(0+0.2/2,1+0.2*) =(0.1, 1.12) =1.1960

f(x,y)=y+x/y-x h=0.2 x0=0 and y0=1

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k4= f(x0+h, y0+hk3) = f(0+0.2,1+0.2*1.196) = (0.2,1.2392) = 1.3848

y1=y0+(k1+2k2+2k3+k4)

=1+(1+2*1.2+2*1.960+1.3848)

= 1+(7.1768) =1.2392

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x1=0.2, y1=1.2392, calculate y2=y1+h/6(k1+2k2+2k3+k4)

k1=f(x1,y1) =f(0.2,1.2392) =(1.4392/1.0392) =1.3849

k2=f(x1+h/2, y1+h) =(0.2+0.1,1.2392+.1384) =(.3,1.3776) =1.5567

k3= f(x1+h/2, y1+h) =f(.2+.2/2,1.2392+.2*) =(0.3, 1.1556) =1.7012

k4= f(x1+h, y1+hk3) = f(.2+.2,1.2392+.3402) = (0.4,1.5794) = 1.6783

Y2= (1.2392+.2/6(1.3849+2*1.5546+2*1.7012+1.6783))

= 1.55836

y2 =y1+(k1+2k2+2k3+k4)

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x2=0.4, y2=1.55836, calculate y3=y2+h/6(k1+2k2+2k3+k4)

k1=f(x2,y2) =f(0.4,1.5583) =(1.9583/1.1583) =1.6906

k2=f(x2+h/2, y2+h) =(0.4+0.1,1.5583+.16906) =(.5,1.7273) =1.8147

k3= f(x2+h/2, y2+h) =f(.4+.2/2,1.5583+.2*) =(0.5, 1.7397) =1.8066

k4= f(x2+h, y2+hk3) = f(.4+.2,1.5583+.18066) = (0.6,1.7389) = 2.0536

Y3= (1.5583+.2/6(1.6906+2*1.8147+2*1.8066+2.0536))

= 1.9245

y3 =y2+(k1+2k2+2k3+k4)

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x3=0.6, y3=1.92452, calculate y4=y3+h/6(k1+2k2+2k3+k4)

k1=f(x3,y3) =f(0.6,1.9245) =(2.5245/1.3245) =1.9060

k2=f(x3+h/2, y3+h) =(0.6+0.1,1.9245+.1906) =(.7,2.1151) =1.9893

k3= f(x3+h/2, y3+h) =f(.6+.2/2,1.9245+.2*) =(0.7, 2.1234) =1.9835

k4= f(x3+h, y3+hk3) = f(.6+.2,1.9245+.1983) = (0.8,2.1228) = 2.2095

Y4= (1.9245+.2/6(1.9060+2*1.9893+2*1.9835+2.2095))

= 2.3265

y4 =y3+(k1+2k2+2k3+k4)

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