Runge-Kutta Method :

Runge-Kutta method here after called as RK method is the generalization of the concept used in Modified Euler's method.

In Modified Eulers method the slope of the solution curve has been approximated with the slopes of the curve at the end points of the each sub interval in computing the solution.  The natural generalization of this concept is computing the slope by taking a weighted average of the slopes taken at more number of points in each sub interval.  However, the implementation of the scheme differes from Modified Eulers method so that the developed algorithm is explicit in nature.  The final form of the scheme is of the form

y_i+1  =  y_i +  (weighted average of the slopes)             for i = 0, 1, 2 . . .
where h is the step length and y_i and y_i+1 are the values of y at x_i and x_i+1 respectively.

In genaral, the slope is computed at various points x_s in each sub interval [x_i, x_i+1] and multiplied them with the step length h and then weighted average of it is then added to y_i to compute y_i+1.  Thus the RK method with v slopes called as v-stage RKmethod can be written as

K₁  =  h f(x_i, y_i)
K₂  =  h f(x_i + c₂h, y_i + a₂₁K₁)
K₃  =  h f(x_i + c₃h, y_i + a₃₁K₁+ a₃₂K₂)
. . .
. . .
. . .

K_v  =  h f(x_i + c_vh, y_i + a_v1K₁+ a_v2K₂ + . . . +a_vv-1K_v-1)
and

y_i+1  =  y_i +  (W₁ K₁ + W₂ K₂ + . . . + W_v K_v )             for i = 0, 1, 2 . . .

To determine the parameters c's, a's and W's in the above equation, y_i+1 defined in the scheme is  expanded interms of steplengh h and the resultant equation is then compared with Taylor series expansion of the solution of the differential equation upto a certain number of terms say p.  Then the v-stage RKmethod will be of order p or is an p^th order RK method.  Here for any v>4 the maximum possible order p of the R Kmethod is always less than v.  However, for any v lessthan or equal to 4, it is possible derive an RK method of order p = v.  Now, consider the case v = 2 to derive the 2-stage RK method.  For this

K₁  =  h f(x_i, y_i)
K₂  =  h f(x_i + c₂h, y_i + a₂₁K₁)
y_i+1  =  y_i +  ( W₁ K₁ + W₂ K₂ )                                             for i = 0, 1, 2 . . .

Now by Taylor series expansion

y(x_i+1) = y(x_i) + h y'(x_i) + h² y''(x_i)/2! + h³ y'''(x_i)/3! + o(h⁴)
            = y(x_i) + h f + h²  ( f_x + f_yf ) / 2! + h³( f_xx + 2f_xyf + f_yy f² + f_y( f_x + f_yf )) / 3! + o(h⁴)

Also

K₁  =  h f_i
K₂  =  h f(x_i + c₂h, y_i + a₂₁K₁)
      =  h( f_i + c₂h f_x +  a₂₁K₁f_y + ( c₂h)² f_xx /2! + (a₂₁K₁)² f_yy /2! + c₂h a₂₁K₁ f_xy + o(h⁴) )
    =  h( f_i + c₂h f_x +  a₂₁h f_i f_y + ( c₂h)² f_xx /2! + (a₂₁h f_i)² f_yy /2! + c₂h a₂₁h f_i f_xy + o(h⁴) )
y_i+1 = y_i + (W₁+W₂) h f_i + h²(W₂c₂f_x + W₂a₂₁f f_y) + h³ W₂(c₂²a₂₁f f_xy + a²₂₁f² f_yy)/2 + o(h⁴)

Now by comparing the equal powers of h iny_i+1and y(x_i+1) we get

W₁ + W₂ = 1                 c₂W₂ = 1/2       and            a₂₁W₂ =  1/2

The solution of this system is

a₂₁ =  c₂,       W₂ = 1/(2c₂)        and        W₁ = 1  -  1/(2c₂)

where c₂ is any arbitrary constant not equal to zero.  For these values of  a₂₁, W₂ , W₁, since 2-stage RK method compares with Taylor series upto h²for any value of c₂ the 2-stage RK method is of order two and hence this scheme is denoted in many text books as a second order RK method.  Now, to give some numerical values to a₂₁, W₂ , W₁ first the valuec₂ of  has to be fixed.  Generally the value of c₂ is fixed such that the values of a₂₁, W₂ , W₁ are integers or some real numbers which easy to remember.  Two of such cases are c₂ = 1/2 and c₂=1.

Case (i):  c₂ = 1/2    Þ    a₂₁= 1/2, W₂ = 1, W₁ = 0.  The corresopnding 2-stage (second order) RK method is

K₁  =  h f(x_i, y_i)
K₂  =  h f(x_i + h/2, y_i + K₁ /2 )
y_i+1  =  y_i +  (  K₂ )                                             for i = 0, 1, 2 . . .

or equivalently

y_i+1  =  y_i +  h f(x_i + h/2, y_i + h f(x_i, y_i) /2)       for i = 0, 1, 2 . . .

Which is knothing but Eulers method with step length h = 1/2.

Case (ii):  c₂ = 1    Þ    a₂₁= 2, W₂ =  W₁ = 1/2.  The corresopnding 2-stage (second order) RK method is

K₁  =  h f(x_i, y_i)
K₂  =  h f(x_i + h, y_i + K₁ )
y_i+1  =  y_i +  (  K₁ + K₂ )/2                                             for i = 0, 1, 2 . . .
or equivalently

y_i+1  =  y_i + .5 h (f(x_i, y_i) + f(x_i + h, y_i + h f(x_i, y_i) ))       for i = 0, 1, 2 . . .

Which is knothing but the Modified Eulers method.

Following the same procedure one can develope the higher order RK methods by giving various values to v and comparing the obtained y_i+1 with the same obtained by Taylor series method.  Classical RK methods of order three and four are
 

1	RK method of order three (v = 3)	K1  =  h f(xi, yi)    K2  =  h f(xi + h/2, yi + K1 /2)    K3  =  h f(xi + h, yi  - K1  + 2K2 )    yi+1  =  yi +  (  K1 + 4K2 + K3 )/6
2	RK method of order three (v = 4)	K1  =  h f(xi, yi)    K2  =  h f(xi + h/2, yi + K1 /2)    K3  =  h f(xi + h/2, yi + K2 /2)    K4  =  h f(xi + h, yi  + K3 )    yi+1  =  yi +  (  K1 + 2K2 + 2K3 + K4 )/6

 

Worked out problems

Example 1	Find   y(1.0)  using RK method of order four by solving the IVP  y' = -2xy2,  y(0) = 1 with step length 0.2. Also compre the solution obtained with RK methods of order three and two.	Solution
Example 2	Find  y  in  [0,3] by solving the initial value problem y' = (x - y)/2,  y(0) = 1 using RK method of order four with  h = 1/2 and 1/4.	Solution
Example 3	Using RK method of order four find y(0.1) for y' = x - y2,  y(0) = 1.	Solution
Example 4	Using RK method of order four find y at x = 1.1 and 1.2 by solving y' = x2 + y2 ,  y(1) = 2.3	Solution

 
 

Problems to workout