Fourth Order

RK-Method
The most commonly used method is
Runge-Kutta fourth order method.

The fourth order RK-method is

1
yi+1 = yi + (k1 + 2k2 + 2k3 + k4 ),
6

Ordinary Differential Equations (ODE) – p.44/89

where
k1 = hf (xi , yi ),
h k1
!
k2 = hf xi + , yi + ,
2 2
h k2
!
k3 = hf xi + , yi + ,
2 2
k4 = hf (xi + h, yi + k3 ).

Ordinary Differential Equations (ODE) – p.45/89

 Example 1
Find the approximate solution of the initial value
problem
dx x
=1+ , 1≤t≤3
dt t
with the initial condition
x(1) = 1,
using the Runge-Kutta second order and fourth
order with step size of h = 1.

Ordinary Differential Equations (ODE) – p.46/89

 Solution
RK 2nd order method. The formula is
h
yi+1 = yi + (k1 + k2 ) ,
2
where k1 = f (xi , ti ), k2 = f (xi + h, ti + hk1 ).

Here, h = 1 and t0 = x0 = 1. Therefore,

x0
k1 = f (t0 , x0 ) = 1 + = 2,
t0
3
k2 = f (t0 + h, x0 + hk1 ) = f (2, 3) = 1 + = 2.5.
2
Ordinary Differential Equations (ODE) – p.47/89
Thus,
1
y1 = 1 + (2 + 2.5) = 3.25.
2
Similary, we calculate y2 .

Ordinary Differential Equations (ODE) – p.48/89

RK 4th order method. The formula is
1
yi+1 = yi + (k1 + 2k2 + 2k3 + k4 ),
6
where
k1 = hf (ti , xi ),
h k1
!
k2 = hf ti + , xi + ,
2 2
h k2
!
k3 = hf ti + , xi + ,
2 2
k4 = hf (ti + h, xi + k3 ). Ordinary Differential Equations (ODE) – p.49/89
k1 = hf (t0 , x0 ) = f (1, 1) = 2,
h k1
!
k2 = hf t0 + , x0 + = f (1.5, 2) = 2.333333,
2 2
h k2
!
k3 = hf t0 + , x0 + = 2.444444,
2 2
k4 = hf (t0 + h, x0 + k3 ) = 2.722222.

Ordinary Differential Equations (ODE) – p.50/89

We find
1
y1 = 1 + (2 + 2 ∗ 2.333333
6
+ 2 ∗ 2.444444 + 2.722222)
= 3.379630.
Similarly, we calculate y2 .

Ordinary Differential Equations (ODE) – p.51/89

 System of First Order
ODE’s
We consider the n system of first order equations
dy1
= f1 (x, y1 , y2 , · · · , yn ), y1 (x0 ) = y10 ,
dx
dy2
= f2 (x, y1 , y2 , · · · , yn ), y2 (x0 ) = y20 ,
dx
···
dyn
= fn (x, y1 , y2 , · · · , yn ), yn (x0 ) = yn0 .
dx
With the help of single step methods, we find the
approximate solution for the system of n
equations. Ordinary Differential Equations (ODE) – p.52/89
We now consider the two first order equations of
the form
dy1
= f1 (x, y1 , y2 ), y1 (x0 ) = y10 ,
dx
dy2
= f2 (x, y1 , y2 ), y2 (x0 ) = y20 .
dx
We use RK second order and fourth order
method to solve the system of equations.

Ordinary Differential Equations (ODE) – p.53/89

 2nd Order RK Method
The formula for second order RK method with
step size h is
1
y1,n+1 = = y1,n + (s1 + k1 ),
2
1
y2,n+1 = y2,n + (s2 + k2 ),
2
where
k1 = hf1 (xn , y1n , y2n ),
k2 = hf2 (xn , y1n , y2n ),
Ordinary Differential Equations (ODE) – p.54/89
 s1 = hf1 (xn + h, y1n + k1 , y2n + k2 ),
s2 = hf2 (xn + h, y1n + k1 , y2n + k2 ).
Similarly, the formula for fourth order RK-method
is
1
y1,n+1 = y1,n + (k1 + 2s1 + 2l1 + p1 ),
6
1
y2,n+1 = y2,n + (k2 + 2s2 + 2l2 + p2 ),
6

Ordinary Differential Equations (ODE) – p.55/89

where
k1 = hf1 (xn , y1n , y2n ),
k2 = hf2 (xn , y1n , y2n ),
h k1 k2
s1 = hf1 (xn + , y1n + , y2n + ),
2 2 2
h k1 k2
s2 = hf2 (xn + , y1n + , y2n + )
2 2 2

Ordinary Differential Equations (ODE) – p.56/89

 h s1 s2
l1 = hf1 (xn + , y1n + , y2n + ),
2 2 2
h s1 s2
l2 = hf2 (xn + , y1n + , y2n + ),
2 2 2
p1 = hf1 (xn + h, y1n + l1 , y2n + l2 ),
p2 = hf2 (xn + h, y1n + l1 , y2n + l2 ).

Ordinary Differential Equations (ODE) – p.57/89

 Example 2
Use RK-method 2nd order and 4th order to find
the approximate solution of y(0.1) and z(0.1) as a
solution of pair of equations
dy
= x + z,
dx
dz
= y−x
dx
with the initial conditions
y(0) = 1, z(0) = −1.
Take step size h = 0.1. Ordinary Differential Equations (ODE) – p.58/89
 Multistep Methods
The general form of a linear m-step multistep
method is
yn+1 − a1 yn − a2 yn−1 − · · · − am yn+1−m
hi
= b0 f (xn+1 , yn+1 ) + b1 f (xn , yn )
+ · · · + bm f (xn+1−m , yn+1−m ).
When b0 = 0, the method is called explicit,
otherwise, it is said to be implicit.

In multistep method, we require multiple starting

values. Ordinary Differential Equations (ODE) – p.64/89
 Adams-Bashforth
Methods
We assume a uniform discretization in the
x-domain, i.e., we define
xi = a + ih,
where h = N ,
b−a
for some positive integer N .

Let the given differential equation is of the form

y 0 (x) = f (x, y).
We now integrate the given differential equation
from x = xi to x = xi+1 .
Ordinary Differential Equations (ODE) – p.65/89
This yields the equation
Z x
i+1
y(xi+1 ) − y(xi ) = f (x, y(x)) dx.
xi

Next, we write
f (x, y(x)) = Pm−1 (x) + Rm−1 (x),

Ordinary Differential Equations (ODE) – p.66/89

where
m
X
Pm−1 (x) = Lm−1,j (x)f (xi+1−j , y(xi+1−j )),
j=1

is the Lagrange form of the polynomial of degree

at most m − 1 that interpolates f at the m points
xi , xi−1 , xi−2 , · · · , xi+1−m .

Ordinary Differential Equations (ODE) – p.67/89

And
f (m) (ξ, y(ξ)) m
Rm−1 (x) = Πj=1 (x − xi+1−j )
m!
is the corresponding remainder term.

Ordinary Differential Equations (ODE) – p.68/89

 Two-Step Adams
Bashforth Method
Since m = 2, we write
x − xi−1
f (x, y(x)) = f (xi , yi )
xi − xi−1
x − xi
+ f (xi−1 , yi−1 )
xi−1 − xi
f 00 (ξ, y(ξ))
+ (x − xi )(x − xi−1 ).
2
Integrating the Lagrange polynomials yields

Ordinary Differential Equations (ODE) – p.69/89

 Z x
i+1 x − xi−1 3h
b1 = dx = ,
xi xi − xi−1 2
Z x
i+1 x − xi h
b2 = dx = − .
xi xi−1 − xi 2

Ordinary Differential Equations (ODE) – p.70/89

Therefore, the two-step Adams-Bashforth
method is
h
yn+1 = yn + [3f (xn , yn ) − f (xn−1 , yn−1 )] .
2
Proceeding in a similar manner, the three step
Adams-bashforth method is
h
yn+1 = yn + [23f (xn , yn ) − 16f (xn−1 , yn−1 )
12
+ 5f (xn−2 , yn−2 )].

Ordinary Differential Equations (ODE) – p.71/89

The four-step Adams-Bashforth method is
h
yn+1 = yn + [55f (xn , yn ) − 59f (xn−1 , yn−1 )
24
+ 37f (xn−2 , yn−2 ) − 9f (xn−3 , yn−3 ].

Ordinary Differential Equations (ODE) – p.72/89

 Example 5
Use two-step Adams-Bashforth method to find
the approximate solution of
dx x
= 1+ ,
dt t
x(1) = 1,
near x(1.5). Take step size h = 0.5.

Ordinary Differential Equations (ODE) – p.73/89

 Solution
The two-step Adams-Bashforth formula is
h
xn+1 = xn + [3f (tn , xn ) − f (tn−1 , xn−1 )] .
2
We note that for finding x2 , we require x1 and x0 .

We calculate x1 with the help of second order

Taylor’s method
x1 = 2.125.

Ordinary Differential Equations (ODE) – p.74/89

Now,
h
x2 = x1 + [3f (t1 , x1 ) − f (t0 , x0 )] ,
2
= 3.4375.

Ordinary Differential Equations (ODE) – p.75/89

 Adams-Moulton
Methods
The derivation of Adams-Moulton methods
follows exactly the same procedure as the
derivation of the Adams-Bashforth method with
one exception.

In addition to interpolating f at
xi , xi−1 , xi−2 , · · · , xi+1−m , we also interpolate at
xi+1 .

Ordinary Differential Equations (ODE) – p.76/89

Hence, we write
f (x, y(x)) = Pm (x) + Rm (x),
where
m
X
Pm (x) = Lm,j (x)f (xi+1−j , y(xi+1−j )),
j=0

and
f (m+1) (ξ, y(ξ)) m
Rm (x) = Πj=0 (x − xi+1−j ).
(m + 1)!
Ordinary Differential Equations (ODE) – p.77/89
Since Pm (x) contains a term involving
f (xi+1 , y(xi+1 )), the resulting method will be
implicit.

Furthermore, by using an additional point in the

interpolating polynomial the degree of the
remainder term is increased by one over the
Adams-Bashforth case.

Ordinary Differential Equations (ODE) – p.78/89

 2-Step Adams-Moulton Method

Since m = 2, we write
(x − xi )(x − xi−1 )
f (x, y(x)) = f (xi+1 , yi+1 )
(xi+1 − xi )(xi+1 − xi−1 )
(x − xi+1 )(x − xi−1 )
+ f (xi , yi )
(xi − xi+1 )(xi − xi−1 )
(x − xi+1 )(x − xi )
+ f (xi−1 , yi−1 )
(xi−1 − xi+1 )(xi−1 − xi )
f 000 (ξ, y(ξ))
+ (x − xi+1 )(x − xi )(x − xi−1 ).
6
Ordinary Differential Equations (ODE) – p.79/89
Integrating the Lagrange polynomials yields
Z x
i+1 (x − xi )(x − xi−1 ) 5h
b0 = dt = ,
xi (xi+1 − xi )(xi+1 − xi−1 ) 12
Z x
i+1 (x − xi+1 )(x − xi−1 ) 2h
b1 = dt = ,
xi (xi − xi+1 )(xi − xi−1 ) 3
Z x
i+1 (x − xi+1 )(x − xi ) h
b2 = dt = − .
xi (xi−1 − xi+1 )(xi−1 − xi ) 12

Ordinary Differential Equations (ODE) – p.80/89

Therefore, the two-step Adams-Moulton method
is
yi+1 − yi 5
= f (xi+1 , yi+1 )
h 12
2 1
+ f (xi , yi ) − f (xi−1 , yi−1 ).
3 12
To find the approximate solution, we required y0
and y1 .

The initial condition give y0 and y1 can be

obtained using any third order one-step method.
Ordinary Differential Equations (ODE) – p.81/89
Similarly, we derive the three-step
Adams-Moulton method is given by
h
yi+1 = yi + [9f (xi+1 , yi+1 )
24
+19f (xi , yi ) − 5f (xi−1 , yi−1 )
+f (xi−2 , yi−2 )].
The required starting values for the three step
method should be obtained using a fourth-order
one-step method.

Ordinary Differential Equations (ODE) – p.82/89

 Predictor-Corrector
Method
The most popular of the predictor-corrector
scheme is the Adams fourth-order
predictor-corrector method.

This uses the four-step, fourth-order

Adams-Bashforth method
(0) h
yn+1 = yn + [55f (xn , yn ) − 59f (xn−1 , yn−1 )
24
+ 37f (xn−2 , yn−2 ) − 9f (xn−3 , yn−3 )],
as a predictor
Ordinary Differential Equations (ODE) – p.83/89
followed by the three-step, fourth-order
Adams-Moulton method
(1) h (0)
yn+1 = yn + [9f (xn+1 , yn+1 ) + 19f (xn , yn )
24
− 5f (xn−1 , yn−1 ) + f (xn−2 , yn−2 )],
as a corrector.

Ordinary Differential Equations (ODE) – p.84/89

Note that the computation of y4 , we require the
(0)

values (x0 , y0 ), (x1 , y1 ), (x2 , y2 ), (x3 , y3 ).

These values should be given with the given

equation or should be computed using single
step methods like, Euler’s method, Taylor’s
method or Runge-Kutta methods.

To compute the value y4 , we require the values

(1)

(x1 , y1 ), (x2 , y2 ), (x3 , y3 ) and

(0)
(x4 , y4 ).
Ordinary Differential Equations (ODE) – p.85/89
 Modifier
Let y(xn+1 ) denotes the true solution of y at xn+1 .
Then
(0) 251 5 (iv)
y(xn+1 ) = yn+1 + h f (ξ1 ), xn−3 < ξ1 < xn+1 ,
720
(1) 19 5 (iv)
y(xn+1 ) = yn+1 − h f (ξ2 ), xn−2 < ξ2 < xn+1 .
720
On subtracting, we get
(0) (1) 270 5 (iv)
0= yn+1 − yn+1 + h f (ξ).
720
Ordinary Differential Equations (ODE) – p.86/89
This implies that
(1) (0) 270 5 (iv)
yn+1 − yn+1 = h f (ξ).
720
We rewrite as
h5 (iv) 1 (1) (0)
f (ξ) = (yn+1 − yn+1 ).
720 270

Ordinary Differential Equations (ODE) – p.87/89

Finally, we find the modifier as
(1) 19 (1) (0)
y(xn+1 ) − yn+1 =− (yn+1 − yn+1 ) = Dn+1 .
270
Here, Dn+1 is called as modifier.

Ordinary Differential Equations (ODE) – p.88/89

 Example
Find y(1.4) by Adams Moulton 4th order
predictor-corrector pair with modifier as a
solution of
dy
= x3 + xy,
dx
y(1) = 2, y(1.1) = 1.6, y(1.2) = 0.34 and
y(1.3) = 0.594 with spacing h = 0.1.

Ordinary Differential Equations (ODE) – p.89/89