This document discusses numerical methods for solving second order ordinary and partial differential equations. It begins by introducing the general form of a second order differential equation and how it can be transformed into a system of first order equations. It then describes the Taylor series and Runge-Kutta methods for solving these equations numerically. Specific Runge-Kutta methods are derived for general and special cases of second order equations. An example demonstrates applying a 4th order Runge-Kutta method.

1. Numerical solution of ordinary and
partial dierential Equations
Module 10: Second order equations
Dr.rer.nat. Narni Nageswara Rao
£
August 2011
1 Second Order Equations
Many dierential equations which appear in practice are systems of the sec-
ond order
y
HH
= f(t; y; y
H
); y(t0) = y0; y
H
(t0) = y
H
0 t0 t b (1)
This is mainly due to the fact that the forces are proportional to acceleration.
i.e, to second derivatives. The equation (1) can be transformed into a

2. rst
order dierential equation of doubled dimension by considering the vector
(y; yH
) as the new variable.
y
yH
!H
=
yH
f(t; y; yH
)
!
y(t0) = y0; (2)
y
H
(t0) = y
H
0
In order to solve (1) numerically, one can for instance apply a Taylor series
method or Runge-Kutta method.
£nnrao maths@yahoo.co.in
1

3. 1.1 Taylor series method of order p
We write the Taylor series method to

4. nd yj+1 and yH
j+1 as
yj+1 = yj + hy
H
j +
h2
2!
y
HH
j + ¡¡¡+
hp
p!
y
(p)
j
y
H
j+1 = y
H
j + hy
HH
j + ¡¡¡+
hp
p!
y
(p+1)
j ; j = 0; 1; ¡¡¡ ; N 1 (3)
The truncation error is y(hp+1
) in both yj+1 and yH
j+1.
The hider order derivatives are calculated as follows:
y
HH
j = f(tj; yj; y
H
j) = fj
y
HHH
j = (ft + y
H
fy + ffy0 )j
=
@
@t
+ y
H @
@y
+ f
@
@yH
fj
= Dfj
y
iv
j = [ftt + (y
H
)2
fyy + f
2
fy0
y0 + 2y
H
fty + 2ffty0
+2y
H
ffyy0 + f
H
y(ft + y
H
fy + ffy0 + ffy]j
= D
2
fj + (fy0 )jDfj + fj(fy)j
where D =
@
@t
+ y
H
j
@
@y
+ fj
@
@yH
and D
2
= D(D) (4)
) yj+1 = yj + hy
H
j +
h2
2
fj +
h3
6
Dfj +
h4
24
¢
D
2
fj + (fy0 )j + fj(fy)j
£
+ ¡¡¡
y
H
j+1 = y
H
j + hfj +
h2
2
Dfj +
h3
6
¢
D
2
fj + (fy0 )jDfj + fj(fy)j
£
+ ¡¡¡ (5)
1.2 Runge-Kutta methods
For the second order initial value problem (1), we de

5. ne the method as
yj+1 = yj + hy
H
j + W1K1 + W2K2
y
H
j+1 = y
H
j +
1
h
(W
£
1 K1 + W
£
2 K2) (6)
K1 =
h2
2!
f(tj; yj; y
H
j)
K2 =
h2
2!
f
tj + c2h; yj + c2hy
H
j + a21K1; y
H
j +
1
h
b21K1
2

6. where c2, a21,b21,W1,W2,W £
1 ,W £
2 are arbitrary constants to be determined.
Note that b21K1=h is an y(h) term, in the argument of f in K2. Because of
the de

7. nition of K1, we choose b21 = 2c2. Expanding K1 and K2 in Taylor
series about tj, we get
K1 =
h2
2
fj
K2 =
h2
2
fj + hc2Dfj +
h2
2
c
2
2D
2
fj + a21fjfy
¡
+ ¡¡¡
!
where D and D2
are as de

8. ned in (4).
Substituting in (6), we obtain
yj+1 = yj + hy
H
j +
h2
2
(W1 + W2)fj +
h3
2
c2W2Dfj +
h4
4
W2c
2
2D
2
fj
+W2a21fjfy) + y(h
5
)
y
H
j+1 = y
H
j +
h
2
(W
£
1 + W
£
2 )fj +
h2
2
c2W
£
2 Dfj +
h3
4
W
£
2 c
2
2D
2
fj
+ W
£
2 a21fjfy) + y(h
4
) (7)
Comparing (7) with (3), we get
W1 + W2 = 1; W
£
1 + W
£
2 = 2
c2W2 =
1
3
; c2W
£
2 = 1 (8)
The coecient of h4
in yj+1 and of h3
in yH
j+1 of equation (7) will not match
with the corresponding coecients in (3) for any choice of c2, a21,W2 and
W £
2 . Therefore,the error is y(h4
) in yj+1 and o(h3
) in yH
j+1. We have four
equations to determine the six parameters c2, a21,W1,W2,W £
1 ,W £
2 . Choose
c2 = 2
3
and a21 = c2. Then, we obtain from (8)
W1 = W2 =
1
2
; c2 = a21 =
2
3
; b21 =
4
3
; W
£
1 =
1
2
; W
£
2 =
3
2
Thus, the Runge-Kutta method (6) becomes
yj+1 = yj + hy
H
j +
1
2
(K1 + K2)
y
H
j+1 = y
H
j +
1
2h
(K1 + 3K2) (9)
K1 =
h2
2
f(tj; yj; y
H
j)
K2 =
h2
2
f
tj +
2
3
h; yj +
2
3
hy
H
j +
2
3
K1; y
H
j +
4
3h
K1
3

9. If we use four evaluations of f, then we get the following Runge-Kutta
method.
yj+1 = yj + hy
H
j +
1
2
(K1 + K2 + K3)
y
H
j+1 = y
H
j +
1
3h
(K1 + 2K2 + 2K3 + K4) (10)
K1 =
h2
2
f(tj; yj; y
H
j)
K2 =
h2
2
f
tj +
h
2
; yj +
h
2
y
H
j +
1
4
K1; y
H
j +
K1
h
K3 =
h2
2
f
tj +
h
2
; yj +
h
2
y
H
j +
1
4
K1; y
H
j +
1
h
K2
K4 =
h2
2
f
tj + h; yj + hy
H
j + K3; y
H
j +
2
h
K3
The error is y(h5
) in yj+1 and y(h4
) in yH
j+1
1.3 Special second order equations
Now, let the function f be independent of yH
. Thus the equation yHH
= f(t; y)
is called the special second order equation. The initial value problem becomes
y
HH
= f(t; y); t0 t b
y(t0) = y0; y
H
(t0) = y
H
0 (11)
Consider the Runge-Kutta method (6) using 2 evaluations of f. Since fy0 = 0,
we have from (4)
y
iv
j = D
2
fj + fj(fy)j
where D = @
@t + yH
j
@
@y and D2
is suitably de

10. ned.
) it is possible to compare y(h3
) terms in yH
j+1 and yH
(tj+1). Comparing
y(h3
) terms in (7) and (3) we get two more equations, besides the four
equations in (8). We now have the equations
W1+W2 = 1; c2W2 =
1
3
; c
2
2W
£
2 =
2
3
; a21W
£
2 =
2
3
; W
£
1 +W
£
2 = 2; c2W
£
2 = 1
This system uniquely determines the six parameters. The solution is
c2 =
2
3
; a21 = c
2
2 =
4
9
; W1 =
1
2
; W2 =
1
2
; W
£
1 =
1
2
; W
£
2 =
3
2
4

11. ) The method is given by
yj+1 = yj + hy
H
j +
1
2
(K1 + K2)
y
H
j+1 = y
H
j +
1
2h
(K1 + 3K2)
K1 =
h2
2
f(tj; yj)
K2 =
h2
2
f
tj +
2
3
h; yj +
2
3
hy
H
j +
4
9
K1
The error in both yj+1 and yH
j+1 is y(h4
).
The Runge-Kutta method using four evaluations for (11) is given by
yj+1 = yj + hy
H
j +
1
96
[23K1 + 75K2 27K3 + 25K4]
y
H
j+1 = y
H
j +
1
96h
(23K1 + 125K2 81K3 + 125K4) (12)
K1 =
h2
2
f(tj; yj)
K2 =
h2
2
f
tj +
2
5
h; yj +
2
5
hy
H
j +
4
25
K1
K3 =
h2
2
f
tj +
2
3
h; yj +
2
3
hy
H
j +
4
9
K1
K4 =
h2
2
f
tj +
4
5
h; yj +
4
5
hy
H
j +
8
25
(K1 + K2)
The error in both yj+1 and yH
j+1 is y(h5
). This method is called the Runge-
Kutta-Nystroem method.
Example Use the Runge-Kutta method (10) for solving the initial value
problem
y
HH
= y
H
y(0) = 1; y
H
(0) = 1
with step length h. Show that
yj+1
yH
j+1
!
= A
yj
yH
j
!
where A is a real 2¢2 matrix. Compute the solution at t = 0:2 with h = 0:1.
Compare with the exact solution y(t) = et
.
5

12. Solution We have
f(t; y; y
H
) = y
H
) K1 =
h2
2
f(tj; yj; y
H
j) =
h2
2
y
H
j
K2 =
h2
2
f
tj +
h
2
; yj +
h
2
y
H
j +
1
h
K1; y
H
j +
K1
h
=
h2
2
y
H
j +
K1
h
=
h2
2
y
H
j +
h
2
y
H
j
=
h2
2
1 +
h
2
y
H
j
K3 =
h2
2
f
tj +
h
2
; yj +
h
2
y
H
j +
1
4
K1; y
H
j +
K2
h
=
h2
2
y
H
j +
K2
h
!
=
h2
2
y
H
j +
h
2
1 +
h
2
y
H
j
!
=
h2
2
1 +
h
2
+
h2
4
!
y
H
j
K4 =
h2
2
f
tj + h; yj + hy
H
j + K3; y
H
j +
2
h
K3
=
h2
2
y
H
j +
2
h
K2
!
=
h2
2
y
H
j + h
1 +
h
2
+
h2
4
y
H
j
!
=
h2
2
1 + h +
h2
2
+
h4
4
y
H
j
yj+1 = yj + hy
H
j +
1
3
(K1 + K2 + K3)
= yj + hy
H
j +
1
3
h2
2
y
H
j +
h2
2
1 +
h
2
y
H
j +
h2
2
1 +
h
2
+
h2
4
y
H
j
!
= yj + h +
h2
2
+
h3
6
+
h4
24
!
y
H
j
y
H
j+1 = y
H
j +
1
3h
(K1 + 2K2 + 2K3 + K4)
= y
H
j +
1
3h
h2
2
y
H
j + h
1 +
h
2
y
H
j + h
2
1 +
h
2
+
h2
4
!
y
H
j
+
h2
2
1 + h +
h2
2
+
h3
4
!
y
H
j
'
= 1 + h +
h2
2
+
h3
6
+
h4
24
!
y
H
j
6

13. ) yj+1
yH
j+1
!
=
1 E(h) 1
0 E(h)
!
yj
yH
j
!
where E(h) = 1 + h +
h2
2
+
h3
6
+
h4
24
Thus, we

14. nd
A =
1 E(h) 1
0 E(h)
!
The matrix A for h = 0:1 becomes
A =
1 0:1051708
0 1:1051708
!
For j = 0, we have
y0 = 1; y
H
0 = 1; t1 = 0:1
y1
yH
1
!
=
1 0:1051708
0 1:1051708
!
1
1
!
=
1:1051708
1:1051708
!
For j = 1, we have
y1 = 1:1051708; y
H
1 = 1:1051708; t2 = 0:2
y2
yH
2
!
=
1 0:1051708
0 1:1051708
!
1:1051708
1:1051708
!
=
1:2214025
1:2214025
!
The exact solution is y(t) = et
and the exact values at t = 0:2 are given by
y2
yH
2
!
=
1:2214028
1:2214028
!
Example Solve the initial value problem
y
HH
= (1 + t
2
)y; y(0) = 1; y
H
(0) = 0; t[0; 0:4]
by the Runge-Kutta-Nystrom method with h = 0:2. Compare with the exact
solution y(t) = et2=2
.
Solution For j = 0, we have
t0 = 0; y0 = 1; y
H
0 = 0
K1 =
h2
2
f(t0; y0) =
h2
2
(1 + t
2
0)y0 =
(0:2)2
2
(1 + 0)1 = 0:02
7

15. K2 =
h2
2
f
t0 +
2
5
h; y0 +
2
5
hy
H
0 +
4
25
K1
=
h2
2
4
1 +
t0 +
2
5
h
2
5
y0 +
2
5
hy
H
0 +
4
25
K1
!
=
(0:2)2
2
¢
1 + (0:08)2
£
1 + 0 +
4
25
¢0:02
!
= (0:02)(1:0064)(1:0032) = 0:0201924
K3 =
h2
2
f
t0 +
2
3
h; y0 +
2
3
hy
H
0 +
4
9
K1
=
h2
2
4
1 +
t0 +
2
3
h
2
5
y0 +
2
3
hy
H
0 +
4
9
K1
!
=
(0:2)2
2
¢
1 + (0:1333333)2
£
1 + 0 +
4
9
(0:0201924)
!
= (0:02)(1:0177778)(1:0089744)
= 0:0205382
K4 =
h2
2
f
t0 +
4
5
h; y0 +
4
5
hy
H
0 +
8
25
(K1 + K2)
=
h2
2
4
1 +
t0 +
4
5
h
2
5
y0 +
4
5
hy
H
0 +
8
25
(K1 + K2)
!
=
(0:2)2
2
¢
1 + (0:16)2
£
1 + 0 +
8
25
(0:0401924)
!
= (0:02)(0:0256)(1:0128616)
= 0:0207758
y1 = y0 + hy
H
0 +
1
96
(23K1 + 75K2 27K3 + 25K4)
= 1 + 0 +
1
96
[23(0:02) + 75(0:0201924) 27(0:0205382) + 25(0:0207758)]
= 1 +
1
96
(1:9392936) = 1:0202010
y
H
1 = y
H
0 +
1
96h
(23K1 + 125K2 81K3 + 125K4)
= 0 +
1
96(0:2)
[23(0:02) + 125(0:0201924) 81(0:0205382) + 125(0:0207758)]
= 0 +
1
19:2
(3:9174308) = 0:2040329
8

16. For j = 1, we have
t1 = 0:2; y2 = 1:0202010; y
H
1 = 0:2040329
K1 =
h2
2
f(t1; y1) =
h2
2
(1 + t
2
1)y2 =
(0:2)2
2
[1 + (0:2)2
](1:0212202)
= (0:02)(1:04)(1:0202010) = 0:0212202
K2 =
h2
2
f
t1 +
2
5
h; y1 +
2
5
hy
H
1 +
4
25
K1
=
h2
2
4
1 +
1 +
2
5
h
2
5
y1 +
2
5
hy
H
1 +
4
25
K1
!
=
(0:2)2
2
[1 + (0:28)2
][1:0202010 + (0:08)(0:2040329) + (0:16)(0:0212202)]
= (0:02)(1:0784)(1:0399189) = 0:0224290
K3 =
h2
2
f
t1 +
2
3
h; y1 +
2
3
hy
H
1 +
4
9
K1
=
h2
2
4
1 +
1 +
2
3
h
2
5
y1 +
2
3
hy
H
1 +
4
9
K1
!
=
(0:2)2
2
[1 + (0:3333333)2
][1:0202010 + (0:133333)(0:2040329)
+(0:4444444)(0:0212202)]
= (0:02)(1:1111111)(1:0568366) = 0:0234853
K4 =
h2
2
f
t1 +
4
5
h; y1 +
4
5
hy
H
1 +
8
25
(K1 + K2)
=
h2
2
1 + (t1 +
4
5
h)2
!
y1 +
4
5
hy
H
1 +
8
25
(K1 + K2)
!
=
(0:2)2
2
[1 + (0:36)2
][1:0202010 + (0:16)(0:2040329) + (0:32)(0:0436492)
= (0:02)(1:1296)(1:0668140)
= 0:0241015
y2 = y1 + hy
H
1 +
1
96
[23K1 + 75K2 27K3 + 25K4)]
= 1:0202010 + (0:2)(0:2040329) +
1
96
[23(0:0212202)
+75(0:0224290) 27(0:0234853) + 25(0:0241015)]
= 1:0202010 + 0:0408066 +
1
96
(2:1386740)
= 1:0832855
9

17. y
H
2 = y
H
1 +
1
96h
(23K1 + 125K2 81K3 + 125K4)
= 0:2040329 +
1
19:2
[23(0:0212202) + 125(0:0224290) 81(0:0234853)
+125(0:0241015)]
= 0:2040329 +
1
19:2
(4:4020678) = 0:4333073
) we have
y(0:2) % y1 = 1:0202010; y
H
(0:2) % y
H
1 = 0:2040329;
y(0:4) % y2 = 1:0832855; y
H
(0:4) % y
H
2 = 0:4333073
The exact solution is y(t) = et2=2
. The exact values are given by
y(0:2) = 1:02020134; y
H
(0:2) = 0:20404027
y(0:4) = 1:0832871; y
H
(0:4) = 0:43331484
10