math

1. 1
Numerical Solution of Ordinary
Differential Equations (ODE):
Boundary Value Problem (BVP)
Lecture-3

2. 2
Specific Aims
 Applications
 Introduction of Boundary value problem (BVP)
 Discuss finite difference method
 Examples of the solution of BVP by using finite difference
method
 Multiple questions
 Exercises

3. 3
Applications
 Science and Engineering
 Introduction of BVP
Problems involving second and higher order differential
equations, we may prescribed the conditions at two or more
points. Such problems are called boundary value problems
(BVP).
 Finite Difference Formula
 Forward difference formula
 Backward difference formula
 Central difference formula

4. 4
Forward difference formula
𝑦′
=
1
ℎ
𝑦𝑛+1 − 𝑦𝑛
For first order
𝑦′′
=
1
ℎ2
𝑦𝑛+2 − 2𝑦𝑛+1 + 𝑦𝑛
For second order
For third order 𝑦′′′
=
1
ℎ3
𝑦𝑛+3 − 3𝑦𝑛+2 + 3𝑦𝑛+1 − 𝑦𝑛

5. 5
Backward difference formulas:
𝑦′ =
1
ℎ
𝑦𝑛 − 𝑦𝑛−1
For first order
𝑦′′
=
1
ℎ2
𝑦𝑛 − 2𝑦𝑛−1 + 𝑦𝑛−2
For second order
For third order 𝑦′′′
=
1
ℎ3
𝑦𝑛 − 3𝑦𝑛−1 + 3𝑦𝑛−2 − 𝑦𝑛−3

6. 6
Central Difference formula
For first order 𝑦′
=
1
2ℎ
𝑦𝑛+1 − 𝑦𝑛−1
For second order 𝑦′′ =
1
ℎ2
𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1
For third order 𝑦′′′ =
1
2ℎ3
𝑦𝑛+2 − 2𝑦𝑛+1 + 2𝑦𝑛−1 − 𝑦𝑛−2

7. 7
Examples
Question 1#: Consider the boundary value problem
𝑦′′ − 𝑦′ = 1, 𝑦 0 = 1, 𝑦(1) = 2 𝑒 − 1
(a) Derive a recurrence relation using three points central
difference approximations with ℎ = 1 3.
(b) Using the above finite difference formula solve the
above BVP.
Solution:
(a) Using central difference formulas, we have
𝑦′
=
1
2ℎ
𝑦𝑛+1 − 𝑦𝑛−1
𝑦′′ =
1
ℎ2
𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1

8. 8
Given ODE is
𝑦′′ − 𝑦′ = 1
Or,
1
ℎ2 𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1 −
1
2ℎ
𝑦𝑛+1 − 𝑦𝑛−1 = 1
Taking ℎ =
1
3
,
Or
1
(
1
3
)2
𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1 −
1
2(
1
3
)
𝑦𝑛+1 − 𝑦𝑛−1 = 1
Or 9 𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1 −
3
2
𝑦𝑛+1 − 𝑦𝑛−1 = 1
Or 21𝑦𝑛−1 − 36𝑦𝑛 + 15𝑦𝑛+1 = 2

9. 9
(b) With ℎ = 1 3, the nodal points are 𝑥0 = 0, 𝑥1 =
1
3
, 𝑥2 =
2
3
,
𝑥3 = 1.
For 𝑛 = 1, 𝑥1 =
1
3
, 𝑦0 = 1,
−36𝑦1 + 15𝑦2 = 2 − 21 = −19, (1)
Solving the equations (1) and (2), we have
𝑦1 = 1.4548 and 𝑦2 = 2.2250
For 𝑛 = 2, 𝑥2 =
2
3
, 𝑦3 = 2(𝑒 − 1) ,
21𝑦1 − 36𝑦2 = 2 − 30(𝑒 − 1) = −49.5485, (2)
[Since 𝑒 = 2.718283333]
21𝑦𝑛−1 − 36𝑦𝑛 + 15𝑦𝑛+1 = 2

10. 10
Question 2#:
(a) Use three point central difference formula for
derivative to derive a recurrence relation for the above
IVP.
Given that 𝑦′ = 2𝑥𝑦2 − 𝑦, where 𝑦 = 1 at 𝑥 = 0.
(b) Estimate the values of y at 𝑥 = 0.4 and 0.6 using this
recurrence relation.
Three point central difference formula for derivatives is as
follows:
𝑦𝑛
′ ≈
𝑦𝑛+1 − 𝑦𝑛−1
2ℎ
Solution:
𝑦′𝑛 = 2𝑥𝑛𝑦𝑛
2 − 𝑦𝑛
(a) Differential equation at (𝑥𝑛, 𝑦𝑛) is

11. 11
Using three point central difference formula for derivatives
with the differential equation at (𝑥𝑛, 𝑦𝑛) we have
𝑦𝑛
′ ≈
𝑦𝑛+1 − 𝑦𝑛−1
2ℎ
= 2𝑥𝑛𝑦𝑛
2 − 𝑦𝑛
𝑦𝑛+1 = 𝑦𝑛−1 + 2ℎ 2𝑥𝑛𝑦𝑛
2
− 𝑦𝑛
(b) For ℎ = 0.2 with the starting values 𝑥0 = 0, 𝑦0 = 1 . For
getting the values of 𝑥1 and 𝑦1, apply Runge-Kutta four order
method. Here same ODE is used with same initial condition.
The formula of second order Runge-Kutta method:
𝑦1 = 𝑦0 +
1
6
[𝑘1 + 2𝑘2 +2𝑘3 +𝑘4]

12. 12
𝑘1 = ℎ 𝑓(𝑥0, 𝑦0)
𝑘2 = ℎ 𝑓(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
)
𝑘3 = ℎ 𝑓(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
)
𝑘4 = ℎ 𝑓(𝑥0 + ℎ, 𝑦0 + 𝑘3)
where
[Since ℎ = 0.2, 𝑥0 =
0, 𝑦0 = 1 ]
Now
𝑘1 = ℎ 𝑓 𝑥0, 𝑦0
= ℎ [2𝑥0𝑦0
2 − 𝑦0]
= 0.2 [2 0 1)2 − 1 = −0.2
[Since 𝑓 𝑥, 𝑦 = 2𝑥𝑦2 − 𝑦 ]

13. 13
𝑘2 = ℎ 𝑓(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
)
= ℎ 𝑓(0 +
0.2
2
, 1 +
−0.2
2
) [Since 𝑥0 = 0, 𝑦0 = 1,
𝑘1= - 0.2, h=0.2]
= 0.2 𝑓(0.1, 0.9)
[Since 𝑓 𝑥, 𝑦 = 2𝑥𝑦2
− 𝑦 ]
= 0.2 [2 0.1 0.9)2
− 0.9
= −0.1476
𝑘3 = ℎ 𝑓(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
)
= ℎ 𝑓(0 +
0.2
2
, 1 +
−0.1476
2
) [Since 𝑥0 = 0, 𝑦0 = 1,
𝑘2= -0.1476, h=0.2]
[Since 𝑓 𝑥, 𝑦 = 2𝑥𝑦2 − 𝑦 ]
= 0.2 𝑓(0.1, 0.926)
= 0.2 [2 0.1 0.926)2 − 0.926
= −0.1509

14. 14
𝑘4 = ℎ 𝑓(𝑥0 + ℎ, 𝑦0 + 𝑘3)
= ℎ 𝑓(0 + 0.2, 1 − 0.1509)
[Since 𝑥0 = 0, 𝑦0 = 1, 𝑘3= - 0.1509, h=0.2]
[Since 𝑓 𝑥, 𝑦 = 2𝑥𝑦2 − 𝑦 ]
= ℎ 𝑓(0.2, 0.849)
= 0.2 [2 0.2 0.849)2 − 0.849
= −0.1121
𝑦1 = 𝑦0 +
1
6
[𝑘1 + 2𝑘2 +2𝑘3 +𝑘4]
Therefore,

15. 15
Substituting the values of 𝑦0, 𝑘1, 𝑘2, 𝑘3 and 𝑘4 , we get
𝑦1 = 𝑦0 +
1
6
[𝑘1 + 2𝑘2 +2𝑘3 +𝑘4]
= 1 +
1
6
[−0.2 + 2 −0.1476 + 2(−0.1509) − 0.1121]
= 1 +
1
6
[−0.2 − 0.2952 − 0.3018 − 0.1121]
= 0.8485
We get, 𝑥1 = 0.2, 𝑦1 = 0.8485
Taking 𝑛 = 1,
𝑦(0.4) ≈ 𝑦2
= 𝑦0 + 2 0.2 2𝑥1𝑦1
2
− 𝑦1
= 1 + 0.4 2 0.2 0.84852 − 0.8485 = 0.7758
𝑦𝑛+1 = 𝑦𝑛−1 + 2ℎ 2𝑥𝑛𝑦𝑛
2 − 𝑦𝑛

16. 16
Taking 𝑛 = 2,
𝑦(0.6) ≈ 𝑦3
= 𝑦1 + 2 0.2 2𝑥2𝑦2
2
− 𝑦2
= 0.8485 + 0.4 2 0.4 0.77582 − 0.7758 = 0.6099
[Since h=0.2, 𝑥1 = 0.4, 𝑦1 = 0.7758]
Outcomes
 Numerically solved ODE with conditions by using Finite
difference method.
𝑦𝑛+1 = 𝑦𝑛−1 + 2ℎ 2𝑥𝑛𝑦𝑛
2 − 𝑦𝑛

17. 17
S.No. Questions
1 Which formula refers to central difference formula of second derivative?
(a) 𝑦′
=
1
2ℎ
𝑦𝑛+1 − 𝑦𝑛−1 ,
(b) 𝑦′′
=
1
ℎ2 𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1
(c) Both of them,
(d) None of them
2 Which formula refers to central difference formula of first derivative?
(a) 𝑦′ =
1
2ℎ
𝑦𝑛+1 − 𝑦𝑛−1 ,
(b) 𝑦′
=
1
ℎ2 𝑦𝑛+1 − 2𝑦𝑛 + 𝑦𝑛−1
3 Which formula refers to forward difference formula of second derivative?
(a) 𝑦′′
=
1
ℎ2 𝑦𝑛+2 − 2𝑦𝑛+1 + 𝑦𝑛 ,
(b) 𝑦′′ =
1
ℎ3 𝑦𝑛+2 − 𝑦𝑛+1 + 𝑦𝑛 ,
(c) Both of them
4 Which formula refers to backward difference formula of second derivative?
(a) 𝑦′′
=
1
ℎ3 𝑦𝑛+2 − 2𝑦𝑛+1 + 𝑦𝑛 ,
(b) 𝑦′′ =
1
ℎ2 𝑦𝑛 − 2𝑦𝑛−1 + 𝑦𝑛−2 ,
(c) None of them
Multiple questions:

18. 18
Try to do yourself
Exercise 2:
(a) Use three point central difference formula for
derivative to derive a recurrence relation for the above
IVP.
Given that 𝑦′ = 𝑦𝑥2 + 2𝑥, where 𝑦 = −1 at 𝑥 = 0,
and h=0.2.
(b) Estimate the values of y at 𝑥 = 0.4 and 0.6 using this
recurrence relation.
Exercise 1: Consider the boundary value problem
(a) Derive a recurrence relation using three points central
difference approximations with ℎ = 1 3.
(b) Using the above finite difference formula solve the
above BVP.
𝑦′′
+ 5𝑦′
+ 3𝑦 = 1, 𝑦 0 = 1, 𝑦 1 = 0