Lecture-1-Mech.pptx

1. SOLUTION OF BOUNDARY VALUE PROBLEMS IN ORDINARY DIFFERENTIAL EQUATIONS
Consider a second order differential equation 𝐹 𝑥, 𝑦, 𝑦′, 𝑦′′ = 0.
Its general solution contains two arbitrary constants. To determine these constants we need to
prescribe two conditions.
The conditions are called initial conditions if 𝑦 and 𝑦′ are specified at a certain value of 𝑥.
The differential equation together with the initial conditions is called the initial value problem.
If 𝑦 or 𝑦′ or their combination is prescribed at two different values of 𝑥, then the conditions
are called boundary conditions.
The differential equation together with the boundary conditions is called a boundary value
problem.
1

2. The solution of an initial value problem, in general, exists and is unique, but the solution of a
boundary value problem may exist or may not.
The following simple examples will illustrate this-
(1) 𝒚′′ + 𝒚 = 𝟎, 𝒚 𝟎 = 𝟎, 𝒚 𝝅 = 𝟏
Auxiliary equation, 𝑚2 + 1 = 0 ⇒ 𝑚 = ±𝑖
The general solution of DF is, 𝑦 = 𝐴 𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥
𝑦 0 = 0 ⇒ A = 0; solution is 𝑦 = 𝐵𝑠𝑖𝑛𝑥
𝑦(𝜋) = 1 ⇒ 1 = 𝐵 0 , which is not possible.
Hence the boundary value problem does not have a solution.
2

3. 𝟐 𝒚′′ + 𝒚 = 𝟎, 𝒚 𝟎 = 𝟎, 𝒚 𝝅 = 𝟎
The general solution of DF is, 𝑦 = 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥
𝑦 0 = 0 ⇒ A = 0; and solution is 𝑦 = 𝐵𝑠𝑖𝑛𝑥
𝑦(𝜋) = 1 ⇒ 0 = 𝐵 0 .
Here 𝐴 = 0 𝑎𝑛𝑑 𝐵 is undeterminable.
Hence the boundary value problem has an infinite number of solutions, 𝒚 = 𝑩𝒔𝒊𝒏𝒙, Where 𝐵
is an arbitrary constant.
3

4. 𝟑 𝒚′′ + 𝒚 = 𝟎, 𝒚 𝟎 = 𝟎, 𝒚 𝝅/𝟐 = 𝟏
The general solution of DF is, 𝑦 = 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥
𝑦 0 = 0 ⇒ A = 0; and solution is 𝑦 = 𝐵𝑠𝑖𝑛𝑥
𝑦
𝜋
2
= 1 ⇒ 𝐵 = 1.
Hence the boundary value problem has the unique solutios, 𝒚 = 𝒔𝒊𝒏𝒙
4

5. To solve the boundary value problem of the form,
𝑝 𝑥 𝑦′′ + 𝑞 𝑥 𝑦′ + 𝑟 𝑥 𝑦 = 𝑓(𝑥) with boundary conditions 𝑦 𝑎 = 𝑦𝑎 and 𝑦 𝑏 = 𝑦𝑏,
we use finite difference method.
Method of finite differences:
Divide the interval 𝑎, 𝑏 into 𝑛 sub-intervals, each of length ℎ =
𝑏−𝑎
𝑛
.
Let 𝑥𝑖 = 𝑎 + 𝑖ℎ, 𝑖 = 0, 1, 2, … , 𝑛, and let us use the following notation:-
𝑦 𝑥𝑖 = 𝑦𝑖, 𝑦′ 𝑥𝑖 = 𝑦𝑖
′
, 𝑦′′ 𝑥𝑖 = 𝑦𝑖
′′
.
𝑝 𝑥𝑖 = 𝑝𝑖, 𝑞 𝑥𝑖 = 𝑞𝑖, 𝑟 𝑥𝑖 = 𝑟𝑖, 𝑓 𝑥𝑖 = 𝑓𝑖
𝑥0 = 𝑎 𝑥1 𝑥2 𝑥𝑖 𝑥𝑖+1 𝑥𝑛−1 𝑥𝑛 = b
5

6. By Taylor’s series expansion,
𝑦𝑖+1 = 𝑦 𝑥𝑖+1 = 𝑦 𝑥𝑖 + ℎ = 𝑦𝑖 + ℎ𝑦𝑖
′
+
ℎ2
2!
𝑦𝑖
′′
+
ℎ3
3!
𝑦𝑖
′′′
+ … ______ (1)
𝑦𝑖−1 = 𝑦 𝑥𝑖−1 = 𝑦 𝑥𝑖 − ℎ = 𝑦𝑖 − ℎ𝑦𝑖
′
+
ℎ2
2!
𝑦𝑖
′′
−
ℎ3
3!
𝑦𝑖
′′′
+ … ______ (2)
Eqn(1) gives 𝑦′
𝑥𝑖 =
1
ℎ
𝑦 𝑥𝑖 + ℎ − 𝑦 𝑥𝑖 −
ℎ
2
𝑦′′
𝑥𝑖 − ⋯
i.e., 𝑦′
𝑥𝑖 =
1
ℎ
𝑦 𝑥𝑖 + ℎ − 𝑦 𝑥𝑖 + 𝑂 ℎ
which is the forward difference approximation of 𝑦′(𝑥𝑖) with an error of the order ℎ.
Similarly eqn(2) gives 𝑦′
𝑥𝑖 =
1
ℎ
𝑦 𝑥𝑖 − 𝑦 𝑥𝑖 − ℎ + 𝑂 ℎ
which is the backward difference approximation of 𝑦′(𝑥𝑖) with an error of the order ℎ.
6

7. Eqn(1)-Eqn(2) ⟹ 𝑦′ 𝑥𝑖 =
1
2ℎ
𝑦 𝑥𝑖 + ℎ − 𝑦 𝑥𝑖 − ℎ + 𝑂(ℎ2)
which is the central-difference approximation of 𝑦′(𝑥𝑖) with an error of the order ℎ2
.
Clearly this central difference approximation to 𝑦′(𝑥𝑖) is better than the forward or
backward difference approximations and hence should be preferred.
Yielding approximation,
𝒚𝒊
′
=
𝒚𝒊+𝟏−𝒚𝒊−𝟏
𝟐𝒉
with an error of the order ℎ2.
(3)
7

8. Eqn(1)+eqn(2) ⟹ 𝑦𝑖+1 + 𝑦𝑖−1 = 2𝑦𝑖 + ℎ2
𝑦𝑖
′′
+
ℎ4
12
𝑦𝑖
′′′′
+ ⋯ ,
We get approximation as
𝒚𝒊
′′
=
𝒚𝒊+𝟏−𝟐𝒚𝒊+𝒚𝒊−𝟏
𝒉𝟐 with an error of the order ℎ2
(4)
which is the central difference approximation of 𝑦′′ 𝑥𝑖 .
In finite difference method 𝑦′ and 𝑦′′ are replaced by the finite differences using eqns (3) and
(4).
8

9. Consider the two cases that arise while solving linear boundary value problems.
Case (1): The boundary conditions dot not involve 𝒚′
To solve the boundary value problem (BVP),
𝑝 𝑥 𝑦′′ + 𝑞 𝑥 𝑦′ + 𝑟 𝑥 𝑦 = 𝑓(𝑥) with boundary conditions 𝑦 𝑎 = 𝑦𝑎 and 𝑦 𝑏 = 𝑦𝑏.
We set 𝑥 = 𝑥𝑖 in BVP and replace 𝑦𝑖′ and 𝑦𝑖′′ by the approximations eqn(3) and (4) and we get
the system of equations
𝑝𝑖 𝑦𝑖
′′
+ 𝑞𝑖 𝑦𝑖
′
+ 𝑟𝑖 𝑦𝑖 = 𝑓𝑖
⟹ 𝑝𝑖
𝑦𝑖+1−2𝑦𝑖+𝑦𝑖−1
ℎ2 + 𝑞𝑖
𝑦𝑖+1−𝑦𝑖−1
2ℎ
+ 𝑟𝑖𝑦𝑖 = 𝑓𝑖 _______ (5)
with 𝑖 = 1, 2, 3, … , 𝑛 − 1 for the 𝑛 − 1 unknowns 𝑦1, 𝑦2, … , 𝑦𝑛−1.
9

10. The system (5) involves 𝑦0 for 𝑖 = 1 and 𝑦𝑛 for 𝑖 = 𝑛 − 1. However these are known from the
boundary conditions as 𝑦0 = 𝑦 𝑎 = 𝑦𝑎 and 𝑦𝑛 = 𝑦 𝑏 = 𝑦𝑏.
Examples:
1) Solve 𝒙 𝒚′′ + 𝒚 =0, 𝒚 𝟏 = 𝟏, 𝒚 𝟐 = 𝟐 with 𝒉 = 𝟎. 𝟐𝟓 by finite difference method.
Answer: With ℎ = 0.25, 𝑛 = 4
i.e., Divide the interval [1, 2] into 4 equal parts
We find the unknowns 𝑦1, 𝑦2, 𝑦3 from the system,
𝑥𝑖
𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
ℎ2
+ 𝑦𝑖 = 0
𝑥0 = 1 𝑥1 = 1.25 𝑥2 = 1.5 𝑥3 = 1.75 𝑥4 = 2
𝑦0 = 1 𝑦4 = 2
𝑦1 =? 𝑦2 =? 𝑦3 =?
10

11. 𝑥𝑖
𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
1
4
2 + 𝑦𝑖 = 0
16𝑥𝑖 𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1 + 𝑦𝑖 = 0, 𝑖 = 1, 2, 3
At 𝑖 = 1,
16𝑥1 𝑦2 − 2𝑦1 + 𝑦0 + 𝑦0 = 0
16 1.25 𝑦2 − 2𝑦1 + 1 + 1 = 0
⟹ 𝟑𝟗𝒚𝟏 − 𝟐𝟎𝒚𝟐 = 𝟐𝟎 _____ (1)
11

12. At 𝑖 = 2,
16𝑥2 𝑦3 − 2𝑦2 + 𝑦1 + 𝑦1 = 0
16 1.5 𝑦3 − 2𝑦2 + 𝑦1 + 𝑦1 = 0
⟹ 𝟐𝟒𝒚𝟏 − 𝟒𝟕𝒚𝟐 + 𝟐𝟒𝒚𝟑 = 𝟎 _____ (2)
At 𝑖 = 3,
16𝑥3 𝑦4 − 2𝑦3 + 𝑦2 + 𝑦2 = 0
16 1.75 2 − 2𝑦3 + 𝑦2 + 𝑦2 = 0
⟹ 𝟐𝟖𝒚𝟐 − 𝟓𝟓𝒚𝟑 = −𝟓𝟔 _____ (3)
Solving equations (1), (2) and (3), We get 𝒚𝟏 = 𝟏. 𝟑𝟓𝟏𝟑, 𝒚𝟐 = 𝟏. 𝟔𝟑𝟒𝟗, 𝒚𝟑 = 𝟏. 𝟖𝟓𝟎𝟖
12

13. 2) Solve
𝟏
𝟒
𝒚′′ − 𝒙𝒚′ + 𝒚 = 𝟏, 𝒚 𝟎 = 𝟎, 𝒚 𝟏 =0 with 𝒉 =
𝟏
𝟑
by finite difference method.
Answer: With ℎ =
1
3
, 𝑛 = 3
i.e., Divide the interval [0, 1] into 3 equal parts
We find the unknowns 𝑦1, 𝑦2 from the system,
1
4
𝑦′′
− 𝑥𝑦′
+ 𝑦 = 1 ⟹ 𝑦𝑖
′′
− 4𝑥𝑖𝑦𝑖
′
+ 4𝑦𝑖 = 4
𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
ℎ2
− 4𝑥𝑖
𝑦𝑖+1 − 𝑦𝑖−1
2ℎ
+ 𝑦𝑖 = 4
𝑥0 = 0
𝑥1 =
1
3
𝑥2 =
2
3
𝑥3 = 1
𝑦0 = 0 𝑦1 =? 𝑦2 =? 𝑦3 = 0
13

14. ⟹
𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
1
9
− 4𝑥𝑖
𝑦𝑖+1 − 𝑦𝑖−1
2
3
+ 𝑦𝑖 = 4
⟹ 9 𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1 − 6𝑥𝑖 𝑦𝑖+1 − 𝑦𝑖−1 + 4𝑦𝑖 = 4, 𝑖 = 1, 2
When 𝑖 = 1,
9 𝑦2 − 2𝑦1 + 𝑦0 − 6
1
3
𝑦2 − 𝑦0 + 4𝑦1 = 4
9 𝑦2 − 2𝑦1 + 0 − 2 𝑦2 − 0 + 4𝑦1 = 4
−𝟏𝟒𝒚𝟏 + 𝟕𝒚𝟐 = 𝟒 ________ (1)
14

15. When 𝑖 = 2,
9 𝑦3 − 2𝑦2 + 𝑦1 − 6
2
3
𝑦3 − 𝑦1 + 4𝑦2 = 4
9 𝑦2 − 2𝑦1 + 0 − 2 𝑦2 − 0 + 4𝑦1 = 4
𝟏𝟑𝒚𝟏 − 𝟏𝟒𝒚𝟐 = 𝟒 ________ (2)
Solving equations (1) and (2), We get 𝒚𝟏 = −
𝟒
𝟓
, 𝒚𝟐 = −
𝟑𝟔
𝟑𝟓
15

16. Case (2): The boundary conditions involve 𝒚′
Examples:
(1) Solve 𝒚′′
+ 𝒙𝒚 = 𝟏, 𝒚 𝟎 = 𝟎, 𝒚′
𝟏 = 𝟏 with 𝒉 = 𝟎. 𝟓 by finite difference method
Answer: With ℎ = 0.5, 𝑛 = 2
i.e., Divide the interval [0, 1] into 2 equal parts
We find the unknowns 𝑦1, 𝑦2 from the system,
𝑦𝑖
′′
+ 𝑥𝑖𝑦𝑖 = 1, 𝑖 = 1, 2
𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
ℎ2
+ 𝑥𝑖𝑦𝑖 = 1
𝑥0 = 0 𝑥1 = 0.5 𝑥2 = 1
𝑦0 = 0 𝑦1 =? 𝑦2 =?
16

17. 𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
1
2
2 + 𝑥𝑖𝑦𝑖 = 1
4 𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1 + 𝑥𝑖𝑦𝑖 = 1, 𝑖 = 1, 2
When 𝑖 = 1
4 𝑦2 − 2𝑦1 + 𝑦0 + 𝑥1𝑦1 = 1
4 𝑦2 − 2𝑦1 + 0 + (0.5)𝑦1 = 1
−7.5𝑦1 + 4𝑦2 = 1 _______ (1)
When 𝑖 = 2
4 𝑦3 − 2𝑦2 + 𝑦1 + 𝑥2𝑦2 = 1
17

18. 4 𝑦3 − 2𝑦2 + 𝑦1 + 1 𝑦2 = 1 _______ (2)
Here the problem involves 𝑦3, the value of 𝑦 outside the given interval.
Equation (2) will enable us to find 𝑦3 in terms of 𝑦1 𝑎𝑛𝑑 𝑦2 as follows,
Given the boundary condition 𝑦′ 1 = 1 ⟹ 𝑦′ 𝑥2 = 1 (∵ 𝑥2 = 1)
⟹ 𝑦2
′
= 1
We know that, 𝑦𝑖
′
=
𝑦𝑖+1−𝑦𝑖−1
2ℎ
⟹ 𝑦2
′
=
𝑦3−𝑦1
2
1
2
⟹ 1 = y3 − y1 or 𝑦3 = 1 + 𝑦1
18

19. Now equation (2)⟹ 4 1 + 𝑦1 − 2𝑦2 + 𝑦1 + 1 𝑦2 = 1
8𝑦1 − 7𝑦2 = −3 _____ (3)
Solving equations (1) and (3), we get, 𝑦1 =
10
41
, 𝑦2 =
29
41
(2) Solve 𝒚′′ + 𝟏 + 𝒙 𝒚′ − 𝒚 = 𝟎, 𝒚 𝟎 = 𝒚′ 𝟎 , 𝒚 𝟏 + 𝒚′ 𝟏 = 𝟏 with 𝒉 = 𝟎. 𝟓 by finite
difference method.
Answer: With ℎ = 0.5, 𝑛 = 2
i.e., Divide the interval [0, 1] into 2 equal parts
𝑥0 = 0 𝑥1 = 0.5 𝑥2 = 1
𝑦0 =? 𝑦1 =? 𝑦2 =?
19

20. We find the unknowns 𝑦1, 𝑦2 from the system,
𝑦𝑖
′′
+ 1 + 𝑥𝑖 𝑦𝑖
′
− 𝑦𝑖 = 0, 𝑖 = 0, 1, 2
𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1
ℎ2
+ 1 + 𝑥𝑖
𝑦𝑖+1 − 𝑦𝑖−1
2ℎ
− 𝑦𝑖 = 0
4(𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1) + (1 + 𝑥𝑖) [𝑦𝑖+1 − 𝑦𝑖−1] − 𝑦𝑖 = 0, 𝑖 = 0, 1, 2
When 𝑖 = 0,
4(𝑦1 − 2𝑦0 + 𝑦−1) + (1 + 𝑥0) [𝑦1 − 𝑦−1] − 𝑦0 = 0 ---- (1)
Here the problem involves 𝑦−1, the value of 𝑦 outside the given interval.
Given boundary condition 𝑦 0 = 𝑦′
0
𝑦 𝑥0 = 𝑦′ 𝑥0 ∵ 𝑥0 = 0
𝑦0 = 𝑦0
′
20

21. We know that, 𝑦𝑖
′
=
𝑦𝑖+1−𝑦𝑖−1
2ℎ
𝑦0 =
𝑦1−𝑦−1
2(0.5)
𝑦0 = 𝑦1 − 𝑦−1 ⟹ 𝑦−1 = 𝑦1 − 𝑦0
Now equation (1) ⟹ 4(𝑦1 − 2𝑦0 + 𝑦1 − 𝑦0) + (1 + 0) [𝑦1 − (𝑦1 − 𝑦0)] − 𝑦0 = 0
𝟐𝒚𝟏 − 𝟑𝒚𝟎 = 𝟎_______ (2)
When 𝑖 = 1,
4(𝑦2 − 2𝑦1 + 𝑦0) + (1 + 𝑥1) [𝑦2 − 𝑦0] − 𝑦1 = 0
4(𝑦2 − 2𝑦1 + 𝑦0) + (1 + 0.5) [𝑦2 − 𝑦0] − 𝑦1 = 0
𝟏𝟏𝒚𝟐 − 𝟏𝟖𝒚𝟏 + 𝟓𝒚𝟎 = 𝟎 ---- (3)
21

22. When 𝑖 = 2,
4(𝑦3 − 2𝑦2 + 𝑦1) + (1 + 𝑥2) [𝑦3 − 𝑦1] − 𝑦2 = 0
4(𝑦3 − 2𝑦2 + 𝑦1) + (1 + 1) [𝑦3 − 𝑦1] − 𝑦2 = 0 ---- (4)
Here the problem involves 𝑦3, the value of 𝑦 outside the given interval.
Given another boundary condition 𝑦 1 + 𝑦′ 1 = 1
𝑦 𝑥2 + 𝑦′ 𝑥2 = 1 ∵ 𝑥2 = 1
𝑦2 + 𝑦2
′
= 1
𝑦2 +
𝑦3 − 𝑦1
2 0.5
= 1
𝑦3 = 1 + 𝑦1 − 𝑦2
∵ 𝑦𝑖
′
=
𝑦𝑖+1 − 𝑦𝑖−1
2ℎ
22

23. Now equation (4) ⟹
4(1 + 𝑦1 − 𝑦2 − 2𝑦2 + 𝑦1) + (1 + 1) [1 + 𝑦1 − 𝑦2 − 𝑦1] − 𝑦2 = 0
𝟏𝟓𝒚𝟐 − 𝟖𝒚𝟏 = 𝟔 −− −(𝟓)
Solving equations (2), (3) and (5), we get,
𝒚𝟎 =
𝟏
𝟑
, 𝒚𝟏 =
𝟏
𝟐
, 𝒚𝟐 =
𝟐
𝟑
23

24. Exercises:
Solve the following boundary value problems by finite difference method.
1. 𝑦′′
− 3𝑦′
+ 2𝑦 = 1 + 2𝑥, 𝑦 1 = 𝑦 2 = 0, ℎ = 0.25
2. 𝑦′′ + 𝑦 = 0, 𝑦 0 = 0, 𝑦
𝜋
2
= 1, ℎ =
𝜋
8
3. 𝑦′′
+ 𝑥𝑦′
− 2𝑦 = 0, 𝑦 1 = 𝑦′
1 , 𝑦 2 = 5, ℎ =
1
2
4. 𝑦′′
= 𝑥𝑦, 𝑦 0 + 𝑦′
0 = 1, 𝑦 1 = 1, ℎ =
1
2
24