Methods of Solving Differential Equations as Boundary Value Problems using the Point Collocation Method.

1. Point Collocation Method
FEM - Introduction - Methods of Solving Differential Equations
Suddhasheel Ghosh, PhD
Department of Civil Engineering
Jawaharlal Nehru Engineering College
N-6 CIDCO, 431003
Advanced Numerical Methods Series
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2. DiffEq1
Introduction to terminology
Given a differential equation
Ψ
d2y
dx2
,
dy
dx
, y, x = 0, (1)
and the initial conditions,
F1
dy
dx
, y, x = a = 0 F2
dy
dx
, y, x = b = 0
So, given the points x1 = a, x2, x3, . . . , xi, . . . , xn, xn+1 = b, it is desired to
find the solution of the differential equation at the points
xj, ∀j = 2, . . . , n. The points xj, j = 2, . . . , n are known as the points of
collocation.
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3. DiffEq1
A second-order Boundary Value Problem
A boundary value problem is given as follows:
d2y
dx2
+ P(x)
dx
dy
+ Q(x)y = R(x)
along with the conditions
y(x1) = A, y(xn+1) = B
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4. Collocation Method
Point collocation Method
Derivative calculation
Assume that
y =
n
i=0
αixi
.
Therefore, we will have
dy
dx
=
n
i=0
i · αixi−1
,
and
d2y
dx2
=
n
i=0
i(i − 1)αixi−2
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5. Collocation Method
Point collocation Method I
Substitution and formulation
Substituting these in the differential equation, we have
n
i=0
i(i − 1)αixi−2
+ P(x)
n
i=0
i · αixi−1
+ Q(x)
n
i=0
αixi
= R(x).
Thus giving,
n
i=0
αi i(i − 1)xi−2
+ ixi−1
P(x) + xi
Q(x) = R(x)
The aim of the interpolation method is to “agree” at the node points,
and therefore, we shall have:
n
i=0
αi i(i − 1)xi−2
j + ixi−1
j P(xj) + xi
jQ(xj) = R(xj), ∀j = 2, . . . , n
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6. Collocation Method
Point collocation Method II
Substitution and formulation
For the nodes x1 and xn+1, we have the following conditions:
n
i=0
αixi
1 = A
n
i=0
αixi
n+1 = B
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7. Collocation Method
Point collocation Method
Matrix formulation of the problem


1 x1 x2
1
. . . xn
1
Q(x2) P(x2) (2 + 2x2P(x2) + x2
2
Q(x2)) . . . [n(n − 1)xn−2
2
+ nP(x2)xn−1
2
+ xn
2
Q(x2)]
...
... . . . . . .
...
Q(xn) P(xn) (2 + 2xnP(xn) + x2
nQ(xn)) . . . [n(n − 1)xn−2
n + nP(xn)xn−1
n + xn
nQ(xn)]
1 xn+1 x2
n+1
. . . xn
n+1




α0
α1
...
αn−1
αn


=


A
R(x2)
...
R(xn)
B


The solution can then be achieved by any of the standard methods like Gauss-Siedel,
Gaussian Elimination or Gauss-Jordan Elimination.
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8. Collocation Method
Point collocation method I
Example
Use the point collocation method to solve the following differential
equation:
d2y
dx2
− y = x
Use the boundary conditions y(x = 0) = 0 and y(x = 1) = 0. Choose
x = 0.25 and x = 0.5 as collocation points. (Desai, Eldho, Shah)
Solution: There are four points where we are considering the solution
for, x = 0, 0.25, 0.5, 1. We label them as x1, x2, x3, x4. Since there are four
points, we will consider a cubic polynomial.
y = α0 + α1x + α2x2
+ α3x3
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9. Collocation Method
Point collocation method II
Example
We have
dy
dx
= α1 + 2α2x + 3α3x2
d2y
dx2
= 2α2 + 6α3x
Substituting these in the given differential equation, we have
2α2 + 6α3x − α0 − α1x − α2x2
− α3x3
= x
−α0 − α1x + (2 − x2
)α2 + (6x − x3
)α3 = x
From the first boundary condition y(x = 0) = 0, we have
α0 + α1(0) + α2(02
) + α3(03
) = 0 =⇒ α0 = 0 (2)
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10. Collocation Method
Point collocation method III
Example
From the second boundary condition y(x = 1) = 0, we have
α0 + α1(1) + α2(12
) + α3(13
) = 0 =⇒ α1 + α2 + α3 = 0 (3)
At the collocation points, we have the following equations:
For x = 0.25, we have
−α1(0.25) + (2 − (0.25)2
)α2 + (6 × 0.25 − (0.25)3
) = 0.25
−0.25α1 + 1.9375α2 + 1.4844α3 = 0.25 (4)
For x = 0.5, we have
−0.5α1 + (2 − 0.52
)α2 + (6 × 0.5 − 0.53
) = 0.5
−0.5α1 + 1.75α2 + 2.875α3 = 0.5 (5)
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11. Collocation Method
Point collocation method IV
Example
Using the equations above, we have the following matrix based
arrangement


1 1 1
−0.25 1.9375 1.4844
−0.5 1.75 2.875




α1
α2
α3


=


0
0.25
0.5


(6)
which gives on the inverse operation,
α1 = −0.1459, α2 = −0.006738, α3 = 0.1526
Therefore the polynomial approximation for y is
y = 0.1459x − 0.006738x2
+ 0.1526x3
(7)
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12. Collocation Method
Thank you!
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