The document discusses solving ordinary differential equations (ODEs) using the least squares finite element method, focusing on minimizing the sum of squares to find solutions. It provides examples of boundary value problems, formulates the least squares method, and demonstrates its application through step-by-step calculations. The content is primarily aimed at civil engineering applications and includes numerical examples to elucidate the method.

1. Solving ODE through least squares
FEM: Introduction
Suddhasheel Ghosh, PhD
Department of Civil Engineering
Jawaharlal Nehru Engineering College
N-6 CIDCO, 431003
Series on Advanced Numerical Methods
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2. DiffEq1
Introduction to terminology
Given a differential equation
Ψ
d2
y
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 a and b, it is desired to find the solution of the
differential equation using the least squares technique.
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3. DiffEq1
A second-order Boundary Value Problem
A boundary value problem is given as follows:
d2
y
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. Least squares method
Recalling least squares
Least squares method
Least squares means that the sum of squares has to be minimised. The
minimization will happen at the internal points.
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5. Least squares method
Formulation I
Least squares method
From the earlier computations, for x = x1,
α0 +α1x1 +α2x2
1 +···+αnxn
1 = A, (2)
and for x = xn+1,
α0 +α1xn+1 +α2x2
n+1 +···+αnxn
n+1 = B, (3)
From an earlier computation, we have obtained:
n
i=0
αi i(i−1)xi−2
+ixi−1
P(x)+xi
Q(x) = R(x)
Thus, for every xj, j = 2,...,n, we have
(x) =
n
i=0
αi i(i−1)xi−2
+ixi−1
P(xj)+xi
Q(x) −R(x) (4)
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6. Least squares method
Formulation II
Least squares method
For the least squares issue, we have to compute
F =
b
a
[ (x)]2
dx =
b
a
n
i=0
αi i(i−1)xi−2
+ixi−1
P(xj)+xi
Q(x) −R(x)
2
dx
Therefore,
∂F
∂αi
=
b
a
n
i=0
αi i(i−1)xi−2
+ixi−1
P(x)+xi
Q(x) −R(x) ×
i(i−1)xi−2
+ixi−1
P(x)+xi
Q(x) dx = 0
This, and the equations generated by the boundary conditions can be
solved by using Linear Algebra.
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7. Least squares method
Example I
Least squares method
Problem: Use the least squares method to solve the following differential
equation:
d2
y
dx2
−y = x
Use the boundary conditions y(x = 0) = 0 and y(x = 1) = 0. (Desai, Eldho,
Shah)
Solution: Let us assume that the solution is in the cubic form
y = α0 +α1x+α2x2
+α3x3
.
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8. Least squares method
Example II
Least squares method
Then, we will have
dy
dx
= α1 +2α2x+3α3x2
d2
y
dx2
= 2α2 +6α3x (5)
Substituting these into the given differential equation, we have
(2α2 +6α3x)−(α0 +α1x+α2x2
+α3x3
) =x
=⇒ −α0 −α1x+(2−x2
)α2 +(6x−x3
)α3 =x
We can therefore frame the error term as follows:
(x) = α0 −α1x+(2−x2
)α2 +(6x−x3
)α3 −x
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9. Least squares method
Example III
Least squares method
And therefore
( (x))2
= α0 −α1x+(2−x2
)α2 +(6x−x3
)α3 −x
2
(6)
From the boundary conditions, we will have the following
α0 = 0 (7)
α0 +α1 +α2 +α3 = 0 (8)
Therefore, we have
α3 = −α2 −α1
Substituting these in (6), we have
F = ( (x))2
= −α1x+(2−x2
)α2 +(x3
−6x)(α1 +α2)−x
2
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10. Least squares method
Example IV
Least squares method
Therefore,
∂F
∂α1
= −α1x+(2−x2
)α2 +(x3
−6x)(α1 +α2)−x x3
−7x
= α1(x3
−7x)+α2(x3
−x2
−6x+2)−x (x3
−7x) (9)
∂F
∂α2
= −α1x+(2−x2
)α2 +(x3
−6x)(α1 +α2)−x x3
−6x−x2
+2
= α1(x3
−7x)+α2(x3
−x2
−6x+2)−x x3
−x2
−6x+2 (10)
Now,
1
0
∂F
∂α1
dx = 0 =⇒ 13.6762α1 +6.6262α2 =−2.1333 (11)
1
0
∂F
∂α2
dx = 0 =⇒ 6.6262α1 +4.2762α2 =−1.05 (12)
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11. Least squares method
Example V
Least squares method
Thus, we can obtain α1 and α2 using the following
13.6762 6.6262
6.6262 4.2762
α1
α2
=
−2.1333
−1.05
(13)
Thus giving α0 = 0,α1 = −0.1485,α2 = −0.0154, and α3 = 0.1639 (since
α1 +α2 +α3 = 0).
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12. Least squares method
Thank you!
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