Solving boundary value problems using the Galerkin's method. This is a weighted residual method, studied as an introduction to the Finite Element Method. This is a part of a series on Advanced Numerical Methods.

1. Solving ODE-BVP through Galerkin’s Method
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 Galerkin’s Method.
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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(x = a) = A, y(x = b) = B
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4. GM
Concept of Linear Independence
In Vector Algebra, n vectors, namely v1,v2,...,vn are linearly independent,
when
n
i=1
aivi = 0 ⇐⇒ ai = 0,∀i = 1,...,n
Linear independence means that no vector can be expressed as a linear
combination of other vectors.
This concept of linear independence is not only limited to vectors, but has
also been extended to the area of functions and various algebraic
polynomials.
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5. GM
Galerkin’s method I
Formulation
The Galerkin’s Method is a “weighted-residual” method. We will try to
solve the following differential equation:
d2
y
dx2
+ P(x)
dy
dx
+ Q(x)y = R(x)
with the following boundary conditions y(x = a) = A and y(x = b) = B.
Let us assume that the solution is in the form
y = α0 + α1x + α2x2
+ ··· + αnxn
=
n
i=0
αixi
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6. GM
Galerkin’s method II
Formulation
Differentiating the above form, with respect to x, we have:
dy
dx
=
n
i=0
iαixi−1
(2)
d2
y
dx2
=
n
i=0
i(i − 1)αixi−2
(3)
Substituting, these into the differential equation we have:
n
i=0
αi i(i − 1)xi−2
+ iP(x)xi−1
+ Q(x)xi
= R(x) (4)
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7. GM
Galerkin’s method III
Formulation
From the boundary conditions, we have
n
i=0
αiai
= A, (5)
n
i=0
αibi
= B (6)
We work out the residual function as follows:
(x) =
n
i=0
αi i(i − 1)xi−2
+ iP(x)xi−1
+ Q(x)xi
− R(x) (7)
If there are n − 1 unknowns, then there should be n − 1 linearly
independent polynomials chosen to be multiplied as weights to the
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8. GM
Galerkin’s method IV
Formulation
residual function. Therefore, for each j = 1,...,n − 1, we should have
n − 1 equations
b
a
Nj(x) (x)dx = 0 (8)
where Nj(x) denotes the jth polynomial.
These equations are solved using linear algebra to obtain the values of
αi, i = 0,...,n
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9. GM
Example I
Galerkin’s Method
Problem: Use the Galerkin’s 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 to the given differential
equation is in the following form, where there are four unknowns:
y = α0 + α1x + α2x2
+ α3x3
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10. GM
Example II
Galerkin’s Method
From the boundary conditions given, we have
α0 + α1(0) + α2(02
) + α3(03
) = 0 =⇒ α0 = 0
α0 + α1(1) + α2(12
) + α3(13
) = 0 =⇒ α1 + α2 + α3 = 0(or)α3 = −(α1 + α2
We calculate the derivatives as follows:
dy
dx
= α1 + 2α2x + 3α3x2
d2
y
dx2
= 2α2 + 6α3x
Substituting these into the differential equation, we have the following:
α1x + α2(2 − x2
) + α3(6x − x3
) = x
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11. GM
Example III
Galerkin’s Method
Since α3 = −(α1 + α2), we will have
−α1x + α2(2 − x2
) + (α1 + α2)(x3
− 6x) = x
=⇒ α1(x3
− 7x) + α2(x3
− x2
− 6x + 2) = x (9)
We can therefore formulate
(x) = α1(x3
− 7x) + α2(x3
− x2
− 6x + 2) − x (10)
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12. GM
Example IV
Galerkin’s Method
Since there are two unknown parameters here, we will consider two
functions N1(x) = x − x2
, and N2(x) = x2
− x3
, as weighting functions.
Therefore,
1
0
N1(x) (x)dx = 0 =⇒ −0.5500α1 − 0.1833α2 = 0.0833
1
0
N2(x) (x)dx = 0 =⇒ −0.3262α1 − 0.1429α2 = 0.0500
We will have this system
−0.5500 −0.1833
−03262 −0.1429
α1
α2
=
0.0833
0.0500
(11)
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13. GM
Example V
Galerkin’s Method
Using this, the relations α3 = −(α1 + α2), and α0 = 0, we have
α0 = 0,α1 = −0.1456,α2 = −0.01743,α3 = 0.1631
Therefore, we can say that for our differential equation, we have the
following solution:
y = −0.1456x − 0.01743x2
+ 0.1631x3
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14. GM
Thank you!
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