1. science
2. mathematics
3. geometry

MEG 800 Finite Element Analysis
Variational Calculus
INSTRUCTOR: OA Fakinlede
oafak@unilag.edu.ng oafak@hotmail.com
Department of Systems Engineering,
University of Lagos
Scope of Instructional Material
Course Schedule:
Slide Title
Variational Calculus Intro
Integral Formulation
1-D Models
Applications
Computer Applications
Slides
70
50
50
60
50
Wks
1
1
1
1
3
Text
Reddy
Reddy
Reddy
Reddy
Wolfram
Read Pages
1-57
58-102
103-154
155-231
The read-ahead materials are from Reddy except the
part marked red. There please read Wolfram Manuals.
Home work assignments will be drawn from the range
of pages in the respective books.
Department of Systems Engineering, University of Lagos
2
oafak@unilag.edu.ng 12/31/2012
Recommended Texts
1. Methods of Mathematical Physics, R. Courant & D.
Hilbert, Vol1, Wiley Interscience, 1975
2. An Introduction to the Finite Element Method, JN
Reddy, Third Edition, McGraw-Hill Education, 2006
3. The Variational Principles of Mechanics, C Lanczos,
Fourth Edition, Dover Publications Inc, 1970
4. Install a copy of Mathematica on your computer to
prepare for the computational part of the course.
The Help System has all the literature you need for
the computational methods
oafak@unilag.edu.ng, University of Lagos
3
Wednesday, April 15, 2015
Purpose of the Course
 Provide the basis of the Finite Element Method for
beginning graduate Students
 Show the formulation of Practical Mechanics
Problems
 Build Finite Element Solutions with a
Symbolics/Numerical Processor
 Introduction to Finite Element Simulations.
oafak@unilag.edu.ng, University of Lagos
4
Wednesday, April 15, 2015
Variations & the Geodesic Problem
 Consider the following problem in Geodesics:
oafak@unilag.edu.ng, University of Lagos
5
Wednesday, April 15, 2015
Shortest Distance
 We are given two fixed points A and B on the surface, 𝑧 =
𝜙(𝑥, 𝑦). We desire to find the shortest distance from A to B.
 Projected to the x-y plane, particular continuous, singlevalued paths may appear like the any of the following.
 It is no longer imperative, for example, that the straight line
is the shortest.
 This is an example of a
problem that can be
solved by the Calculus
of Variations
oafak@unilag.edu.ng, University of Lagos
6
Wednesday, April 15, 2015
Shortest Distance
 Minimize the integral,
𝑏
𝐽=
𝑏
𝑑𝑥 2 + 𝑑𝑦 2 + 𝑑𝑧 2
𝑑𝑠 =
𝑎
𝑏
=
𝑎
𝑎
𝑑𝑦
1+
𝑑𝑥
2
𝑑𝑧
+
𝑑𝑥
2
𝑑𝑥
 But we are given that, along the surface, 𝑧 = 𝜙(𝑥, 𝑦) . Therefore,
𝜕𝜙
𝜕𝜙
𝑑𝑧 =
𝑑𝑥 +
𝑑𝑦
𝜕𝑥
𝜕𝑦
 Let the image of the path on the 𝑥 − 𝑦 plane be 𝑦 = 𝜓(𝑥) so
that
𝑑𝑦
= 𝜓′ 𝑥
𝑑𝑥
oafak@unilag.edu.ng, University of Lagos
7
Wednesday, April 15, 2015
Shortest Distance
 We therefore find that
𝜕𝜙
𝜕𝜙
𝑑𝑥 +
𝑑𝑦
𝜕𝑥
𝜕𝑦
𝜕𝜙
𝜕𝜙 𝑑𝑦
=
𝑑𝑥 +
𝑑𝑥.
𝜕𝑥
𝜕𝑦 𝑑𝑥
So that, we are to minimize the integral
𝑑𝑧 =
𝑏
𝐽=
𝑎
𝑏
=
𝑎
𝑑𝜓
1+
𝑑𝑥
2
𝜕𝜙 𝜕𝜙 𝑑𝑦
+
+
𝜕𝑥 𝜕𝑦 𝑑𝑥
𝜕𝜙 𝑥, 𝜓 𝑥
2
′
1+ 𝜓 𝑥
+
𝜕𝑥
2
𝑑𝑥
𝜕𝜙 𝑥, 𝜓 𝑥
+
𝜕𝑦
2
𝜓′ 𝑥
𝑑𝑥
𝑏
𝐹 𝑥, 𝜓 𝑥 , 𝜓 ′ 𝑥
=
𝑑𝑥
𝑎
by finding the best function, 𝑦 = 𝜓(𝑥), among the set of admissible functions.
Recall that we already know the function 𝑧 = 𝜙 𝑥, 𝑦 .
oafak@unilag.edu.ng, University of Lagos
8
Wednesday, April 15, 2015
Other Historical Problems
There have been a large number of problems that can
be formulated in a similar way. Variational Calculus has a
rich history of problems that have engaged the most
fertile mathematical geniuses over the years. Examples
include:
 Dido’s Problem, or the Isoperimetric problem,
 The brachistochrone,
 Minimum Surface,
 Fermat’s principle of least time,
 Lagrangian Principle of least action and other
minimization principles in Mechanics.
oafak@unilag.edu.ng, University of Lagos
9
Wednesday, April 15, 2015
Dido’s Problem
 Enclose the largest possible area of land within a
fence with a predetermined length. The problem
becomes more interesting if we allow a weight
function 𝜌 𝑥 𝑦 to be defined at each point. Such a
function may represent things such as fertility or the
concentration of some valuable quantity.
Mathematically, this is equivalent to finding the
closed curve of a given length that extremises the
integral 𝜌 𝑥, 𝑦 𝑑𝑥𝑑𝑦.
oafak@unilag.edu.ng, University of Lagos
10
Wednesday, April 15, 2015
The Brachistochrone
 The Brachistochrone. A frictionless ring attached to a
fixed smooth rod is falling under the influence of
gravity from point A to point B. What shape of the rod
will make the travel time smallest? We show later that
this curve is a catenary.
oafak@unilag.edu.ng, University of Lagos
11
Wednesday, April 15, 2015
Minimal Surface
 Minimal Surface. A plane curve joining given points A
and B is to be rotated about an axis. We desire to
know the particular curve out of all the possible
curves that can be made to pass these two points,
that will produce the smallest surface area.
oafak@unilag.edu.ng, University of Lagos
12
Wednesday, April 15, 2015
Functions and Functionals
 From differential and integral Calculus, we are used
the concept of functions.
 Given a domain 𝑎, 𝑏 , we can define a function that take
values such as 𝑦 = 𝑥 2 in the domain.
 We may use the methods of calculus to find the maxima
or minima of such a function, if any exists, in the
domain.
 We can define multi-valued functions in the same way.
 In each case, we are mapping from the real space or a
product real space to the real space.
 The geodesic problem we saw earlier has something
fundamentally different. It is called a functional.
oafak@unilag.edu.ng, University of Lagos
13
Wednesday, April 15, 2015
What is a Functional?
 In the minimization of
𝑏
𝐽=
𝐹
𝑎
𝑥, 𝜓 ′
𝜕𝜙 𝑥, 𝜓 𝑥
𝑥 ,
𝜕𝑥
𝜕𝜙 𝑥, 𝜓 𝑥
,
𝜕𝑦
𝑑𝑥
 We are trying to find a function that minimizes the integral.
Unlike our example in calculus, it is a function that will
minimize our integral rather than a particular point! In this
case, we are really looking for the function 𝑦 = 𝜓(𝑥) that
minimizes the functional which is a function of the function
𝜓(𝑥) and some of its derivatives.
 Variational Calculus deals with the extremal values of
functionals.
oafak@unilag.edu.ng, University of Lagos
14
Wednesday, April 15, 2015
The Simplest Problem
 The simplest problem of Calculus of Variations is to find the
function, 𝑦 = 𝜓 𝑥 , that extremizes the Integral,
𝑏
𝐹 𝑥, 𝑦 𝑥 , 𝑦 ′ 𝑥
𝐽[𝑦 𝑥 ] =
𝑑𝑥
𝑎
𝑥 ∈ 𝑎, 𝑏 , 𝑦 𝑎 = 𝑦𝑎 , and 𝑦 𝑏 = 𝑦𝑏
 We talk of extremization because sometimes we are looking for
minimum and sometimes we are looking for maximum.
 Such extreme values always occur at the stationary points of
the functional under examination.
oafak@unilag.edu.ng, University of Lagos
15
Wednesday, April 15, 2015
Admissible functions
 We assume that the functions that will qualify satisfy certain
basic smoothness conditions, are single-valued.
 Each function must necessarily satisfy, a priori, certain end
conditions such that 𝑦 𝑎 = 𝑦𝑎 , and 𝑦 𝑏 = 𝑦𝑏 with the
values of 𝑦𝑎 , and 𝑦𝑏 supplied in advance.
 Such explicit conditions are called “Essential Boundary
Conditions” as they must be satisfied by any trial functions
we shall choose.
 Other boundary conditions will arise later. They will be called
“Natural Boundary Conditions”. Suffice it is to say that any
boundary conditions that cannot be classified as “Essential”
will be in the second category.
oafak@unilag.edu.ng, University of Lagos
16
Wednesday, April 15, 2015
Inadmissible Functions
 The following functions are NOT admissible: 1 and 3 are
not smooth, 2 is discontinuous, 4 is not single-valued
and five does not satisfy the essential boundary
conditions.
oafak@unilag.edu.ng, University of Lagos
17
Wednesday, April 15, 2015
The First Variation
 In order to solve the problem, we consider the rate of
change in the functional resulting in a variation in the
admissible trial function 𝑦 = 𝑦(𝑥).
 We shall be specific and ensure that each varied path
satisfies the admissibility conditions as stated earlier.
 In particular, assume that 𝑦 = 𝑦(𝑥) is admissible,
Then introduce the function 𝜙 𝑥 which is such that
𝜙 𝑎 = 𝜙 𝑏 = 0, then, it follows that, the function
𝑦 𝑥 + 𝜖𝜙 𝑥 ≡ 𝑦 + 𝛿𝑦
oafak@unilag.edu.ng, University of Lagos
18
Wednesday, April 15, 2015
The First Variation
𝑦 𝑥 + 𝜖𝜙 𝑥 ≡ 𝑦 + 𝛿𝑦
Is also admissible
provided 𝜙 𝑥 also
satisfies the
smoothness conditions
stated earlier.
Moreover, we can make
𝜖 as small as we desire
and in the limit, as 𝜖 →
0, this function
approaches 𝑦(𝑥).
oafak@unilag.edu.ng, University of Lagos
19
Wednesday, April 15, 2015
The First Variation
The variation in the functional as a result of variation in
the paths from the extremizing path is,
𝛿𝐹 𝑥, 𝑦, 𝑦 ′ ≡ 𝐹 𝑥, 𝑦 + 𝛿𝑦, 𝑦 ′ + 𝛿𝑦 ′ − 𝐹 𝑥, 𝑦, 𝑦 ′
Note that there is no variation in the independent
variable; the above variations are only in 𝑦 and 𝑦 ′ . We
can therefore write,
𝜕𝐹
𝜕𝐹 ′
𝜕𝐹
𝜕𝐹
′
𝛿𝐹 𝑥, 𝑦, 𝑦 =
𝛿𝑦 +
𝛿𝑦 = 𝜖
𝜙+
𝜙′
𝜕𝑦
𝜕𝑦′
𝜕𝑦
𝜕𝑦′
oafak@unilag.edu.ng, University of Lagos
20
Wednesday, April 15, 2015
Taylor Series
Where we have only taken the first two terms of the
Taylor Series,
𝐹 𝑥 + Δ𝑥, 𝑦 + Δ𝑦
𝜕𝐹
𝜕𝐹
= 𝐹 𝑥, 𝑦 +
Δ𝑥 +
Δ𝑦
𝜕𝑥
𝜕𝑦
2𝐹
2𝐹
1 𝜕2𝐹
2𝜕
𝜕
2
2
+
Δ𝑥
+
Δ𝑥Δ𝑦
+
𝛥𝑦
+⋯
2
2
2! 𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
oafak@unilag.edu.ng, University of Lagos
21
Wednesday, April 15, 2015
First Variation
Upon Integrating by parts, we can write,
𝑏
𝛿𝐽
𝜕𝐹
𝜕𝐹
=
𝜙+
𝜙′ 𝑑𝑥
𝜖
𝜕𝑦
𝜕𝑦′
𝑎
𝑏
=
𝑎
𝜕𝐹
𝑑 𝜕𝐹
𝜕𝐹
−
𝜙𝑑𝑥 +
𝜙
𝜕𝑦 𝑑𝑥 𝜕𝑦′
𝜕𝑦 ′
Coming from the fact that,
𝑏
𝜕𝐹
𝜕𝐹
𝜙′ 𝑑𝑥 =
′𝜙
𝜕𝑦′
𝜕𝑦
𝑎
oafak@unilag.edu.ng, University of Lagos
22
𝑏
𝑏
−
𝑎
𝑎
𝑏
𝑎
𝑑 𝜕𝐹
𝜙𝑑𝑥
𝑑𝑥 𝜕𝑦′
Wednesday, April 15, 2015
Euler-Lagrange Equations
 We arrived at the expression of this variation by
ignoring higher terms than the second in the Taylor
Series. It is the first variation that must vanish in order
for an extremum to exist at 𝑦 = 𝑦(𝑥). It can be
shown, by an invocation of the fundamental theorem
of integral calculus that this implies that,
𝜕𝐹
𝑑 𝜕𝐹
−
=0
′
𝜕𝑦 𝑑𝑥 𝜕𝑦
The essential boundary conditions, that 𝜙 𝑎 = 𝜙 𝑏 =
0, ensures that
oafak@unilag.edu.ng, University of Lagos
𝑏
𝜕𝐹
𝜙
𝜕𝑦 ′
𝑎
= 0 automatically.
23
Wednesday, April 15, 2015
Integral Theorems
The divergence theorem is central to several other results in
Continuum Mechanics. We present here a generalized form
[Ogden] which states that,
Gauss Divergence Theorem
For a tensor field 𝚵, The volume integral in the region Ω ⊂ ℰ,
grad 𝚵 𝑑𝑣 =
Ω
𝚵 ⊗ 𝐧 𝑑𝑎
𝜕Ω
where 𝐧 is the outward drawn normal to 𝜕Ω – the boundary
of Ω.
Dept of Systems Engineering, University of Lagos
24
oafak@unilag.edu.ng 12/27/2012
2-D Version
Consider a cylindrical volumetric domain such that
the volume element lies between two surfaces of a
cylinder as shown. It is assumed that the depth of
the cylinder 𝑏 → 0. We can therefore write, 𝑑𝑣 =
𝑏𝑑𝐴, and 𝑑𝑎 = 𝑏𝑑𝑠
Clearly,
grad 𝜩 𝑑𝐴 =
𝜩 ⊗ 𝒏 𝑑𝑠
𝜕C
Where we have assumed that the area integral on
the two lateral surfaces will cancel out as 𝑏 → 0 and
the outwardly drawn normals oppose.
Dept of Systems Engineering, University of Lagos
25
oafak@unilag.edu.ng 12/27/2012
Special Cases: Vector Field
Vector field. Replacing the tensor with the vector field 𝐟
and contracting, we have,
div 𝐟 𝑑𝑣 =
Ω
𝐟 ⋅ 𝐧 𝑑𝑎
𝜕Ω
Which is the usual form of the Gauss theorem. In 2-D,
this becomes,
div 𝐟 𝑑𝐴 =
𝐟 ⋅ 𝐧 𝑑𝑠
Ω
Dept of Systems Engineering, University of Lagos
26
oafak@unilag.edu.ng 12/27/2012
Scalar Field
For a scalar field 𝜙, the divergence becomes a gradient
and the scalar product on the RHS becomes a simple
multiplication. Hence the divergence theorem becomes,
grad 𝜙 𝑑𝑣 =
Ω
𝜙𝐧 𝑑𝑎
𝜕Ω
The procedure here is valid and will become obvious if
we write, 𝐟 = 𝜙𝒃 where 𝒃 is an arbitrary constant
vector.
Dept of Systems Engineering, University of Lagos
27
oafak@unilag.edu.ng 12/27/2012
div 𝜙𝒃 𝑑𝑣 =
Ω
𝜙𝒃 ⋅ 𝐧 𝑑𝑎 = 𝒃 ⋅
𝜙𝐧 𝑑𝑎
𝜕Ω
𝜕Ω
For the LHS, note that, div 𝜙𝒃 = tr grad 𝜙𝒃
grad 𝜙𝒃 = 𝜙𝑏 𝑖 ,𝑗 𝐠 𝑖 ⊗ 𝐠𝑗
= 𝑏 𝑖 𝜙,𝑗 𝐠 𝑖 ⊗ 𝐠𝑗
The trace of which is,
𝑗
𝑏 𝑖 𝜙,𝑗 𝐠 𝑖 ⋅ 𝐠𝑗 = 𝑏 𝑖 𝜙,𝑗 𝛿𝑖
= 𝑏 𝑖 𝜙,𝑖 = 𝒃 ⋅ grad 𝜙
For the arbitrary constant vector 𝒃, we therefore have that,
div 𝑑𝑣 = 𝒃 ⋅
Ω
Ω
grad 𝜙 𝑑𝑣 =
Dept of Systems Engineering, University of Lagos
grad 𝜙 𝑑𝑣 = 𝒃 ⋅
Ω
𝜕Ω
𝜙𝐧 𝑑𝑎 or
𝜙𝐧 𝑑𝑎
𝜕Ω
Ω
28
grad 𝜙 𝑑𝐴 =
𝜙𝒏 𝑑𝑠
oafak@unilag.edu.ng 12/27/2012
Further Results (Scalar)
grad 𝜙 𝑑𝐴 =
𝜙𝒏 𝑑𝑠
Ω
From here we can write that,
grad 𝜙𝜓 𝑑𝐴 =
Ω
𝜙 grad 𝜓 𝑑𝐴 +
Ω
𝜓 grad 𝜙 𝑑𝐴 =
𝜙𝜓 𝒏 𝑑𝑠
Ω
So that,
𝜓 grad 𝜙 𝑑𝐴 =
𝜙𝜓 𝒏 𝑑𝑠 −
Ω
𝜙 grad 𝜓 𝑑𝐴.
Ω
We can break this, in Cartesian coordinates, to its two components:
𝜕𝜙
𝜕𝜓
𝜓
𝑑𝐴 = 𝜙𝜓 𝑛𝑥 𝑑𝑠 − 𝜙
𝑑𝐴
𝜕𝑥
𝜕𝑥
Ω
Ω
𝜓
Ω
Dept of Systems Engineering, University of Lagos
𝜕𝜙
𝑑𝐴 =
𝜕𝑦
𝜙𝜓 𝑛𝑦 𝑑𝑠 −
𝜙
Ω
29
𝜕𝜓
𝑑𝐴
𝜕𝑦
oafak@unilag.edu.ng 12/27/2012
Further Results (Vector)
Beginning from the standard Gauss theorem (slide 26) in 2-D,
div 𝐟 𝑑𝐴 =
𝐟 ⋅ 𝐧 𝑑𝑠
Ω
Now, let 𝐟 = 𝑤𝐅 where 𝐅 is a differentiable vector field and 𝑤 is a scalar field. Then we
can write,
div 𝑤𝐅 𝑑𝐴 =
𝑤𝐅 ⋅ 𝐧 𝑑𝑠
Ω
But div 𝑤𝐅 = grad 𝑤 ⋅ 𝐅 + 𝑤 div 𝐅. If we are also given that the field 𝐅 is derived from
a scalar potential 𝜙, so that 𝐅 = grad 𝜙, then we can write that,
grad 𝑤 ⋅ grad 𝜙 𝑑𝐴 +
Ω
𝑤 div grad 𝜙 𝑑𝐴 =
𝑤 grad 𝜙 ⋅ 𝐧 𝑑𝑠
Ω
Which can be rearranged to obtain,
−
𝑤 grad2 𝜙 𝑑𝐴 =
Ω
Dept of Systems Engineering, University of Lagos
grad 𝑤 ⋅ grad 𝜙 𝑑𝐴 −
𝑤 grad 𝜙 ⋅ 𝐧 𝑑𝑠
Ω
30
oafak@unilag.edu.ng 12/27/2012
Second-Order Tensor field
For a second-order tensor 𝐓, the Gauss Theorem
becomes,
div𝐓 𝑑𝑣 =
Ω
𝐓𝐧 𝑑𝑎
𝜕Ω
The original outer product under the integral can be
expressed in dyadic form:
𝑇 𝑖𝑗 ,𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 ⊗ 𝐠 𝑘 𝑑𝑣
grad 𝐓 𝑑𝑣 =
Ω
Ω
=
𝐓 ⊗ 𝐧 𝑑𝑎
𝜕Ω
𝑇 𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 ⊗ 𝑛𝑘 𝐠 𝑘 𝑑𝑎
=
𝜕Ω
Dept of Systems Engineering, University of Lagos
31
oafak@unilag.edu.ng 12/27/2012
Second-Order Tensor field
Or
𝑇 𝑖𝑗 ,𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 ⊗ 𝐠 𝑘 𝑑𝑣 =
Ω
𝑇 𝑖𝑗 𝑛𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 ⊗ 𝐠 𝑘 𝑑𝑎
𝜕Ω
Contracting, we have
𝑇 𝑖𝑗 ,𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 𝐠 𝑘 𝑑𝑣 =
Ω
𝑇 𝑖𝑗 𝑛𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 𝐠 𝑘 𝑑𝑎
𝜕Ω
Ω
𝑇 𝑖𝑗 ,𝑘 𝛿𝑗𝑘 𝐠 𝑖 𝑑𝑣 =
𝜕Ω
𝑇 𝑖𝑗 ,𝑗 𝐠 𝑖 𝑑𝑣 =
Ω
𝑇 𝑖𝑗 𝑛𝑘 𝛿𝑗𝑘 𝐠 𝑖 𝑑𝑎
𝑇 𝑖𝑗 𝑛𝑗 𝐠 𝑖 𝑑𝑎
𝜕Ω
Which is the same as,
div𝐓 𝑑𝑣 =
Ω
Dept of Systems Engineering, University of Lagos
𝐓𝐧 𝑑𝑎
𝜕Ω
32
oafak@unilag.edu.ng 12/27/2012
Stokes Theorem
Consider the Euclidean Point Space ℰ. A curve 𝒞 is defined by
the parametrization (Gurtin):
𝐱 = 𝐱 𝜆 where 𝜆0 ≤ 𝜆 ∈ ℛ ≤ 𝜆1
𝒞 is said to be a closed curve if
𝐱 𝜆0 = 𝐱 𝜆1
Define 𝐭 𝜆 ≡
𝑑𝐱 𝜆
𝑑𝜆
.
For any vector point function defined everywhere along 𝒞,
the line integral,
𝜆1
𝜆1
𝑑𝐱 𝜆
𝐯 ⋅ 𝒅𝐱 =
𝐯 𝐱 𝜆0 ⋅
𝑑𝜆 =
𝐯 𝐱 𝜆0 ⋅ 𝐭 𝜆 𝑑𝜆
𝑑𝜆
𝒞
𝜆0
𝜆0
Dept of Systems Engineering, University of Lagos
33
oafak@unilag.edu.ng 12/27/2012
Stokes Theorem
Let 𝜙 be a scalar field on ℰ, the
Chain rule immediately implies
that,
𝜆1
𝜆1
grad 𝜙 ⋅ 𝒅𝐱 =
𝜆0
𝜆1
=
𝜆0
𝜆0
𝜕𝜙 𝐱 𝜆
𝜕𝜆
𝑑𝐱 𝜆
grad 𝜙 ⋅
𝑑𝜆
𝑑𝜆
𝑑𝜆 = 𝜙 𝐱 𝜆1
− 𝜙 𝐱 𝜆0
So that for a close curve 𝒞 grad 𝜙 ⋅ 𝒅𝐱 = 0
For a positively oriented surface bounded by a closed curve 𝒞
(Gurtin), Stokes theorem is stated as follows:
Dept of Systems Engineering, University of Lagos
34
oafak@unilag.edu.ng 12/27/2012
Stokes Theorem
Stokes’ Theorem Let 𝜙, 𝐯, and 𝐓 be scalar, vector, and tensor
fields with common domain a region ℛ. Then given any
positively oriented surface 𝒮, with boundary 𝒞 closed curve,
in ℛ
𝜙𝑑𝐱 =
𝒞
𝐧 × grad 𝜙 𝑑𝑎
𝒮
𝐯 ⋅ 𝑑𝐱 =
𝒞
𝐧 × curl 𝐯 𝑑𝑎
𝒮
curl 𝐓 T 𝑑𝑎
𝐓𝑑𝐱 =
𝒞
Dept of Systems Engineering, University of Lagos
𝒮
35
oafak@unilag.edu.ng 12/27/2012
Simplest Problem Explicit
𝜕𝐹 𝑑 𝜕𝐹
−
=0
𝜕𝑦 𝑑𝑥 𝜕𝑦′
Noting the fact that,
𝜕𝐹
= Ψ 𝑥, 𝑦 𝑥 , 𝑦′(𝑥) ,
𝜕𝑦′
Explicitly showing it is a function of the three variables. The total
derivative of
𝜕Ψ
𝜕Ψ
𝜕Ψ
𝑑Ψ =
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑦′
𝜕𝑥
𝜕𝑦
𝜕𝑦′
So that,
𝑑 𝜕𝐹
𝜕Ψ 𝜕Ψ ′ 𝜕Ψ ′′
𝜕2𝐹
𝜕2𝐹 ′
𝜕 2 𝐹 ′′
=
+
𝑦 + ′𝑦 =
+
𝑦 + ′
𝑦
𝑑𝑥 𝜕𝑦′
𝜕𝑥 𝜕𝑦
𝜕𝑦
𝜕𝑥𝜕𝑦′ 𝜕𝑦𝜕𝑦′
𝜕𝑦 𝜕𝑦′
oafak@unilag.edu.ng, University of Lagos
36
Wednesday, April 15, 2015
Simplest Problem
So that the minimization condition becomes,
𝜕2𝐹
𝜕2𝐹 ′
𝜕 2 𝐹 ′′ 𝜕𝐹
+
𝑦 + ′
𝑦 −
= 0.
𝜕𝑥𝜕𝑦′ 𝜕𝑦𝜕𝑦′
𝜕𝑦 𝜕𝑦′
𝜕𝑦
This expression is also called the “variational derivative” of
the functional:
𝜕𝐹
𝑑 𝜕𝐹
𝐹𝑦𝑥 =
−
𝜕𝑦 𝑑𝑥 𝜕𝑦 ′
𝜕2𝐹
𝜕2𝐹 ′
𝜕 2 𝐹 ′′ 𝜕𝐹
=−
−
𝑦 − ′ ′𝑦 +
=0
𝜕𝑥𝜕𝑦 ′ 𝜕𝑦𝜕𝑦 ′
𝜕𝑦 𝜕𝑦
𝜕𝑦
oafak@unilag.edu.ng, University of Lagos
37
Wednesday, April 15, 2015
Plane Geodesic
 Recall that the plane geodesic problem is simply the extremization of
the integral,
𝑏
𝐽=
𝑎
𝑑𝑦
1+
𝑑𝑥
2
𝑑𝑥
Where the functional 𝐹 = 𝐹(𝑦 ′ 𝑥 ) which is the Simplest Problems
with only the derivative of 𝑦 appearing in the functional. Hence, all
other terms in the variational derivative vanishes and the
extremizing condition is simply,
𝜕 2 𝐹 ′′
𝑦 =0
𝜕𝑦 ′ 𝜕𝑦 ′
From which we easily recover the obvious fact that only a straight
line can minimize the distance on a flat plane.
oafak@unilag.edu.ng, University of Lagos
38
Wednesday, April 15, 2015
Several Independent Functions
 The foregoing can be extended to include the case of
functionals with two or more functional arguments.
Consider the case where the function
𝐽 𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥)
𝑥1
=
𝐹 𝑥, 𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥), 𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥) 𝑑𝑥
𝑥0
Where we have used dots to denote derivatives with respect
to the independent variable 𝑥 so that the dependent variable
arguments of the functional are 𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥) and their
first derivatives 𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥). We assume that the
continuity requirements for the single function case remain
valid in this case for all the independent and function
arguments shown. We now suppose that functions
oafak@unilag.edu.ng, University of Lagos
39
Wednesday, April 15, 2015
𝑦 𝑖 𝑥 = 𝑓 𝑖 𝑥 , 𝑖 = 1,2, … 𝑛
exist in the interval 𝑥0 ≤ 𝑥 ≤ 𝑥1 with prescribed values at the
endpoints of this interval which are such that the above functional
assumes an extreme value in comparison with other admissible sets
of functions in a predetermined neighbourhood of the set of
functions in 𝑓 𝑖 𝑥 , 𝑖 = 1,2, … 𝑛 satisfying the same prescribed
conditions at the endpoints. For this to happen, the first variation of
the definite integral must vanish. This immediately leads to the set
of equations,
𝑥1
𝜕𝐹
𝜕𝐹
𝑖
𝑖 𝑑𝑥 = 0
𝛿𝐽 =
𝛿𝑦
+
𝛿
𝑦
𝜕𝑦 𝑖
𝜕𝑦 𝑖
𝑥0
oafak@unilag.edu.ng, University of Lagos
40
Wednesday, April 15, 2015
Note that each term in the brackets is actually a sum of 𝑛 terms by
the Einstein summation convention. Integrating the second sum by
parts, we have
𝑥1
𝑥1
𝑥1
𝜕𝐹
𝜕𝐹
𝑑 𝜕𝐹
𝑖
𝑖
𝑖 𝑑𝑥
𝛿
𝑦
𝑑𝑥
=
𝛿𝑦
−
𝛿𝑦
𝑖
𝑑𝑥
𝜕𝑦 𝑖
𝜕𝑦 𝑖
𝜕
𝑦
𝑥0
𝑥
0
𝑥
0
The first term vanishes as usual on account of the vanishing of the
variations in the dependent variables at the endpoints. Substituting
this back into the minimization condition, we can write,
𝑥1
𝜕𝐹
𝑑 𝜕𝐹
𝑖 𝑑𝑥 = 0
𝛿𝐽 =
−
𝛿𝑦
𝜕𝑦 𝑖 𝑑𝑥 𝜕𝑦 𝑖
𝑥0
But the variations in the functional arguments are all applied
independently.
oafak@unilag.edu.ng, University of Lagos
41
Wednesday, April 15, 2015
The fundamental lemma of variational calculus therefore requires
the vanishing of the terms in the square brackets for each 𝑖 =
1, 2, . . . , 𝑛 so that each variational derivative vanishes:
𝜕𝐹
𝜕 𝜕𝐹
−
=0
𝜕𝑦 𝑖 𝜕𝑥 𝜕𝑦 𝑖
The functional 𝐹 here is a function of 2𝑛 + 1 variables
𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥) and their first derivatives 𝑦1 𝑥 , … , 𝑦 𝑛 (𝑥). In
analytical mechanics, if the independent variable 𝑥 is the elapsed
time 𝑡, 𝐹 represents the work function, the 𝑦 𝑖 𝑥 𝑠 ’s are the
generalized coordinates and the 𝑦 𝑖 𝑥 𝑠 are the generalized
velocities. These equations are then are the celebrated EulerLagrange equations of motion.
oafak@unilag.edu.ng, University of Lagos
42
Wednesday, April 15, 2015
Higher-Order Derivatives
Consider the case where the functional has another term of
the second order:
𝑥1
𝐽𝑦 𝑥
=
𝐹 𝑥, 𝑦 𝑥 , 𝑦′(𝑥), 𝑦′′(𝑥) 𝑑𝑥.
𝑥0
A simple extension of the previous arguments with similar
conditions lead us to the extremization condition,
𝑥1
𝜕𝐹
𝜕𝐹 ′ 𝜕𝐹
𝛿𝐽 =
𝛿𝑦 +
𝛿𝑦 +
𝛿𝑦 ′ ′ 𝑑𝑥 = 0
𝜕𝑦
𝜕𝑦′
𝜕𝑦′′
𝑥0
Integrating the second term by parts, we have,
𝑥1
𝑥1
𝑥1
𝜕𝐹 ′
𝜕𝐹
𝑑 𝜕𝐹
𝛿𝑦 𝑑𝑥 =
𝛿𝑦
−
𝛿𝑦𝑑𝑥
′
𝜕𝑦
𝑥0 𝜕𝑦′
𝑥0 𝑑𝑥 𝜕𝑦′
𝑥
0
oafak@unilag.edu.ng, University of Lagos
43
Wednesday, April 15, 2015
For the third term, we have,
𝑥1
𝜕𝐹
𝜕𝐹
′
𝛿𝑦′ 𝑑𝑥 =
𝛿𝑦′
𝜕𝑦′′
𝑥0 𝜕𝑦′′
𝜕𝐹
=
𝛿𝑦′
𝜕𝑦′′
𝑥1
𝑥1
−
𝑥0
𝑥1
𝑥0
𝑥0
𝑑 𝜕𝐹
𝛿𝑦′𝑑𝑥
𝑑𝑥 𝜕𝑦′′
𝑑 𝜕𝐹
−
𝛿𝑦
𝑑𝑥 𝜕𝑦 ′′
𝑥1
𝑥0
Substituting back into the extremizing condition,
𝑥1
𝛿𝐽 =
𝑥0
𝜕𝐹 𝑑 𝜕𝐹
𝑑 2 𝜕𝐹
𝜕𝐹
𝜕𝐹
𝑑 𝜕𝐹
′
−
+
𝛿𝑦𝑑𝑥 +
𝛿𝑦 +
𝛿𝑦 −
𝛿𝑦
𝜕𝑦 𝑑𝑥 𝜕𝑦′ 𝑑𝑥 2 𝜕𝑦′′
𝜕𝑦 ′
𝜕𝑦′′
𝑑𝑥 𝜕𝑦 ′′
𝑥1
𝑥0
=0
oafak@unilag.edu.ng, University of Lagos
44
Wednesday, April 15, 2015
Substituting back into the extremizing condition,
𝛿𝐽
𝑥1
𝜕𝐹 𝑑 𝜕𝐹
𝑑 2 𝜕𝐹
𝜕𝐹
𝜕𝐹
𝑑 𝜕𝐹
′
=
−
+ 2 ′′ 𝛿𝑦𝑑𝑥 +
𝛿𝑦 + ′′ 𝛿𝑦 −
𝛿𝑦
′
′
′′
𝜕𝑦
𝑑𝑥
𝜕𝑦
𝑑𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑦
𝑑𝑥
𝜕𝑦
𝑥0
𝑥1
𝑥0
=0
Which contains the appropriate Euler’s equations
𝑥1
𝜕𝐹 𝑑 𝜕𝐹
𝑑 2 𝜕𝐹
𝛿𝐽 =
−
+ 2
𝛿𝑦𝑑𝑥 = 0
𝜕𝑦
𝑑𝑥
𝜕𝑦′
𝑑𝑥
𝜕𝑦′′
𝑥0
And the boundary conditions,
𝑥1
𝜕𝐹
𝜕𝐹
𝑑
𝜕𝐹
′
𝛿𝑦
+
𝛿𝑦
−
𝛿𝑦
𝜕𝑦 ′
𝜕𝑦′′
𝑑𝑥 𝜕𝑦 ′′
𝑥
0
that normally vanish identically when there is no variation at the endpoints –
that is, when the end conditions are fixed.
oafak@unilag.edu.ng, University of Lagos
45
Wednesday, April 15, 2015
Several Independent Variables
The case of two variables 𝑥 and 𝑦 sufficiently illustrates
what happens here:
𝐽𝑢 𝑥
=
𝐹 𝑥, 𝑦, 𝑢 𝑥, 𝑦 , 𝑢𝑥 𝑥, 𝑦 , 𝑢𝑦 (𝑥, 𝑦) 𝑑𝑥𝑑𝑦
Ω
Without any further ado, there will be no variations in
the independent variables 𝑥 and 𝑦. The first variation
immediately leads to the vanishing of the integral,
𝜕𝐹
𝜕𝐹
𝜕𝐹
𝛿𝑢 +
𝛿𝑢𝑥 +
𝛿𝑢𝑦 𝑑𝑥𝑑𝑦 = 0
𝜕𝑢
𝜕𝑢𝑥
𝜕𝑢𝑦
Ω
oafak@unilag.edu.ng, University of Lagos
46
Wednesday, April 15, 2015
We change the order of integration in the extremizing condition, and obtain,
𝜕𝐹
𝜕𝐹 𝜕𝛿𝑢 𝜕𝐹 𝜕𝛿𝑢
𝛿𝑢 +
+
𝑑𝑥𝑑𝑦 = 0
𝜕𝑢
𝜕𝑢𝑥 𝜕𝑥
𝜕𝑢𝑦 𝜕𝑦
Ω
We now appeal to the Gauss Divergence Theorem (Slide 29) to obtain the
necessary integration by parts here; recall,
𝜕𝜙
𝜕𝜓
𝜓
𝑑𝐴 = 𝜙𝜓 𝑛𝑥 𝑑𝑠 − 𝜙
𝑑𝐴
𝜕𝑥
𝜕𝑥
Ω
Ω
In the second term, substituting 𝜓 for
Ω
𝜕𝐹 𝜕𝛿𝑢
𝑑𝑥𝑑𝑦 =
𝜕𝑢𝑥 𝜕𝑥
𝜕𝐹
𝜕𝑢𝑥
and 𝜙 for 𝛿𝑢, it follows that,
𝜕𝐹
𝑛 𝛿𝑢𝑑𝑠 −
𝜕𝑢𝑥 𝑥
Ω
𝜕 𝜕𝐹
𝛿𝑢𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑢𝑥
Applying the second part of the Gauss theorem to the third term leads to,
Ω
𝜕𝐹 𝜕𝛿𝑢
𝑑𝑥𝑑𝑦 =
𝜕𝑢𝑦 𝜕𝑦
oafak@unilag.edu.ng, University of Lagos
𝜕𝐹
𝑛 𝛿𝑢𝑑𝑠 −
𝜕𝑢𝑦 𝑦
47
Ω
𝜕 𝜕𝐹
𝛿𝑢𝑑𝑥𝑑𝑦
𝜕𝑦 𝜕𝑢𝑦
Wednesday, April 15, 2015
Finally, we can write that,
𝜕𝐹
𝜕 𝜕𝐹
𝜕 𝜕𝐹
𝛿𝐽 =
−
−
𝜕𝑢 𝜕𝑥 𝜕𝑢𝑥
𝜕𝑦 𝜕𝑢𝑦
Ω
𝛿𝑢𝑑𝑥𝑑𝑦 +
Γ
𝜕𝐹
𝜕𝐹
𝑛𝑥 +
𝑛 𝛿𝑢𝑑𝑠
𝜕𝑢𝑥
𝜕𝑢𝑦 𝑦
(Notational inconsistency here: the subscript on 𝑛 is the component. On 𝑢, it is partial
derivative wrt subscript)
Let the green line denote the boundary and
Consider an outward drawn normal 𝐧 = cos 𝛼 𝐢 + sin 𝛼 𝐣 in the figure below:
𝑑𝑦
𝑑𝑥
cos 𝛼 =
, sin 𝛼 = −
𝑑𝑠
𝑑𝑠
so that the above boundary becomes,
Γ
𝜕𝐹
𝜕𝐹
𝑛 +
𝑛 𝛿𝑢𝑑𝑠 =
𝜕𝑢𝑥 𝑥 𝜕𝑢𝑦 𝑦
Γ
=
𝜕𝐹 𝑑𝑦 𝜕𝐹 𝑑𝑥
−
𝛿𝑢𝑑𝑠
𝜕𝑢𝑥 𝑑𝑠 𝜕𝑢𝑦 𝑑𝑠
=
𝜕𝐹
𝜕𝐹
𝑑𝑦 −
𝑑𝑥 𝛿𝑢
𝜕𝑢𝑥
𝜕𝑢𝑦
=0
oafak@unilag.edu.ng, University of Lagos
48
Wednesday, April 15, 2015
so that the above boundary becomes,
Γ
𝜕𝐹
𝜕𝐹
𝑛 +
𝑛 𝛿𝑢𝑑𝑠 =
𝜕𝑢𝑥 𝑥 𝜕𝑢𝑦 𝑦
=
𝜕𝐹 𝑑𝑦 𝜕𝐹 𝑑𝑥
−
𝛿𝑢𝑑𝑠
𝜕𝑢𝑥 𝑑𝑠 𝜕𝑢𝑦 𝑑𝑠
Γ
𝜕𝐹
𝑑𝑦
𝜕𝑢𝑥
Γ
Finally, we can write that,
𝛿𝐽
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
=
−
−
𝜕𝑢 𝜕𝑥 𝜕𝑢𝑥
𝜕𝑦 𝜕𝑢𝑦
Ω
−
𝜕𝐹
𝑑𝑥
𝜕𝑢𝑦
𝛿𝑢𝑑𝑥𝑑𝑦 +
Γ
𝛿𝑢 = 0
𝜕𝐹
𝜕𝐹
𝑑𝑦 −
𝑑𝑥 𝛿𝑢
𝜕𝑢𝑥
𝜕𝑢𝑦
=0
oafak@unilag.edu.ng, University of Lagos
49
Wednesday, April 15, 2015
Multi-Dimensional Systems
Following the same logic, a functional with two multivariable functions
𝐽 𝑢 𝑥, 𝑦 , 𝑣(𝑥, 𝑦) =
𝐹 𝑥, 𝑦, 𝑢 𝑥, 𝑦 , 𝑣 𝑥, 𝑦 , 𝑢𝑥 , 𝑢𝑦 , 𝑣𝑥 , 𝑣𝑦 ) 𝑑𝑥𝑑𝑦.
With the primary variables 𝑢 𝑥, 𝑦 , 𝑣(𝑥, 𝑦) will lead to the equation,
𝛿𝐽 =
Ω
+
Γ
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
𝜕𝑢 𝜕𝑥 𝜕𝑢𝑥
𝜕𝑦 𝜕𝑢𝑦
𝛿𝑢 +
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
𝜕𝑣 𝜕𝑥 𝜕𝑣𝑥
𝜕𝑦 𝜕𝑣𝑦
𝛿𝑣 𝑑𝑥𝑑𝑦
𝜕𝐹
𝜕𝐹
𝜕𝐹
𝜕𝐹
𝑛 +
𝑛 𝛿𝑢 +
𝑛 +
𝑛 𝛿𝑣 𝑑𝑠
𝜕𝑢𝑥 𝑥 𝜕𝑢𝑦 𝑦
𝜕𝑣𝑥 𝑥 𝜕𝑣𝑦 𝑦
oafak@unilag.edu.ng, University of Lagos
50
Wednesday, April 15, 2015
And the fact that the variations in the two functions 𝑢 𝑥, 𝑦 , 𝑣(𝑥, 𝑦) are
applied independently implies that the coefficients of 𝛿𝑢 and 𝛿𝑣 must
vanish:
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
=0
𝜕𝑢 𝜕𝑥 𝜕𝑢𝑥
𝜕𝑦 𝜕𝑢𝑦
And
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
=0
𝜕𝑣 𝜕𝑥 𝜕𝑣𝑥
𝜕𝑦 𝜕𝑣𝑦
And the boundary term,
𝜕𝐹
𝜕𝐹
𝜕𝐹
𝜕𝐹
𝑛𝑥 +
𝑛𝑦 𝛿𝑢 +
𝑛𝑥 +
𝑛𝑦 𝛿𝑣 𝑑𝑠
𝜕𝑢𝑥
𝜕𝑢𝑦
𝜕𝑣𝑥
𝜕𝑣𝑦
Γ
Must also vanish.
oafak@unilag.edu.ng, University of Lagos
51
Wednesday, April 15, 2015
Primary & Secondary Variables
 Integration by parts, achieved here by using the 2-D
version of the Gauss Divergence theorem achieved
two purposes:
 1. Creates a “weak” form of the problem. Weak in the
sense that the level of differentiation is transferred
from one variable (weakening the smoothness
requirements) to another.
 2. Exposing another set of variables apart from the
primary variables of interest.
oafak@unilag.edu.ng, University of Lagos
52
Wednesday, April 15, 2015
Primary & Secondary Variables
 In the original problem, the variables are the functions
𝑢 𝑥, 𝑦 𝑎𝑛𝑑 𝑣(𝑥, 𝑦). As a result of the Integration by parts, we
now have new functions 𝑄 𝑥, 𝑦 𝑎𝑛𝑑 𝑅 𝑥, 𝑦 defined by the
integrals,
𝑄 𝑥, 𝑦 ≡
𝜕𝐹
𝜕𝑢𝑥
𝑛𝑥 +
𝜕𝐹
𝜕𝑢𝑦
𝑛𝑦
and
𝑅 𝑥, 𝑦 ≡
𝜕𝐹
𝜕𝑣𝑥
𝑛𝑥 +
𝜕𝐹
𝜕𝑣𝑦
𝑛𝑦
The prescription of the primary variables on the boundary are the
Dirichlet, rigid, kinematic or Essential Boundary Conditions; the
specification of the secondary variables are the Newman or Natural
Boundary Conditions.
oafak@unilag.edu.ng, University of Lagos
53
Wednesday, April 15, 2015
We now rewrite the variational extremization in terms of primary and secondary
variables as follows:
1.
First, observe that in the interior, the primary variables must vanish; This
leads naturally to the set of Euler-Lagrange Equations:
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
𝜕𝐹 𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
= 0, and
−
−
=0
𝜕𝑢 𝜕𝑥 𝜕𝑢𝑥
𝜕𝑦 𝜕𝑢𝑦
𝜕𝑣 𝜕𝑥 𝜕𝑣𝑥
𝜕𝑦 𝜕𝑣𝑦
2. And the boundary term,
𝑄 𝑥, 𝑦 𝛿𝑢 + 𝑅 𝑥, 𝑦 𝛿𝑣 𝑑𝑠
Γ
Now looks clearer in terms of the primary and secondary variables. If the
primary variables are not specified, their variations will be non-zero. In
order for the extremization to remain valid, their coeficients, - the
secondary variables will automatically vanish.
It is clear here that, on the boundary, the vanishing coefficient of the
variations in the primary variables (secondary variables) give the natural
boundary conditions.
oafak@unilag.edu.ng, University of Lagos
54
Wednesday, April 15, 2015
Boundary Conditions
 One of the big confusions in the application of
variational, and consequently, finite element,
methods is in the practitioners’ inability to correctly
articulate the issues of Boundary conditions.
 In early elucidations on this issue, (See For example
Courant & Hilbert I (pg 208-214), the matter, for the
patient reader, is well spelt out.
 We use the (Lanczos’) demonstration of a
transversely-loaded beam to concretize these ideas.
oafak@unilag.edu.ng, University of Lagos
55
Wednesday, April 15, 2015
 Instead of finding the optimal path from A to B,
imagine we are now finding the best path from one
end to the other with no restrictions.
oafak@unilag.edu.ng, University of Lagos
56
Wednesday, April 15, 2015
Unrestricted Varied Paths
 The Extremized and varied paths may now look like
the following:
The end values are no
longer prescribed. The
varied paths still
satisfy previous
conditions for
extremisation.
We consider a concrete example of this first before we present the
theoretical treatment of the boundary conditions and its practical
implications. Ref: Lanczos (68-73), Courant & Hilbert (208-214)
oafak@unilag.edu.ng, University of Lagos
57
Wednesday, April 15, 2015
Concrete Example
 Consider the loaded beam as shown in different
support configurations:
oafak@unilag.edu.ng, University of Lagos
58
Wednesday, April 15, 2015
The bending elastica of the above beam can be found by solving the ODE: Finding
𝑦(𝑥) that satisfies,
𝑑 4 𝑦 𝑤(𝑥)
=
4
𝑑𝑥
𝑘
where 𝑘 is a constant depending on the elastic property of the material and
geometry. We derive this same equation from a variational principle:
The potential energy, 𝑉1 due to elastic forces and the that due to gravitational
forces 𝑉2 are given by,
𝑘
𝑉1 =
2
𝑥1
𝑦 ′′ 𝑥
2
𝑥1
𝑑𝑥 and 𝑉2 = −
𝑥0
𝑤 𝑥 𝑦(𝑥)𝑑𝑥
𝑥0
The functional to minimize in this case is therefore,
𝑥1
𝐽𝑦 𝑥
=
𝑥0
𝑥1
=
𝑘 ′′
𝑦 𝑥
2
2
− 𝑤 𝑥 𝑦 𝑥 𝑑𝑥
′
𝐹(𝑥, 𝑦 𝑥 , 𝑦 ′ 𝑥 )𝑑𝑥
𝑥0
The functional depending explicitly only on 𝑦(𝑥) and 𝑦 ′′ 𝑥 .
oafak@unilag.edu.ng, University of Lagos
59
Wednesday, April 15, 2015
The variational derivative (Slide 41) of this functional is,
𝜕𝐹 𝑑 𝜕𝐹
𝑑 2 𝜕𝐹
−
+ 2
𝜕𝑦 𝑑𝑥 𝜕𝑦′ 𝑑𝑥 𝜕𝑦′′
And the boundary conditions,
𝜕𝐹
𝜕𝐹
𝑑 𝜕𝐹
′
𝛿𝑦 +
𝛿𝑦 −
𝛿𝑦
𝜕𝑦 ′
𝜕𝑦′′
𝑑𝑥 𝜕𝑦 ′′
𝑥1
𝑥0
where
𝑘 ′′
2
𝐹 𝑥, 𝑦 𝑥
𝑥 = 𝑦 𝑥
− 𝑤 𝑥 𝑦(𝑥)
2
A rather tedious expansion of the variational derivative, using the explicit
expression, immediately leads to the ODE,
, 𝑦 ′′
𝑑4 𝑦 𝑤 𝑥
=
.
𝑑𝑥 4
𝑘
Once this is satisfied, we only need to bother about the BCs. First
𝜕𝐹
recall that, ′ = 0 as, 𝐹 𝑥, 𝑦 𝑥 , 𝑦 ′′ 𝑥
𝜕𝑦
′′
and 𝑦 𝑥 .
oafak@unilag.edu.ng, University of Lagos
60
depends explicitly only on 𝑦(𝑥)
Wednesday, April 15, 2015
The boundary conditions are now given as

𝜕𝐹
𝛿𝑦 ′
′′
𝜕𝑦
−
= 𝑘𝑦 ′′
= 𝑘𝑦 ′′
𝑥1
𝑑 𝜕𝐹
𝛿𝑦
𝑑𝑥 𝜕𝑦 ′′
𝑥0
𝑑
′
𝑥 𝛿𝑦 − 𝑘
𝑦 ′′ 𝑥 𝛿𝑦
𝑑𝑥
𝑥1
𝑥0
𝜕
𝜕
𝜕
𝜕
′
′
′′
′′′
𝑥 𝛿𝑦 − 𝑘
+𝑦
+𝑦
+𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑦 ′
𝜕𝑦 ′′
𝑥
𝑥1
𝑦 ′′ 𝑥 𝛿𝑦
𝑥0
= 𝑘𝑦 ′′ 𝑥 𝛿𝑦 ′ − 𝑘 𝑦 ′′′ + 0 + 0 + 𝑦 ′′′ 𝛿𝑦 𝑥01
= 𝑘𝑦 ′′ 𝑥1 𝛿𝑦 ′ 𝑥1 − 2𝑘𝑦 ′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦 ′′ 𝑥0 𝛿𝑦 ′ 𝑥0
−2𝑘𝑦 ′′′ 𝑥0 𝛿𝑦 𝑥0
Let us now examine how these conditions are satisfied for the different loading
conditions.
Before we proceed, first recall that a fourth-order ODE must have four boundary
conditions to give a unique solution.
oafak@unilag.edu.ng, University of Lagos
61
Wednesday, April 15, 2015
𝜕𝐹
First recall that, ′ = 0 as, 𝐹 𝑥, 𝑦 𝑥 , 𝑦 ′′ 𝑥 depends explicitly only on 𝑦(𝑥)
𝜕𝑦
′′
and 𝑦 𝑥 . The boundary conditions are now given as

𝜕𝐹
𝛿𝑦 ′
′′
𝜕𝑦
−
= 𝑘𝑦 ′′
= 𝑘𝑦 ′′
𝑥1
𝑑 𝜕𝐹
𝛿𝑦
𝑑𝑥 𝜕𝑦 ′′
𝑥0
𝑑
′
𝑥 𝛿𝑦 − 𝑘
𝑦 ′′ 𝑥 𝛿𝑦
𝑑𝑥
𝑥1
𝑥0
𝜕
𝜕
𝜕
𝜕
′
′
′′
′′′
𝑥 𝛿𝑦 − 𝑘
+𝑦
+𝑦
+𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑦 ′
𝜕𝑦 ′′
𝑥1
𝑥0
𝑥1
𝑦 ′′ 𝑥 𝛿𝑦
𝑥0
= 𝑘𝑦 ′′ 𝑥 𝛿𝑦 ′ − 𝑘 𝑦 ′′′ + 0 + 0 + 𝑦 ′′′ 𝛿𝑦
= 𝑘𝑦 ′′ 𝑥1 𝛿𝑦 ′ 𝑥1 − 2𝑘𝑦 ′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦 ′′ 𝑥0 𝛿𝑦 ′ 𝑥0
+2𝑘𝑦 ′′′ 𝑥0 𝛿𝑦 𝑥0
Let us now examine how these conditions are satisfied for the different loading
conditions.
Before we proceed, first recall that a fourth-order ODE must have four boundary
conditions to give a unique solution.
oafak@unilag.edu.ng, University of Lagos
62
Wednesday, April 15, 2015
Special Cases
Let us now examine how these conditions are satisfied for the different loading
conditions. Before we proceed, first recall that a fourth-order ODE must have
four boundary conditions to give a unique solution.
Case 1. Both ends fixed.
Here, the displacements, 𝑦 𝑥0 as well as 𝑦 𝑥1 at both ends are given (zero for
both in this case; what really matters is not that they are zero, but that they are
prespecified). Furthermore, the slopes at the two ends, 𝑦′ 𝑥0 as well as 𝑦′ 𝑥1
are also given (again, they are zero-prescribed is what matters).
Clearly, these conditions mean that
𝛿𝑦 ′
𝛿𝑦
𝛿𝑦 ′
𝛿𝑦
oafak@unilag.edu.ng, University of Lagos
𝑥1
𝑥1
𝑥0
𝑥0
63
=0
=0
= 0,
=0
Wednesday, April 15, 2015
Case 1 Continued
It is essential that these conditions, prescriptions of no variation of the
function and its partial derivatives at the endpoints be satisfied a-priori
by any extremizing function to be admissible. They are called “Essential
Boundary conditions”. These are the conditions we need to be able to
get a unique solution to our ODE! With the essential boundary
conditions satisfied, it is clear that our extremizing conditions are now
satisfied:
𝑘𝑦 ′′ 𝑥1 𝛿𝑦 ′ 𝑥1 − 2𝑘𝑦 ′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦 ′′ 𝑥0 𝛿𝑦 ′ 𝑥0 + 2𝑘𝑦 ′′′ 𝑥0 𝛿𝑦 𝑥0
= 𝑘𝑦 ′′ 𝑥1 0 − 2𝑘𝑦 ′′′ 𝑥1 0 − 𝑘𝑦 ′′ 𝑥0 0 + 2𝑘𝑦 ′′′ 𝑥0 0 = 0.
oafak@unilag.edu.ng, University of Lagos
64
Wednesday, April 15, 2015
Case 2. One end (𝑥0 ), fixed,
the other end free.
Clearly, these conditions mean that 𝛿𝑦 ′ 𝑥1 =?; 𝛿𝑦 𝑥1 =?; 𝛿𝑦 ′ 𝑥0 = 0; 𝛿𝑦 𝑥0 = 0
Only two of the parameters are prescribed. We cannot get a unique solution to our ODE
now because we do not have a sufficient number of boundary conditions. The variation at
the end 𝑥1 can no longer be assumed to vanish. The power of the variational method is
that, in this case, it carries along with the extremization, the remaining boundary
conditions needed to still obtain a unique solution. This conditions, occurring “naturally”
with the first variation, are called “Natural Boundary Conditions”. In order to still obtain a
vanishing of the boundary expression,
𝑘𝑦 ′′ 𝑥1 𝛿𝑦 ′ 𝑥1 − 2𝑘𝑦 ′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦 ′′ 𝑥0 𝛿𝑦 ′ 𝑥0 + 2𝑘𝑦 ′′′ 𝑥0 𝛿𝑦 𝑥0
when 𝛿𝑦 𝑥1 ≠ 0 and 𝛿𝑦 ′ 𝑥1 ≠ 0 we MUST have that,
𝑦 ′′ 𝑥1 = 0,
𝑎𝑛𝑑
𝑦 ′′′ 𝑥1 = 0.
Which in this case are proportional to the bending moment and shear forces at these
points. The variational formulation yields these conditions naturally!
oafak@unilag.edu.ng, University of Lagos
65
Wednesday, April 15, 2015
Case 3. Both ends simply supported
Here for each end, the displacement function is prescribed
𝛿𝑦 ′
𝛿𝑦
𝛿𝑦 ′
𝛿𝑦
𝑥1
𝑥1
𝑥0
𝑥0
=?
=0
=? ,
=0
(zero actually, but bear in mind, it is the prescription that matters), we still want
the expression,
𝑘𝑦 ′′ 𝑥1 𝛿𝑦 ′ 𝑥1 − 2𝑘𝑦 ′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦 ′′ 𝑥0 𝛿𝑦 ′ 𝑥0 + 2𝑘𝑦 ′′′ 𝑥0 𝛿𝑦 𝑥0
To vanish. Again, as an ODE problem, we need two more boundary conditions.
At this point, it is easy to see that the Natural boundary conditions, following
the previous development, are:
𝑦 ′′ 𝑥0 = 0 𝑎𝑛𝑑 𝑦 ′′ 𝑥1 = 0
Which has the physical interpretation that the shear forces vanish at these
ends.
oafak@unilag.edu.ng, University of Lagos
66
Wednesday, April 15, 2015
Case 4. One end fixed,
The other Simply Supported
𝛿𝑦 ′ 𝑥1
𝛿𝑦 𝑥1
𝛿𝑦 ′ 𝑥0
𝛿𝑦 𝑥0
=?
=0
= 0,
=0
The remaining condition needed here is supplied by the
natural boundary condition, 𝑦 ′′ 𝑥1 .
oafak@unilag.edu.ng, University of Lagos
67
Wednesday, April 15, 2015
Small Elastic Deformations
For a homogeneous, isotropic elastic material undergoing small
deformations given by the displacement vector 𝐮, assuming that
𝜆 and 𝜇 are the Lame’s coefficients, the energy functional,
𝜇
𝐹𝐮 =
grad 𝐮 2 + 𝜆 + 𝜇 div 2 𝐮
2
can be minimized to find the equilibrium conditions. In Cartesian
coordinates, we can write this functional in terms of the two
multivariable functions by writing, 𝐮 = 𝑢 𝑥, 𝑦 𝐢 + 𝑣 𝑥, 𝑦 𝐣. Recall
that
2
grad 𝐮
𝜕𝑢
=
𝜕𝑥
and div 𝐮 =
𝜕𝑢
𝜕𝑥
oafak@unilag.edu.ng, University of Lagos
+
2
= tr gradT 𝐮 grad 𝐮
𝜕𝑢
+
𝜕𝑦
2
𝜕𝑣
+
𝜕𝑥
2
𝜕𝑣
+
𝜕𝑦
2
𝜕𝑣
𝜕𝑦
68
Wednesday, April 15, 2015
Elastic Deformations
We therefore need to extremize the functional,
 𝐽 𝑢, 𝑣 =
=
Ω
Ω
𝜇
2
𝜆
+𝜇
2
2
grad 𝐮
𝜕𝑢
𝜕𝑥
2
+ 𝜆 + 𝜇 div 2 𝐮 𝑑𝐴
𝜕𝑣
+
𝜕𝑦
2
𝜇
+
2
𝜕𝑢
𝜕𝑦
2
𝜕𝑣
+
𝜕𝑥
2
+ 𝜆+𝜇
𝜕𝑢 𝜕𝑣
𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
Clearly,
𝐹 𝑢, 𝑣 = 𝐹 𝑢𝑥 , 𝑢𝑦 , 𝑣𝑥 , 𝑣𝑦
Which corresponds to the case of multiple functions of the independent variable as we have
in slide 45. There is no direct dependency on the functions 𝑢 and 𝑣 themselves. Note that,
𝜕𝐹(𝑢, 𝑣)
𝜕𝑢
𝜕𝑣
= 𝜆 + 2𝜇
+ 𝜆+𝜇
𝜕𝑢𝑥
𝜕𝑥
𝜕𝑦
oafak@unilag.edu.ng, University of Lagos
69
Wednesday, April 15, 2015
The Variational derivatives corresponding to this problem are,
𝜕𝐹
𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
=0
𝜕𝑢 𝜕𝑥 𝜕𝑢𝑥
𝜕𝑦 𝜕𝑢𝑦
and
𝜕𝐹
𝜕 𝜕𝐹
𝜕 𝜕𝐹
−
−
=0
𝜕𝑣 𝜕𝑥 𝜕𝑣𝑥
𝜕𝑦 𝜕𝑣𝑦
After differentiating and ignoring the vanishing first term in each equation, we have,
𝜕2𝑢
𝜕 2𝑢
𝜕2𝑣
𝜆 + 2𝜇
+𝜇 2+ 𝜆+𝜇
=0
𝜕𝑥 2
𝜕𝑦
𝜕𝑦 2
and
𝜕2𝑢
𝜕2𝑣
𝜕 2𝑣
𝜆+𝜇
+ 𝜇 2 + 𝜆 + 2𝜇
=0
𝜕𝑥 2
𝜕𝑥
𝜕𝑦 2
Which are the Navier Equilibrium equations of elasticity. With the boundary terms,
𝑄 𝑥, 𝑦 𝛿𝑢 + 𝑅 𝑥, 𝑦 𝛿𝑣 𝑑𝑠
Γ
where 𝑄 𝑥, 𝑦 ≡
𝜕𝐹
𝑛
𝜕𝑢𝑥 𝑥
oafak@unilag.edu.ng, University of Lagos
+
𝜕𝐹
𝑛
𝜕𝑢𝑦 𝑦
and 𝑅 𝑥, 𝑦 ≡
70
𝜕𝐹
𝑛
𝜕𝑣𝑥 𝑥
+
𝜕𝐹
𝑛
𝜕𝑣𝑦 𝑦
Wednesday, April 15, 2015