SSG 311 Calculus of variations _Tensors.pdf

SSG 311 Calculus of Multivariable Functions
FACULTY OF ENGINEERING
Synopsis
. Calculus of variations
. Tensor analysis

Calculus of Variations
In the static optimization problems - a point or points
that would maximize or minimize a given function at a
particular point or period of time can be sought for from
a given function, i.e.
 
x
y
y 
For optimal, we set
  0
*
' 
 x
y
dx
dy
Such that the optimal point is  
y
x*,

In dynamic optimization we seek a curve x*(t) which will
maximize or minimize a given integral expression, such that;
An integral of such which assumes a numerical value for each
of the class of functions x(t) is called a functional. A curve
that maximizes or minimizes the value of a functional is
called an extremal. Acceptable candidates for an extremal
are the class of functions x(t) which are continuously
differentiable on the defined interval and which typically
satisfy some fixed endpoint conditions.

Distance Between Two Points on a Plane
The length S of any nonlinear curve connecting two points
on a plane, such as the curve connecting the points (t0, x0)
and (t1, x1) (Fig. (a)) can be approximated mathematically by
subdividing the curve mentally into subintervals, as in Fig.
(b), and applying the Pythagorean theorem

Dividing through by dt,
And more compactly written as,
then integrating,

EULER’S EQUATION: The necessary condition for dynamic optimization
For a curve x*=x*(t) connecting points (t0, x0) and (t1, x1) to
be an extremal for (i.e., to optimize) a functional
the necessary condition, called Euler’s equation, is
Being equivalent of the first-order necessary conditions in static
optimization, Euler’s equation is actually a second-order differential
equation which can be written as

To prove that Euler’s equation is a necessary condition for an extremal,
let X*=x*(t) be the curve connecting points (t0, x0) and (t1, x1) in Fig.2
which optimizes the functional (i.e., posits the optimizing function for)
Fig. 2

Let X^= x*(t) + mh(t) be a neighbouring curve joining these points,
where m is an arbitrary constant and h(t) is an arbitrary function. In
order for the curve X^ to also pass through the points (t0, x0) and (t1,
x1), that is, for X^ to also satisfy the endpoint conditions, it is
necessary that
By holding both x*(t) and h(t) fixed, the value of the integral becomes a
function of m alone and can be written
the function g(m) can be optimized only when m=0 and

Using Leibnitz rule, which states that
Then,
Since the boundaries of integration t0 and t1 are fixed in the present
example, t0/m t1/m 0, and we have to consider only the first term in
Leibnitz’s rule. Applying the chain rule we find F/m, because F is a
function of x and x · , which in turn are functions of m, and substituting,
we have

Leaving the first term in the brackets untouched and integrating the
second term by means of parts,
Since h(t) is an arbitrary function that need not equal zero, it follows
that a necessary condition for an extremal is that the integrand within
the brackets equal zero, namely,
which is the Euler’s equation

EXAMPLE 1. Given the functional
Let
Then
Substitute in Euler’s equation

EXAMPLE 2. The functional
subject to
is to be optimized.

Solving for x,
Applying bc’s
Exercises
1. Minimize subject to
2. Optimize subject to

Example 3. For example 1, the sufficiency conditions are as illustrated

EXAMPLE 5. The sufficiency conditions are illustrated below for Example
1 where the functional was

Dynamic Optimization Subject To Functional Constraints
To form the Lagrangian function:
The necessary, but not sufficient, condition to have an extremal for
dynamic optimization is the Euler equation

EXAMPLE 6. Constrained optimization of functionals is commonly used in problems
to determine a curve with a given perimeter that encloses the largest area. Such
problems are called isoperimetric problems and are usually expressed in the
functional notation of y(x) rather than x(t). Adjusting for this notation, to find the
curve Y of given length k which encloses a maximum area A, where
Length of the curve is
Lagrangian function
Letting H equal the integrand , the Euler equation is

Substituting in Euler’s equation

1. Find the curve connecting (t0, x0) and (t1, x1) which will generate the surface of
minimal area when revolved around the t axis, as in Fig. Q1. That is,
Calculus of Variations - Exercises
2. Minimize
3. (i)
(ii) Explain the significance.
3. (i)
(ii) Explain the significance.

Tensor Analysis
Einstein summation convention

Tensor Analysis
Example 3.
SUBSTITUTIONS
Example 4.

Tensor Analysis
Example 5.
Example 6.

Tensor Analysis
Matrices, Vectors and Determinants
Examples:
Matrix sum
Scalar multiplication

Tensor Analysis
Matrix multiplication
Also, can be written as
where i and j are not summed on.
Identity matrix
Transpose of a matrix

Tensor Analysis
Determinant of matrix
m x n system of equations
A quadratic form Q in the n variables x1, x2, ..., xn
Example: Write the quadratic equation
using a symmetric matrix.

Tensor Analysis
Solution: quadratic form is given in terms of nonsymmetric matrix
Symmetric equivalent is replacing the off-diagonal elements by half of
sum of its value and its mirror image

Vectors
Tensor Analysis
Column vectors: Components:
Or simply:
Scalar product of vectors
Norm of vectors
Angle between two vectors

Tensor Analysis
Vector product in R3

Tensor Analysis
Dot product
Cross product
Also,

SSG 311 Calculus of variations _Tensors.pdf

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