Winzer said:
Well I guess this question will have to wait till I cover Mathematical Physics my junoir year.
Here is the classic Calculus of Variations problem.
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In
Calculus you study functions (real), basically a function is something that takes a number and turns it into another number. For instance, [tex]f(x)=x^2[/tex] means for each number you give in [tex]x[/tex] the output number is [tex]x^2[/tex].
In
Integral Trasforms you study
transforms, for example the http://en.wikipedia.org/wiki/Laplace_Transform" . Basically what the Laplace/Fourier transforms are is something that transforms a
function into another
function. So for example, the Laplace transform on [tex]f(x)=x[/tex] is [tex]F(s)=s^{-2}[/tex]. Which means the function gets turned into another funtion.
In
Calculus of Variations you study a similar situation called a
functional. And basically a functional is something that turns a
function into a
number. I will give you an example below and hopefully you will understand.
Here is a classical application of Calulcus of Varations I promised in the beginning of this thread.
Given two distinct points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]. What is the shortest path between them? You certainly know the answer is a straight line. But look at how this problem is approach. First we set us an equation involving a
functional. Here is what we do. Let [tex]y=f(x)[/tex] be any curve between those two points. And the length of that curve is [tex]\int_{x_1}^{x_2} \sqrt{1+[f'(x)]^2} \ dx[/tex], this is simply the arc length formula from Calculus II. So we write,
[tex]I[f(x)] = \int_{x_1}^{x_2}\sqrt{1+[f'(x)]^2} \ dx[/tex].
This means if [tex]f(x)[/tex] is any curve passing through those two points then [tex]I[f(x)][/tex] is the value of this functional. Note, the input is a
function but the output is a
number.
So the way the problem is stated is:
Given [tex]I[f(x)]=\int_{x_1}^{x_2}\sqrt{1+[f'(x)]^2} \ dx \mbox{ with }f(x_1)=y_1 \mbox{ and }f(x_2)=y_2[/tex]
minimize the functional.
The conditions [tex]f(x_1)=y_1 \ f(x_2)=y_2[/tex] are referred to as the
boundary condition, because that assures us the curve passes through those two points.
But the above problem is a typical application of Calculus of Variations which is easily solved with
Euler-Lagrange Equation if you read the Wikipedia article.
. But I can't wait to learn more about those, they sound really interesting.