Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus of variations

  1. Apr 5, 2013 #1
    My first question is with regards to the "status" of calculus of variations. Because I read in wolfram alpha that it was a generalization of calculus? Is that right?

    Anyway; my main question has to do with the process of getting the answer you're looking for. Is every problem in calculus of variations possible to be set as a solvable differential equation? And by solvable I don't mean keep guessing the answer until you eventually get it and then prove it was the right one using some method. Which to me looks like a general trend in calculus of variations.

    If you can set at least most problems of calculus of variations as differential equations, how do you go about doing it? Should I look deeper into functional analysis, banach spaces, metric spaces and so on?

    Could anyone show me an "algorithmic" way to find, for instance, the solution to the problem of finding the shortest path between two point in the xy plane? Without assuming you already know the answer and is just trying to prove it's the right one?
  2. jcsd
  3. Apr 5, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

  4. Apr 5, 2013 #3
    If you have around 15 bucks or so I recommend: "Mathematics of Classical and Quantum Physics" from Byron and Fuller, Dover Publishing. Forget about deeper functional analysis for a first understanding (<- my opinion).
    What do you mean with "guessing"?

    I cite the book: "One of the oldest problems in mathematical physics is that of trying to minimize expressions which do not depend on some simple continuous variable, but rather on a function".
    That's it. Why should this be something different than "analysis"? Where does "analysis" end? Is "vector analysis" a new one? Or "theory of functions"?

    You simply set up an expression (or a couple of them) like

    ds = sqrt(1 + y'(x)^2) dx

    which you would like to minimize. Here you simply describe the length of the little path pieces. And now you think: Fine, I would like to minimize the path. So you are writing down on a piece of paper the integral. Now you realize: Nice, I have an integral I can't integrate because I don't know the function y(x). Now you think hmmm if I want to have the minimum then the function y(x,epsilon) can only be the minimum if epsilon is zero, meaning you are adding simply something which has to be gone in the end.

    Example: Integral from 0 to 1. y(x) unknown. So y(x) + epsilon * sin(x / PI) is an idea. But the test function is not important.

    So the brilliant idea is to think about a test function which disappears in the minimum and that function has nothing to do but fulfilling some conditions.

    Doing that you will get the Euler-Lagrange equation(s). This means that you have a (set of) differential equation(s) to be solved.

    You have a deep lack in understanding if you think that the test functions are the problem. I would say: Have fun with solving the system of differential equations you will get! ;)
  5. Apr 5, 2013 #4


    User Avatar
    Science Advisor
    Gold Member

    For the optimization problem you essentially follow the same steps as in single and multivariable calculus.

    In single variable calculus you express the quantity to be optimized as a function of the variable and solve for where the derivative is zero. After taking the derivative this is a single variable algebra problem.

    In multi-variable calculus you express the quantity to be optimized as a function of the many variables (which you can call a single vector variable). The derivative generalizes to the "gradient" here i.e. the vector of partial derivatives. You then solve the vector equation you get setting this general derivative to (the) zero (vector). The vector equation is then a system of equations.

    In functional analysis a function can be thought of as a vector (infinite dimensional) and you do the same as above except now the "function" of your vector is a functional, i.e. a mapping from the function to a number. This functional has a derivative in the generalized setting and setting it to (the) zero (functional) gives you a continuum system of equations typically expressed as a differential equation on the function at which optimization occurs.

    As an example of shortest path. You would express an arbitrary path by giving position as a function of some parameter, i.e. x(t), y(t), and z(t) for 0<t<1. You then need to express how to get the thing to be optimized form the functions, e.g. for distance you have to do a path integral to get arclength...
    [tex] s[x,y,z] = \int_{0}^1 \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2}dt[/tex]
    Note here that the arclength s is not a function of the values of x,y, and z but rather a functional of x,y,z as functions of t. (We'd be better off using different names for the variables x,y,z and their functional dependence on t... something like:
    [tex]x=f(t), y=g(t), z=h(t)[/tex]
    and similarly arclength s as some functional, say Lambda of these functions (I use Lambda to stand for the "L"ength functional.)
    [tex] s=\Lambda[f,g,h] = \int_{0}^1 \sqrt{\dot{f}(t)^2 + \dot{g}(t)^2 + \dot{h}(t)^2}dt[/tex]
    Now we have this mapping from functions to a number s. We take its generalized derivative which is a linear functional depending on the function (including its derivatives). So setting it to the zero function gives you typically a differential equation.

    In summary you in all cases express the quantity to be optimized as a mapping from the variable entity to a number, you in some sense take the derivative of this mapping and set that to the zero object for its type and solve.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Calculus of variations
  1. Variational Principles (Replies: 8)

  2. Variational Equation (Replies: 0)