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Calculus of Variation Problem

  1. Jun 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the Lagrange-Euler equation for the functional J[y] = ∫F(x,y,y')dx = ∫F(x(y),y,1/x')*x' dx = I[x(y)]

    2. Relevant equations
    The Lagrange-Euler equations ∂F/∂y = d(∂F/∂y')/dx

    3. The attempt at a solution
    I'm new to the subject, so I don't really know what to do, I've tried to use taylor theorem and finite differences method but in vain, maybe I've done it wrong, my last attempt was ∂F/∂y = ∂F/∂x * ∂x/∂y, ∂F/∂y' = ∂F/∂x*∂x/∂y', but i didn't come to express ∂x/∂y' in simpler terms, can anyone help ?
     
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  3. Jun 28, 2015 #2

    Zondrina

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    Just to have the notation clear, you want to find the Euler-Lagrange equation for the functional:

    $$J(y(x)) = \int_a^b F(x, y(x), y'(x)) \space dx$$

    The limits are important because they impose boundary conditions on ##y(x)##. Namely, ##y(a) = A## and ##y(b) = B## for some ##A, B \in \mathbb{R}##.

    I am unsure why you have written the functional is equal to ∫F(x(y),y,1/x')*x' dx in your original post. This doesn't make sense due to a few reasons, one being ##y## is a function of ##x##.

    You have somehow written the answer to your problem in the relevant equations, i.e the Euler-Lagrange equation is given by:

    $$\frac{\partial F}{\partial y} - \frac{d}{dx} \frac{\partial F}{\partial y'} = 0$$

    You have provided no explanation how you have obtained it though.
     
  4. Jun 28, 2015 #3

    Fredrik

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    Was the function F defined in the problem statement? If so, then all you have to do is to compute these derivatives. F is usually something very simple, like a 3-variable polynomial, and you know how to compute partial derivatives of those. Don't let the notation make you think that you need to do something more complicated.

    If F wasn't specified in the problem, then I guess the task is to derive the Euler-Lagrange equation for an arbitrary F. That's a fundamental and non-trivial theorem, so there has to be a proof of that in your book.

    The proper notation is ##J[y]=\int_{t_1}^{t_2} F(x,y(x),y'(x))\mathrm dx##. In problems like these, I think it would be very confusing to not be very careful about the notations for functions and numbers. In particular, x denotes a number, y denotes a function, y(x) denotes a number in the range of y.
     
  5. Jun 28, 2015 #4
    Guys, I'm supposed to find Euler-Lagrange equations of the inverse of y, I mean x(y) that minimize I[x(y)] = ∫F(x(y),y,1/x')*x'*dx, I found this in the book, but no correction were available, I've tried over and over but yet i didn't figure it out, I don't even why F is multiplied by x' = dx/dy, Sorry :/
     
  6. Jun 28, 2015 #5

    Zondrina

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    So you're saying the functional is actually this:

    $$J[x(y)] = \int_a^b F(y, x(y), \frac{1}{x'(y)}) \times x'(y) \space dy$$

    If so, then see this wiki article, it contains two proofs (one more "loose" than the other) showing how to derive the equation you seek:

    https://en.wikipedia.org/wiki/Euler–Lagrange_equation

    Take some time to understand the derivation, it will help you with this problem.

    Otherwise, if ##F## is ever given explicitly, then you have the main theorem you can use, which was mentioned prior.

    @Fredrik I like the square brackets on the left hand side after thinking about it a bit. It distinguishes the functional from a regular function I suppose.
     
    Last edited: Jun 29, 2015
  7. Jun 29, 2015 #6

    Fredrik

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    The argument for [] would be that it emphasizes that this is not a function from ℝ into ℝ. The argument for () would be that it's still a function, it's just a function with a domain that isn't ℝ. Since we don't use [] for such functions outside of this very narrow context (action functionals), I think I actually prefer the () notation. But I don't mind seeing [] either.

    So I didn't mean to object to the () in your notation. I just used [] because the OP did. I did however mean to object to the y(x) on the left-hand side. The domain of J is a set of functions, but y(x) is a number, an element of the codomain of the function y. So I strongly prefer the notations J(y) or J[y] over the notations J(y(x)) or J[y(x)].

    The notation F(x,y(x),y'(x)) is appropriate on the right-hand side, since the domain of F is a subset of ℝ3.
     
  8. Jun 29, 2015 #7

    Fredrik

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    OK, then I think I understand the problem. That rewrite is just the change of variables s=y(x) in the integral, combined with a typo (or sloppy notation) and the awful choice to denote the inverse of y by x. What they wrote as dx should be d(the new variable), so I will write it as ds. The definition s=y(x) gives us ##x=y^{-1}(s)##. To figure out how to rewrite y'(x), you need to use the rule for derivatives of inverse functions:
    $$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$
    Use ds=y'(x)dx to rewrite dx. Don't forget to change the integration interval.

    You will end up with something of the form
    $$\int_{s_1}^{s_2} G(s,y^{-1}(s),(y^{-1})'(s))\mathrm ds.$$ You will see what the function G is when you get there.

    If you're given an action functional I defined by
    $$I[g]=\int_{s_1}^{s_2} G(s,g(s),g'(s))\mathrm ds$$ for all ##g##, you know how to find the associated Euler-Lagrange equation. When you get to this point, you will have to figure out how to rewrite the partial derivatives of G as partial derivatives of F. I haven't tried to do this part, but I expect it to be fairly easy if you're comfortable with the chain rule and the product rule.
     
    Last edited: Jun 29, 2015
  9. Jun 29, 2015 #8
    Waow thanks, Here's what I've done so far, If g extremizes then functional ∂G/∂g = d(∂G/∂g')/ds
    g(s) = x bcz g = y-1 and s = y, d/ds = d/(ds/dx * dx) = dx/dy * d/dx
    g'(s) = dg/ds = dx/dy, ∂G/∂g' = ∂G/∂s * ∂s/∂g' = ∂G/∂s * ∂s/(∂(dx/dy)/∂y * ∂y), ,since x is a function of x only, ∂G/∂s*∂s/∂y*d2y/(dx)2 = ∂G/∂s*y'' now can i change G byF ?
     
  10. Jun 29, 2015 #9

    Zondrina

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    Yes, I agree with saying ##J[y]## instead of ##J[y(x)]##. People could potentially be confused by the functional argument ##y(x) \in \mathbb{R}[x]##. They could read it as ##y(x) \in \mathbb{R}## instead of reading it as a function.

    Both notations seemingly work, but one may cause a little less confusion.
     
  11. Jun 29, 2015 #10

    Fredrik

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    I didn't check everything you wrote, but this can't be right: ∂G/∂g' = ∂G/∂s * ∂s/∂g'.

    Did you do the variable change? Did you find the relationship between G and F?
     
  12. Jun 29, 2015 #11
    I'm really confused, sorry !
     
  13. Jun 29, 2015 #12

    Fredrik

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    About how to change variables in an integral? What we do when we change variables is just to use the fundamental theorem of calculus and the chain rule in the following way: (F is a primitive function of f):
    \begin{align*}
    &\int_{g(a)}^{g(b)} f(x)\mathrm dx = F(g(b))-F(g(a)) =(F\circ g)(b)-(F\circ g)(a) =\int_a^b (F\circ g)'(x) \mathrm dx\\
    &=\int_a^b F'(g(x))g'(x)\mathrm dx = \int_a^b f(g(x))g'(x)\mathrm dx.
    \end{align*}
    Typically, you'd start with the right-hand side above, and write things like the following in order to end up with the left-hand side.
    \begin{align*}
    &y=g(x),\quad dy =\frac{dy}{dx}dx =g'(x)dx\\
    &x=a\Rightarrow y=g(a)\\
    &x=b\Rightarrow y=g(b).
    \end{align*}
    But this is more of a mnemonic than anything else. The calculation I did above is the rigorous way to do it.

    Example: y=sin x, dy=cos x dx.
    $$\int_a^b\sin^2x\cos x\mathrm dx =\int_{\sin a}^{\sin b} y^2\mathrm dy =\left[\frac{y^3}{3}\right]_a^b =\frac 1 3(\sin^3b-\sin^3a).$$
     
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