1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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. Oct 14, 2005 #1
    I am facing a difficult integral here for calculus of variations. The question reads:
    Find the extremum to the integral:
    [tex]
    I[y(x)] = \int_{Q}^{P} (dy/dx)^2(1+dy/dx)^2 dx [/tex]
    [tex] where [/tex][tex] P = (0,0) [/tex] [tex] and [/tex] [tex]Q = (1,2)[/tex]
     
    Last edited: Oct 14, 2005
  2. jcsd
  3. Oct 14, 2005 #2

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    How about using Euler's equation:

    [tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0[/tex]

    where:

    [tex]F(x,y,y^{'})=\left(y^{'}\right)^2(1+y^{'})^2[/tex]

    However, that's usually on an interval and not from point to point as you have it defined above.

    This of course results in a highly non-linear ODE. Push come to shove, I'd solve it numerically.
     
    Last edited: Oct 14, 2005
  4. Oct 14, 2005 #3
    Since this is about extremum, we clearly have to start with the Euler-Lagrange equation. But, you are right. We will obtain a non-linear ODE. It is this ODE I am having headache with.
     
  5. Oct 14, 2005 #4

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Can you post the resulting ODE? I'd like to check it against mine which is really not too bad: second-degree, first order ODE.
     
  6. Oct 14, 2005 #5
    Because the integrand is independent of y, hence, when I use Euler equation, it gives me,

    [tex] 2y'^3 + 3y'^2 + y' = constant [/tex]
     
  7. Oct 14, 2005 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?
     
  8. Oct 14, 2005 #7

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get:

    [tex]6(y^{'})^2+6y^{'}+1=0[/tex]

    Can someone rectify this discrepancy please?
     
  9. Oct 14, 2005 #8
    Errr.... I guess there is a way to solve this non-linear differential equation which I am not aware of its existence. What should you do after you factorise a differential equation?
    Let say for example,

    [tex] (y'+1)(y'+2) = 0 [/tex]

    Then, [tex] y'+1 = 0 [/tex] or [tex] y'+2 = 0 [/tex]

    That means to say that y is a function whereby its first derivative can take only the value -1 or -2 ?

    What's the next step?
     
  10. Oct 14, 2005 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, there aren't many functions whose derivative is always either -1 or -2, are there? (Even fewer if you assume the derivative is continuous...)
     
  11. Oct 14, 2005 #10
    How can the derivative be continuous when the derivative takes on discrete values of either -1 or -2 ?
    IMO, the only function whose derivative is either -1 or -2, are somewhat like stepwise functions with straight lines. I suppose you are saying that this applies to the original question. Are there any other functions?
     
  12. Oct 15, 2005 #11

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    What's wrong with:

    [tex]y(x)=-x+b[/tex]

    or:

    [tex]y(x)=-2x+b[/tex]

    Thus, being a variational problem, I'd substitute them both into the integral, evaluate it, and see which one yields the extremum I seek.
    Or should I say, "is that what I SHOULD do?"
     
    Last edited: Oct 15, 2005
  13. Oct 15, 2005 #12
    I think there is something wrong if you factorise the equation and let y' be equal to a constant. If y' does equal a constant, then, y is a straight line having a slope equals to that constant.
    But in the original question, there are boundary conditions, ie,
    P=(0,0) and Q=(1,2)

    If it is a straight line, obviously we can get the function y straight away from just the end points P and Q.
    What is your opinion?
     
  14. Oct 15, 2005 #13

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Straight lines, even splits, et cetera are often solutions to extremal problems, so I'm not surprised. :smile:

    I'm pretty sure my approach is valid -- if you've proven that the derivative must satisfy that cubic, then at all points it must be equal to one of its roots, and if the derivative is continuous, it must be a constant.

    I don't know if you should assume the derivative is continuous or not.

    If it's not, then the simpler solutions for y are piecewise-linear... I'm not sure if there are pathological continuous solutions.
     
  15. Oct 15, 2005 #14

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    For the record I wish to correct a conclusion I made above:

    Solving the Euler equation, I obtained:

    [tex]2(1+6y^{'}+6(y^{'})^2)y^{''}=0[/tex]

    Now, either the quantity in paranthesis is zero or the second derivative. In the former case:

    [tex]1+6y^{'}+6(y^{'})^2=0[/tex]

    I obtain:

    [tex]y^{'}=\frac{-1\pm\sqrt{1-2/3}}{2}[/tex]

    Note that the slope is negative in both instances so this solution cannot meet the boundary conditions above. In the second case:

    [tex]y^{''}=0[/tex]

    The solution is:

    [tex]y(x)=ax+b[/tex]

    Using the boundary conditions: y(0)=0 and y(1)=2, I obtain:

    [tex]y(x)=2x[/tex]

    Toqura, I assume you're finished with this problem by now right?

    Edit: Oh yea, can someone tell me how to prove y(x) is a solution to this functional equation?
     
    Last edited: Oct 15, 2005
  16. Oct 16, 2005 #15
    Thanks.

    I think I will write assumptions whether the first derivative is continuous or not, first case and second case.
    And then, go on to solve it by factorising like what saltydog did.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



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

  2. Calculus of variations (Replies: 0)

  3. Calculus by variations (Replies: 2)

  4. Variational Calculus (Replies: 0)

  5. Calculus of variations (Replies: 3)

Loading...