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Determine minimum value of integral

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine minimum value of integral ##B(y)=\int_{0}^{2}({y}')^2dx## for function ##y\in C^1(\mathbb{R})## and ##y(0)=y(2)=0## and ##\int_{0}^{2}y^2dx=4##


    2. Relevant equations



    3. The attempt at a solution

    IF I am not mistaken, the idea is to first find the function ##y## for given conditions and than calculate ##B(y)##.

    My idea:

    Since ##y\in C^1(\mathbb{R})## I simply started with polynomial ##y(x)=ax^2+bx+c##. Of course ##y(0)=0## so immediately ##c=0## than ##y(2)=0=4a+2b=0## so ##b=-2a##.

    Third condition says ##\int_{0}^{2}y^2dx=\int_{0}^{2}(ax^2+bx)^2dx=4##

    ##\frac{32a^2}{5}+\frac{ab16}{2}+8b^2=4## using ##b=-2a## gives me ##a=\sqrt{\frac{5}{28}}## and ##b=-\sqrt{\frac{5}{7}}##.

    So using ##y(x)=\sqrt{\frac{5}{28}}x^2-\sqrt{\frac{5}{7}}x## the ##B(y)=\int_{0}^{2}({y}')^2dx=\frac{5}{3}##.

    Or is it not?
     
  2. jcsd
  3. Mar 6, 2014 #2

    LCKurtz

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    I'm not sure how to work that problem, but I'm pretty sure that isn't it. There are lots of functions that satisfy ##y(0)=y(2)=0## such that ##\int_0^2 y^2~dx = 4##. You are to find the one that minimizes ##\int_0^2 y'^2~dx## over all such functions. You must be pretty lucky if the one function you found is it. This looks like a calculus of variations problem. I'm not very familiar with such problems, but if that's what you are studying, surely your text must give a method for solving them.
     
  4. Mar 6, 2014 #3

    Ray Vickson

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    You made some errors: after solving ##y(2) = 0## to get ##b = -2a##, your other condition should be
    [tex]\int_0^2 (ax^2 - 2ax)^2 \, dx = 4[/tex]

    More seriously: why would you assume a quadratic polynomial? How do you know that a polynomial of degree 1000 won't produce better results? How do you know that you could not do better using a non-polynomial?
    You need to use a Lagrange multiplier method:
    [tex] \min \left( \int_0^2 y'^2 \, dx + \lambda \int_0^2 y^2 \, dx \right)[/tex]
    where ##\lambda## is a constant. This gives a Lagrangian of ##L = y'^2 + \lambda y^2##. Now apply the Euler-Lagrange equations.
     
  5. Mar 6, 2014 #4
    Hmmm, I completely agree with you. I guess I should go the opposite way and firstly find a general function that minimizes the integral and than find one that satisfies given conditions.

    And yes I am dealing with variations problems.
     
  6. Mar 6, 2014 #5

    HallsofIvy

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    It's not clear to me what course this is for or what knowledge you have. If this we in a "Calculus of Variations" class then I would think that a good "Relevant Equation" would be the Euler-Lagrange equation: if f minimizes
    [tex]\int_{x_1}^{x_2} L(x, y, y')dx[/tex]
    then
    [tex]\frac{\partial L}{\partial y}- \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)= 0[/tex]

    But you seem to be saying this is just for a Calculus class. My question then is do you have any reason to assume the solution can be written as a polynomial? And even so, why quadratic? There appear to be four conditions to be met:
    (1)y(0)= 0.
    (2)y(2)= 0.
    (3)[tex]\int_0^2 y^2 dx= 4[/tex] and
    (4)[tex]\int_0^2 (y')^2dy[/tex] is a minimum.

    If I had reason to believe that y was a polynomial function of x, seeing that there were four conditions to be met, I would expect a polynomial with four constants to be determined, a cubic polynomial.
     
  7. Mar 6, 2014 #6
    I integrated here [tex]\int_0^2 y(x)^2 \, dx = 4[/tex] where I used ##y(x)=ax^2+bx+c##, however one of the conditions already gave me ##c=0## so ##\int_0^2 (ax^2+bx)^2 dx = 4##. After integrating I used the fact that ##b=-2a##.

    Since ##b## and ##a## are not functions of ##x##, I don't think this actually plays a huge role here. Unless I am mistaken.

    Haha, I know that simply by having only two conditions. :D Thanks for the hint. Will do that and publish it here.

    EDIT: This is in a way also answer to Hallsoflvy. Why two and not four as you stated?... Yup, that's what I am asking myself at this very moment when you are making it obvious. :D
     
    Last edited: Mar 6, 2014
  8. Mar 6, 2014 #7
    Version 2.0:

    ##F(y,{y}')=\int_{0}^{2}({y}')^2dx+\lambda \int_{0}^{2}y^2dx=\int_{0}^{2}(({y}')^2+\lambda y^2)dx##

    Let's say that ##L=({y}')^2+\lambda y^2##, than Euler-Lagrange equation tells me that

    ##2\lambda y-2{y}''=0## which is a differential equation with solutions ##y(x)=Ae^{\sqrt{\lambda }x}+Be^{-\sqrt{\lambda }x}##.

    Condition ##y(0)=0## gives me ##A+B=0## and from ##y(2)=0## I get

    ##Ae^{\sqrt{\lambda }2}+Be^{-\sqrt{\lambda }2}=0##

    ##Ae^{4\sqrt{\lambda }}=-B##

    ##-Be^{4\sqrt{\lambda }}=-B##

    ##4\sqrt{\lambda }=ln(1)=0## therefore ##\lambda =0##.


    This has to be wrong but I can't find a mistake...
     
  9. Mar 6, 2014 #8

    pasmith

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    You've made the error of assuming that [itex]\lambda > 0[/itex]. If [itex]\lambda = -k^2 < 0[/itex] you get a solution in terms of sines and cosines, and you can choose [itex]k > 0[/itex] such that [itex]\sin(2k) = 0[/itex].

    (There are a countable infinity of such [itex]k[/itex], so you'll have to decide which of them minimizes [itex]B(y)[/itex].)

    Finding [itex]\lambda[/itex] such that [itex]y'' = \lambda y[/itex] has a non-zero solution with [itex]y(0) = y(L) = 0[/itex] is a standard eigenvalue problem which you will probably see again in other contexts.
     
    Last edited: Mar 6, 2014
  10. Mar 6, 2014 #9
    Ok, so for ##\lambda >0## nothing interesting happens.

    Now let's assume that ##\lambda <0##, than ##y(x)=Ae^{i\sqrt{\lambda}x}+Be^{-i\sqrt{\lambda}x}=Asin(\sqrt{\lambda}x)+Bcos(\sqrt{\lambda}x)##.

    Due to ##y(0)=0## we find out that ##B=0##.

    Also ##y(2)=0=Asin(2\sqrt{\lambda})##. Since we are looking for non trivial solutions, let's assume that ##A\neq 0##, than ##2\sqrt{\lambda}=n\pi ##.

    ##\sqrt{\lambda}=\frac{n\pi }{2}## so ##y(x)=Asin(\frac{n\pi }{2}x)##.

    Using condition ##\int_{0}^{2}y^2dx=4## we get value for ##A## which is ##A=2\sqrt{2}##.

    Now finally ##y(x)=\left\{\begin{matrix}
    0 ;& \lambda \geq 0\\
    2\sqrt{2}sin(\frac{n\pi }{2}x) ; &\lambda <0
    \end{matrix}\right.##

    Now ##B(y)=\int_{0}^{2}({y}')^2dx## to determine ##n## and finally the right ##y(x)##.

    ##B(y)=2n^2\pi ^2## and obviously ##n=1## if we want ##B(y)## to be minimzed but not zero.

    Therefore ##y(x)=2\sqrt{2}sin(\frac{\pi }{2}x)##.


    Right or wrong... You guys are being very helpful! Thanks to all of you!
     
  11. Mar 8, 2014 #10
    Since nobody had any complaints, I assume everything is more or less ok and the next step is to say thanks to you.

    So, thanks!
     
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