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Variational expression: a demonstration

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello!
    My problem is with a variational expression. I have a quantity, say L, which could be determined by the ratio:

    [itex]K = \displaystyle \frac{\left[\int_a^b E(x)e_n(x)dx\right]^2}{\left[\int_a^b E(x)e_1(x)dx\right]^2}[/itex]

    Where [itex]e_1(x), e_n(x)[/itex] are known functions and [itex]n[/itex] is an integer, with [itex]n > 1[/itex].
    If I couldn't know exactly [itex]E(x)[/itex] and I substitute it with [itex]E_0(x) + \delta E(x)[/itex], I have to demonstrate that the corresponding [itex]\delta K[/itex] is proportional to [itex][\delta E(x)]^2[/itex].
    So an error of 10 % made during the estimation of [itex]E(x)[/itex] is an error of only 1 % for the corresponding estimation of [itex]K[/itex].

    2. Relevant equations



    3. The attempt at a solution

    I tried to write the square of numerator and denominator:

    [itex]K + \delta K = \displaystyle \frac{\left[\int_a^b [E(x) + \delta E(x)] e_n(x)dx\right]^2}{\left[\int_a^b [E(x) + \delta E(x)] e_1(x)dx\right]^2} = [/itex]
    [itex] \displaystyle = \frac{\left[\int_a^b E(x)e_n(x)dx\right]^2 + 2\int_a^b E(x)e_n(x)dx \int_a^b \delta E(x)e_n(x)dx + \left[\int_a^b \delta E(x)e_n(x)dx\right]^2}{\left[\int_a^b E(x)e_1(x)dx\right]^2 + 2\int_a^b E(x)e_1(x)dx \int_a^b \delta E(x)e_1(x)dx + \left[\int_a^b \delta E(x)e_1(x)dx\right]^2}[/itex]

    We can easily substitute the integral expression with letters and write:

    [itex]K + \delta K = \displaystyle \frac{a^2 + 2ab + b^2}{c^2 + 2cd + d^2}[/itex]

    How could I do now?
    Could I neglect the [itex]b^2[/itex] and [itex]d^2[/itex] terms because they are small? If I do this, then I could divide both numerator and denominator by [itex]c^2[/itex] and take the Taylor series expansion of the denominator truncated to the first order, so

    [itex]K + \delta K \simeq \frac{\displaystyle \frac{a^2}{c^2} + \displaystyle \frac{2ab}{c^2}}{1 + \displaystyle \frac{2d}{c}} \simeq \left(\displaystyle \frac{a^2}{c^2} + \frac{2ab}{c^2}\right)\left(1 + \displaystyle \frac{2d}{c}\right)[/itex]

    But in this way I can't separate yet [itex]a[/itex] and [itex]c[/itex] from [itex]b[/itex] and [itex]d[/itex], and I can't write a term with only [itex]c^2[/itex] or [itex]d^2[/itex]. How can I proceed?
    Thank you if you read this post,

    Emily
     
  2. jcsd
  3. May 16, 2012 #2

    Ray Vickson

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    The result you want is false, in general. Let us write [itex] \delta E(x) [/itex] as [itex] r h(x),[/itex] where [itex] h(x) [/itex] is a function (not necessarily small) and [itex] r[/itex] is the perturbation parameter (which we do think of a small). We have
    [tex] K = K(r) = \left(\frac{a + br}{c + dr}\right)^2, [/tex] where
    [tex] \begin{array}{l}a = \int_a^b e_0(x) E_0(x) \, dx \\
    b = \int_a^b e_0(x) h(x) \, dx \\
    c = \int_a^b e_n(x) E_0(x) \, dx \\
    d = \int_a^b e_n(x) h(x) \, dx \end{array}
    [/tex]
    Your result requires that the Taylor expansion of K(r) have the form [itex]K(0) + K_2 r^2, [/itex] with no [itex] r^1[/itex] term. This is false: we have
    [tex] K'(0) = dK(r)/dr|_{r=0} = \frac{2a}{c^3}(bc - ad), [/tex]
    which is nonzero in general.

    RGV
     
  4. May 17, 2012 #3
    First of all, thank you for your reply!
    Maybe you wanted to write:

    [tex] \begin{array}{l}a = \int_a^b E_0(x) e_n(x) \, dx \\
    b = \int_a^b h(x) e_n(x) \, dx \\
    c = \int_a^b E_0(x) e_1(x) \, dx \\
    d = \int_a^b h(x) e_1(x) \, dx \end{array}
    [/tex]

    Substituting:

    [tex]bc - ad = 0 \Rightarrow \int_a^b h(x) e_n(x) \, dx \int_a^b E_0(x) e_1(x) \, dx - \int_a^b E_0(x) e_n(x) \, dx \int_a^b h(x) e_1(x) \, dx = 0[/tex]

    It seems to be true just if the product of the integrals equals the integrals of the products. Isn't it?
    According to my notes, the result about [tex][\delta E(x)]^2[/tex] should be true in general: my professor and another one said that in their lessons, without explaining why :frown:.

    Emily
     
  5. May 17, 2012 #4

    Ray Vickson

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    Sorry, no. The product b*c involves the integral of e_n * h, while the product a*d involves the integral of e_1 * h. There is no reason why they should cancel. Look at it this way:
    [tex]bc-ad = \int_a^b h(x) e_n(x) \, dx \int_a^b E_0(w) e_1(w) \, dw - \int_a^b E_0(w) e_n(w) \, dw \int_a^b h(x) e_1(x) \, dx .[/tex]
    Does that look like 0 to you?

    RGV
     
    Last edited: May 17, 2012
  6. May 17, 2012 #5

    Ray Vickson

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    To convince you the result is false, let's look at a numerical example. (Of course, the functions you are using in your study may not resemble the ones I use below; it may be that for the special functions you are using, the result could, conceivably, hold---just not for the reasons you claim.) Let's take [itex] E_0(x) = x,\: h(x) = x^2,\: e_n(x) = \sin(\pi x/2), \: e_1(x) = \cos(\pi x/2), \: a = 0, \:b = 1.[/itex] Then
    [tex] \begin{array}{rcl}K &=& \frac{\left( \int_0^1 (x + r x^2) \sin(\pi x/2) \, dx \right)^2}{\left( \int_0^1 (x + r x^2) \cos(\pi x/2) \, dx \right)^2}\\
    &\doteq& \frac{4(2.283185308\: r + 3.141592654)^2}{(3.586419096+1.869604404\: r)^2} \\
    &=&3.069288133 + 1.261227604\: r - 0.527913928\: r^2 + O(r^3).
    \end{array}
    [/tex]
    The term linear in r is not zero.

    RGV
     
  7. May 21, 2012 #6
    Sorry if I'm late (I didn't receive notifications by e-mail).
    I tried to do some numerical examples too, with functions very similar to your ones. In the particular problem, it should be

    [itex]e_n(x) = \sin(n \pi x/w)\\
    e_1(x) = \sin(\pi x/w)\\
    E(x) = \displaystyle \sqrt{(x - a)(b - x)}[/itex]

    with [itex]h(x)[/itex] whatever error function and [itex]w[/itex] a numerical positive constant; in my example I chose a sort of Gaussian distribution, [itex]h(x) = e^{-x^2}[/itex]. [itex]E(x)[/itex] should be the estimated function, so it is [itex]E(x) = E_0(x) + rh(x)[/itex] and this is all we know about the function. Don't worry if somewhere [itex]E(x)[/itex] becomes imaginary, it could happen.
    But now I wonder why the professor said that :surprised. I agree with you, the first order error is nonzero!
    Anyway, thank you so much for your calculations and observations, which are all right and useful!!!

    Emily
     
  8. May 21, 2012 #7
    With a more accurate example, I tried these realistic values for each integral:

    [itex]a = 1.9;\\
    b=2.1;\\
    w = 4[/itex]

    and I had to consider [itex]E_0(x) = \sqrt{(x - a)(b - x)}[/itex] instead of [itex]E(x)[/itex] to make this example (I need an expression for [itex]E_0(x)[/itex]!).

    [itex]n = 3, 5[/itex] gave a difference [itex]bc - ad[/itex] which is nonzero only if we consider a [itex]10^{-7}[/itex] precision (now a and b are the integrals, not their extremes!).
    [itex]n = 9[/itex] gave a difference [itex]bc - ad[/itex] which is nonzero only if we consider a [itex]10^{-6}[/itex] precision and so on for increasing values of n.
    But the contribution of the nth term for increasing [itex]n[/itex] is decreasing, so maybe we can consider in this case [itex]bc - ad \simeq 0[/itex].
    I can't say which are the mathematical reasons for this result, but - strictly numerically - it seems to work.

    Emily
     
  9. May 22, 2012 #8
    Even if in this textbook the calculation are quite different, the results that I should demonstrate are the same: the statement is:
    K «is stationary for small arbitrary variations in the electric field distribution about its correct value. It, therefore, follows that a first-order approximation to the electric field distribution will yield a second-order approximation to» the value of K.
    (Robert E. Collin, Field Theory of Guided Waves, Ch. 8)
     
  10. May 22, 2012 #9

    Ray Vickson

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    I don't know why you are arguing; we are talking about two different things. The thing you wrote down in your original post was not, in general, stationary, so that is why the first-order differential was nonzero. If you did have a stationary thing then, of course, the first-order differential would be zero BY DEFINITION, and that would mean your deviations would be of second or higher order---no argument there.

    RGV
     
  11. May 22, 2012 #10
    Oh, so it had been a misunderstanding. My professor presented the first expression,

    [tex]K = \displaystyle \frac{\left[\int_a^b E(x)e_n(x)dx\right]^2}{\left[\int_a^b E(x)e_1(x)dx\right]^2}[/tex]

    as a variational, stationary expression. This is why I wrote the post.

    So, the [tex]K[/tex] expression is not stationary?
    How can I recognize a stationary expression and where I can see that it has zero first-order differential?

    Emily
     
  12. May 22, 2012 #11

    Ray Vickson

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    I don't know the background or the context, so I don't know what K is supposed to be, or where it came from. I just took your word for it that K was supposed to be stationary at E = E_0, and I disputed that. However, maybe what you wrote is not what was meant, etc. At this point I give up.

    RGV
     
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