Homework Help: Variational expression: a demonstration

1. May 16, 2012

EmilyRuck

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:

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

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

2. Relevant equations

3. The attempt at a solution

I tried to write the square of numerator and denominator:

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

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

$K + \delta K = \displaystyle \frac{a^2 + 2ab + b^2}{c^2 + 2cd + d^2}$

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

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

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

Emily

2. May 16, 2012

Ray Vickson

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

RGV

3. May 17, 2012

EmilyRuck

Maybe you wanted to write:

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

Substituting:

$$bc - ad = 0 \Rightarrow \int_a^b h(x) e_n(x) \, dx \int_a^b E_0(x) e_1(x) \, dx - \int_a^b E_0(x) e_n(x) \, dx \int_a^b h(x) e_1(x) \, dx = 0$$

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 $$[\delta E(x)]^2$$ should be true in general: my professor and another one said that in their lessons, without explaining why .

Emily

4. May 17, 2012

Ray Vickson

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:
$$bc-ad = \int_a^b h(x) e_n(x) \, dx \int_a^b E_0(w) e_1(w) \, dw - \int_a^b E_0(w) e_n(w) \, dw \int_a^b h(x) e_1(x) \, dx .$$
Does that look like 0 to you?

RGV

Last edited: May 17, 2012
5. May 17, 2012

Ray Vickson

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 $E_0(x) = x,\: h(x) = x^2,\: e_n(x) = \sin(\pi x/2), \: e_1(x) = \cos(\pi x/2), \: a = 0, \:b = 1.$ Then
$$\begin{array}{rcl}K &=& \frac{\left( \int_0^1 (x + r x^2) \sin(\pi x/2) \, dx \right)^2}{\left( \int_0^1 (x + r x^2) \cos(\pi x/2) \, dx \right)^2}\\ &\doteq& \frac{4(2.283185308\: r + 3.141592654)^2}{(3.586419096+1.869604404\: r)^2} \\ &=&3.069288133 + 1.261227604\: r - 0.527913928\: r^2 + O(r^3). \end{array}$$
The term linear in r is not zero.

RGV

6. May 21, 2012

EmilyRuck

I tried to do some numerical examples too, with functions very similar to your ones. In the particular problem, it should be

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

with $h(x)$ whatever error function and $w$ a numerical positive constant; in my example I chose a sort of Gaussian distribution, $h(x) = e^{-x^2}$. $E(x)$ should be the estimated function, so it is $E(x) = E_0(x) + rh(x)$ and this is all we know about the function. Don't worry if somewhere $E(x)$ 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

7. May 21, 2012

EmilyRuck

With a more accurate example, I tried these realistic values for each integral:

$a = 1.9;\\ b=2.1;\\ w = 4$

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

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

Emily

8. May 22, 2012

EmilyRuck

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)

9. May 22, 2012

Ray Vickson

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

10. May 22, 2012

EmilyRuck

Oh, so it had been a misunderstanding. My professor presented the first expression,

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

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

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

Emily

11. May 22, 2012

Ray Vickson

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