Variational expression: a demonstration

[EmilyRuck]

Homework Statement

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.

Homework Equations

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

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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

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 \\<br /> b = \int_a^b e_0(x) h(x) \, dx \\<br /> c = \int_a^b e_n(x) E_0(x) \, dx \\<br /> d = \int_a^b e_n(x) h(x) \, dx \end{array}<br />
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&#039;(0) = dK(r)/dr|_{r=0} = \frac{2a}{c^3}(bc - ad),
which is nonzero in general.

RGV

 

[EmilyRuck]

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 \\<br /> b = \int_a^b e_0(x) h(x) \, dx \\<br /> c = \int_a^b e_n(x) E_0(x) \, dx \\<br /> d = \int_a^b e_n(x) h(x) \, dx \end{array}<br />
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&#039;(0) = dK(r)/dr|_{r=0} = \frac{2a}{c^3}(bc - ad),
which is nonzero in general.

RGV

First of all, thank you for your reply!
Maybe you wanted to write:

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

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

 

[Ray Vickson]

First of all, thank you for your reply!
Maybe you wanted to write:

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

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

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

 

[Ray Vickson]

First of all, thank you for your reply!
Maybe you wanted to write:

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

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

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 &amp;=&amp; \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}\\<br /> &amp;\doteq&amp; \frac{4(2.283185308\: r + 3.141592654)^2}{(3.586419096+1.869604404\: r)^2} \\<br /> &amp;=&amp;3.069288133 + 1.261227604\: r - 0.527913928\: r^2 + O(r^3).<br /> \end{array}<br />
The term linear in r is not zero.

RGV

 

[EmilyRuck]

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.

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

e_n(x) = \sin(n \pi x/w)\\<br /> e_1(x) = \sin(\pi x/w)\\<br /> 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 . 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

 

[EmilyRuck]

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

a = 1.9;\\<br /> b=2.1;\\<br /> 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

 

[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)

 

[Ray Vickson]

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)

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

 

[EmilyRuck]

we are talking about two different things. The thing you wrote down in your original post was not, in general, stationary

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.

If you did have a stationary thing then, of course, the first-order differential would be zero BY DEFINITION

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

 

[Ray Vickson]

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

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