Modes orthogonality in a dielectric slab

[EmilyRuck]

A typical mode in a dielectric slab like this, with propagation along x, uniformity along z and refractive index variation along y, is represented by the following function:

f (y) = \begin{cases} \displaystyle \frac{\cos (k_1 y)}{\cos (k_1 d)} && |y| \leq d \\ e^{-j k_2 (y - d)} && |y| \geq d \end{cases}

The slab is symmetrical with respect to the (x,z) plane and its thickness is 2d. k_2 = - j \gamma is purely imaginary and it causes attenuation out of the core. Another acceptable mode can have \sin (k_1 y).
If f(y) represents the electric field E_z and the field is maximum for y = 0, the mode is said to be "even" and it has a cosine behaviour inside the core; if the electric field is zero for y = 0, the mode is said to be "odd" and it has a sine behaviour inside the core.

According to the solutions (k_1, k_2) of the system, modes can have one or more zeros in the |x| \leq d zone.

I would like to prove that two different modes f_1 (y) and f_2 (y) (f_1 \neq f_2) are orthogonal. This is a very important proof because it would mean that modes can't bring power jointly and can't exchange their power.

Each mode satisfies the wave equation

\displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} + [n_1^2 k_0^2 - \beta_1^2] f_1 (y) = 0
\displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} + [n_1^2 k_0^2 - \beta_2^2] f_2 (y) = 0

(n_1 is not related to the mode 1 or mode 2: it is the refractive index of the core). \beta is the propagation constant along the propagation direction x. Suppose that \beta_1 \neq \beta_2. Let's multiply the first equation by f_2 and the second by f_1; then, let's subtract the second equation from the first. The result is

f_2 (y) \displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} - f_1 (y) \displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} = [\beta_1^2 - \beta_2^2] f_1 (y) f_2 (y)

Now, let's integrate between 0 and + \infty (if the mode is odd or even, we can integrate on just half of the y axis and the result for the remaining half is the same - if even - or the opposite - if odd):

\displaystyle \int_{0}^{+\infty} \left( f_2 (y) \displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} - f_1 (y) \displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} \right) dy = \displaystyle \int_{0}^{+\infty} \left[ (\beta_1^2 - \beta_2^2) f_1 (y) f_2 (y)\right] dy

The LHS can be integrated by parts:

\displaystyle \int_{0}^{+\infty} \left( f_2 (y) \displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} - f_1 (y) \displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} \right) dy =
= \left[ f_2 (y) \displaystyle \frac{\partial f_1 (y)}{\partial y} \right]_0^{+\infty} - \displaystyle \int_{0}^{+\infty} \left( \frac{\partial f_1 (y)}{\partial y} \frac{\partial f_2 (y)}{\partial y}\right) dy - \left[f_1 (y) \displaystyle \frac{\partial f_2 (y)}{\partial y} \right]_0^{+\infty} + \displaystyle \int_{0}^{+\infty} \left( \frac{\partial f_1 (y)}{\partial y} \frac{\partial f_2 (y)}{\partial y}\right) dy =
= \left[ f_2 (y) \displaystyle \frac{\partial f_1 (y)}{\partial y} \right]_0^{+\infty} - \left[f_1 (y) \displaystyle \frac{\partial f_2 (y)}{\partial y} \right]_0^{+\infty}

Both f_1 and f_2 and their derivatives are 0 for y \to +\infty because of the decreasing exponential. So, just the following term has to be evaluated:

- f_2 (0) \displaystyle \frac{\partial f_1 (0)}{\partial y} + f_1 (0) \displaystyle \frac{\partial f_2 (0)}{\partial y}

For y = 0 several cases can be distinguished. If f_1 and f_2 are both even or both odd, the expression vanished as desired. But if f_1 is odd and f_2 is even, or viceversa, respectively the first and the second addend are not zero.

How can be proved in these last two cases that both the addends (and so all the LHS) anyway vanish? Any hint or any references? I did not find something useful on Google.

 

[RUber]

As you mentioned, if $f_1$ and $f_2$ are both even or both odd, you have been able to show what you want to show.
If one is odd and another is even, then they are orthogonal from the start.

 

[EmilyRuck]

As you mentioned, if $f_1$ and $f_2$ are both even or both odd, you have been able to show what you want to show.
If one is odd and another is even, then they are orthogonal from the start.

Are you meaning that the integral in the RHS

\displaystyle \int_{0}^{+ \infty} f_1(y) f_2(y)dy

is 0 by itself? I know that sine and cosine are orthogonal functions, but it is difficult to figure out how this integral can vanish when the k_1, k_2 in f_1 are different from the k_1, k_2 in f_2.

Moreover,

\displaystyle \int_{d}^{+ \infty} e^{- k_2^{(1)}(y - d)}e^{- k_2^{(2)}(y - d)}dy = \displaystyle - \frac{1}{\left( k_2^{(1)} + k_2^{(2)} \right)}\left[ e^{- \left( k_2^{(1)} + k_2^{(2)} \right) (y - d)} \right]_d^{+\infty} = \frac{1}{\left( k_2^{(1)} + k_2^{(2)} \right)} \neq 0

Note: there was an error in my original post, where f(y) should be e^{-k_2(y-d)} for |y| \geq d, with k_2 \geq 0, k_2 \in \mathbb{R}, and not e^{-jk_2(y-d)}.

 

[RUber]

When you say:

Now, let's integrate between 0 and +∞ (if the mode is odd or even, we can integrate on just half of the y axis and the result for the remaining half is the same - if even - or the opposite - if odd):

you complicate the integral.
If $f$ is odd, then $\int_{-d}^d f(y) dy =0$ and $\int_d^\infty f(y) dy + \int_{-\infty}^{-d} f(y) dy = 0$.
And note that, as with integers, an odd function times an even function is an odd function.