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I am currently studying the textbook Microwave Engineering, fourth edition, by David Pozar. Chapter 1.4 THE WAVE EQUATION AND BASIC PLANE WAVE SOLUTIONS says the following:

Plane Waves in a Lossless Medium

In a lossless medium, $\epsilon$ and $\mu$ are real numbers, and so $k$ is real. A basic plane wave solution to the above wave equation can be found by considering an electric field with only an $\hat{x}$ component and uniform (no variation) in the $x$ and $y$ directions. Then, $\partial/\partial{x} = \partial/\partial{y} = 0$, and the Helmholtz equation of (1.42) reduces to $$\dfrac{\partial^2{E_x}}{\partial{z}^2} + k^2 E_x = 0. \tag{1.44}$$ The two independent solutions to this equation are easily seen, by substitution, to be of the form $$E_x(z) = E^+e^{-jkz} + E^-e^{jkz}, \tag{1.45}$$ where $E^+$ and $E^-$ are arbitrary amplitude constants.

I calculated the following:

$$\dfrac{\partial^2}{\partial{z}^2}E_x(z) = -k^2E^+ e^{-jkz} - k^2 E^- e^{jkz} = -k^2\left( E^+ e^{-jkz} + E^- e^{jkz} \right)$$

After substitution, we have

$$-k^2\left( E^+ e^{-jkz} + E^- e^{jkz} \right) + k^2 E_x = -k^2\left( E^+ e^{-jkz} + E^- e^{jkz} - E_x \right) = 0$$

How does $-k^2\left( E^+ e^{-jkz} + E^- e^{jkz} - E_x \right) = 0$? Specifically, how does $E^+ e^{-jkz} + E^- e^{jkz} = E_x$? I would like to better understand the components of this equation, such as $E^+$ and $E^-$.

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You're almost there! $E_x=E^+e^{-jkz}+E^-e^{jkz}$ so just sub that into your final expression and everything cancels.

Note you should learn to recognise that the equation $$\frac{\text{d}^2f(z)}{\text{d}z^2}+k^2f(z)=0$$ has general solution $$f(z)=Ae^{jkz}+Be^{-jkz}$$ Possibly one of the most important differential equations in physics!

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  • $\begingroup$ Thanks for the clarification. Would you please explain why $E_x=E^+e^{-jkz}+E^-e^{jkz}$? $\endgroup$
    – The Pointer
    Commented Apr 23, 2022 at 13:17
  • $\begingroup$ @ThePointer Isn't this is exactly eq 1.45? $\endgroup$
    – garyp
    Commented Apr 23, 2022 at 13:52
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    $\begingroup$ @ThePointer Aha! The author condensed the notation too much. It's a delicate balance to convey enough information without decorating the variables with subscripts, superscripts, accsents, arrows, etc. $E^+$ and $E^-$ are the field amplitudes for the wave solutions that travel in opposite directions. One propagates in the positive $z$ direction, the other in the negative $z$ direction. I now notice that in your example you use a function in $x$ where the author is writing a function in $z$. Is the fact that the variables are different causing confusion? $\endgroup$
    – garyp
    Commented Apr 23, 2022 at 14:09
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    $\begingroup$ @ThePointer BTW, which of the author's terms are going in the positive direction and which is going negative can't be determined from what's posted. It's a matter of convention set earlier. The author might have mentioned that explicitly earlier in the presentation. $\endgroup$
    – garyp
    Commented Apr 23, 2022 at 14:12
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    $\begingroup$ I've edited my answer to use $z$ instead of $x$, but what the variable is called doesn't matter. As for the symbols $E^+$ and $E^-$, or $A$ and $B$, you're right we don't know what they are. They have to be specified by extra information (called "boundary conditions"). Right now, they are just general constants which could be anything. The precise value of these constants isn't important here though. What is important is the functional form of $E_x(z)$ or my $f(z)$ -- this function, regardless of the coefficients of the exponentials, satisfies the differential equation given. $\endgroup$
    – Garf
    Commented Apr 23, 2022 at 14:17

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