Wave equation boundary conditions for an engineer versus physicist

Question

The wave equation is: $$\nabla^2 \mathbf{E} + k^2 \mathbf{E} = 0$$

Using separation of variables, we get a solution of $E = a(x)b(y)c(z)d(t)$. Say for the $x$-direction we get a solution of:

$$ a(x) = C_1 e^{-j k_x x} + C_2 e^{+j k_x x}$$

and the time solution is

$$ d(t) = C_3 e^{-j \omega t} + C_4 e^{+j \omega t}$$

But I noticed that for physics books, they apply boundary conditions that yield the following scenario

$$ a(x) = 0 + C_2 e^{+j k_x x}$$

and

$$ d(t) = C_3 e^{-j \omega t} + 0$$

Hence, we get

$$ a(x)d(t) = C_5 e^{\mathbf{+}j k_x x \mathbf{-} j \omega t}$$

For engineering books, they apply boundary conditions that yield the following scenario

$$ a(x) = C_1 e^{-j k_x x} + 0$$

and

$$ d(t) = 0 + C_4 e^{+j \omega t}$$

Hence, we get

$$ a(x)d(t) = C_5 e^{\mathbf{-}j k_x x \mathbf{+} j \omega t}$$

Question: which set of boundary conditions is the "correct" one for wave propagation? Or what does the engineering book assume over the physics book (and vice-versa)

Background:

1. The physics book (e.g. Griffith, chapter 9) claims $+k_x$ is for the incident/transmitted wave, while $-k_x$ is for the reflected wave. Moreover, He states $\pm \omega$ controls the wave direction (i.e. moving to the left or right).
2. The engineering book (Pozar, chapter 1) states "terms with + signs result in waves traveling in the negative $x,y,z$ direction, while terms with - signs result in waves traveling in the positive direction" when talking about the sign of $k_x$ and Pozar slaps a $e^{+j \omega t}$ onto the final solution.

Answer 1

It all goes back to the ancient habit of defining what a Fourier transform is. By tradition long lost in the mist of history, physicists define a Fourier transform of a function $f(t)$ as $$F_p(\omega)=\int_{-\infty}^{+\infty}dt f(t) e^{\mathfrak i \omega t}$$ but engineers define it as $$F_e(\omega)=\int_{-\infty}^{+\infty}dt f(t) e^{-\mathfrak j \omega t}$$ so that we have $F_p(-\omega) = F_e(\omega)$. There is some reason in the engineers definition because we, at least I do, intuitively think in terms of positive frequencies and then since the Inverse transforms is $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}d\omega F_e(\omega) e^{\mathfrak j \omega t}$$ you can say that you assemble an arbitrary $f(t)$ from its frequency components represented at $\omega$ by the complex sinusoids $e^{\mathfrak j \omega t}$ with amplitudes $F_e(\omega)$. Of course a physicist would say his frequency contents are equally well represented at positive frequencies by the sinusoids $e^{-\mathfrak i \omega t}$. Don't worry, you get used to it, just mentally change every $\mathfrak j$ to $-\mathfrak i$ or $\mathfrak i$ to $-\mathfrak j$.