Solution to electromagnetic wave equation

Question

The Helmholtz wave equation is given as :

$$\nabla^2 \vec E =\mu\epsilon \frac{\partial^2 \vec E}{\partial t^2}$$

Considering $\vec E=E_x(z) e^{j \omega t}$ the Helmholtz wave equation now takes the form

$$\frac{\partial^2 E_x}{\partial z^2} + \omega^2 \mu \epsilon E_x=0$$

I know the solution to this equation can take many forms.

But which solution is generally used?

I mean some books use $$E_x=(E_x^+ e^{-jkz}+E_x^- e^{jkz})e^{j \omega t} = E_x^+ e^{j(wt-kz)}+E_x^- e^{j(wt+kz)}$$

while some other books use $$E_x=E_{0x}\cos(wt-kz)$$ where $E_{0x}$ is some value obtained from initial conditions.

Am I at liberty to use any of these solutions considering what is applicable in the problem in front of me?

I see that for formulation equations of polarization the cosine solution is used and not the exponential solution.

Answer 1

Your first solution refers to the two waves propagating in the opposite directions ($+z$ and $-z$), while the second one has a wave propagating in one direction only. If propagation direction is specifically stated in your problem, or you don't need two-directional propagation to illustrate a concept, you can use the 2nd one of course.