My electromagnetism (Maxwell's equations) textbook gives the following wave equation for free space:
$$\nabla^2 \mathbf{h}(\mathbf{r}, t) - \dfrac{1}{c^2} \dfrac{\partial^2{\mathbf{h}(\mathbf{r}, t)}}{\partial{t}^2} = 0,$$
where $c^2$ is the square of the velocity of light, $\mathbf{h}(\mathbf{r}, t) = \mathcal{R}e[\mathbf{R}(\mathbf{r})e^{j \omega t}]$ is the phasor representation of sinusoidal variations of the field, $\mathcal{R}e$ is the real part, $\mathbf{r} = x \mathbf{a}_x + y \mathbf{a}_y + \dots + z \mathbf{a}_z$, $\mathbf{a}_i$ is the unit vector in the $i$th direction, and the capital letter $\mathbf{H}$ is a complex vector quantity depending on space coordinates but not on time.
The author then says that
... any function of the form $f \left(t - \mathbf{a}_n \cdot \dfrac{\mathbf{r}}{c} \right)$ is a solution, where $\mathbf{a}_n$ is a unit vector. It is easy to show this in one dimension and only slightly more complicated to do so for the general case. Physically, it merely means that the wave propagates in the direction of $\mathbf{a}_n$ with a velocity of $c$.
I want to prove that this is indeed a solution for the general case.
My (as of yet incomplete) proof begins as follows:
$$\nabla^2 \mathbf{h}(\mathbf{r}, t) - \dfrac{1}{c^2} \dfrac{\partial^2{\mathbf{h}(\mathbf{r}, t)}}{\partial{t}^2} = 0,$$
where everything is the same as above, except for $\mathbf{r} = x_1 a_{x_1} + x_2 a_{x_2} + \dots + x_n a_{x_n}$.
This is the $n$-dimensional wave equation -- $n$-dimensional because the function $\mathbf{h}(\mathbf{r}, t)$ may depend in general on $\mathbf{r} = x_1 a_{x_1} + x_2 a_{x_2} + \dots + x_n a_{x_n}$.
This is made explicit if we write out the terms of the Laplace operator:
$$\dfrac{\partial^2{\mathbf{h}}(\mathbf{r}, t)}{\partial{x_1}^2} + \dfrac{\partial^2{\mathbf{h}}(\mathbf{r}, t)}{\partial{x_2}^2} + \dots + \dfrac{\partial^2{\mathbf{h}}(\mathbf{r}, t)}{\partial{x_n}^2} - \dfrac{1}{c^2} \dfrac{\partial^2{\mathbf{h}}(\mathbf{r}, t)}{\partial{t}^2} = 0$$
Note: This is one of my problem areas. Since $\mathbf{h}(\mathbf{r}, t) = \mathcal{R}e[\mathbf{R}(\mathbf{r})e^{j \omega t}]$ is a vector function of a vector and a variable, I am not sure how the Laplace operator $\nabla^2$ affects it and, therefore, how it should be written. I would appreciate it if someone could please clarify this.
I then make some simplifications. I conclude that any function of the form
$$f \left( t - \mathbf{a}_n \cdot \dfrac{\mathbf{r}}{c} \right) = f \left[ \dfrac{-1}{c} \left( \mathbf{a}_n \cdot \mathbf{r} - ct \right)\right] = f \left( \mathbf{a}_n \cdot \mathbf{r} - ct \right)$$
is a solution.
I am unsure of how to proceed with the proof from here. This is primarily due to the same problem I mentioned above: I am not sure how I should be taking the derivatives of $f \left( \mathbf{a}_n \cdot \mathbf{r} - ct \right)$.
I would then substitute the differentiated values into the equation
$$\dfrac{\partial^2{f}}{\partial{x_1}^2} + \dfrac{\partial^2{f}}{\partial{x_2}^2} + \dots + \dfrac{\partial^2{f}}{\partial{x_n}^2} - \dfrac{1}{c^2} \dfrac{\partial^2{f}}{\partial{t}^2} = 0,$$
which should then equal $0$, completing the proof.
I would greatly appreciate it if people could please take the time to carefully explain this to me.
We need to use the chain rule: If we let $$g(r, t)=f\left(t-a\cdot \frac{r}{c}\right)$$ Then we have that \begin{align} \partial_t g(r,t) &=\frac{\partial f\left(t-a\cdot \frac{r}{c}\right)}{\partial \left(t-a\cdot \frac{r}{c}\right)} \frac{\partial \left(t-a\cdot \frac{r}{c}\right)}{\partial t}\\ &=f'\left(t-a\cdot \frac{r}{c}\right) \end{align} And \begin{align} \partial_ig(r,t) &=\frac{\partial f\left(t-a\cdot \frac{r}{c}\right)}{\partial \left(t-a\cdot \frac{r}{c}\right)} \frac{\partial \left(t-a\cdot \frac{r}{c}\right)}{\partial x_i}\\ &=f'\left(t-a\cdot \frac{r}{c}\right) \frac{-1}{c}\partial_i \sum_{j}a_jx_j\\ &=-f'\left(t-a\cdot \frac{r}{c}\right)\frac{1}{c}a_i \end{align}