Step 1: Try a plane wave trial solution directly.
Instead of deriving the general solution from scratch, guess a traveling wave trial form $\varphi(x,t) = A\sin(kx - \omega t)$, where $A$, $k$, and $\omega$ are constants, and check what condition the given PDE forces on this guess.
Step 2: Substitute the trial solution into the PDE.
Compute the two derivatives needed:
\[ \frac{\partial \varphi}{\partial t} = -A\omega\cos(kx-\omega t), \qquad \frac{\partial \varphi}{\partial x} = Ak\cos(kx-\omega t) \]
Plugging into $\varphi_t + a\varphi_x = 0$:
\[ -A\omega\cos(kx-\omega t) + aAk\cos(kx-\omega t) = 0 \]
\[ A\cos(kx-\omega t)\,(ak - \omega) = 0 \]
Step 3: Read off the condition this forces.
For this to hold at every $x$ and $t$, not just at special points, and since $A \ne 0$ and $\cos(kx-\omega t)$ is not always zero, we need $ak - \omega = 0$, that is $\omega = ak$. This is a linear relationship between frequency $\omega$ and wavenumber $k$, called a dispersion relation, and a linear dispersion relation $\omega = ak$, with constant $a$ not depending on $k$, is the signature of non-dispersive wave motion, all wavenumbers travel at exactly the same speed $a = \omega/k$.
Step 4: Confirm this describes wave propagation, not diffusion or a simple decay law.
The trial solution becomes $\varphi = A\sin(k(x-at))$, a sinusoidal shape moving rigidly at speed $a$, exactly like a traveling wave on a string, not spreading out as heat conduction would, and not decaying in place as Newton's cooling would, since there is no $x$ dependence in cooling at all. Because the coefficients in the PDE are plain constants, not depending on $\varphi$ itself, the wave motion described is linear.
Final Answer:
The plane wave trial solution only works when $\omega = ak$, a non-dispersive relation that describes a rigidly translating wave, confirming linear wave propagation.
\[ \boxed{\text{linear wave propagation}} \]