Does a wave have inertia? In de Broglie hypothesis, particles have wave nature. The question is does this wave have inertia? If so is it represented in the corresponding wave equation?
 A: Let's first address the general question--do "waves" of any kind have (or can they have) inertia? I suppose here by "inertia" you mean "resistance to changes in velocity." This is certainly true of waves in, say, water--you've certainly felt resistance to your hand if you sweep it through water to make a wave; the destructive force of a tsunami is a more extreme example of the force a wave can transmit.
Even in classical electromagnetism, where the waves (light) are massless, they still have inertia. Consider, for a very direct example, a solar sail which reflects light from a source and acquires momentum from the rebounding light--we wouldn't expect this to happen if light waves had no inertia, since in that case presumably we could easily change their direction with no resistance.
Now, do quantum "matter waves" have inertia? This a more subtle question, because the wave should not be viewed as a "spread out" particle, but rather a probability amplitude for the particle to be measured as present at a particular point in space. 
However, I think it should be clear that these waves do have inertia, by considering collision problems, just as we did for the previous two cases. We know that when two particles collide, they must conserve momentum. For instance, it is not possible that two particles of equal mass collide head-on and then both move in the same direction afterwards, and furthermore we know that this collision is extremely well-modeled quantum mechanically by Schrodinger's wave equations.
To get a more direct and mathematical answer, let's consider another definition of inertia, which is that an object has inertia if its velocity (and thus its momentum) remains unchanged unless acted on by an external force. For a massive, non-relativistic, particle, the Hamiltonian is given by:
\begin{equation}
H = \frac{p^2}{2m} + V(x)
\end{equation}
Here V(x) is our external potential ("force" is a difficult quantity to work with in quantum mechanics), $p$ is the momentum operator, and $m$ is the mass of the particle. Suppose there is no external force, so $V(x) = 0$ (or any constant really). Then the rate of change of the momentum operator using the Heisenberg picture (equivalent to the more traditional Schrodinger wave picture) is:
\begin{equation}
\frac{ \mathrm{d}{p}}{\mathrm{d} t} = \frac{i}{\hbar} \left[ H , p \right] = 0
\end{equation}
This is because the momentum operator is only commuting with $p^2$, and operators always commute with their own squares. Therefore, momentum is unchanging, and we can see that the "principle of inertia" is indeed respected by the equations of quantum mechanics. This occurs in the equations essentially by the fact that the free-particle Hamiltonian commutes with momentum, and therefore if the particle is subject to this Hamiltonian the momentum will not change.
A: TDSE
TISE
Here you have the time dependent and independent Schrodinger wave equations, respectively.  These relate to the energy of particles, but the trident symbol, Psi is representative of the actual wave equation I believe you are referring to.  
While De Broglie and schrodinger and others like them do describe particles as behaving like waves, they are referring primarily to probability functions, or wave equations describing where a particle is likely to be found.  Meaning, in a given space, the particles and the "paths" they move in can be represented by waves, where you have crests (areas of high probability) and troughs (areas of low probability).  
In short, particles CAN have mass and thus COULD have inertia, but the wave equation themselves do not as they are merely a representation of probability.
Hope this helps!  
