Editing Acoustic theory (section)

===For Irrotational Fluids===
In the case that the fluid is irrotational, that is <math>\nabla\times\mathbf{v} = 0</math>, we can then write <math>\mathbf{v} = -\nabla\phi</math> and thus write our equations of motion as
: <math>
  \begin{align}
     \frac{\partial p'}{\partial t} -\rho_0c^2\nabla^2\phi & = 0 \\
     -\nabla\frac{\partial\phi}{\partial t} + \frac{1}{\rho_0}\nabla p' & = 0
  \end{align}
</math>

The second equation tells us that
: <math>
  p' = \rho_0 \frac{\partial \phi}{\partial t}
</math>

And the use of this equation in the continuity equation tells us that
: <math>
     \rho_0\frac{\partial^2 \phi}{\partial t} -\rho_0c^2\nabla^2\phi = 0
</math>

This simplifies to
: <math>
     \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} -\nabla^2\phi = 0
</math>

Thus the velocity potential <math>\phi</math> obeys the wave equation in the limit of small disturbances. The boundary conditions required to solve for the potential come from the fact that the velocity of the fluid must be 0 normal to the fixed surfaces of the system.

Taking the time derivative of this wave equation and multiplying all sides by the unperturbed density, and then using the fact that <math>p' = \rho_0 \frac{\partial \phi}{\partial t}</math> tells us that
: <math>
     \frac{1}{c^2}\frac{\partial^2 p'}{\partial t^2} -\nabla^2p' = 0
</math>

Similarly, we saw that <math>p' = \left(\frac{\partial p}{\partial \rho_{0}}\right)_{s}\rho' = c^{2}\rho'</math>. Thus we can multiply the above equation appropriately and see that
: <math>
     \frac{1}{c^2}\frac{\partial^2 \rho'}{\partial t^2} -\nabla^2\rho' = 0
</math>

Thus, the velocity potential, pressure, and density all obey the wave equation. Moreover, we only need to solve one such equation to determine all other three. In particular, we have
: <math>
  \begin{align}
     \mathbf{v} & = -\nabla \phi \\
     p' & = \rho_0 \frac{\partial \phi}{\partial t}\\
\rho' & = \frac{\rho_0}{c^2}\frac{\partial\phi}{\partial t}
  \end{align}
</math>