Derivation of the acoustic wave equation

The acoustic wave equation describes sound waves in a liquid or gas. Another more complicated set of equations describes elastic waves in solids. Begin with the acoustic case. Define

$\rho$  =  mass per unit volume of the fluid             
u  		 = 		 velocity flow of fluid in the x-direction                   
w  		 =  		 velocity flow of fluid in the z-direction      
P  		 =  		 pressure in the fluid

Newton's law of momentum conservation says that a small volume within a gas will accelerate if there is an applied force. The force arises from pressure differences at opposite sides of the small volume. Newton's law says

$$\ \rm{mass} \ \times \ \rm{acceleration}\ \ =\ \ \rm{force}\ \ =\ \ -\ \rm{pressure\ gradient}$$

$$\begin{eqnarray} \rho\ {\partial u \over \partial t } \ \ \ &=&\ \ \ - \ { \pa... ...r \partial t } \ \ \ &=&\ \ \ -\ { \partial P \over \partial z }\end{eqnarray}$$ (9) (10)

The second physical process is energy storage by compression and volume change. If the velocity vector u at $x \ +\ \Delta x$exceeds that at x, then the flow is said to be diverging. In other words, the small volume between x and $x \ +\ \Delta x$ is expanding. This expansion must lead to a pressure drop. The amount of the pressure drop is in proportion to a property of the fluid called its incompressibility K. In one dimension the equation is

$$\ \rm{pressure\ drop} \ \ =\ \ \ \rm{(incompressibility)} \ \times \ \ \rm{(divergence\ of\ velocity)}$$ (11)

 

$$-\ {\partial P \over \partial t } \ \ \ = \ \ \ K \ { \partial u \over \partial x }$$ (12)

In two dimensions it is  

$$-\ { \partial P \over \partial t } \eq K \ \left( { \partial... ...ver \partial x } \ +\ { \partial w \over \partial z } \right)$$ (13)

To arrive at the one-dimensional wave equation from (9) and (12), first divide (9) by $\rho$ and take its x-derivative:  

$${\partial\ \over \partial x }\ {\partial \ \over \partial t ... ...\partial x } \ {1 \over \rho }\ { \partial P \over \partial x }$$ (14)

Second, take the time-derivatives of (12) and (13). In the solid-earth sciences we are fortunate that the material in question does not change during our experiments. This means that K is a constant function of time:  

$${ \partial^2 P \over \partial t^2 } \eq -\ K\ { \partial \ \over \partial t } \ { \partial \ \over \partial x } \ u$$ (15)

Inserting (14) into (15), the one-dimensional scalar wave equation appears.  

$${ \partial^2 P \over \partial t^2 } \eq K \ { \partial \ \o... ...partial x } \ {1 \over \rho }\ { \partial P \over \partial x }$$ (16)

In two space dimensions, the exact, acoustic scalar wave equation is  

$${ \partial^2 P \over \partial t^2 } \eq K \ \left( { \partia... ...\ {1 \over \rho }\ { \partial \ \over \partial z } \ \right) P$$ (17)

You will often see the scalar wave equation in a simplified form, in which it is assumed that $\rho$ is not a function of x and z. Two reasons are often given for this approximation. First, observations are generally unable to determine density, so density may as well be taken as constant. Second, we will soon see that Fourier methods of solution do not work for space variable coefficients. Before examining the validity of this approximation, its consequences will be examined. It immediately reduces (17) to the usual form of the scalar wave equation:  

$$\begin{tabular} {\vert c\vert} \hline \\ $ \displaystyle {\... ...ver\partial z^2} \right)\ P $\space \\ \\ \hline\end{tabular}$$ (18)

To see that this equation is a restatement of the geometrical concepts of previous sections, insert the trial solution  

$$P \eq \exp ( -\,i \omega t \ +\ i\,k_x x \ +\ i\,k_z z )$$ (19)

What is obtained is the dispersion relation of the two-dimensional scalar wave equation:  

$${ \omega^2 \over K / \rho } \eq k_x^2 \ +\ k_z^2$$ (20)

Later an equation like (20) will be developed by considering only the geometrical behavior of waves. In that development the wave velocity squared is found where $K / \rho$ stands in equation (20). Thus physics and geometry will be reconciled by the association  

$$v^2 \eq {K \over \rho }$$ (21)

Last, let us see why Fourier methods fail when the velocity is space variable. Assume that $\omega$, k_x, and k_z are constant functions of space. Substitute (19) into (18) and you get the contradiction that $\omega$, k_x, and k_z must be space variable if the velocity is space variable. Try again assuming space variability, and the resulting equation is still a differential equation, not an algebraic equation like (20).