Cavity Solution of the Non-linear Equations
In this section we develop the sound wave solution to the non-linear equations for a standing wave inside a cavity. The cavity extends from x=0 to x=a meters, and has a uniform cross-section parallel to the y-z plane. We consider wave solutions that are independent of y and z. The primary boundary condition is that the mean molecular velocity u(x,t) is zero at x=0 and x=a. The solution method used for a traveling wave doesn't work for a standing wave, so here the solution is derived as an exact correction to the first order solution. The equations are then solved numerically. Formulating the equations in terms of a correction to the 1st order solution enhances the accuracy of the numerical solution, compared to numerically calculating the complete solution. A gas consisting of diatomic molecules is assumed in the analysis, but numerical results for a monotonic gas are also discussed.
Without loss of generality we can write the solution as
The subscript zero denotes the static equilibrium values, and the subscript one denotes the first order solution, where, for a cavity
The objective is then to solve for the exact correction terms denoted by the subscript two.
Equations for the Correction Terms
Using equation (1) the equation of continuity becomes
Euler's equation becomes
and the energy equation
These are the three equations that are solved numerically.
Form of the Correction Solution
Consider the following form for the solution of the correction terms
It is not difficult to show that equations (4) and (6) then contain only cosnkx terms, and equation (5) contains only sinnkx terms, so equation (7) is a general form of the solution.
The initial excitation of the cavity is arbitrary; we consider the case where the gas begins at equilibrium, u=0, ρ=ρ0, p=P0. The kinetic energy density (Joules per cubic meter) in the gas is
Now imagine a piston at x=0 that begins vibrating with velocity (u1/2)sinωt; assume that it generates the first order wave in the cavity. It delivers power to the cavity for 2a/c seconds, the time it takes the wave to travel to, and reflect from, the boundary at x=a, and then return to x=0. At that point, which we call t=0, the piston stops, and the wave continues to reflect back and forth in the cavity. During the initial excitation the piston will do work on the gas, as given by equation (37) in the section on plane waves. The energy density in the cavity will increase due to this work by
Energy Density vs. Time
Subsequently, the correction terms evolve according to equations (4) - (6). The kinetic energy density inside the cavity can be divided into two parts: KEu due to the mean velocity u(x,t) and the KEσ due to the velocity variance σ 2(x,t). Using equation (2) we have
Substituting equation (7) into equation (1) and integrating over the cavity length we get
Similarly, for t -> 0 we get
Comparing equations (9), (11), and (12) we must have
Numerical Solution of the Non-linear Equations
Given equation (7), the wave behavior as a function of x is analytic. The only approximation involving the x variation is the truncation of the infinite series. The differential equations are numerically integrated with respect to time. A Matlab program has been written to solve the equations, and accuracies on the order of one part in 105 have been obtained. This is sufficient to accurately determine the wave momentum. Many program runs with a wide range of wave amplitudes, frequency, length of time integration, and size of the time increment, yield very stable numerical results: the bottom line is that the ratio of wave energy flux to wave momentum flux is (5/3)c. In general, the result is equal to two times the number of degrees of freedom, divided by the number of degrees of freedom plus one, times the velocity of propagation c.
In the calculation of this ratio the standing wave is viewed as the combination of two waves traveling in opposite directions, each with amplitude u1/2. The wave energy flux is computed from equation (37) in the plane wave section. The wave momentum flux is computed as 1/2 of the additional force produced by the presence of the wave - i.e. the increase in pressure compared to equilibrium conditions. This is completely equivalent to the basic flux definitions given elsewhere (Section 1, table 2).
It is interesting to note that the presence of the sound wave slightly increases the average temperature in the cavity, according to equation (9); this causes part of the increase in pressure. This part is reflected in the n=0 term in the pressure correction term in equation (7). It turns out that the n=0 term represents 1/3 of the total pressure increase for a diatomic gas. The remaining 2/3 of the pressure increase is due to the n>0 terms, which have no effect on average temperature. For a monatomic gas the n=0 term represents 1/2 of the total pressure increase. At first it may seem odd to identify an increase in pressure due to an increase in average temperature as "sound wave momentum." However, as discussed elsewhere, pressure of a gas is always a manifestation of momentum flux in the gas, and from this perspective it is quite natural.
Equations (12) and (13) are derived in the limit of t→0. The actual numerical behavior as a function of time diverges from these equations, but quite slowly, so the equations are very good approximations for many cycles, for small wave amplitudes.
As noted elsewhere for the first order solution the momentum flux has a time-independent component. If this component were a function of x, the momentum in some regions would increase steadily without limit. Clearly this is physically impossible. On the other hand, imposing the condition of x-independence leads to an inconsistency with the coefficients defined by equation (3). For the non-linear numerical solution the momentum flux is also x-dependent, and initially it is the same as the first order solution. However as time increases the x-dependency for the non-linear solution decreases. Obviously the numerical solution can only be computed for a finite time period, so this doesn't prove that it will go to zero, but after 30 cycles the problematic flux component is nearly gone. I am satisfied that the solution is valid.
As explained in a note in the Plane Wave Section I subsequently learned of analytical solutions which confirm my numerical result.To the physics table of contents