Since acoustic wave is longitudinal, its equation is exhibited nonlinear. In particular, I derived an acoustic wave equation within uneven, varying pipe. It goes:
$$ m_{tt} (m_x)^2 -2 m_{tx}m_tm_x + m_{xx}(m_t)^2=0 $$
where $m$ is accumulated mass of air along $x$-axis.
Since I am attempting speech synthesis, the initial/boundary conditions are:
$$ m(0,x) = \rho_\text{atm}(V_\text{lungs} + \int_0^x A(0,\lambda) \space \text{d}\lambda) \\ m_t(0,x) = 0 \\ m(t,0) = \rho_\text{atm}V_\text{lungs}-m_\text{exhale}(t) \\ m_x(t,0) = \rho_\text{atm}A(t,0) $$ where $\rho_\text{atm}$ is the atmospheric density, $V_\text{lungs}$ is the air capacity of lungs, $A$ is the cross-sectional area of the pipe (which models the space between articulators), and $m_\text{exhale}$ is the exhaled air mass.
The actual quantity to be picked up is the air density at the right end of the pipe. That is, $m_x(t,L) / A(t,L)$. $L$ is the length from the glottis to the lips.
I attempted solving this PDE numerically using finite difference method. I tried few stencils, for all which the solution blew up. Yet to think about it, denoting the grid coefficients by $h_t$ and $h_x$, each term of the equation has $h_t^{-2}h_x^{-2}$ coefficients, which will be factored out. As a result, the amplification factor (regardless along $t$ or $x$) will be independent to both $h_t$ and $h_x$. Does this mean the numerical solution will blow up anyway, regardless of the choice of stencils? Does this mean finite difference method is unusable here?