Acoustic theory

In the following we are going to derive the linear acoustic wave equation starting from the continuity and momentum equations on non-conservation differential form. The equations are repeated here for convenience.

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0$

(Eq. 6.63)

$\rho \frac{D𝐯}{Dt}+\mathrm{\nabla }p=0$

(Eq. 6.64)

Remember that $D/Dt$ denotes the substantial derivative operator defined as follows

$\frac{D}{Dt}=\frac{\partial }{\partial t}+𝐯\cdot \mathrm{\nabla }$

(Eq. 6.65)

where $\partial /\partial t$ is the local temporal derivative and $𝐯\cdot \mathrm{\nabla }$ is the convective derivative.

We are going to analyze acoustic waves in one dimension, which means that the equations above reduces to

$\frac{\partial \rho }{\partial t}+u\frac{\partial \rho }{\partial x}+\rho \frac{\partial u}{\partial x}=0$

(Eq. 6.66)

$\rho \frac{\partial u}{\partial t}+\rho u\frac{\partial u}{\partial x}+\frac{\partial p}{\partial x}=0$

(Eq. 6.67)

Pressure is a thermodynamic property and thus it can be expressed as a function of two other thermodynamic properties. Let's express pressure as a function of density ($\rho$) and entropy ($s$).

$p=p(\rho ,s)\Rightarrow dp={\left(\frac{\partial p}{\partial \rho }\right)}_{s}d\rho +{\left(\frac{\partial p}{\partial s}\right)}_{\rho }ds$

(Eq. 6.68)

Since weak acoustic waves are considered, entropy will be constant and thus $ds=0$, which means that

$dp={\left(\frac{\partial p}{\partial \rho }\right)}_{s}d\rho =a^{2}d\rho$

(Eq. 6.69)

$\rho \frac{\partial u}{\partial t}+\rho u\frac{\partial u}{\partial x}+a^2\frac{\partial \rho }{\partial x}=0$

(Eq. 6.70)

The acoustic perturbations can be described as small deviations around a reference state

$\begin{array}{cc}& \rho ={\rho }_{\mathrm{\infty }}+\Delta \rho \\ & p=p_{\mathrm{\infty }}+\Delta p\\ & T=T_{\mathrm{\infty }}+\Delta T\\ & u=u_{\mathrm{\infty }}+\Delta u=\{u_{\mathrm{\infty }}=0\}=\Delta u\\ \end{array}$

Inserted in Eqns.~\ref{eq:unstady:acoustic:wave:cont} and \ref{eq:unstady:acoustic:wave:mom:b} and using the fact that derivatives of the constant reference state flow quantities are zero, we get

$\frac{\partial }{\partial t}(\Delta \rho )+\Delta u\frac{\partial }{\partial x}(\Delta \rho )+({\rho }_{\mathrm{\infty }}+\Delta \rho )\frac{\partial }{\partial x}(\Delta u)=0$

(Eq. 6.71)

$({\rho }_{\mathrm{\infty }}+\Delta \rho )\frac{\partial }{\partial t}(\Delta u)+({\rho }_{\mathrm{\infty }}+\Delta \rho )\Delta u\frac{\partial }{\partial x}(\Delta u)+a^2\frac{\partial }{\partial x}(\Delta \rho )=0$

(Eq. 6.72)

In the same way as pressure, being a thermodynamic variable, can be expressed as a function of two other thermodynamic variables, so can the speed of sound. Once again we will select density and entropy as the two thermodynamic variables

$a^2=a^2(\rho ,s)$

(Eq. 6.73)

and since entropy is constant

$a^2=a^2(\rho )$

(Eq. 6.74)

Taylor expansion of $a^2$ around the reference state $a_{\mathrm{\infty }}$ with $\Delta \rho =\rho -{\rho }_{\mathrm{\infty }}$ gives

$a^2=a_{\mathrm{\infty }}^2+{(\frac{\partial }{\partial \rho }(a^2))}_{\mathrm{\infty }}\Delta \rho +{(\frac{{\partial }^2}{\partial {\rho }^2}(a^2))}_{\mathrm{\infty }}(\Delta \rho {)}^2+\cdots$

(Eq. 6.75)

Inserted in Eqn.~\ref{eq:unstady:acoustic:wave:mom:pert}, we get

$({\rho }_{\mathrm{\infty }}+\Delta \rho )\frac{\partial }{\partial t}(\Delta u)+({\rho }_{\mathrm{\infty }}+\Delta \rho )\Delta u\frac{\partial }{\partial x}(\Delta u)+[a_{\mathrm{\infty }}^2+{(\frac{\partial }{\partial \rho }(a^2))}_{\mathrm{\infty }}\Delta \rho +\cdots ]\frac{\partial }{\partial x}(\Delta \rho )=0$

(Eq. 6.76)

The perturbations $\Delta u$ and $\Delta \rho$ are small, which implies that $\Delta u\ll a_{\mathrm{\infty }}$ and $\Delta \rho \ll {\rho }_{\mathrm{\infty }}$. This means that products of perturbations can be canceled and so can higher-order terms in the Taylor expansion of $a^2$. This means that the continuity and momentum equations reduces to

$\frac{\partial }{\partial t}(\Delta \rho )+{\rho }_{\mathrm{\infty }}\frac{\partial }{\partial x}(\Delta u)=0$

(Eq. 6.77)

${\rho }_{\mathrm{\infty }}\frac{\partial }{\partial t}(\Delta u)+a_{\mathrm{\infty }}^2\frac{\partial }{\partial x}(\Delta \rho )=0$

(Eq. 6.78)

Before making the assumption that the perturbations are small compared to the corresponding reference state flow quantities and thus justifying the cancelation of products of perturbations from the equations, the flow equations were still the exact fully non-linear equations. Eqns.~\ref{eq:unstady:acoustic:wave:cont:linear}. and \ref{eq:unstady:acoustic:wave:mom:linear}, however, are approximations as several terms has been removed. The equations are linear and are good approximations as long as the perturbations are small. The smaller the perturbations, the better the approximation are the linear equations. Eqns.~\ref{eq:unstady:acoustic:wave:cont:linear} and \ref{eq:unstady:acoustic:wave:mom:linear} describes the motion induced in a gas by the passage of a sound wave. By combining the temporal derivative of Eqn.~\ref{eq:unstady:acoustic:wave:cont:linear} with the divergence of Eqn.~\ref{eq:unstady:acoustic:wave:mom:linear}, it is possible to obtain a wave equation describing the propagation of acoustic waves in a quiescent surrounding.

The temporal derivative of the continuity equation:

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )+{\rho }_{\mathrm{\infty }}\frac{{\partial }^2}{\partial x\partial t}(\Delta u)=0$

(Eq. 6.79)

The divergence of the momentum equation:

${\rho }_{\mathrm{\infty }}\frac{{\partial }^2}{\partial x\partial t}(\Delta u)+a_{\mathrm{\infty }}^2\frac{{\partial }^2}{\partial x^2}(\Delta \rho )=0$

(Eq. 6.80)

The second term in the first equation is the same as the first term in the second equation. Substituting the term, the two equations reduces to one single equation

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )=a_{\mathrm{\infty }}^2\frac{{\partial }^2}{\partial x^2}(\Delta \rho )$

(Eq. 6.81)

which is a one-dimensional form of the classic wave equation with the general solution

$\Delta \rho =F(x-a_{\mathrm{\infty }}t)+G(x+a_{\mathrm{\infty }}t)$

(Eq. 6.82)

$F$ and $G$ are arbitrary functions. The function $F$ describes the shape of a wave traveling in the positive $x$-direction at the speed of sound of the ambient gas and the function $G$ describes the shape of a wave traveling in the negative $x$-direction at the same speed. In Eqn.~\ref{eq:wave} $\Delta \rho$ appears with second derivatives in space and time. Let's differentiate the proposed solution (Eqn.~\ref{eq:wave:solution}) two times in time and space, respectively, and check that it is actually a valid solution to Eqn.~\ref{eq:wave}.

$\frac{\partial }{\partial t}(\Delta \rho )=\frac{\partial F}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial t}+\frac{\partial G}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial t}$

(Eq. 6.83)

$\frac{\partial }{\partial t}(\Delta \rho )=-a_{\mathrm{\infty }}{F}^{\prime }+a_{\mathrm{\infty }}{G}^{\prime }$

(Eq. 6.84)

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )=-a_{\mathrm{\infty }}\frac{\partial F^{\prime }}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial t}+a_{\mathrm{\infty }}\frac{\partial G^{\prime }}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial t}$

(Eq. 6.85)

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )=a_{\mathrm{\infty }}^2{F}^{″}+a_{\mathrm{\infty }}^2{G}^{″}$

(Eq. 6.86)

$\frac{\partial }{\partial x}(\Delta \rho )=\frac{\partial F}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial x}+\frac{\partial G}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial x}$

(Eq. 6.87)

$\frac{\partial }{\partial x}(\Delta \rho )=F^{\prime }+G^{\prime }$

(Eq. 6.88)

$\frac{{\partial }^2}{\partial x^2}(\Delta \rho )=\frac{\partial F^{\prime }}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial x}+\frac{\partial G^{\prime }}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial x}$

(Eq. 6.89)

$\frac{{\partial }^2}{\partial x^2}(\Delta \rho )=F^{″}+G^{″}$

(Eq. 6.90)

Eqns. \ref{eq:wave:ddt} and \ref{eq:wave:ddx} inserted Eqn. \ref{eq:wave} gives

$a_{\mathrm{\infty }}^2{F}^{″}+a_{\mathrm{\infty }}^2{G}^{″}=a_{\mathrm{\infty }}^2(F^{″}+G^{″})$

(Eq. 6.91)

which shows that Eqn. \ref{eq:wave:solution} is a valid solution to the wave equation.

$F$ and $G$ are arbitrary functions and thus $G=0$ is a valid solution, which gives

$\Delta \rho (x,t)=F(x-a_{\mathrm{\infty }}t)$

(Eq. 6.92)

If $\Delta \rho$ is constant, i.e. a wave with constant amplitude, we see from Eqn.~\ref{eq:wave:solution:F} that $(x-a_{\mathrm{\infty }}t)$ is constant and thus

$x=a_{\mathrm{\infty }}t+c\Rightarrow \frac{dx}{dt}=a_{\mathrm{\infty }}$

(Eq. 6.93)

From Eqn.~\ref{eq:wave:solution:F}, we get

$\frac{\partial }{\partial t}(\Delta \rho )=-a_{\mathrm{\infty }}{F}^{\prime }$

(Eq. 6.94)

$\frac{\partial }{\partial x}(\Delta \rho )=F^{\prime }$

(Eq. 6.95)

and thus

$\frac{\partial }{\partial x}(\Delta \rho )=-\frac{1}{a_{\mathrm{\infty }}}\frac{\partial }{\partial t}(\Delta \rho )$

(Eq. 6.96)

which gives a relation between the temporal derivative of $\Delta \rho$ and the spatial derivative of $\Delta \rho$. With Eqn.~\ref{eq:wave:solution:F:b}, the linearized momentum equation Eqn.~\ref{eq:unstady:acoustic:wave:mom:linear} can be rewritten as follows

$\frac{\partial }{\partial t}(\Delta u)=-\frac{a_{\mathrm{\infty }}^2}{{\rho }_{\mathrm{\infty }}}\frac{\partial }{\partial x}(\Delta \rho )=\{\frac{\partial }{\partial x}(\Delta \rho )=-\frac{1}{a_{\mathrm{\infty }}}\frac{\partial }{\partial t}(\Delta \rho )\}=\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\frac{\partial }{\partial t}(\Delta \rho )\Rightarrow$

(Eq. 6.97)

$\frac{\partial }{\partial t}(\Delta u-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho )=0\Rightarrow \Delta u-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =const$

(Eq. 6.98)

In an undisturbed gas $\Delta u=\Delta \rho =0$ and thus

$\Delta u-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =0$

(Eq. 6.99)

or

$\Delta u=\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho$

(Eq. 6.100)

If instead $F$ is set to zero and $G$ is non-zero, we get

$\Delta u=-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho$

(Eq. 6.101)

${\left(\frac{\partial p}{\partial \rho }\right)}_s=a^2\Rightarrow \Delta p=a_{\mathrm{\infty }}^2\Delta \rho$

(Eq. 6.102)

Acoustic wave traveling in the positive $x$-direction:

$\Delta u=\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =\frac{1}{a_{\mathrm{\infty }}{\rho }_{\mathrm{\infty }}}\Delta p$

(Eq. 6.103)

Acoustic wave traveling in the negative $x$-direction:

$\Delta u=-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =-\frac{1}{a_{\mathrm{\infty }}{\rho }_{\mathrm{\infty }}}\Delta p$

(Eq. 6.104)