Finite non-linear waves

Starting point: the governing flow equations on partial differential form

Continuity equation:

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

(Eq. 6.105)

Momentum equation:

$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho }\frac{\partial p}{\partial x}=0$

(Eq. 6.106)

Any thermodynamic property can be expressed as a function of two other thermodynamic properties. This means that we can get density as a function of pressure and entropy: $\rho =\rho (p,s)$ and therefore

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

(Eq. 6.107)

Assuming isentropic flow $ds=0$ gives

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

(Eq. 6.108)

$\begin{array}{cc}& \frac{\partial \rho }{\partial t}={\left(\frac{\partial \rho }{\partial p}\right)}_s\frac{\partial p}{\partial t}=\frac{1}{a^2}\frac{\partial p}{\partial t}\\ & \\ & \frac{\partial \rho }{\partial x}={\left(\frac{\partial \rho }{\partial p}\right)}_s\frac{\partial p}{\partial x}=\frac{1}{a^2}\frac{\partial p}{\partial x}\end{array}$

(Eq. 6.109)

Now, insert \ref{eq:rhotop} in \ref{eq:pde:cont} gives

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

(Eq. 6.110)

Dividing \ref{eq:pde:cont:b} by $\rho a$ gives

$\frac{1}{\rho a}(\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x})+a\frac{\partial u}{\partial x}=0$

(Eq. 6.111)

A slightly modified form of the momentum equation is obtained by multiplying and dividing the last term by $a$

$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho a}\left(a\frac{\partial p}{\partial x}\right)=0$

(Eq. 6.112)

If the continuity equation on the form \ref{eq:pde:cont:c} is added to the momentum equation on the form \ref{eq:pde:mom:c}, we get

$[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}]+\frac{1}{\rho a}[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}]=0$

(Eq. 6.113)

If, instead, the continuity equation on the form \ref{eq:pde:cont:c} is subtracted from the momentum equation on the form \ref{eq:pde:mom:c}, we get

$[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}]+\frac{1}{\rho a}[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}]=0$

(Eq. 6.114)

Since $u=u(x,t)$, we have from the definition of a differential

$du=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}dx=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}\frac{dx}{dt}dt$

(Eq. 6.115)

Now, let $dx/dt=u+a$

$du=\frac{\partial u}{\partial t}dt+(u+a)\frac{\partial u}{\partial x}dt=[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}]dt$

(Eq. 6.116)

which is the change of $u$ in the direction $dx/dt=u+a$

In the same way

$dp=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}dx=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}\frac{dx}{dt}dt$

(Eq. 6.117)

and thus, in the direction $dx/dt=u+a$

$dp=\frac{\partial p}{\partial t}dt+(u+a)\frac{\partial p}{\partial x}dt=[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}]dt$

(Eq. 6.118)

If we go back and examine Eqn. \ref{eq:nonlin:a}, we see that Eqns. \ref{eq:du:b} and \ref{eq:dp:b} appear in the equation and thus it can now be rewritten as follows

$\frac{du}{dt}+\frac{1}{\rho a}\frac{dp}{dt}=0\Rightarrow du+\frac{dp}{\rho a}=0$

(Eq. 6.119)

Eqn. \ref{eq:nonlin:a:ode} applies along a $C^{+}$ characteristic, i.e., a line in the direction $dx/dt=u+a$ in $xt$-space and is called the compatibility equation along the $C^{+}$ characteristic. If we instead chose a $C^{-}$ characteristic, i.e., a line in the direction $dx/dt=u-a$ in $xt$-space, we get

$du=[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}]dt$

(Eq. 6.120)

$dp=[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}]dt$

(Eq. 6.121)

which can be identified as subsets of Eqn. \ref{eq:nonlin:b} and thus

$\frac{du}{dt}-\frac{1}{\rho a}\frac{dp}{dt}=0$

(Eq. 6.122)

In order to fulfil the relation above, either $du=dp=0$ or

$du-\frac{dp}{\rho a}=0$

(Eq. 6.123)

Eqn. \ref{eq:nonlin:b:ode} applies along a $C^{-}$ characteristic, i.e., a line in the direction $dx/dt=u-a$ in $xt$-space and is called the compatibility equation along the $C^{-}$ characteristic.

So, what we have done now is that we have have found paths through a point ($x_1$, $t_1$) along which the governing partial differential equations Eqns. \ref{eq:nonlin:a} and \ref{eq:nonlin:b} reduces to the ordinary differential equations \ref{eq:nonlin:a:ode} and \ref{eq:nonlin:b:ode}. The $C^{+}$ and $C^{-}$ characteristic lines are physically the paths of right- and left-running sound waves in the $xt$-plane.

Riemann Invariants

If the compatibility equations are integrated along respective characteristic line, i.e., integrate \ref{eq:nonlin:a:ode} along the $C^{+}$ characteristic and \ref{eq:nonlin:b:ode} along the $C^{-}$ characteristic, we get the Riemann invariants $J^{+}$ and $J^{-}$.

$J^{+}=u+\int \frac{dp}{\rho a}=const$

(Eq. 6.124)

$J^{-}=u-\int \frac{dp}{\rho a}=const$

(Eq. 6.125)

The Riemann invariants are constants along the associated characteristic line.

We have assumed isentropic flow and thus we may use the isentropic relations

$p=C_1{T}^{\gamma /(\gamma -1)}=C_2{a}^{2\gamma /(\gamma -1)}$

(Eq. 6.126)

where $C_1$ and $C_2$ are constants. Differentiating Eqn. \ref{eq:isentropic:a} gives

$dp=C_2\left(\frac{2\gamma }{\gamma -1}\right)a^{[2\gamma /(\gamma -1)-1]}da$

(Eq. 6.127)

Now, if we further assume the gas to be calorically perfect

$a^2=\gamma RT=\frac{\gamma p}{\rho }\Rightarrow \rho =\frac{\gamma p}{a^2}$

(Eq. 6.128)

Eqn. \ref{eq:isentropic:a} in \ref{eq:isentropic:c} gives

$\rho =C_2\gamma a^{[2\gamma /(\gamma -1)-2]}$

(Eq. 6.129)

and thus

$J^{+}=u+\int \frac{C_2\left(\frac{2\gamma }{\gamma -1}\right)a^{[2\gamma /(\gamma -1)-1]}}{C_2\gamma a^{[2\gamma /(\gamma -1)-2]}a}da=u+\left(\frac{2}{\gamma -1}\right)\int da$

(Eq. 6.130)

$J^{+}=u+\frac{2a}{\gamma -1}$

(Eq. 6.131)

$J^{-}=u-\frac{2a}{\gamma -1}$

(Eq. 6.132)

Eqns. \ref{eq:riemann:a:b} and \ref{eq:riemann:b:b} are the Riemann invariants for a calorically perfect gas. The Riemann invariants are constants along $C^{+}$ and $C^{-}$ characteristics and if the situation shown in Fig. \ref{fig:characteristics} appears, that fact can be used to calculate the flow velocity and speed of sound in the location ($x_1$, $t_1$).

$J^{+}+J^{-}=u+\frac{2a}{\gamma -1}+u-\frac{2a}{\gamma -1}=2u\Rightarrow u=\frac{1}{2}(J^{+}+J^{-})$

(Eq. 6.133)

$J^{+}=u+\frac{2a}{\gamma -1}=\frac{1}{2}(J^{+}+J^{-})+\frac{2a}{\gamma -1}\Rightarrow a=\frac{\gamma -1}{4}(J^{+}-J^{-})$

(Eq. 6.134)