Finite non-linear waves

\section{Finite Nonlinear Waves}

\noindent Starting point: the governing flow equations on partial differential form\\

\noindent Continuity equation:

\begin{equation} \frac{\partial \rho}{\partial t}+u\frac{\partial \rho}{\partial x}+\rho\frac{\partial u}{\partial x}=0 \label{eq:pde:cont} \end{equation}\\

\noindent Momentum equation:

\begin{equation} \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho}\frac{\partial p}{\partial x}=0 \label{eq:pde:mom} \end{equation}\\

\noindent 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\\

\begin{equation*} d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp+\left(\frac{\partial \rho}{\partial s}\right)_p ds \end{equation*}\\

\noindent Assuming isentropic flow $ds=0$ gives\\

\begin{equation*} d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp \end{equation*}\\

\begin{equation} \begin{aligned} &\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{aligned} \label{eq:rhotop} \end{equation}\\

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

\begin{equation} \frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}+\rho a^2\frac{\partial u}{\partial x}=0 \label{eq:pde:cont:b} \end{equation}\\

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

\begin{equation} \frac{1}{\rho a}\left(\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}\right)+a\frac{\partial u}{\partial x}=0 \label{eq:pde:cont:c} \end{equation}\\

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

\begin{equation} \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho a}\left(a\frac{\partial p}{\partial x}\right)=0 \label{eq:pde:mom:c} \end{equation}\\

\noindent 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\\

\begin{equation} \left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]=0 \label{eq:nonlin:a} \end{equation}\\

\noindent 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\\

\begin{equation} \left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]=0 \label{eq:nonlin:b} \end{equation}\\

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

\begin{equation} 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 \label{eq:du} \end{equation}\\

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

\begin{equation} du=\frac{\partial u}{\partial t}dt+(u+a)\frac{\partial u}{\partial x}dt=\left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]dt \label{eq:du:b} \end{equation}\\

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

\noindent In the same way\\

\begin{equation} 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 \label{eq:dp} \end{equation}\\

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

\begin{equation} dp=\frac{\partial p}{\partial t}dt+(u+a)\frac{\partial p}{\partial x}dt=\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]dt \label{eq:dp:b} \end{equation}\\

\noindent 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\\

\begin{equation} \frac{du}{dt}+\frac{1}{\rho a}\frac{dp}{dt}=0\Rightarrow du+\frac{dp}{\rho a}=0 \label{eq:nonlin:a:ode} \end{equation}\\

\noindent 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\\

\begin{equation} du=\left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]dt \label{eq:du:c} \end{equation}\\

\begin{equation} dp=\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]dt \label{eq:dp:c} \end{equation}\\

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

\begin{equation*} \frac{du}{dt}-\frac{1}{\rho a}\frac{dp}{dt}=0 \end{equation*}\\

\noindent In order to fulfill the relation above, either $du=dp=0$ or\\

\begin{equation} du-\frac{dp}{\rho a}=0 \label{eq:nonlin:b:ode} \end{equation}\\

\noindent 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.\\

\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter07/pdf/characteristic-lines.pdf} \caption{Characteristic lines through a point ($x_1$,$t_1$)} \label{fig:characteristics} \end{center} \end{figure}

\noindent 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.\\

\subsection{Riemann Invariants}

\noindent 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 $J^-$.\\

\begin{equation} J^+=u+\int\frac{dp}{\rho a}=const \label{eq:riemann:a} \end{equation}\\

\begin{equation} J^-=u-\int\frac{dp}{\rho a}=const \label{eq:riemann:b} \end{equation}\\

\noindent The Riemann invariants are constants along the associated characteristic line.\\

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

\begin{equation} p=C_1T^{\gamma/(\gamma-1)}=C_2a^{2\gamma/(\gamma-1)} \label{eq:isentropic:a} \end{equation}\\

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

\begin{equation} dp=C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}da \label{eq:isentropic:b} \end{equation}\\

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

\begin{equation} a^2=\gamma RT=\frac{\gamma p}{\rho}\Rightarrow \rho=\frac{\gamma p}{a^2} \label{eq:isentropic:c} \end{equation}\\

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

\begin{equation} \rho=C_2\gamma a^{[2\gamma/(\gamma-1)-2]} \label{eq:isentropic:d} \end{equation}\\

\noindent and thus\\

\begin{equation*} 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 \end{equation*}\\

\begin{equation} J^+=u+\frac{2a}{\gamma-1} \label{eq:riemann:a:b} \end{equation}\\

\begin{equation} J^-=u-\frac{2a}{\gamma-1} \label{eq:riemann:b:b} \end{equation}\\

\noindent 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 ($x_1$,$t_1$).\\

\begin{equation} J^++J^-=u+\frac{2a}{\gamma-1}+u-\frac{2a}{\gamma-1}=2u\Rightarrow u=\frac{1}{2}(J^++J^-) \label{eq:riemann:evaluation:a} \end{equation}\\

\begin{equation} J^+=u+\frac{2a}{\gamma-1}=\frac{1}{2}(J^++J^-)+\frac{2a}{\gamma-1}\Rightarrow a=\frac{\gamma-1}{4}(J^+-J^-) \label{eq:riemann:evaluation:b} \end{equation}\\