# Nonlinear wave equation meeting notes

## Governing equation

The perfect fluid in $\mathbb{R}^{(3+1)}$ space-time is descibed by the relativistic Euler equation:
$$
\require{ams}\begin{equation}
\tag{1}\label{eq:governing}
\displaystyle \frac{\partial}{\partial t}\left(\frac{\rho+Pu^2/c^4}{1-u^2/c^2}\right)+\sum_{k=1}^3\frac{\partial}{\partial x_k}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u_k\right) = 0,\\
\displaystyle \frac{\partial}{\partial t}\left(\frac{\rho+Pu/c^2}{1-u^2/c^2}u_j\right)+\sum_{k=1}^3\frac{\partial}{\partial x_k}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u_ju_k+P\delta_{ij}\right) = 0, \hspace{5pt}j = 1,2,3,
\end{equation}$$

where $\rho$ is the density, $P = \rho \sigma^2$ is the pressure, $c$ is the speed of light, and $u = \sqrt{u_1^2+u_2^2+u_3^2}$ is the magnitude of the velocity vector $(u_i)_{i=1}^3$ in space,
Consider the $\rho = \rho(r, t), u = u(r,t)$, $r = |x|$, $u_j = \displaystyle\frac{x_j}{r}u(r, t)$

The spatial derivative can be written as :
$\begin{equation}
\displaystyle\frac{\partial}{\partial
x_k}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u_k\right) = \frac{\partial }{\partial x_k}\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)u_k+\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)\frac{\partial }{\partial x_k}u_k
\end{equation}$
With the property of sphereical symmetry, we have
$$
\frac{\partial }{\partial x_k}\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)u_k = \frac{\partial }{\partial r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)\frac{\partial r}{\partial x_k}u_k
$$
On the other hand, $\displaystyle\frac{\partial r}{\partial x_k} = \frac{x_k}{r}$, $\displaystyle\frac{\partial u_k}{\partial x_k} = \left(\frac{1}{r}-\frac{x_k^2}{r^3}\right)u+\frac{x_k^2}{r^2}\frac{\partial u}{\partial r}$ , therefore the equation can be written as
$\displaystyle \frac{\partial}{\partial
x_k}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u_k\right)$=$\displaystyle\frac{\partial }{\partial r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)\frac{x_k^2}{r^2}u+\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)\left(\frac{-x_k^2}{r^3}u+\frac{x_k^2}{r^2}\frac{\partial u}{\partial r }+\frac{1}{r}u\right)$.

Furthermore, $\eqref{eq:governing}$ becomes
$$\begin{equation}
\displaystyle \frac{\partial}{\partial t}\left(\frac{\rho+Pu^2/c^4}{1-u^2/c^2}\right)+\frac{\partial}{\partial r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u\right)+\frac{2}{r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u\right)= 0.
\end{equation}$$

In addition, with the property of spherical symmetry, the second equation of $\eqref{eq:governing}$ can also be rewritten.

The time derivative is
$\displaystyle\frac{\partial }{\partial t}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u_j\right) = \frac{\partial}{\partial t}\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)\frac{x_j}{r}u+\left(\frac{\rho+P/c^2}{1-u^2/c^2}\right)\frac{x_j}{r}\frac{\partial u}{\partial t}$.

The spatial derivative of $\frac{\partial u_j u_k}{\partial x_k}$follows from
$$
\left\{
\begin{array}{l}
\displaystyle\left(\frac{2x_k}{r^2}-2\frac{x_k^3}{r^4}\right)u^2+2\frac{x_k^3}{r^3}u\frac{\partial u}{\partial r}\mbox{, if }k = j\\
\displaystyle\left(\frac{x_j}{r^2}-2\frac{x_k^2x_j}{r^4}\right)u^2+2\frac{x_k^2x_j}{r^3}u\frac{\partial u}{\partial r}\mbox{, if }k \neq j
\end{array}
\right.
$$

Therefore we have the second equation of $\eqref{eq:governing}$ be written as
$$
\displaystyle\frac{\partial }{\partial t}\left(\displaystyle\frac{\rho+P/c^2}{1-u^2/c^2}u\right)+\frac{\partial }{\partial r}\left(\frac{\rho u^2+P}{1-u^2/c^2}\right)+\frac{2}{r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u^2\right) = 0
$$

Finally, the governing equation within sphreically symmtry
$$\require{ams}\tag{2}\label{eq:relEulerSph}\begin{equation}
\left\{
\begin{array}{cc}
\displaystyle \frac{\partial}{\partial t}\left(\frac{\rho+Pu^2/c^4}{1-u^2/c^2}\right)+\frac{\partial}{\partial r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u\right)-\frac{1}{r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u\right)= 0,\\
\displaystyle\frac{\partial }{\partial t}\left(\displaystyle\frac{\rho+P/c^2}{1-u^2/c^2}u\right)+\frac{\partial }{\partial r}\left(\frac{\rho u^2+P}{1-u^2/c^2}\right)-\frac{2}{r}\left(\frac{\rho+P/c^2}{1-u^2/c^2}u^2\right) = 0
\end{array}
\right.
\end{equation}
$$

## Steady State Solution
In this section we will consider the steady-state solution of $\eqref{eq:relEulerSph}$

First af all, we consider the solution of the first equation
$$
\frac{\rho+P/c^2}{1-u^2/c^2}u = \alpha(r_0) r,
$$
where $\displaystyle\alpha(r_0) = \frac{\rho(r_0)+P(r_0)/c^2}{1-u^2(r_0)/c^2}u(r_0).$

Next, after we substitute the solution to the second equation, we have
$$
\begin{equation}
\displaystyle \frac{\partial }{\partial r}\left(\alpha r u+P\right) = \frac{2}{r}(\alpha r u)\\
\Rightarrow\alpha r u_r+P_r = \alpha u.
\end{equation}
$$

Therefore
$$
\begin{equation}
\displaystyle\int_{r_0}^r\alpha s\frac{\partial u}{\partial s}ds+\int_{r_0}^r \frac{\partial P}{\partial s} ds = \alpha \int_{r_0}^ru(s)ds\\
\Rightarrow \alpha\left(ru(r)-r_0u(r_0)-\int_{r_0}^ru(s)ds\right)+P(r)-P(r_0) = \alpha \int_{r_0}^ru(s)ds\\
\end{equation}
$$
Let $\beta = \alpha r_0u(r_0)+P(r_0)$, we have
$$
\displaystyle 2\alpha \int_{r_0}^ru(s)ds = \alpha\left(ru(r)\right)+P(r)-\beta,
$$

which shows that the pressure can be written as a function of velocity $u$.

Furthermore, $\displaystyle\frac{\rho+P/c^2}{1-u^2/c^2}u = \alpha r$, we have
$$
\displaystyle\frac{\rho+\left(2\alpha \displaystyle\int_{r_0}^ru(s)ds -\alpha\left(ru(r)\right)+\beta\right)/c^2}{1-u^2/c^2}u = \alpha r\\
\Rightarrow
\rho u+ 2\alpha u/c\displaystyle\int_{r_0}^ru(s)/cds-\alpha(ru^2(r))/c^2+\beta u/c^2 = \alpha r-\alpha r u^2(r)/c^2\\
\Rightarrow
\rho u+ 2\alpha u/c\displaystyle\int_{r_0}^ru(s)/cds+\beta u/c^2 = \alpha r.
$$
Let $v = u/c$ a dimensionless variable, we have
$$\displaystyle\rho vc+2\alpha v\int_{r_0}^rvds+\beta v/c = \alpha r$$.
Since $v$ is never zero, we have
$$\displaystyle\rho c = \alpha r/v - 2\alpha\int_{r_0}^r v ds-\beta/c
$$

---
* Rmk:
Is there a physical meaning of $\rho c^2$ ?
If yes, can we use the physical meaning to interpreate the relation between density and pressure and velocity?
---

## Smoller/Temple
[Global Solutions of the Relativistic Euler Equations ](https://link.springer.com/content/pdf/10.1007/BF02096733.pdf)

Authors consider the 1+1 dimensional case with the case $P = \sigma^2\rho$, therefore $\eqref{eq:governing}$ have the following form
$$
\require{ams}
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\frac{\partial }{\partial t}\left(\rho\bigg[\left(\frac{\sigma^2+c^2}{c^2}\right)\frac{v^2}{c^2-v^2}+1\bigg]\right)+\frac{\partial }{\partial x}\left(\rho\bigg[(\sigma^2+c^2)\frac{v^2}{c^2-v^2} \bigg] \right) = 0\\
\displaystyle\frac{\partial }{\partial t}\left(\rho\bigg[(\sigma^2+c^2)\frac{v}{c^2-v^2} \bigg] \right)+\frac{\partial }{\partial x}\left(\rho\bigg[(\sigma^2+c^2)\frac{v^2}{c^2-v^2} +\sigma^2 \bigg] \right) = 0
\end{array}
\right.
\end{equation}
$$

### Proposition 1
Considering a map $F:(\rho, v) \rightarrow(U_1, U_2)$, where $\displaystyle(U_1, U_2) = \left(\rho\bigg[\left(\frac{\sigma^2+c^2}{c^2}\right)\frac{v^2}{c^2-v^2}+1\bigg], \rho\bigg[(\sigma^2+c^2)\frac{v}{c^2-v^2} \bigg] \right)$
is 1-1, and the determinant of its Jacobian is nonzero in the region $\rho>0, |v|<c$.

$Proof.$
(1) 1-1
Suppose not, let $(\rho_1, v_1) \neq (\rho_2, v_2)$, since $\frac{\partial U_i}{\partial \rho} \neq 0$ therefore, we may assume $v_1\neq v_2.$

We have
$$
\rho_1\bigg[\left(\frac{\sigma^2+c^2}{c^2}\right)\frac{v_1^2}{c^2-v_1^2}+1\bigg] = \rho_2\bigg[\left(\frac{\sigma^2+c^2}{c^2}\right)\frac{v_2^2}{c^2-v_2^2}+1\bigg]
$$
$$
\rho_1\bigg[(\sigma^2+c^2)\frac{v_1}{c^2-v_1^2} \bigg] = \rho_2\bigg[(\sigma^2+c^2)\frac{v_2}{c^2-v_2^2} \bigg]
$$

With the two relationships, we have
$$
\displaystyle\frac{\bigg[\left(\frac{\sigma^2+c^2}{c^2}\right)\frac{v_1^2}{c^2-v_1^2}+1\bigg]}{\bigg[\left(\frac{\sigma^2+c^2}{c^2}\right)\frac{v_2^2}{c^2-v_2^2}+1\bigg]} = \frac{\bigg[(\sigma^2+c^2)\frac{v_1}{c^2-v_1^2} \bigg]}{\bigg[(\sigma^2+c^2)\frac{v_2}{c^2-v_2^2} \bigg]}
$$

We have
$$
\frac{(\sigma^2+c^2)v_1^2+c^4-v_1^2c^2}{(\sigma^2+c^2)v_2^2+c^4-v_2^2c^2} = \frac{v_1}{v_2}
$$
and
$$
(\sigma^2+c^2)v_1^2v_2+c^4v_2-v_1^2c^2v_2 = (\sigma^2+c^2)v_2^2v_1+c^4v_1-v_2^2c^2v_1.
$$
, which leads $(v_1-v_2)(\sigma^2 v_1v_2-c^4) = 0$.
In our assumption, $(v_1-v_2)\neq 0$, which forces $\sigma^2 v_1v_2-c^4 = 0$.

Finally, we have $v_1v_2 = c^4/\sigma^2$, base on geometry inequality, $\frac{v_1+v_2}{2}\geq c^2/\sigma>c$, which is impossible, since neither $v_1$ nor $v_2$ can exceed $c$.

---
#### Some results in Smoller and Temple's paper

First, in the paper, they applied Lax's result to determine the speed of shock wave and behaviors of rarefection waves.

$\underline{\mbox{Theorem 1}}:$

If we assume $v = v(\rho)$, then we have the following differential equation:
$$
\displaystyle\frac{dv}{d\rho} = \pm\frac{\sqrt{p'}(c^2-v^2)}{p+\rho c^2},
$$
where $'$ represents the derivative w.r.t $\rho$.

$\underline{\mbox{Theorem 2}}$
Assume $(\rho_L, v_L)$ and $(\rho, v)$ satisfy jump condition. Then we have the following relation:
$$
\displaystyle\frac{\rho}{\rho_L} = 1+\beta+\left\{1+\pm\sqrt{1+\frac{2}{\beta}}\right\},
$$
where $\beta = \beta(v, v_L) = \displaystyle\frac{(\sigma^2+c^2)^2}{2\sigma^2}\frac{(v-v_L)^2}{(c^2-v^2)(c^2-v_L)^2}$

The result of Theorem 1 provides us some information w.r.t rarefection waves:
|Family|sign|riemann invariants|eigenvalue(determine label)|eigenvectors|
|--|--|--|--|--|
|$R_1$|negative|$\displaystyle \frac{1}{2}\ln\left(\frac{c+v}{c-v}\right)+c\int_1^\rho\frac{\sqrt{p'(s)}}{p(s)+c^2s} ds$|$\lambda_1 = \displaystyle\frac{v-\sqrt{p'}}{1-\frac{v\sqrt{p'}}{c^2}}$|$\left(\displaystyle\frac{-c}{c^2-v^2}, \frac{c\sqrt{p'}}{p+\rho c^2}\right)$|
|$R_2$|positive|$\displaystyle \frac{1}{2}\ln\left(\frac{c+v}{c-v}\right)-c\int_1^\rho\frac{\sqrt{p'(s)}}{p(s)+c^2s} ds$|$\lambda_2 = \displaystyle\frac{v+\sqrt{p'}}{1+\frac{v\sqrt{p'}}{c^2}}$|$\left(\displaystyle\frac{c}{c^2-v^2}, \frac{c\sqrt{p'}}{p+\rho c^2}\right)$|

|Shock Family|equation|
|--|--|
|$S_1$|$\displaystyle\frac{\rho}{\rho_L} = 1+\beta+\left\{1+\sqrt{1+\frac{2}{\beta}}\right\}, \rho>\rho_L$|
|$S_2$|$\displaystyle\frac{\rho}{\rho_L} = 1+\beta+\left\{1-\sqrt{1+\frac{2}{\beta}}\right\}, \rho<\rho_L$|

The following figure shows the rarefection family and shock family where $(\rho_L, v_L) = (3,0).$

In addition, we show some states as follows: