My online note for the Euler equations in the useful form.
Governing Equations
The Euler equations consist of the mass conservation
(1)\[ \begin{align}\begin{aligned}\newcommand{\bvec}[1]{\mathbf{#1}}
\newcommand{\defeq}{\buildrel{\text{def}}\over{=}}
\newcommand{\dpd}[3][]{\mathinner{
\dfrac{\partial{^{#1}}#2}{\partial{#3^{#1}}}
}}\\\frac{\partial\rho}{\partial t} + \frac{\partial\rho v_j}{\partial x_j}
= 0\end{aligned}\end{align} \]
momentum conservation
(2)\[\frac{\partial\rho v_i}{\partial t}
+ \frac{\partial\rho v_iv_j}{\partial x_j}
= \frac{\partial p}{\partial x_j} + \rho b_i\]
and energy conservation
(3)\[\frac{\partial}{\partial t}
\left[\rho\left( e + \frac{v_k^2}{2} \right)\right]
+ \frac{\partial}{\partial x_j}
\left[\rho\left( e + \frac{v_k^2}{2} \right)v_j\right]
= \rho \dot{q} - \frac{\partial pv_j}{\partial x_j} + \rho b_jv_j\]
Einstein’s index summation convention was used.
Equations (1), (2), and (3) aren’t
closed even if we choose \(\bvec{b}\) and \(\dot{q}\) as given. We
have 5 equtions but 6 unknowns (\(\rho\), \(\bvec{v}\), \(p\), and
\(e\)). To close the system of equations, I use the equation of state:
Internal energy is related to temperature:
(5)\[e = c_vT = \frac{RT}{\gamma-1} = \frac{1}{\gamma-1}\frac{p}{\rho}\]
With the additional two equations (Eqs. (4) and
(5)) and one variable \(T\), the equations are closed.
Vector Flux Function
Define the conservation variables:
(6)\[\begin{split}\bvec{u} \defeq \left(\begin{array}{c}
u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5
\end{array}\right) = \left(\begin{array}{c}
\rho \\ \rho v_1 \\ \rho v_2 \\ \rho v_3 \\
\rho\left(e+\frac{v_k^2}{2}\right)
\end{array}\right)\end{split}\]
Aided by writing the pressure with \(\bvec{u}\):
\[p = (\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)\]
the conservation equations (Eqs. (1), (2), and
(3)) can be cast to use only \(\bvec{u}\):
(7)\[\frac{\partial u_1}{\partial t}
+ \frac{\partial u_2}{\partial x_1}
+ \frac{\partial u_3}{\partial x_2}
+ \frac{\partial u_4}{\partial x_3} = 0\]
(8)\[\begin{split}\begin{aligned} &\frac{\partial u_2}{\partial t}
+ \frac{\partial}{\partial x_1}\left(\frac{u_2^2}{u_1}\right)
+ \frac{\partial}{\partial x_2}\left(\frac{u_2u_3}{u_1}\right)
+ \frac{\partial}{\partial x_3}\left(\frac{u_2u_4}{u_1}\right) = \\
&\quad -\frac{\partial}{\partial x_1}\left[
(\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)
\right] + b_1u_1
\end{aligned}\end{split}\]
(9)\[\begin{split}\begin{aligned} &\frac{\partial u_3}{\partial t}
+ \frac{\partial}{\partial x_1}\left(\frac{u_2u_3}{u_1}\right)
+ \frac{\partial}{\partial x_2}\left(\frac{u_3^2}{u_1}\right)
+ \frac{\partial}{\partial x_3}\left(\frac{u_3u_4}{u_1}\right) = \\
&\quad -\frac{\partial}{\partial x_2}\left[
(\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)
\right] + b_2u_1
\end{aligned}\end{split}\]
(10)\[\begin{split}\begin{aligned} &\frac{\partial u_4}{\partial t}
+ \frac{\partial}{\partial x_1}\left(\frac{u_2u_4}{u_1}\right)
+ \frac{\partial}{\partial x_2}\left(\frac{u_3u_4}{u_1}\right)
+ \frac{\partial}{\partial x_3}\left(\frac{u_4^2}{u_1}\right) = \\
&\quad -\frac{\partial}{\partial x_3}\left[
(\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)
\right] + b_3u_1
\end{aligned}\end{split}\]
(11)\[\begin{split}\begin{aligned} &\frac{\partial u_5}{\partial t}
+ \frac{\partial}{\partial x_1}\left(\frac{u_2u_5}{u_1}\right)
+ \frac{\partial}{\partial x_2}\left(\frac{u_3u_5}{u_1}\right)
+ \frac{\partial}{\partial x_3}\left(\frac{u_4u_5}{u_1}\right) = \\
&\quad - \frac{\partial}{\partial x_1}\left[
(\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)
\frac{u_2}{u_1}
\right] \\
&\quad - \frac{\partial}{\partial x_2}\left[
(\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)
\frac{u_3}{u_1}
\right] \\
&\quad - \frac{\partial}{\partial x_3}\left[
(\gamma-1)\left(u_5 - \frac{u_2^2+u_3^2+u_4^2}{2u_1}\right)
\frac{u_4}{u_1}
\right]
+ \rho\dot{q} + b_1u_2 + b_2u_3 + b_3u_4
\end{aligned}\end{split}\]
Then organize Eqs. (7) – (11) into a vector form:
(12)\[\frac{\partial\bvec{u}}{\partial t}
+ \sum_{\mu=1}^3 \frac{\partial\bvec{f}^{(\mu)}}{\partial x_{\mu}}
= \bvec{s}\]
The flux functions are defined as:
(13)\[\begin{split}\bvec{f}^{(1)} &= \left(\begin{array}{c}
f^{(1)}_1 \\ f^{(1)}_2 \\ f^{(1)}_3 \\ f^{(1)}_4 \\ f^{(1)}_5
\end{array}\right) \defeq \left(\begin{array}{l}
u_2 \\
(\gamma-1)u_5 - \frac{\gamma-3}{2}\frac{u_2^2}{u_1}
- \frac{\gamma-1}{2}\frac{u_3^2}{u_1}
- \frac{\gamma-1}{2}\frac{u_4^2}{u_1} \\
\frac{u_2u_3}{u_1} \\
\frac{u_2u_4}{u_1} \\
\gamma\frac{u_2u_5}{u_1}
- \frac{\gamma-1}{2}\frac{u_2^2+u_3^2+u_4^2}{u_1}\frac{u_2}{u_1}
\end{array}\right)\end{split}\]
(14)\[\begin{split}\bvec{f}^{(2)} &= \left(\begin{array}{c}
f^{(2)}_1 \\ f^{(2)}_2 \\ f^{(2)}_3 \\ f^{(2)}_4 \\ f^{(2)}_5
\end{array}\right) \defeq \left(\begin{array}{l}
u_3 \\
\frac{u_2u_3}{u_1} \\
(\gamma-1)u_5 - \frac{\gamma-1}{2}\frac{u_2^2}{u_1}
- \frac{\gamma-3}{2}\frac{u_3^2}{u_1}
- \frac{\gamma-1}{2}\frac{u_4^2}{u_1} \\
\frac{u_3u_4}{u_1} \\
\gamma\frac{u_3u_5}{u_1}
- \frac{\gamma-1}{2}\frac{u_2^2+u_3^2+u_4^2}{u_1}\frac{u_3}{u_1}
\end{array}\right)\end{split}\]
(15)\[\begin{split}\bvec{f}^{(3)} &= \left(\begin{array}{c}
f^{(3)}_1 \\ f^{(3)}_2 \\ f^{(3)}_3 \\ f^{(3)}_4 \\ f^{(3)}_5
\end{array}\right) \defeq \left(\begin{array}{l}
u_4 \\
\frac{u_2u_4}{u_1} \\
\frac{u_3u_4}{u_1} \\
(\gamma-1)u_5 - \frac{\gamma-1}{2}\frac{u_2^2}{u_1}
- \frac{\gamma-1}{2}\frac{u_3^2}{u_1}
- \frac{\gamma-3}{2}\frac{u_4^2}{u_1} \\
\gamma\frac{u_4u_5}{u_1}
- \frac{\gamma-1}{2}\frac{u_2^2+u_3^2+u_4^2}{u_1}\frac{u_4}{u_1}
\end{array}\right)\end{split}\]
At the right-hand side, the source term is
(16)\[\begin{split}\bvec{s} = \left(\begin{array}{c}
s_1 \\ s_2 \\ s_3 \\ s_4 \\ s_5
\end{array}\right) \defeq \left(\begin{array}{l}
0 \\ b_1u_1 \\ b_2u_1 \\ b_3u_3 \\ \dot{q}u_1 + b_1u_2 + b_2u_3 + b_3u_4
\end{array}\right)\end{split}\]
Quasi-linear System Equation
Expand Eq. (12) to an index form:
(17)\[\frac{\partial u_m}{\partial t}
+ \sum_{\mu=1}^3 \frac{\partial f^{(\mu)}_m}{\partial x_{\mu}}
= s_m, \quad m = 1, \ldots, 5\]
Because we want to construct an inviscid baseline solver, later we will drop
the source term from Eq. (17).
Define
\[\begin{split}u_{mt} &\defeq \dpd{u_m}{t}, \\
u_{mx_{\mu}} &\defeq \dpd{u_m}{x_{\mu}}, \\
f^{(\mu)}_{m,l} &\defeq \dpd{f^{(\mu)}_m}{u_l}\end{split}\]
where \(\mu = 1, 2, 3,\) and \(m, l = 1, 2, \ldots, 5\).
Aided by the above definition, we rewrite the equation to a matrix-vector form:
(18)\[\dpd{\bvec{u}}{t} + \sum_{\mu=1}^3
\mathrm{A}^{(\mu)} \dpd{\bvec{u}}{x_{\mu}} = 0\]
where \(\mathrm{A}^{(1)}\), \(\mathrm{A}^{(2)}\), and
\(\mathrm{A}^{(3)}\) are Jacobian matrices
(\(\left[\mathrm{A}^{(\mu)}\right]_{ml} \defeq f^{(\mu)}_{m,l}\)).
Components of the Jacobian matrices are tabulated.
Constant components:
(19)\[\begin{split}f^{(1)}_{1,1} &= f^{(1)}_{1,3} = f^{(1)}_{1,4} = f^{(1)}_{1,5} = \\
f^{(2)}_{1,1} &= f^{(2)}_{1,2} = f^{(2)}_{1,4} = f^{(2)}_{1,5} = \\
f^{(3)}_{1,1} &= f^{(3)}_{1,2} = f^{(3)}_{1,3} = f^{(3)}_{1,5} = 0, \\
f^{(1)}_{1,2} &= f^{(2)}_{1,3} = f^{(3)}_{1,4} = 1\end{split}\]
Non-constant components of \(A^{(1)}\):
(20)\[\begin{split}f^{(1)}_{2,1} &= \frac{\gamma-3}{2}\frac{u_2^2}{u_1^2}
+ \frac{\gamma-1}{2}\frac{u_3^2}{u_1^2}
+ \frac{\gamma-1}{2}\frac{u_4^2}{u_1^2}, \\
f^{(1)}_{2,2} &= -(\gamma-3)\frac{u_2}{u_1}, \quad
f^{(1)}_{2,3} = -(\gamma-1)\frac{u_3}{u_1}, \quad
f^{(1)}_{2,4} = -(\gamma-1)\frac{u_4}{u_1}, \quad
f^{(1)}_{2,5} = \gamma-1, \\
f^{(1)}_{3,1} &= -\frac{u_2u_3}{u_1^2}, \quad
f^{(1)}_{3,2} = \frac{u_3}{u_1}, \quad
f^{(1)}_{3,3} = \frac{u_2}{u_1}, \quad
f^{(1)}_{3,4} = f^{(1)}_{3,5} = 0, \\
f^{(1)}_{4,1} &= -\frac{u_2u_4}{u_1^2}, \quad
f^{(1)}_{4,2} = \frac{u_4}{u_1}, \quad
f^{(1)}_{4,4} = \frac{u_2}{u_1}, \quad
f^{(1)}_{4,3} = f^{(1)}_{4,5} = 0, \\
f^{(1)}_{5,1} &= -\gamma\frac{u_2u_5}{u_1^2}
+ (\gamma-1)\frac{u_2^2+u_3^2+u_4^2}{u_1^2}\frac{u_2}{u_1}, \quad
f^{(1)}_{5,2} = \gamma\frac{u_5}{u_1}
- \frac{\gamma-1}{2}\frac{3u_2^2 + u_3^2 + u_4^2}{u_1^2}, \\
f^{(1)}_{5,3} &= -(\gamma-1)\frac{u_2u_3}{u_1^2}, \quad
f^{(1)}_{5,4} = -(\gamma-1)\frac{u_2u_4}{u_1^2}, \quad
f^{(1)}_{5,5} = \gamma\frac{u_2}{u_1}\end{split}\]
Non-constant components of \(A^{(2)}\):
(21)\[\begin{split}f^{(2)}_{2,1} &= -\frac{u_2u_3}{u_1^2}, \quad
f^{(2)}_{2,2} = \frac{u_3}{u_1}, \quad
f^{(2)}_{2,3} = \frac{u_2}{u_1}, \quad
f^{(2)}_{2,4} = f^{(2)}_{2,5} = 0, \\
f^{(2)}_{3,1} &= \frac{\gamma-1}{2}\frac{u_2^2}{u_1^2}
+ \frac{\gamma-3}{2}\frac{u_3^2}{u_1^2}
+ \frac{\gamma-1}{2}\frac{u_4^2}{u_1^2}, \\
f^{(2)}_{3,2} &= -(\gamma-1)\frac{u_2}{u_1}, \quad
f^{(2)}_{3,3} = -(\gamma-3)\frac{u_3}{u_1}, \quad
f^{(2)}_{3,4} = -(\gamma-1)\frac{u_4}{u_1}, \quad
f^{(2)}_{3,5} = \gamma-1, \\
f^{(2)}_{4,1} &= -\frac{u_3u_4}{u_1^2}, \quad
f^{(2)}_{4,3} = \frac{u_4}{u_1}, \quad
f^{(2)}_{4,4} = \frac{u_3}{u_1}, \quad
f^{(2)}_{4,2} = f^{(2)}_{4,5} = 0, \\
f^{(2)}_{5,1} &= -\gamma\frac{u_3u_5}{u_1^2}
+ (\gamma-1)\frac{u_2^2+u_3^2+u_4^2}{u_1^2}\frac{u_3}{u_1}, \quad
f^{(2)}_{5,3} = \gamma\frac{u_5}{u_1}
- \frac{\gamma-1}{2}\frac{u_2^2 + 3u_3^2 + u_4^2}{u_1^2}, \\
f^{(2)}_{5,2} &= -(\gamma-1)\frac{u_2u_3}{u_1^2}, \quad
f^{(2)}_{5,4} = -(\gamma-1)\frac{u_3u_4}{u_1^2}, \quad
f^{(2)}_{5,5} = \gamma\frac{u_3}{u_1}\end{split}\]
Non-constant components of \(A^{(3)}\):
(22)\[\begin{split}f^{(3)}_{2,1} &= -\frac{u_2u_4}{u_1^2}, \quad
f^{(3)}_{2,2} = \frac{u_4}{u_1}, \quad
f^{(3)}_{2,4} = \frac{u_2}{u_1}, \quad
f^{(3)}_{2,3} = f^{(3)}_{2,5} = 0, \\
f^{(3)}_{3,1} &= -\frac{u_3u_4}{u_1^2}, \quad
f^{(3)}_{3,3} = \frac{u_4}{u_1}, \quad
f^{(3)}_{3,4} = \frac{u_3}{u_1}, \quad
f^{(3)}_{3,2} = f^{(3)}_{3,5} = 0, \\
f^{(3)}_{4,1} &= \frac{\gamma-1}{2}\frac{u_2^2}{u_1^2}
+ \frac{\gamma-1}{2}\frac{u_3^2}{u_1^2}
+ \frac{\gamma-3}{2}\frac{u_4^2}{u_1^2}, \\
f^{(3)}_{4,2} &= -(\gamma-1)\frac{u_2}{u_1}, \quad
f^{(3)}_{4,3} = -(\gamma-1)\frac{u_3}{u_1}, \quad
f^{(3)}_{4,4} = -(\gamma-3)\frac{u_4}{u_1}, \quad
f^{(3)}_{4,5} = \gamma-1, \\
f^{(3)}_{5,1} &= -\gamma\frac{u_4u_5}{u_1^2}
+ (\gamma-1)\frac{u_2^2+u_3^2+u_4^2}{u_1^2}\frac{u_4}{u_1}, \quad
f^{(3)}_{5,4} = \gamma\frac{u_5}{u_1}
- \frac{\gamma-1}{2}\frac{u_2^2 + u_3^2 + 3u_4^2}{u_1^2}, \\
f^{(3)}_{5,2} &= -(\gamma-1)\frac{u_2u_4}{u_1^2}, \quad
f^{(3)}_{5,3} = -(\gamma-1)\frac{u_3u_4}{u_1^2}, \quad
f^{(3)}_{5,5} = \gamma\frac{u_4}{u_1}\end{split}\]