28.6.2.2.1. Isothermal compressible flow - 2d
Conservative variables: \((\rho, \vec{m})\)
Physical variables: e.g. \((\rho, \vec{u})\), \((P, \vec{u})\),…
In isothermal compressible flow model, pressure is a function of density only and
\[d p = a^2 d \rho \ ,\]
with constant \(a\), speed of sound.
Conservative form reads
\[\begin{split}\begin{cases}
\partial_t \rho + \nabla \cdot \mathbf{m} = 0 \\
\partial_t \mathbf{m} + \nabla \cdot \left[ \frac{\mathbf{m}\otimes\mathbf{m}}{\rho} + \rho a^2 \mathbf{I} \right] = 0 \\
\end{cases}\end{split}\]
Convective form in a 2-dimensional domain reads
\[\begin{split}
\partial_t \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} +
\begin{bmatrix}
\cdot & 1 & \cdot \\
a^2 - \frac{m_x m_x}{\rho^2} & \frac{2 m_x}{\rho} & \cdot \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} \\
\end{bmatrix} \partial_x \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} +
\begin{bmatrix}
\cdot & \cdot & 1 \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} \\
a^2 - \frac{m_y m_y}{\rho^2} & \cdot & \frac{2 m_y}{\rho} \\
\end{bmatrix} \partial_y \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} = \mathbf{0}
\end{split}\]
\[\begin{split}\begin{aligned}
0 & = \rho_{/t} + m_{x/x} + m_{y/y} \\
0 & = m_{i/t} + \partial_{j} \left( \frac{m_j m_i}{\rho} + \rho a^2 \delta_{ij} \right) \ .
\end{aligned}\end{split}\]
Thus,
\[\begin{split}\begin{aligned}
\mathbf{A}_{\hat{\mathbf{n}}}
& =
\begin{bmatrix}
\cdot & n_x & n_y \\
a^2 n_x - u_n u_x & u_x n_x + u_n & u_x n_y \\
a^2 n_y - u_n u_y & u_y n_x & u_n + u_y n_y \\
\end{bmatrix} = \\
& =
\begin{bmatrix}
0 & \mathbf{n}^T \\
a^2 \mathbf{n} - ( \mathbf{n}^T \mathbf{u} ) \mathbf{u} & \mathbf{n}^T \mathbf{u} \mathbf{I} + \mathbf{u} \mathbf{n}^T
\end{bmatrix} = \\
& =
\begin{bmatrix}
0 & \hat{\mathbf{n}}^T \\
a^2 \mathbf{n} - ( \mathbf{n} \cdot \mathbf{u} ) \mathbf{u} & ( \mathbf{n} \cdot \mathbf{u} ) \mathbb{I} + \mathbf{u} \otimes \mathbf{n}
\end{bmatrix} \ .
\end{aligned}\end{split}\]
The eigenvalue decomposition of the matrix \(\mathbf{A}_{\hat{\mathbf{n}}}\) follows from
\[\begin{split}\begin{aligned}
0 & = \left| \mathbf{A}_{\hat{\mathbf{n}}} - s \mathbf{I} \right| =
\left|
\begin{bmatrix}
-s & \mathbf{n}^T \\
a^2 \mathbf{n} - u_n \mathbf{u} & ( u_n - s ) \mathbf{I} + \mathbf{u} \mathbf{n}^T
\end{bmatrix}
\right| = \\
& = -s ( u_n + u_x n_x - s ) ( u_n + u_y n_y - s ) + n_x u_x n_y ( a^2 n_y - u_n u_y ) + n_y u_y n_x ( a^2 n_x - u_n u_x ) + \\
& - ( a^2 n_y - u_n u_y ) ( u_x n_x + u_n - s) n_y
- ( a^2 n_x - u_n u_x ) ( u_y n_y + u_n - s) n_x
- s u_x u_y n_x n_y = \\
& =
s^3 ( - 1 ) + \\
& \quad + s^2 ( u_n + u_y n_y + u_n + u_x n_x ) + \\
& \quad + s ( - ( u_n + u_x n_x ) ( u_n + u_y n_y ) + n_y ( a^2 n_y - u_n u_y ) + n_x ( a^2 n_x - u_n u_x ) - u_x n_x u_y n_y ) + \\
& \quad + 1 \cdot ( n_x u_x n_y ( a^2 n_y - u_n u_y ) + n_y u_y n_x ( a^2 n_x - u_n u_x ) - n_y ( a^2 n_y - u_n u_y )( u_x n_x + u_n ) - n_x ( a^2 n_x - u_n u_x )( u_y n_y + u_n ) ) = \\
& = - s^3 + 3 u_n s^2 + s ( - u_n^2 - u_n (u_x n_x + u_y n_y) - u_x n_x u_y n_y - u_n u_y n_y - u_n u_x n_x + a^2 - u_x n_x u_y n_y ) + \\
& \quad + ( a^2 \left( n_y^2 u_x n_x + n_x^2 n_y u_y - n_y^2 u_x n_x - n_y^2 u_n - n_x^2 u_y n_y - u_n n_x^2 \right) - u_n u_x n_x u_y n_y - u_n u_x n_x u_y n_y +u_n u_x n_x u_y n_y + u_n^2 u_y n_y + u_n u_x n_x u_y n_y + u_n^2 u_x n_x ) = \\
& = - s^3 + 3 u_n s^2 + s ( a^2 - 3 u_n^2 ) - u_n ( a^2 - u_n^2 ) = \\
& = - ( s - u_n )^3 + ( s - u_n ) a^2 = \\
& = - ( s - u_n ) ( ( s - u_n )^2 - a^2 ) \ .
\end{aligned}\end{split}\]
Thus the eigenvalues of the matrix \(\mathbf{A}_{\hat{\mathbf{n}}}\) are
\[s_{1,3} = u_n \mp a \quad , \quad s_2 = u_n \ ,\]
and the right and left eigenvectors follow
\[\begin{split}\begin{aligned}
\mathbf{R} & = \\
\mathbf{L} & = \\
\end{aligned}\end{split}\]
For \(s_{1,3} = u_n \mp a\),
\[\begin{split}
\mathbf{0} = \begin{bmatrix} -u_n \pm a & \mathbf{n}^T \\ a^2 \mathbf{n} - u_n \mathbf{u} & \pm a \mathbf{I} + \mathbf{u} \mathbf{n}^T \end{bmatrix}
\begin{bmatrix} \hat{\rho} \\ \hat{\mathbf{m}} \end{bmatrix}
\end{split}\]
From the first equation it follows \(\hat{m}_n = ( u_n \mp a ) \hat{\rho}\), and thus the momentum equation gives
\[\begin{split}\begin{aligned}
\mathbf{0}
& = \hat{\rho} ( a^2 \mathbf{n} - u_n \mathbf{u} ) \pm a \hat{\mathbf{m}} + \mathbf{u} \hat{m}_n = \\
& = \hat{\rho} ( a^2 \mathbf{n} - u_n \mathbf{u} ) \pm a \hat{\mathbf{m}} + \mathbf{u} ( u_n \mp a ) \hat{\rho} = \\
& = \hat{\rho} ( a^2 \mathbf{n} \mp a \mathbf{u} ) \pm a \hat{\mathbf{m}} \ ,
\end{aligned}\end{split}\]
and thus
\[\hat{\mathbf{m}}_{1,3} = \hat{\rho}_{1,3} ( \mathbf{u} \mp a \mathbf{n} ) \ . \]
For \(s_{2} = u_n\),
\[\begin{split}
\mathbf{0} = \begin{bmatrix} -u_n & \mathbf{n}^T \\ a^2 \mathbf{n} - u_n \mathbf{u} & \mathbf{u} \mathbf{n}^T \end{bmatrix}
\begin{bmatrix} \hat{\rho} \\ \hat{\mathbf{m}} \end{bmatrix}
\end{split}\]
From the first equation it follows \(\hat{m}_n = u_n \hat{\rho}\), and thus the momentum equation gives
\[\begin{split}\begin{aligned}
\mathbf{0}
& = \hat{\rho} ( a^2 \mathbf{n} - u_n \mathbf{u} ) + \mathbf{u} \, \hat{m}_n = \\
& = \hat{\rho} ( a^2 \mathbf{n} - u_n \mathbf{u} ) + \mathbf{u} \, u_n \, \hat{\rho} = \\
& = \hat{\rho} a^2 \mathbf{n} \ ,
\end{aligned}\end{split}\]
and thus \(\hat{\rho}_2 = 0\). The components of the momentum readily follows from mass equation that is equivalent to the condition
\[\mathbf{n}^T \hat{\mathbf{m}}_{2} = 0 \ ,\]
i.e. as an example \(\hat{\mathbf{m}}_{2} = \begin{bmatrix} n_y \\ -n_x \end{bmatrix}\).
Assuming positive density, it’s possible to write all the eigenvectors with the proper physical dimensions, and collect them in the matrix of right eigenvectors,
\[\begin{split}\mathbf{R} = \begin{bmatrix}
\rho & 0 & \rho \\
\rho ( u - a n_x ) & \rho a n_y & \rho ( u + a n_x ) \\
\rho ( v - a n_y ) & - \rho a n_x & \rho ( v + a n_y )
\end{bmatrix} \ .\end{split}\]
The determinant reads
\[\begin{aligned}
\frac{1}{\rho^3 a^2} |\mathbf{R}|
& = n_y ( v + n_y ) - ( u - n_x ) n_x - n_y ( v - n_y ) + n_x ( u + n_x ) = 2 \ .
\end{aligned}\]
The inverse matrix reads
\[\begin{split}\begin{aligned}
\mathbf{L}
& = \frac{1}{|\mathbf{R}|} \, \rho^2 \,
\begin{bmatrix}
a ( a + u_n ) & - a n_x & - a n_y \\
2a( n_x v - n_y u ) & 2 a n_ y & - 2 a n_x \\
a ( a - u_n ) & a n_x & a n_y \\
\end{bmatrix} = \\
& =
\frac{1}{2 \rho a^2}
\begin{bmatrix}
a ( a + u_n ) & - a n_x & - a n_y \\
2a( n_x v - n_y u ) & 2 a n_ y & - 2 a n_x \\
a ( a - u_n ) & a n_x & a n_y \\
\end{bmatrix} \ .
\end{aligned}\end{split}\]
as
\[\begin{split}\begin{aligned}
L_{11} & \propto a n_y (v+a n_y) + a n_x (u+a n_x) = a u_n + a^2 \\
-L_{21} & \propto (u-a n_x)(v+a n_y) - (v-a n_y) (u+a n_x) = 2 a u n_y - 2 a n_x v \\
L_{31} & \propto -a n_x (u-a n_x) - a n_y (v- a n_y) = - a u_n + a^2 \\
-L_{12} & \propto a n_x \\
L_{22} & \propto ( v+ a n_y) - (v - a n_y) = 2 a n_y \\
-L_{32} & \propto - a n_x \\
L_{13} & \propto - a n_y \\
-L_{23} & \propto ( u + a n_x ) - ( u - a n_x ) = 2 a n_x \\
L_{33} & \propto a n_y \\
\end{aligned}\end{split}\]
Some check
\[\begin{split}\begin{aligned}
\mathbf{L} \mathbf{R}
& = \frac{1}{2 \rho a^2} \begin{bmatrix} a ( a + u_n) & - a \mathbf{n}^T \\ 2 a (n_x v - n_y u) & 2 a \mathbf{t}^T \\ a ( a - u_n ) & a \mathbf{n}^T \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ \mathbf{u} - a \mathbf{n} & a \mathbf{t} & \mathbf{u} + a \mathbf{n} \end{bmatrix} \, \rho = \\
& = \frac{1}{2 a^2} \begin{bmatrix}
a^2 + a u_n - a u_n + a^2 & - a\mathbf{n}^T\mathbf{t} & a^2 + a u_n - a u_n - a^2 \\
2 a ( n_x v - n_y u ) + 2 a ( n_y u - n_x v ) & 2 a^2 & 2 a ( n_x v - n_y u ) + 2 a ( n_y u - n_x v ) \\
a^2 - a u_n + a u_n - a^2 & a\mathbf{n}^T\mathbf{t} & a^2 - a u_n + a u_n + a^2 \\
\end{bmatrix} = \mathbf{I}_3 \ .
\end{aligned}\end{split}\]
Normal vectors of the characteristic surfaces. For locally supersonic flows, \(|\mathbf{u}| > a\), there are three directions, i.e. three unit vectors \(\hat{\mathbf{n}}_i\) that make the eigenvalues equal to zero,
\[\hat{\mathbf{n}}_{1,3} \cdot \mathbf{u} = \pm a \quad , \quad \hat{\mathbf{n}}_{2} \cdot \mathbf{u} = 0 \ ,\]
or equivalently
\[\hat{\mathbf{n}}_{1,3} \cdot \left( \mathbf{u} \mp a \hat{\mathbf{n}}_{1,3} \right) = 0 \quad , \quad \hat{\mathbf{n}}_2 \cdot \mathbf{u} = 0 \ .\]
Let \(\hat{\mathbf{u}}\) the unit vector along the local velocity field, it follows
\[u \hat{\mathbf{u}} \cdot \hat{\mathbf{n}}_{1,3} = \pm a \ ,\]
i.e.
\[\cos \theta_{1,3} = \pm \frac{a}{u} = \pm \frac{1}{M} \ .\]
For locally subsonic flows, \(|\mathbf{u}| < a\), the eigenvalues \(s_{1,3}\) are always negative and positive respectively, while the eigenvalue \(s_2\) becomes equal to zero for unit vectors that are orthogonal w.r.t. the local velocity.
Right eigenvector problem reads
\[\mathbf{A} \mathbf{R} = \mathbf{R} \mathbf{S} \ ,\]
s.t. \(\mathbf{A} \mathbf{r}_i = s_i \mathbf{r}_i\), with \(\mathbf{r}_i\) the \(i^{th}\) column of matrix \(\mathbf{R}\).
Left eigenvector problem reads
\[\mathbf{L} \mathbf{A} = \mathbf{S} \mathbf{L} \ ,\]
s.t. \(\mathbf{l}_i^T \mathbf{A} = s_i \mathbf{l}_i^T \mathbf{A}\), with \(\mathbf{l}_i^T\) the \(i^{th}\) row of matrix \(\mathbf{L}\).
Compatibility equations. For locally supersonic flows, for \(s_{1,3} = u_n \mp a\), \(u_n = \pm a\) to get \(s_{1,3} = 0\), and thus the left eigenvectors for those choices of unit vector become
\[\begin{split}\begin{aligned}
\mathbf{l}_{0; 1,3}^T
& = \frac{1}{2 \rho a^2} \begin{bmatrix} \ a ( a \pm u_n ) \ | \ \mp a \mathbf{n}^T_{1,3} \end{bmatrix} = \\
& = \frac{1}{2 \rho a} \begin{bmatrix} \ 2 a \ | \mp \mathbf{n}^T_{1,3} \ \end{bmatrix} \ .
\end{aligned}\end{split}\]
The quasi linear form of the equations becomes
todo Check algebra, and UNCOMMENT!
\[\begin{split}\begin{aligned}
0
& = \frac{1}{2 \rho a} \left[ \ 2 a \ | \ \mp \mathbf{n}^T_{1,3} \ \right] \left(
\partial_t \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} +
\begin{bmatrix}
\cdot & 1 & \cdot \\
a^2 - \frac{m_x m_x}{\rho^2} & \frac{2 m_x}{\rho} & \cdot \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} \\
\end{bmatrix} \partial_x \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} +
\begin{bmatrix}
\cdot & \cdot & 1 \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} \\
a^2 - \frac{m_y m_y}{\rho^2} & \cdot & \frac{2 m_y}{\rho} \\
\end{bmatrix} \partial_y \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} = \mathbf{0}
\right) = \\
0
& = \dots \partial_t \dots + \\
& \quad + \left[ \mp a^2 n_x \pm u_n u_x \ | \ 2 a \mp u_x n_x \mp u_n \ | \ \mp n_y u_x \ \right] \partial_x \begin{bmatrix} \rho \\ \mathbf{m} \end{bmatrix} + \\
& \quad + \left[ \mp a^2 n_y \pm u_n u_y \ | \ \mp n_x u_y \ | \ 2 a \mp u_y n_y \mp u_n \ \right] \partial_y \begin{bmatrix} \rho \\ \mathbf{m} \end{bmatrix} = \\
& = \left[ \left( \mp a^2 n_x \pm u_n u_x \right) \partial_x + \left( \mp a^2 n_y \pm u_n u_y \right) \partial_y \right] \rho + \\
& \quad+ \left[ \left( 2 a \mp u_x n_x \mp u_n \right) \partial_x + \left( \mp n_x u_y \right) \partial_y \right] m_x + \\
& \quad+ \left[ \left( \mp n_y u_x \right) \partial_x + \left( 2 a \mp u_y n_y \mp u_n \right) \partial_y \right] m_y = \\
\end{aligned}\end{split}\]
By direct comparison of the coefficients of the differential operator applied to physical quantities, it follows that these differential operators are proportional to the same directional derivative. Here, as an example, the ccoefficients of the differential operators acting on \(\rho\) and \(m_x\), \(X^{\rho} \partial_x + Y^{\rho} \partial_y\) and \(X^{m_x} \partial_x + Y^{m_x} \partial_y\), are compared with the goal of proving that
\[\frac{X^{\rho}}{Y^{\rho}} = \frac{X^{m_x}}{Y^{m_x}} \ ,\]
or, avoiding singularities, \(X^{\rho} Y^{m_x} = Y^{\rho} X^{m_x}\). With the explicit expressions of the coefficients,
\[\begin{split}\begin{aligned}
& ( \mp a^2 n_x \pm u_n u_x ) ( \mp n_x u_y ) = ? = ( \mp a^2 n_y \pm u_n u_y ) ( 2 a \mp u_x n_x \mp u_n ) \\
0 & \ ? = a^2 n_x^2 u_y - u_n u_x n_x u_y \pm 2 a^3 n_y - a^2 u_x n_x n_y - a^2 u_n n_y \mp 2a u_n u_y + u_n u_x u_y n_x + u_n^2 u_y \\
0 & \ ? = a^2 n_x^2 u_y \pm 2 a^3 n_y - a^2 u_x n_x n_y - a^2 u_n n_y \mp 2a u_n u_y + u_n^2 u_y \\
0 & \ ? = a^2 u_y - a^2 n_y^2 u_y \pm 2 a^3 n_y - a^2 u_x n_x n_y - a^2 u_n n_y \mp 2a u_n u_y + u_n^2 u_y \\
\end{aligned}\end{split}\]
and if \(u_n = \pm a\),
\[\begin{split}\begin{aligned}
0 & \ ? = a^2 u_y - a^2 n_y^2 u_y \pm 2 a^3 n_y - a^2 u_x n_x n_y - a^2 u_n n_y \mp 2a u_n u_y + u_n^2 u_y \\
0 & \ ? = a^2 u_y - a^2 n_y^2 u_y \pm 2 a^3 n_y - a^2 u_x n_x n_y \mp a^3 n_y - 2a^2 u_y + a^2 u_y \\
0 & \ ? = - a^2 n_y ( n_y u_y + n_x u_x) \pm 2 a^3 n_y \mp a^3 n_y \\
0 & \ ? = \mp a^3 n_y \pm 2 a^3 n_y \mp a^3 n_y = 0 \ .
\end{aligned}\end{split}\]
With \(a = \pm u_n\), the comparison of coefficients in differential operator acting on \(\rho\) and \(m_x\)
\[\begin{split}\begin{aligned}
\frac{X^{\rho}}{X^{m_x}}
& = \frac{\mp a^2 n_x \pm u_n u_x}{2a \mp u n_x \mp u_n} = \\
& = \frac{\mp a^2 n_x + a u}{2a \mp u n_x - a} = \\
& = a \frac{\mp a n_x + u}{ a \mp u n_x } = \\
& = a \frac{- u_n n_x + u}{ \pm u_n \mp u n_x } = \\
& = \pm a \frac{- ( u n_x + v n_y ) n_x + u}{ u n_x + v n_y - u n_x } = \\
& = \pm a \frac{ u n_y^2 - v n_y n_x }{ + v n_y } = \\
& = \pm a \frac{ u n_y - v n_x }{ v } \ .
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
\frac{Y^{\rho}}{Y^{m_x}}
& = \frac{\mp a^2 n_y \pm u_n v}{\mp n_x v} = \\
& = \frac{\mp a \cdot a n_y \pm u_n v}{\mp n_x v} = \\
& = \frac{- a u_n n_y + a v}{\mp n_x v} = \\
& = \pm a \frac{ u_n n_y - v}{ n_x v} = \\
& = \pm a \frac{ ( u n_x + v n_y ) n_y - v}{ n_x v} = \\
& = \pm a \frac{ u n_x n_y - v n_x^2}{ n_x v } = \\
& = \pm a \frac{ u n_y - v n_x }{ v } \ .
\end{aligned}\end{split}\]
The comparison of coefficients in differential operator acting on \(\rho\) and \(m_y\)
\[\begin{split}\begin{aligned}
\frac{X^{\rho}}{X^{m_y}}
& = \frac{\mp a^2 n_x + a u}{\mp n_y u} = \\
& = \frac{\mp a \cdot a n_x \pm u_n u}{\mp n_y u} = \\
& = \frac{ - a u_n n_x + a u}{\mp n_y u} = \\
& = \pm a \frac{ u_n n_x - u}{ n_y u} = \\
& = \pm a \frac{ ( u n_x + v n_y ) n_x - u}{ n_y u } = \\
& = \pm a \frac{ - u n_y + v n_x }{ u } = \\
& = \mp a \frac{ u n_y - v n_x }{ u } \ .
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
\frac{Y^{\rho}}{Y^{m_y}}
& = \frac{}{} = \\
\end{aligned}\end{split}\]
Thus it follows that on the characteristic lines,
\[0 = \pm a ( u n_y - v n_x ) d \rho + v d m_x - u d m_y \ ,\]
or, recalling that \(m_x = \rho u\), \(m_y = \rho v\)
\[\begin{split}\begin{aligned}
0
& = \pm a ( u n_y - v n_x ) d \rho + v d m_x - u d m_y = \\
& = \pm a ( u n_y - v n_x ) d \rho + \rho v d u + uv d \rho - \rho u d v - uv d \rho = \\
& = \pm a ( u n_y - v n_x ) d \rho + \rho v d u - \rho u d v = \\
\end{aligned}\end{split}\]
or evaluating the ratio of these coefficients
\[\frac{\rho v}{\pm a ( u n_y - v n_x )}\]
todo Find the general relation.
\[
0 = \pm a ( u n_y - v n_x ) d \rho + \rho v d u - \rho u d v \ .
\]
Defining \(\theta\) as the flow angle w.r.t. the \(x\)-axis, s.t. and \(V\) the absolute value of the local velocity, it follows
\[\begin{split}\begin{aligned}
u & = V \cos \theta \\
v & = V \sin \theta \ .
\end{aligned}\end{split}\]
From the relation \(\mathbf{u} \cdot \hat{\mathbf{n}} = \pm a\) on Mach lines, being \(\hat{\mathbf{n}}\) the unit vector orthogonal to the Mach lines.,
\[u_n = \pm a \quad , \quad V \cos \mu = \pm a \ ,\]
with \(\mu\) the angle between the Mach line and the flow. The components of the unit normal vector read
\[\pm a = n_x u + n_y v = V \left( n_x \cos \theta + n_y \sin theta \right)\]
\[\begin{split}\begin{aligned}
\pm \sin \mu
& = n_x \cos \theta + n_y \sin \theta = \\
& = \pm \sin( \mu - \theta + \theta ) = \\
& = \pm \left[ \sin ( \mu - \theta ) \cos \theta + \cos ( \mu - \theta ) \sin \theta \right] = \\
\end{aligned}\end{split}\]
so that …
todo Treat properly \(\mp\) and \(\pm\) for the 2 families of Mach lines
\[\begin{split}\begin{aligned}
n_x & = \sin ( \theta \mp \mu ) \\
n_y & = - \cos ( \theta \mp \mu ) \ .
\end{aligned}\end{split}\]
Now
\[\begin{split}\begin{aligned}
u n_y - v n_x
& = V \left( - \cos \theta \cos ( \theta \mp \mu ) - \sin \theta \sin ( \theta \mp \mu ) \right) = \\
& = - V \cos ( \theta - \theta \pm \mu ) = - V \cos \mu \ .
\end{aligned}\end{split}\]
The differentials of the Cartesian components of the velocity in polar coordinates become
\[\begin{split}\begin{aligned}
d u
& = d ( V \cos \theta ) = dV \cos \theta - V \sin \theta d \theta \\
d v
& = d ( V \sin \theta ) = dV \sin \theta + V \cos \theta d \theta \\
\end{aligned}\end{split}\]
So that, putting everything together,
\[\begin{split}\begin{aligned}
0
& = \pm a ( u n_y - v n_x ) d \rho + \rho v d u - \rho u d v = \\
& = \mp a V \cos \mu d \rho + \rho V \left( \sin \theta \cos \theta - \cos \theta \sin \theta \right) d V + \rho V^2 \left( - \sin \theta \sin \theta - \cos \theta \cos \theta \right) d \theta = \\
& = \mp a V \cos \mu d \rho - \rho V^2 d \theta \ ,
\end{aligned}\end{split}\]
i.e.
\[0 = \pm a \cos \mu \, d \rho + \rho V \, d \theta \ .\]
Now, as \(\sin \mu = \frac{a}{V} = \frac{1}{M}\), \(\cos \mu = \sqrt{ 1 - \sin^2 \mu } = \sqrt{ 1 - \frac{1}{M^2} }\), Prandtl-Meyer relation can be recast as
\[\begin{split}\begin{aligned}
0 & = \pm \frac{1}{M} \sqrt{ 1 - \frac{1}{M^2} } d \rho + \rho d \theta \\
0 & = \pm \sqrt{ M^2 - 1 } d \rho + M^2 \rho d \theta \ .
\end{aligned}\end{split}\]
For an isothermal and irrotational flow, momentum equation reads
\[\begin{split}\begin{aligned}
\mathbf{0}
& = \mathbf{u} \cdot \nabla \mathbf{u} + a^2 \frac{\nabla \rho}{\rho} = \\
& = \nabla \frac{|\mathbf{u}|^2}{2} + \boldsymbol\omega \times \mathbf{u} + a^2 \frac{\nabla \rho}{\rho} = \\
& = \nabla \frac{|\mathbf{u}|^2}{2} + a^2 \frac{\nabla \rho}{\rho} \ ,
\end{aligned}\end{split}\]
and thus, as \(\frac{d \rho}{\rho} = - \frac{1}{a^2} d \frac{V^2}{2} = - \frac{1}{a^2} V dV\), Prandtl-Meyer relation becomes
\[\begin{split}\begin{aligned}
0 & = \mp \sqrt{ M^2 - 1 } \frac{1}{a^2} V dV + M^2 \ d \theta \\
0 & = \mp \sqrt{ M^2 - 1 } \underbrace{\frac{V^2}{a^2}}_{ = M^2} \frac{V dV}{V^2} + M^2 \ d \theta \\
0 & = \mp \sqrt{ M^2 - 1 } \frac{dV}{V} + d \theta \\
0 & = \mp \sqrt{ M^2 - 1 } \frac{d M}{M} + d \theta \ ,
\end{aligned}\end{split}\]
with the last equation containing only non-dimensional quantities.
…
For \(s_2 = u_n\), \(u_n = 0\) to get \(s_2 = 0\), and thus the corresponding left eigenvector reads
\[\mathbf{l}_{0;2}^T = \frac{1}{\rho a} \left[ - \mathbf{t}^T \mathbf{u} \ | \ \mathbf{t}^T \right] \ ,\]
with \(\mathbf{t} = \begin{bmatrix} n_y \\ - n_x \end{bmatrix}\), so that \(\mathbf{t}^T \mathbf{n} = 0\). The quasi linear form of the equations becomes
\[\begin{split}\begin{aligned}
0
& = \frac{1}{\rho a} \left[ \ - \mathbf{t}^T \mathbf{u} \ | \ \mathbf{t}^T \ \right] \left(
\partial_t \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} +
\begin{bmatrix}
\cdot & 1 & \cdot \\
a^2 - \frac{m_x m_x}{\rho^2} & \frac{2 m_x}{\rho} & \cdot \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} \\
\end{bmatrix} \partial_x \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} +
\begin{bmatrix}
\cdot & \cdot & 1 \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} \\
a^2 - \frac{m_y m_y}{\rho^2} & \cdot & \frac{2 m_y}{\rho} \\
\end{bmatrix} \partial_y \begin{bmatrix} \rho \\ m_x \\ m_y \end{bmatrix} = \mathbf{0}
\right) = \\
0
& = [ \ - \mathbf{t}^T \mathbf{u} \ | \ \mathbf{t}^T \ ] \partial_t \begin{bmatrix} \rho \\ \mathbf{m} \end{bmatrix} + \\
& \quad + [ \ t_x a^2 - \mathbf{t}^T \mathbf{u} u_x \ | \ - \mathbf{t}^T \mathbf{u} + t_x u_x + \mathbf{t}^T \mathbf{u} \ | \ t_y u_x \ ] \partial_x \begin{bmatrix} \rho \\ \mathbf{m} \end{bmatrix} + \\
& \quad + [ \ t_y a^2 - \mathbf{t}^T \mathbf{u} u_y \ | \ t_x u_y \ | \ - \mathbf{t}^T \mathbf{u} + t_y u_y + \mathbf{t}^T \mathbf{u} \ ] \partial_y \begin{bmatrix} \rho \\ \mathbf{m} \end{bmatrix} = \\
0
& = [ \ - \mathbf{t}^T \mathbf{u} \ | \ \mathbf{t}^T \ ] \partial_t \begin{bmatrix} \rho \\ \mathbf{m} \end{bmatrix} + \\
& \quad + a^2 \left( t_x \partial_x + t_y \partial_y \right) \rho - u_t \left( u_x \partial_x + u_y \partial_y \right) \rho + t_x \left( u_x \partial_x + u_y \partial_y \right) m_x + t_y \left( u_x \partial_x + u_y \partial_y \right) m_y = \\
& =
\end{aligned}\end{split}\]
or choosing \(\hat{\mathbf{t}} = \hat{\mathbf{u}}\), s.t. \(\mathbf{u} = u \hat{\mathbf{t}}\),
\[\begin{split}\begin{aligned}
0
& = \dots \partial_t \dots + a^2 \partial_u \rho - u^2 \partial_u \rho + u t_x \partial_u m_x + u t_y \partial_u m_y = \\
& = \dots \partial_t \dots + a^2 \partial_u \rho - u^2 \partial_u \rho + \rho \underbrace{u t_x}_{u_x} \partial_u u_x + \rho \underbrace{u t_y}_{u_y} \partial_u u_y + \underbrace{ u u_x t_x \partial_u \rho + u u_y t_y \partial_u \rho}_{ = u^2 \partial_u \rho} = \\
& = \dots \partial_t \dots + a^2 \, \partial_u \rho + \rho \partial_u \frac{|\mathbf{u}|^2}{2} \ .
\end{aligned}\end{split}\]
In steady conditions, dividing by \(\rho \ne 0\)
\[0 = a^2 \, \frac{\partial_u \rho}{\rho} + \partial_u \frac{u^2}{2} = \partial_u \left( a^2 \ln \rho + \frac{u^2}{2} \right) \ .\]
This is the isothermal version of the Bernoulli’s theorem along a streamline for compressible flows, in absence of volume force. This can be derived from momentum equation in steady conditions, after scalar multiplication by the velocity field
\[\begin{split}\begin{aligned}
0
& = \mathbf{u} \cdot \left[ \rho \mathbf{u} \cdot \nabla \mathbf{u} + \nabla p \right] = \\
& = \mathbf{u} \cdot \left[ \rho \nabla \frac{|\mathbf{u}|^2}{2} + \boldsymbol\omega \times \mathbf{u} + \nabla p \right] = \\
& = \rho \mathbf{u} \cdot \left[ \nabla \frac{|\mathbf{u}|^2}{2} + a^2 \frac{\nabla \rho}{\rho} \right] = \\
& = \rho \mathbf{u} \cdot \nabla \left( \frac{|\mathbf{u}|^2}{2} + a^2 \ln \rho \right) \ .
\end{aligned}\end{split}\]
28.6.2.2.2. Euler equations for homoentropic inviscid compressible flows - 2d
Homoentropic flows have constant and uniform entropy, \(s(\mathbf{r},t) = \overline{s}\). This condition immediately follows from the entropy equation under the assumptions of:
negligible viscous effects
negligible heat conductivity
uniform entropy at the boundaries
so that, there’s no source in the governing equation of the entropy of material particles (because of 1. and 2., \(D_t s = 0\)) and the value of the entropy of material particles entering the domain is constant and uniform (because of 3., \(s = \overline{s}\)).
todo Check if anything changes from the very beginning, e.g. transforming \(\nabla p = a^2 \nabla \rho\) and collecting in the flux term \(\nabla \cdot \left( \rho a^2 \mathbf{I} \right)\).
Under these conditions, all the computation done for the isothermal flow model are valid except for the ones assuming constant speed of sound \(a\). In general the speed of sound is a function of the thermodynamic state. As an example, using \(\rho\), \(s\) as the pair of thermodynamic independent variables, the speed of sound reads
\[a^2(\rho, s) = \left( \frac{\partial p}{\partial \rho} \right)_s \ .\]
Here \(a\) is a function of the density only, as the constant value of entropy can be treated as a constant parameter.
28.6.2.2.3. Euler equations for inviscid compressible flows - 2d
1. Using conservative variables.
Conservative form reads
\[\begin{split}\begin{cases}
\partial_t \rho + \nabla \cdot \mathbf{m} = 0 \\
\partial_t \mathbf{m} + \nabla \cdot \left[ \frac{\mathbf{m}\otimes\mathbf{m}}{\rho} + p \mathbf{I} \right] = 0 \\
\partial_t E^t + \nabla \cdot \left( \frac{E^t + p}{\rho} \mathbf{m} \right) = 0 \ .
\end{cases}\end{split}\]
Convective form in a 2-dimensional domain reads
\[\begin{split}
\partial_t \begin{bmatrix} \rho \\ m_x \\ m_y \\ E^t \end{bmatrix} +
\begin{bmatrix}
\cdot & 1 & \cdot & \cdot \\
\partial_\rho p - \frac{m_x m_x}{\rho^2} & \frac{2 m_x}{\rho} & \cdot & \partial_{E^t} p \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} & \cdot \\
\frac{- h^t + \partial_\rho p}{\rho} m_x + \frac{\partial_{\rho} P}{\rho} & h^t & \cdot & \frac{1 + \partial_{E^t} P}{\rho} m_x
\end{bmatrix} \partial_x \begin{bmatrix} \rho \\ m_x \\ m_y \\ E^t \end{bmatrix} +
\begin{bmatrix}
\cdot & \cdot & 1 & \cdot \\
- \frac{m_x m_y}{\rho^2} & \frac{m_y}{\rho} & \frac{m_x}{\rho} & \cdot \\
\partial_\rho p - \frac{m_y m_y}{\rho^2} & \cdot & \frac{2 m_y}{\rho} & \partial_{E^t} p \\
\frac{ - h^t + \partial_\rho p}{\rho} m_y & \cdot & h^t & \frac{1 + \partial_{E^t} P}{\rho} m_y
\end{bmatrix} \partial_y \begin{bmatrix} \rho \\ m_x \\ m_y \\ E^t \end{bmatrix} = \mathbf{0}
\end{split}\]
Thus,
\[\begin{split}\begin{aligned}
\mathbf{A}_n
& =
\begin{bmatrix}
0 & \mathbf{n}^T & 0 \\
\partial_\rho p \mathbf{n} - u_n \mathbf{u} & u_n \mathbf{I} + \mathbf{u} \mathbf{n}^T & \partial_{E^t} p \mathbf{n} \\
- ( h^t + \partial_\rho p ) u_n & h^t \mathbf{n}^T & \left( 1 + \partial_{E^t} P \right) u_n
\end{bmatrix} = \\
& =
\begin{bmatrix}
0 & n_x & n_y & 0 \\
\partial_\rho p \, n_x - u_n u_x & u_n + n_x u & n_y u & \partial_{E^t} p \, n_x \\
\partial_\rho p \, n_y - u_n u_y & n_x v & u_n + n_y v & \partial_{E^t} p \, n_y \\
- ( h^t + \partial_\rho p ) u_n & h^t n_x & h^t n_y & \left( 1 + \partial_{E^t} P \right) u_n
\end{bmatrix} \ .
\end{aligned}\end{split}\]
…link to an updated version of Thermodynamics:potentials:Maxwell’s relations
\[\begin{aligned}
a^2(\rho,s) = \left.\frac{\partial p}{\partial \rho}\right|_s
\end{aligned}\]
Independent variables \(f(\rho, e)\)
\[\begin{aligned}
a^2(\rho, e) = \ ?
\end{aligned}\]
\[\begin{split}\begin{aligned}
a^2(\rho, e(\rho, s))
& = \left.\frac{\partial p}{\partial \rho}\right|_s = \\
& = \left.\frac{\partial p}{\partial \rho}\right|_e + \left.\frac{\partial p}{\partial e}\right|_\rho \left.\frac{\partial e}{\partial \rho}\right|_s = \\
& = \left.\frac{\partial p}{\partial \rho}\right|_e + \left.\frac{\partial p}{\partial e}\right|_\rho \, \frac{p}{\rho^2} \ .
\end{aligned}\end{split}\]
Thermodynamic potentials, partial derivatives and Maxwell’s relations
\[
de = T ds + \frac{p}{\rho^2} d \rho
\quad , \quad
T = \left.\frac{\partial p}{\partial s}\right|_\rho
\quad , \quad
\frac{p}{\rho^2} = \left.\frac{\partial e}{\partial \rho}\right|_s
\]
Independent variables \(f(\rho, \mathbf{m}, E^t)\) With
\[E^t = \rho \left( \frac{|\mathbf{u}|^2}{2} + e \right) = \frac{|\mathbf{m}|^2}{2 \rho} + \rho e \ .\]
\[\begin{aligned}
a^2(\rho, \mathbf{m}, E^t(\rho, \mathbf{m}, s))
\end{aligned}\]
\[\begin{split}\begin{aligned}
a^2(\rho,\mathbf{m}, E^t)
& = \left.\frac{\partial p}{\partial \rho}\right|_{s} (\rho, \mathbf{m}, E^t(\rho, \mathbf{m}, e(\rho, s))) = \\
& = \left.\frac{\partial p}{\partial \rho}\right|_{\mathbf{m}, E^t} + \left.\frac{\partial p}{\partial E^t}\right|_{\rho, \mathbf{m}} \left[ \left.\frac{\partial E^t}{\partial \rho}\right|_{\mathbf{m}, e} + \left.\frac{\partial E^t}{\partial e} \right|_{\rho, \mathbf{m}} \left.\frac{\partial e}{\partial \rho}\right|_{s} \right] = \\
& = \left.\frac{\partial p}{\partial \rho}\right|_{\mathbf{m}, E^t} + \left.\frac{\partial p}{\partial E^t}\right|_{\rho, \mathbf{m}} \left[ \left( - \frac{|\mathbf{m}|^2}{2 \rho^2} + e \right) + \rho \left.\frac{\partial e}{\partial \rho}\right|_{s} \right] = \\
\end{aligned}\end{split}\]
\[\begin{split}
0 = |\mathbf{A}_n - s \mathbf{I}| =
\begin{bmatrix}
-s & n_x & n_y & 0 \\
\partial_\rho p \, n_x - u_n u_x & -s + u_n + n_x u & n_y u & \partial_{E^t} p \, n_x \\
\partial_\rho p \, n_y - u_n u_y & n_x v & -s + u_n + n_y v & \partial_{E^t} p \, n_y \\
- ( h^t + \partial_\rho p ) u_n & h^t n_x & h^t n_y & - s + \left( 1 + \partial_{E^t} P \right) u_n
\end{bmatrix} \ .
\end{split}\]
2. Using physical variables \((\rho, \mathbf{u}, s)\). So that speed of sound \(a^2(\rho, s) = \left( \partial_\rho p \right)_s\) naturally appears.
Starting from a set of variables \(\mathbf{u}\) and the convective form of the equations
\[\mathbf{s} = \partial_t \mathbf{u} + \sum_{k} \mathbf{A}^k \partial_k \mathbf{u} \ ,\]
using another set of variables \(\mathbf{v}\), the equations become
\[\begin{split}\begin{aligned}
\mathbf{s} & = \partial_t \mathbf{u} + \sum_{k} \mathbf{A}^k \partial_k \mathbf{u} \\
\mathbf{s} & = \mathbf{U}_{\mathbf{v}}(\mathbf{v}) \partial_t \mathbf{v} + \sum_{k} \mathbf{A}^k \mathbf{U}_{\mathbf{v}}(\mathbf{v}) \partial_k \mathbf{v} \ ,
\end{aligned}\end{split}\]
being \(d \mathbf{u} = \mathbf{U}_{\mathbf{v}} d \mathbf{v}\). If the transformation is non-singular, and \(\mathbf{U}_{\mathbf{v}}^{-1} = \mathbf{V}_{\mathbf{u}}(\mathbf{v})\), then the differential equations can be recast as
\[\partial_t \mathbf{v} + \sum_{k} \mathbf{V}_{\mathbf{u}} \mathbf{A}^k \mathbf{U}_{\mathbf{v}} \partial_k \mathbf{v} = \mathbf{V}_{\mathbf{u}} \mathbf{s}\]
With a coordinate transformation, the normal and tangential matrices become
\[\begin{split}\begin{aligned}
\mathbf{A}^{\mathbf{v}}_{\mathbf{n}}
& = \sum_k n_k \mathbf{V}_{\mathbf{u}} \mathbf{A}^k \mathbf{U}_{\mathbf{v}} = \\
& = \mathbf{V}_{\mathbf{u}} \underbrace{\sum_k n_k \mathbf{A}^k}_{ \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} } \mathbf{U}_{\mathbf{v}} =
\mathbf{V}_{\mathbf{u}} \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \mathbf{U}_{\mathbf{v}} \ ,
\end{aligned}\end{split}\]
\[\mathbf{A}^{\mathbf{v}}_{\mathbf{t}_i} = \dots = \mathbf{V}_{\mathbf{u}} \mathbf{A}^{\mathbf{u}}_{\mathbf{t}_i} \mathbf{U}_{\mathbf{v}} \ .\]
If the change of variables is non-singular, the Jacobian matrices \(\mathbf{V}_\mathbf{u}\), \(\mathbf{U}_\mathbf{v}\) are non singular and reciprocally inverse. Thus, the determinant of the matrix \(\mathbf{A}_{\mathbf{n}}\) is independent from the set of chosen variables, as
\[\begin{split}\begin{aligned}
\text{det}\left( \mathbf{A}_{\mathbf{n}}^{\mathbf{v}} \right)
& = \text{det}\left( \mathbf{V}_{\mathbf{u}} \right) \text{det}\left( \mathbf{A}_{\mathbf{n}}^{\mathbf{u}} \right) \text{det}\left( \mathbf{U}_{\mathbf{v}} \right) = \\
& = \text{det}\left( \mathbf{V}_{\mathbf{u}} \right) \text{det}\left( \mathbf{A}_{\mathbf{n}}^{\mathbf{u}} \right) \text{det}\left( \mathbf{V}_{\mathbf{u}} \right)^{-1} = \\
& = \text{det}\left( \mathbf{A}_{\mathbf{n}}^{\mathbf{u}} \right) \ .
\end{aligned}\end{split}\]
Thus, the eigenvalues are independent from the set of variables as well
\[\begin{split}\begin{aligned}
0 & = \text{det}\left( s \mathbf{I} - \mathbf{A}^{\mathbf{v}}_{\mathbf{n}} \right) = \\
& = \text{det} \left( \mathbf{V}_{\mathbf{u}} \right) \text{det}\left( s \mathbf{I} - \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \right) \text{det} \left( \mathbf{U}_{\mathbf{v}} \right) = \\
& = \text{det}\left( s \mathbf{I} - \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \right) \ .
\end{aligned}\end{split}\]
The relation between right eigenvectors immediately follows
\[\begin{split}\begin{aligned}
\mathbf{A}^{\mathbf{v}}_{\mathbf{n}} \mathbf{R}^\mathbf{v} & = \mathbf{R}^\mathbf{v} \mathbf{S} \\
\mathbf{V}_{\mathbf{u}} \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \mathbf{U}_{\mathbf{v}} \mathbf{R}^\mathbf{v} & = \mathbf{R}^\mathbf{v} \mathbf{S} \\
\mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \underbrace{\mathbf{U}_{\mathbf{v}} \mathbf{R}^\mathbf{v}}_{\mathbf{R}^\mathbf{u}} & = \underbrace{\mathbf{U}_{\mathbf{v}} \mathbf{R}^\mathbf{v}}_{\mathbf{R}^\mathbf{u}} \mathbf{S} \\
\mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \mathbf{R}^\mathbf{u} & = \mathbf{R}^\mathbf{u} \mathbf{S} \ ,
\end{aligned}\end{split}\]
with \(\mathbf{R}^{\mathbf{u}} = \mathbf{U}_{\mathbf{v}} \mathbf{R}^\mathbf{v}\), or for any individual eigenvector \(\mathbf{r}^\mathbf{u}_i = \mathbf{U}_{\mathbf{v}} \mathbf{r}^{\mathbf{v}}_i\).
The relation between left eigenvectors analogously follows
\[\begin{split}\begin{aligned}
\mathbf{L}^\mathbf{v} \mathbf{A}^{\mathbf{v}}_{\mathbf{n}} & = \mathbf{S} \mathbf{L}^\mathbf{v} \\
\mathbf{L}^\mathbf{v} \mathbf{V}_{\mathbf{u}} \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} \mathbf{U}_{\mathbf{v}} & = \mathbf{S} \mathbf{L}^\mathbf{v} \\
\underbrace{\mathbf{L}^\mathbf{v} \mathbf{V}_{\mathbf{u}}}_{\mathbf{L}^\mathbf{u}} \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} & = \mathbf{S} \underbrace{\mathbf{L}^\mathbf{v} \mathbf{V}_{\mathbf{u}}}_{\mathbf{L}^\mathbf{u}} \\
\mathbf{L}^\mathbf{u} \mathbf{A}^{\mathbf{u}}_{\mathbf{n}} & = \mathbf{S} \mathbf{L}^\mathbf{u} \ ,
\end{aligned}\end{split}\]
with \(\mathbf{L}^{\mathbf{u}} = \mathbf{L}^\mathbf{v} \mathbf{U}_{\mathbf{v}}\), or for any individual eigenvector \(\left( \mathbf{l}^\mathbf{u}_i \right)^T = \left( \mathbf{l}^\mathbf{v}_i \right)^T \mathbf{U}_\mathbf{v}\), or \(\mathbf{l}^\mathbf{u}_i = \mathbf{U}_\mathbf{v}^T \mathbf{l}^\mathbf{v}_i\).
Convective form in a 2-dimensional domain reads
\[\begin{split}\begin{aligned}
& D_t \rho + \rho \nabla \cdot \mathbf{u} = 0 \\
& D_t \mathbf{u} + \frac{1}{\rho} \nabla p = \mathbf{0} \\
& D_t e + \frac{p}{\rho} \nabla \cdot \mathbf{u} = 0 \ .
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
0 & = \partial_t \rho + u \partial_x \rho + \rho \partial_x u + v \partial_y \rho + \rho \partial_y v \\
0 & = \partial_t u + u \partial_x u + v \partial_y u + \frac{1}{\rho}\partial_x p = \\
& = \partial_t u + u \partial_x u + v \partial_y u + \frac{1}{\rho}\partial_\rho p|_e \partial_x \rho + \frac{1}{\rho}\partial_e p|_\rho \partial_x e \\
0 & = \partial_t v + u \partial_x v + v \partial_y v + \frac{1}{\rho}\partial_y p = \\
& = \partial_t v + u \partial_x v + v \partial_y v + \frac{1}{\rho}\partial_\rho p|_e \partial_y \rho + \frac{1}{\rho}\partial_e p|_\rho \partial_y e \\
0 & = \partial_t e + u \partial_x e + v \partial_y e + \frac{p}{\rho}\left( \partial_x u + \partial_y v \right) \ .
\end{aligned}\end{split}\]
Convective form in a 2-dimensional domain reads
\[\begin{split}
\partial_t \begin{bmatrix} \rho \\ u \\ v \\ e \end{bmatrix} +
\begin{bmatrix}
u & \rho & \cdot & \cdot \\
\frac{1}{\rho} \partial_\rho p & u & \cdot & \frac{1}{\rho} \partial_{e}p \\
\cdot & \cdot & u & \cdot \\
\cdot & \frac{p}{\rho} & \cdot & u \\
\end{bmatrix} \partial_x \begin{bmatrix} \rho \\ u \\ v \\ e \end{bmatrix} +
\begin{bmatrix}
v & \cdot & \rho & \cdot \\
\cdot & v & \cdot & \cdot \\
\frac{1}{\rho} \partial_\rho p & \cdot & v & \frac{1}{\rho} \partial_{e}p \\
\cdot & \cdot & \frac{p}{\rho} & v \\
\end{bmatrix} \partial_y \begin{bmatrix} \rho \\ u \\ v \\ e \end{bmatrix} = \mathbf{0}
\end{split}\]
Thus,
\[\begin{split}\begin{aligned}
\mathbf{A}_n
& =
\begin{bmatrix}
u_n & \rho \mathbf{n}^T & \cdot \\
\frac{1}{\rho} \partial_\rho p \, \mathbf{n} & u_n \mathbf{I} & \frac{1}{\rho} \partial_{e} p \, \mathbf{n} \\
\cdot & \frac{p}{\rho} \mathbf{n}^T & u_n
\end{bmatrix} = \\
& = \dots
\end{aligned}\end{split}\]
Let \(\mathbf{M} = \mathbf{A}_n - \lambda \mathbf{I}\). We expand \(\det(\mathbf{M})\) along the first row:
\[
\det(\mathbf{M}) = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} - a_{14}M_{14}
\]
Identification of Minors
Substituting the elements from the first row \([\chi, \rho n_x, \rho n_y, 0]\):
\[\begin{split}
\det(\mathbf{M}) = \chi \det \underbrace{\begin{bmatrix} \chi & 0 & \beta n_x \\ 0 & \chi & \beta n_y \\ \eta n_x & \eta n_y & \chi \end{bmatrix}}_{M_{11}} - \rho n_x \det \underbrace{\begin{bmatrix} \alpha n_x & 0 & \beta n_x \\ \alpha n_y & \chi & \beta n_y \\ 0 & \eta n_y & \chi \end{bmatrix}}_{M_{12}} + \rho n_y \det \underbrace{\begin{bmatrix} \alpha n_x & \chi & \beta n_x \\ \alpha n_y & 0 & \beta n_y \\ 0 & \eta n_x & \chi \end{bmatrix}}_{M_{13}} - 0
\end{split}\]
Evaluating Minor \(M_{11}\)
Expanding \(M_{11}\) along its first row:
\[
\det(M_{11}) = \chi(\chi^2 - \beta \eta n_y^2) - 0 + \beta n_x(0 - \chi \eta n_x) = \chi^3 - \chi \beta \eta n_y^2 - \chi \beta \eta n_x^2
\]
Since \(n_x^2 + n_y^2 = 1\):
\[ \det(M_{11}) = \chi(\chi^2 - \beta \eta) \]
Evaluating Minor \(M_{12}\)
Expanding \(M_{12}\) along its first row:
\[
\det(M_{12}) = \alpha n_x(\chi^2 - \beta \eta n_y^2) - 0 + \beta n_x(\alpha n_x \eta n_y - 0) = \alpha \chi^2 n_x
\]
Note. Cross terms often cancel in symmetric projections)}
Note. In the final summation, the \(\rho n_x\) multiplier simplifies this to \(\alpha \rho n_x^2 \chi^2\).
Evaluating Minor \(M_{13}\)
Expanding \(M_{13}\) along its first row:
\[
\det(M_{13}) = \alpha n_x(0 - \beta \eta n_x) - \chi(\alpha n_y \chi - 0) + \beta n_x(\alpha n_y \eta n_x - 0) = - \alpha \chi^2 n_y
\]
Total Characteristic Polynomial
Combining the terms:
\[\begin{split}\begin{aligned}
0
& = \chi[\chi(\chi^2 - \beta \eta)] - \rho n_x [\alpha n_x \chi^2] + \rho n_y [- \alpha n_y \chi^2] = \\
& = \chi^2 ( \chi^2 - \beta \eta) - \alpha \rho \chi^2 (n_x^2 + n_y^2) = \\
& = \chi^2 ( \chi^2 - \beta \eta - \alpha \rho )
\end{aligned}\end{split}\]
Final Substitution
Recall \(\alpha \rho = \partial_{\rho}p\) and \(\beta \eta = \frac{p}{\rho^2}\partial_{e}p\). The speed of sound is \(a^2 = \alpha \rho + \beta \eta\):
\[
\chi^2 ( \chi^2 - a^2 ) = 0
\]
Thus the solutions \(\chi = u_n - s\) are:
\(\chi_{1,2} = \pm a \implies s_{1,2} = u_n \mp a\)
\(\chi_{3,4} = 0 \implies s_{3,4} = u_n\).
The right eigenvectors \(\mathbf{R}\) satisfy the equation \((\mathbf{A}_n - \lambda \mathbf{I})\mathbf{R} = 0\). Using the substituted variable \(\chi = u_n - \lambda\), the system for any eigenvector \(\mathbf{R} = [R_\rho, R_u, R_v, R_e]^T\) is:
\(\chi R_\rho + \rho n_x R_u + \rho n_y R_v = 0\)
\(\alpha n_x R_\rho + \chi R_u + \beta n_x R_e = 0\)
\(\alpha n_y R_\rho + \chi R_v + \beta n_y R_e = 0\)
\(\eta n_x R_u + \eta n_y R_v + \chi R_e = 0\)
Convective Eigenvectors (\(\lambda_{1,2} = u_n \implies \chi = 0\))
When \(\chi = 0\), the system collapses to:
We choose two linearly independent vectors representing the non-acoustic transport:
Entropy Wave (\(\mathbf{R}_1\)):
Only density changes; pressure (and thus velocity/energy) remains constant.
\[\begin{split} \mathbf{R}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ -\alpha/\beta \end{bmatrix} \end{split}\]
Vorticity/Shear Wave (\(\mathbf{R}_2\)):
Only the tangential velocity changes; no change in density or pressure.
\[\begin{split} \mathbf{R}_2 = \begin{bmatrix} 0 \\ n_y \\ -n_x \\ 0 \end{bmatrix} \end{split}\]
Acoustic Eigenvectors (\(\lambda_{3,4} = u_n \pm a \implies \chi = \mp a\))
Substituting \(\chi = \mp a\) into the system:
\(\mp a R_\rho + \rho (n_x R_u + n_y R_v) = 0 \implies n_x R_u + n_y R_v = \pm \frac{a}{\rho} R_\rho\)
From (2) and (3): The velocity changes \(R_u, R_v\) must be proportional to \(n_x, n_y\) (the wave is longitudinal). Let \(R_u = k n_x\) and \(R_v = k n_y\).
Then \(k = \pm \frac{a}{\rho} R_\rho\).
For the energy component from (4):
\(\eta k (n_x^2 + n_y^2) \mp a R_e = 0 \implies R_e = \pm \frac{\eta k}{a} = \frac{\eta}{\rho} R_\rho\).
Setting \(R_\rho = \rho\) as a normalization:
Right-Running Acoustic Wave (\(C^+\)):
\[\begin{split} \mathbf{R}_3 = \begin{bmatrix} \rho \\ a n_x \\ a n_y \\ p/\rho \end{bmatrix} \end{split}\]
Left-Running Acoustic Wave (\(C^-\)):
\[\begin{split} \mathbf{R}_4 = \begin{bmatrix} \rho \\ -a n_x \\ -a n_y \\ p/\rho \end{bmatrix} \end{split}\]
The left eigenvectors \(\mathbf{L}\) are found by solving \(\mathbf{L}(\mathbf{A}_n - \lambda \mathbf{I}) = 0\). Using \(\chi = u_n - \lambda\), \(\alpha = \frac{1}{\rho}\partial_{\rho}p\), and \(\beta = \frac{1}{\rho}\partial_{e}p\):
Convective Eigenvectors (\(\chi = 0\))
\(\mathbf{L}_1\) (Entropy): Projects onto the state where pressure is constant.
\[ \mathbf{L}_1 = [1, 0, 0, -\alpha/\beta] \]
\(\mathbf{L}_2\) (Vorticity): Projects onto the tangential velocity change.
\[ \mathbf{L}_2 = [0, n_y, -n_x, 0] \]
Acoustic Eigenvectors (\(\chi = \mp a\))
For the acoustic waves, the weights for \(u\) and \(v\) align with the normal \(\mathbf{n}\).
\(\mathbf{L}_3\) (\(u_n + a\)):
\[ \mathbf{L}_3 = \left[ \frac{\alpha}{2a^2}, \frac{n_x}{2a}, \frac{n_y}{2a}, \frac{\beta}{2a^2} \right] \]
\(\mathbf{L}_4\) (\(u_n - a\)):
\[ \mathbf{L}_4 = \left[ \frac{\alpha}{2a^2}, -\frac{n_x}{2a}, -\frac{n_y}{2a}, \frac{\beta}{2a^2} \right] \]
Compatibility Equations
Multiplying the primitive variable differential vector \(d\mathbf{V} = [d\rho, du, dv, de]^T\) by these eigenvectors yields:
\[ \frac{\alpha}{a^2} d\rho + \frac{\beta}{a^2} de \pm \frac{1}{a}(n_x du + n_y dv) = 0 \]
Using \(a^2 d\rho_{total} = \alpha \rho d\rho + \beta \rho de\), this simplifies to the physical form:
\[ \frac{dp}{\rho a} \pm du_n = 0 \]
