hyperbolic PDEs:
$U_{,t} + F_{,x} = S$
eigen-decompose:
chain rule:
$U_{,t} + \frac{\partial F}{\partial U} U_{,x} = S$
eigen-decompose the flux derivative:
$U_{,t} + R \Lambda L U_{,x} = S$
has wavespeeds $\lambda_i$ for $\Lambda_{ij} = \delta_{ij} \lambda_i$
Simplest case: 1D system:
$u_{,t} + a u_{,x} = 0$
has a single wavespeed of $a$.
now comes the wave equation:
$u_{,tt} - a^2 u_{,xx} = f$
recast as a 1st order PDE:
$\Pi = u_{,t}$
$\Psi = u_{,x}$
$\left[\begin{matrix}
\Pi \\ \Psi
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
0 & -a^2 \\
-1 & 0
\end{matrix}\right]
\left[\begin{matrix}
\Pi \\ \Psi
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
f \\ 0
\end{matrix}\right]$
has eigensystem:
$\left[\begin{matrix}
\Pi \\ \Psi
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
-a & a \\
1 & 1
\end{matrix}\right]
\left[\begin{matrix}
a & 0 \\
0 & -a
\end{matrix}\right]
\left[\begin{matrix}
-\frac{1}{2a} & \frac{1}{2} \\
\frac{1}{2a} & \frac{1}{2}
\end{matrix}\right]
\left[\begin{matrix}
\Pi \\ \Psi
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
f \\ 0
\end{matrix}\right]$
what about all 2x2 eigensystem classifications?
maybe later.
how about specific to our problem at hand?
where we have to convert 1st derivatives to state variables in order to make our system hyperbolic:
$u_{,t} + a u_{,x} = s$
$u_{,xt} + a_{,x} u_{,x} + a u_{,xx} = s_{,x}$
$u_{,t} + a u_{,x} = s$
$u_{,xt} + a u_{,xx} = s_{,x} - a_{,x} u_{,x}$
$\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
a & 0 \\
0 & a
\end{matrix}\right]
\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s \\ s_{,x} - a_{,x} u_{,x}
\end{matrix}\right]$
has eigensystem
$\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}\right]
\left[\begin{matrix}
a & 0 \\
0 & a
\end{matrix}\right]
\left[\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}\right]
\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s \\ s_{,x} - a_{,x} u_{,x}
\end{matrix}\right]$
and here we have wavespeeds of $\{ a \}$ mult. 2.
or since we can treat the $u_{,x}$ as a state variable, it doesn't need to be in $u$'s flux:
$\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
0 & 0 \\
0 & a
\end{matrix}\right]
\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s - a u_{,x} \\
s_{,x} - a_{,x} u_{,x}
\end{matrix}\right]$
has eigensystem
$\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}\right]
\left[\begin{matrix}
0 & 0 \\
0 & a
\end{matrix}\right]
\left[\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}\right]
\left[\begin{matrix}
u \\ u_{,x}
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s \\ s_{,x} - a_{,x} u_{,x}
\end{matrix}\right]$
so instead we have wavespeeds of $\{ 0, a \}$.
but what if we throw in a 3rd variable dependent on the 1st two? How to rescale it, by $u$ perhaps, and when to convert the rescaling derivative to a state variable or not?
ok before considering 1st derivs as state vars, just look at a scalar system with new var included:
$u_{,t} + a_{11} u_{,x} = s_1$
$v_{,t} + a_{21} u_{,x} + a_{22} v_{,x} = s_2$
$\left[\begin{matrix}
u \\
v
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
a & 0 \\
c & d
\end{matrix}\right]
\left[\begin{matrix}
u \\ v
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s_1 \\
s_2
\end{matrix}\right]$
I almost want to make $u_{,t}$ flux depend on $v_{,x}$ so that my matrix will be a full 2x2, and this problem can turn into a problem of classifying all 2x2 eigensystem behaviors.
but without $a_{12}$:
eigensystem
$\left[\begin{matrix}
u \\ v
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
\frac{a - d}{c} & 0 \\
1 & 1
\end{matrix}\right]
\left[\begin{matrix}
a & 0 \\
0 & d
\end{matrix}\right]
\left[\begin{matrix}
\frac{c}{a - d} & 0 \\
\frac{c}{d - a} & 1
\end{matrix}\right]
\left[\begin{matrix}
u \\ v
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s_1 \\ s_2
\end{matrix}\right]$
So, as long as our new variable doesn't play into the original variable's system, the wavespeeds are determined solely by the diagonals of the flux jacobian.
I guess that means we can adjust the wavespeed to our heart's content -- so long as $A_{22}$ is a constant.
What if it is a function of $u$?
let $v = v(f)$
$u_{,t} + a_{11} u_{,x} = s_1$
$v_{,t} + a_{21} u_{,x} + a_{22} v_{,x} = s_2$
$\frac{\partial v}{\partial f} f_{,t} + a_{21} u_{,x} + a_{22} \frac{\partial v}{\partial f} f_{,x} = s_2$
$f_{,t} + a_{21} (\frac{\partial v}{\partial f})^{-1} u_{,x} + a_{22} f_{,x} = s_2 (\frac{\partial v}{\partial f})^{-1}$
So this is interesting, looks like after transforming $v$ into $v(f)$, the wavespeeds don't change, but only in absence of a $A_{12}$ term.
if $u_{,t}$ does depend on $v_{,x}$ then does this change wavespeeds?
$u_{,t} + a_{11} u_{,x} + a_{12} v_{,x} = s_1$
$v_{,t} + a_{21} u_{,x} + a_{22} v_{,x} = s_2$
let $A = R \Lambda L$ with $\Lambda = diag( \lambda_1, \lambda_2 )$
Let $v = v(f)$
$u_{,t} + a_{11} u_{,x} + a_{12} \frac{\partial v}{\partial f} f_{,x} = s_1$
$\frac{\partial v}{\partial f} f_{,t} + a_{21} u_{,x} + a_{22} \frac{\partial v}{\partial f} f_{,x} = s_2$
$f_{,t} + a_{21} (\frac{\partial v}{\partial f})^{-1} u_{,x} + a_{22} f_{,x} = s_2 (\frac{\partial v}{\partial f})^{-1}$
$\left[\begin{matrix}
u \\
f
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
a_{11} & a_{12} \frac{\partial v}{\partial f} \\
a_{21} (\frac{\partial v}{\partial f})^{-1} & a_{22}
\end{matrix}\right]
\left[\begin{matrix}
u \\ v
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s_1 \\
s_2 (\frac{\partial v}{\partial f})^{-1}
\end{matrix}\right]$
So the determinant is going to still be $det(A)$, the char poly is still going to be the same, and the wavespeed is still going to be the same.
How about with 1st derivative state variables?
$u_{,t} + a_{11} u_{,x} = s_1$
$u_{x,t} + a_{11,x} u_{,x} + a_{11} u_{,xx} = s_{1,x}$
$v_{,t} + a_{21} u_{,x} + a_{22} v_{,x} = s_2$
so now we have a situation where we can put some variables in arbitrary places, as derivs or as state vars:
favoring flux:
$\left[\begin{matrix}
u \\
u_{,x} \\
v
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
a_{11} & 0 & 0 \\
a_{11,x} & a_{11} & 0 \\
a_{21} & 0 & a_{22}
\end{matrix}\right]
\left[\begin{matrix}
u \\ u_{,x} \\ v
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s_1 \\
s_{1,x} \\
s_2
\end{matrix}\right]$
favoring source:
$\left[\begin{matrix}
u \\
u_{,x} \\
v
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
a_{11} & 0 & 0 \\
0 & a_{11} & 0 \\
0 & 0 & a_{22}
\end{matrix}\right]
\left[\begin{matrix}
u \\
u_{,x} \\
v
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s_1 \\
s_{1,x} - a_{11,x} u_{,x} \\
s_2 - a_{21} u_{,x}
\end{matrix}\right]$
both with eigenvalues of $\{a_{11} \times 2, a_{22} \}$
what about a remapping of v that includes u? like $v \rightarrow u v$ ?
$u_{,t} + a_{11} u_{,x} = s_1$
$v_{,t} + a_{21} u_{,x} + a_{22} v_{,x} = s_2$
$u_{,t} = -a_{11} u_{,x} + s_1$
$v_{,t} = -a_{21} u_{,x} - a_{22} v_{,x}+ s_2$
$(uv)_{,x} = u_{,x} v + u v_{,x}$
$(uv)_{,x} - u_{,x} v = u v_{,x}$
$v_{,x} = \frac{1}{u} ((uv)_{,x} - u_{,x} v)$
$(uv)_{,t} = u_{,t} v + u v_{,t}$
$(uv)_{,t} = (
-a_{11} u_{,x} + s_1
) v
+ u (
-a_{21} u_{,x} - a_{22} v_{,x} + s_2
)
$
$(uv)_{,t} =
- (a_{11} v + a_{21} u) u_{,x}
- a_{22} u v_{,x}
+ s_1 v + s_2 u
$
$(uv)_{,t} =
- (a_{11} v + a_{21} u) u_{,x}
- a_{22} u \frac{1}{u} ((uv)_{,x} - u_{,x} v)
+ s_1 v + s_2 u
$
$(uv)_{,t} =
- (a_{11} v + a_{21} u) u_{,x}
- a_{22} (uv)_{,x}
+ a_{22} u_{,x} v
+ s_1 v + s_2 u
$
$(uv)_{,t} =
- (
(a_{11} - a_{22}) v
+ a_{21} u
) u_{,x}
- a_{22} (uv)_{,x}
+ s_1 v + s_2 u
$
So our new linear PDE system looks like:
$\left[\begin{matrix}
u \\
uv
\end{matrix}\right]_{,t}
+
\left[\begin{matrix}
a_{11} & 0 \\
- (
(a_{11} - a_{22}) v
+ a_{21} u
) &
- a_{22}
\end{matrix}\right]
\left[\begin{matrix}
u \\ uv
\end{matrix}\right]_{,x}
=
\left[\begin{matrix}
s_1 \\
s_2 u + s_1 v
\end{matrix}\right]$
I wonder if there's a 1-1 between state variable changes and row operations?
how about a generic linear system change-of-variables?
$\frac{\partial u^i}{\partial t} + {a^i}_j \frac{\partial u^j}{\partial x} = s^i$
$\frac{\partial u^i}{\partial t} = -{a^i}_j \frac{\partial u^j}{\partial x} + s^i$
$v^i = v(u)$
$\frac{\partial v^i}{\partial x} = \frac{\partial v^i}{\partial u^j} \frac{\partial u^j}{\partial x}$
$\frac{\partial u^i}{\partial v^j} \frac{\partial v^j}{\partial x} = \frac{\partial u^i}{\partial x}$
$\frac{\partial v^i}{\partial t} = \frac{\partial v^i}{\partial u^j} \frac{\partial u^j}{\partial t}$
$\frac{\partial v^i}{\partial t} = \frac{\partial v^i}{\partial u^j} (-{a^j}_k \frac{\partial u^k}{\partial x} + s^j)$
$\frac{\partial v^i}{\partial t} = -\frac{\partial v^i}{\partial u^j} {a^j}_k \frac{\partial u^k}{\partial x} + \frac{\partial v^i}{\partial u^j} s^j$
$\frac{\partial v^i}{\partial t} = -\frac{\partial v^i}{\partial u^j} {a^j}_k \frac{\partial u^k}{\partial v^l} \frac{\partial v^l}{\partial x} + \frac{\partial v^i}{\partial u^j} s^j$
$\frac{\partial v^i}{\partial t} + \frac{\partial v^i}{\partial u^j} {a^j}_k \frac{\partial u^k}{\partial v^l} \frac{\partial v^l}{\partial x} = \frac{\partial v^i}{\partial u^j} s^j$
So if $A = R \Lambda L$, and if $\frac{\partial U}{\partial V} = (\frac{\partial V}{\partial U})^{-1}$, then the change-of-variables from $\frac{\partial u^i}{\partial t}$ to $\frac{\partial v^i}{\partial t}$
can be seen as a left-transform of the right-eigenvectors of $A$ by $\frac{\partial V}{\partial U}$, but the eigenvalues are preserved.