I read in a article dealing with a hyperbolic partial differential equations this statement :

For any system of hyperbolic partial differential equations (pde), expressed as

(1) $\partial_t W + A \partial_x W = 0$,

with $A$ a diagonalisable matrix, the eigensystem of A plays a prominent role.

The eigenvalues of A are the wavespeeds, the right eigenvectors define the paths taken in phase space by simple waves, and the left eigenvectors define the characteristic equations.

I usually deal with the hyperbolic pde but here it's the first time i encounter this signification of the right and the left eigenvectors :

  1. right eigenvectors define the paths taken in phase space by simple waves
  2. left eigenvectors define the characteristic equations

I will be grateful if someone can help me to better understand this


I find Pulliam's notes for the Euler equations to be a pretty good introduction to this topic using the equations of fluid motion. The idea is that you start with a conservation law:

$$ \frac{\partial \vec{Q}}{\partial t} + \frac{\partial \vec{F}\left(\vec{Q}\right)}{\partial x} = 0$$

where $Q$ is your variable vector and $F$ is your flux function. You can introduce a flux jacobian and assume the jacobian is constant to get a quasi-linear equation:

$$ \frac{\partial \vec{Q}}{\partial t} + \frac{\partial \vec{F}}{\partial \vec{Q}}\frac{\partial \vec{Q}}{\partial x} = 0$$

where we can call $\partial \vec{F}/\partial \vec{Q} = A^\prime$. Now, we want to seek a new vector of variables related to the first set, $\vec{W} = (\partial \vec{Q}/\partial \vec{W}) \vec{Q} = J \vec{Q}$. We introduce this to our equation and we get:

$$ J \frac{\partial \vec{W}}{\partial t} + A^\prime J \frac{\partial \vec{W}}{\partial x} = 0 \\ \rightarrow \frac{\partial \vec{W}}{\partial t} + J^{-1} A^\prime J \frac{\partial \vec{W}}{\partial x} = 0 $$

And we choose our new variable set $\vec{W}$ such that $J^{-1} A^\prime J = A$ is "nice". How we choose "nice" is up to you. You do want to make it something that you can diagonalize so you can break it apart into $A = T^{-1} \Lambda T$ where $\Lambda$ is a diagonal matrix with the eigenvalues along the diagonal. If you can do that, you will get:

$$ \frac{\partial \vec{W}}{\partial t} + T^{-1} \Lambda T \frac{\partial \vec{W}}{\partial x}$$

Then multiply through by $T$ and you get a new set of variables $T \vec{W} = \vec{S}$:

$$ \frac{\partial \vec{S}}{\partial t} + \Lambda \frac{\partial \vec{S}}{\partial x} $$

If you recall, $\Lambda$ is diagonal! This means you have now decoupled your equations so they can be solved independently -- a huge improvement in finding the solution! The vector $\vec{S}$ is the characteristic variable vector.

Only one component of the vector may change along each characteristic line, the others are constant along the characteristic line. Each vector in the right eigenvector matrix $T$ gives you the characteristic vector (direction of the characteristic line) and the element in $\Lambda$ gives you the characteristic speed along that vector. The vector in the left eigenvector matrix multiplied by your original variable vector gives you the characteristic variable for that equation.

The last thing to be really careful of is notation. Some people choose $A = T \Lambda T^{-1}$ while others choose $A = T^{-1} \Lambda T$ and if the matrix is symmetric it can be $A = T^T \Lambda T$ or $ A = T \Lambda T^T$

| cite | improve this answer | |
  • $\begingroup$ thank you for your answer. If i understand well your explanation works only in the linear case, what about the non-linear one? And what is it the same with the 2D case ? $\endgroup$ – Amine HANINI Feb 8 '16 at 15:32
  • $\begingroup$ if it's possible can you illustrate me your point on the Sallow Water Equation example in the conservative form ? Thanks $\endgroup$ – Amine HANINI Feb 8 '16 at 15:54
  • 1
    $\begingroup$ @AmineHANINI Doing things with non-linear equations throws it entirely out the window. What we do in the Euler equations is assume the Jacobian is fixed and this makes the equation "linear", or sometimes called quasi-linear. It's only true in a small neighborhood around a particular operating point, but that's enough to do the analysis needed. Regarding your question about the Shallow Water Equation -- I'm guessing that is the equation you need to do this for? I'll happily answer specific questions about particular steps, but I won't do the full derivation -- that is an exercise for you to do. $\endgroup$ – tpg2114 Feb 8 '16 at 16:19
  • 1
    $\begingroup$ And it works in any dimension. But, in higher dimension you cannot (at least for the Euler equations) diagonalize both dimensions at the same time. Read through the linked PDF, it explains pretty well what happens and it provides all of the equations needed to work through and verify the answers for 1D and 2D in both conservative and non-conservative form of the Euler equations. $\endgroup$ – tpg2114 Feb 8 '16 at 16:21
  • $\begingroup$ I don't have any problem with the mathematical analysis of the hyperbolic pde, I am just a little confused with the physical interpretation. With my specific problem, namely Shallow Water Equation : $W=(h,hu)$, $ \mathcal A = \begin{pmatrix} 0 & 1 \\ - u^2 + c^2 & 2 u \\ \end{pmatrix}$ we have : $ \mathcal \Lambda = \begin{pmatrix} u-c & 0 \\ 0 & u+c \\ \end{pmatrix}$ and $ \mathcal T = \begin{pmatrix} 1 & 1 \\ u-c & u+c \\ \end{pmatrix}$ $\endgroup$ – Amine HANINI Feb 8 '16 at 17:17

Here some details \begin{equation*} \partial _{t}\mathbf{w}(x,t)+\mathbf{A}\cdot \partial _{x}\mathbf{w}(x,t)=0 \end{equation*} Let $\mathbf{A}$ be an $n\times n$ matrix. Then $\mathbf{w}$ must be $n$ -dimensional. Let us assume that $\mathbf{A}$ has real entries and $\mathbf{w }$ has real components.

\begin{eqnarray*} \mathbf{w}(x,t) &=&\exp [t\mathbf{A}\partial _{x}]\cdot \mathbf{w}(x,0) \\ \mathbf{A} &=&\mathbf{V}^{-1}\mathbf{\cdot B\cdot V} \\ \mathbf{w}(x,t) &=&\exp [t\mathbf{V}^{-1}\mathbf{\cdot B\cdot V}\partial _{x}]\cdot \mathbf{w}(x,0)=\mathbf{V}^{-1}\mathbf{\cdot }\exp [t\mathbf{B} \partial _{x}]\mathbf{\cdot V}\cdot \mathbf{w}(x,0) \\ \mathbf{y}(x,t) &=&\mathbf{V\cdot w}(x,t)=\exp [t\mathbf{B}\partial _{x}] \mathbf{\cdot y}(x,0) \\ \mathbf{B} &=&\sum_{j}b_{j}\mathbf{u}_{j}\mathbf{v}_{j}^{T},\;\mathbf{v}% _{j}^{T}\cdot \mathbf{u}_{h}=\delta _{jh} \\ \exp [t\mathbf{B}\partial _{x}] &=&\sum_{j}\exp [b_{j}t\partial _{x}]\mathbf{ u}_{j}\mathbf{v}_{j}^{T} \\ \mathbf{y}(x,t) &=&\sum_{j}\exp [b_{j}t\partial _{x}]\mathbf{u}_{j}\mathbf{v} _{j}^{T}\cdot \mathbf{y}(x,0) \\ y_{j}(x,t) &=&\mathbf{v}_{j}^{T}\cdot \mathbf{y}(x,t)=\exp [b_{j}t\partial _{x}]\mathbf{v}_{j}^{T}\cdot \mathbf{y}(x,0) \\ &=&\exp [b_{j}t\partial _{x}]y_{j}(x,0)=y_{j}(x+b_{j}t,0) \end{eqnarray*} Note that $b_{j}$ must be real for this to be meaningful. It determines the wave speed in channel $j$. Note further that in higher dimensions the situation becomes more complicated.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.