Characteristic form of the Euler equations

When I first learnt about eigenvalues and eigenvectors, I never thought that it would actually be useful. Nowadays I can’t seem to avoid them even if I wanted to – Linear solvers? Don’t forget about the spectral radius! Control systems? Can’t go without the natural frequencies! Euler equations? Guess where do the wavespeeds come from!

Eigenvalues and eigenvectors intuitively

In order to understand the role of eigenvalues and eigenvectors in fluid dynamics, lets turn to it’s sibling structural dynamics for awhile. The canonical introductory example in structural dynamics is the problem of 2 weights connected by springs as shown in the diagram I pulled from here.

The system can be described with constant mass and stiffness matrices by: \mathbf{M}\ddot{\vec{x}} + \mathbf{K} \vec{x} = 0 when no forces are applied. \mathbf{K} has off diagonal terms so the system is coupled. Rearranging, we get \ddot{\vec{x}} + \mathbf{M}^{-1}\mathbf{K} \vec{x} = 0. By diagonalizing the matrix, one gets:

    \begin{align*} \ddot{\vec{x}} + \mathbf{Q} \mathbf{\Lambda} \mathbf{Q}^{-1} \vec{x} &= 0 \\ \mathbf{Q}^{-1} \ddot{\vec{x}} + \mathbf{\Lambda} \mathbf{Q}^{-1} \vec{x} &= 0 \\ \ddot{\vec{y}} + \mathbf{\Lambda} \vec{y} &= 0 . \end{align*}

We have made the transformation \vec{y} = \mathbf{Q}^{-1}\vec{x}. Since \mathbf{\Lambda} is a diagonal matrix, we get an uncoupled system of equations which is now trivial to solve. By projecting the state vector into the eigenvectors, we are working in components that are independent from each other. Collectively, these independent components give the system its characteristics, like mode shapes in beams or principle components of a reduced system. Note that I am using the term characteristic very loosely here since we are just trying to get an intuitive feel.

Fluid dynamics

So now we’ve seen how to uncouple a system of equations, we can simply apply the same method to the fluid equations and we’re done! Simple, no? Okay, let’s see where does that lead us with the Euler equations. Consider the one-dimensional case with conserved variables \vec{W}:

    \begin{gather*} \frac{\partial \vec{W}}{\partial t} + \frac{d \vec{F}}{d \vec{W}} \frac{\partial \vec{W}}{\partial x} = \frac{\partial \vec{W}}{\partial t} + \mathbf{A} \frac{\partial \vec{W}}{\partial x} = 0 \quad , \quad \mathbf{A} = \mathbf{T}\mathbf{\Lambda}\mathbf{T}^{-1} \\ \Rightarrow \frac{ \mathbf{T}^{-1} \partial \vec{W} }{\partial t} + \mathbf{\Lambda} \frac{ \mathbf{T}^{-1} \partial \vec{W} }{\partial x} = 0. \end{gather*}

So this leads us to the characteristic form with d\vec{C} = \mathbf{T}^{-1} d\vec{W}. Note that \mathbf{T} is not a constant matrix so we can’t bring it into the differential operator. Then the characteristic form looks like:

    \begin{align*} \frac{\partial \vec{C} }{\partial t} + \mathbf{\Lambda} \frac{\partial \vec{C} }{\partial x} = 0. \end{align*}

Despite the simple looking form of the equation, the set of equation in full is:

    \begin{align*} \left(\frac{\partial \rho}{\partial t} - \frac{1}{a^2}\frac{\partial P}{\partial t}\right) + u \left( \frac{\partial \rho}{\partial x} - \frac{1}{a^2}\frac{\partial P}{\partial x} \right) &= 0 \\ \left( \frac{\partial u}{\partial t} + \frac{1}{\rho a}\frac{\partial P}{\partial t} \right) + \left(u+a\right)\left(\frac{\partial u}{\partial x} + \frac{1}{\rho a}\frac{\partial P}{\partial x} \right)&= 0 \\ \left( \frac{\partial u}{\partial t} - \frac{1}{\rho a}\frac{\partial P}{\partial t} \right) + \left(u-a\right)\left(\frac{\partial u}{\partial x} - \frac{1}{\rho a}\frac{\partial P}{\partial x} \right)&= 0 . \end{align*}

It’s still not too bad, we now have 3 advection equations which obey:

    \begin{align*} d\rho - \frac{dP}{a^2} = 0 \quad &for \quad \frac{dx}{dt} = u \\ du + \frac{dP}{\rho a} = 0 \quad &for \quad \frac{dx}{dt} = u+a \\ du - \frac{dP}{\rho a} = 0 \quad &for \quad \frac{dx}{dt} = u-a . \end{align*}

Integrating the left hand side of each equation gives us the characteristic variables which remain constant along the curve dx/dt = \lambda where \lambda is either u, (u+a) or (u-a). The curves are called wave fronts since they are the path where the variables propagate. Let’s try to get the characteristic variable for the middle equation for example,

    \begin{align*} \int du + \int \frac{dP}{\rho a} = const \quad &for \quad \frac{dx}{dt} = u+a \\ u + \int \frac{dP}{\rho a} = const \quad &for \quad \frac{dx}{dt} = u+a . \end{align*}

It seems like we’ve run into some trouble with the integration for the second term on the left. What about applying the perfect gas relation?

    \begin{align*} \int \frac{dP}{\rho a} = \sqrt{\frac{R}{\gamma}} \int \frac{dP}{P / \sqrt{T}}. \end{align*}

No dice. In fact, for the characteristic form of the Euler equation, only the first equation can be integrated to entropy. The other characteristic variables do not have a physical manifestation. In addition, the wave speeds u, (u+a) and (u-a) are also dependent on the solution, which does not lend a simple solution like we had previously thought.

So we’ve gone through all these efforts for nothing? Well, not really. The characteristic variables may not be integratable into something physical, but at least the wave fronts tell us something. For this one-dimensional problem, we see that information in the flow moves with u, (u+a) and (u-a) wave speeds.

Now imagine for a moment that you are swimming alone in a large pool. As you pull your strokes, you generate ripples in the water. Looking ahead, you have a wave moving forward away from you and if you were to look back, you would also see a wave moving away behind you. Now outside of the ripple, the water is still, but within the ripple, the water has been influenced by your presence. Suppose a friend were to join you a little later and he’s caught up with your backward propagating ripple. You can imagine that the shape of his ripples will be affected by your presence, it is dependent on the state of the water surrounding him before he pulls his strokes.

This is very roughly what the characteristics tell us about the fluid. The characteristic variables are like the ripples, and they are broadcast at the speed of sound relative to the flow (i.e. (u+a) and (u-a)), as well as the actual flow speed u. For a subsonic forward flow (u < a), we have 2 characteristics moving forward and one moving back. For example when we have change in area for 1D flow, the air particles where the change occurs sends 2 characteristics downwind and one upwind. It is because of the one characteristic being sent at (u-a) that allows the flow upwind of the area change to adjust gradually, accommodating for the change in area.

Another point to note here, as was already pointed out, is that \mathbf{T} is not constant. Hence we run into difficulties integrating the characteristic variables. If we can approximate a constant \mathbf{T}_0, we have for the second term in \vec{C}:

    \begin{gather*} c_2 = \int du + \frac{1}{\rho_0 a_0} \int dP = u + \frac{P}{\rho_0 a_0} = const \\ \Rightarrow \Delta c_2 = \Delta u + \frac{\Delta P}{\rho_0 a_0} . \end{gather*}

This would come in handy when we look at flux calculation later.

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