13
$\begingroup$

What's really going on as regards characterstics in the theory of PDE's?

(Hopefully you wont have to check the links provided below, included for those who are interested really)

For second order PDE's, in this video in which characteristics were defined as lines in the domain along which the highest order partial derivatives were discontinuous, then using this idea, one can naturally use matrices & determinants to derive the conditions under which the second order terms are equal to zero - deriving the $B^2 - 4AC$ classification of partial differential equations.

  • For those who are interested, the method is that given

    $$\mathcal{L}[\phi] = A \phi_{xx} + B \phi_{xy} + C \phi_{yy} + H(x,y,\phi, \phi_x, \phi_y) = 0$$

     we basically want $ \vec{v} = (\frac{\partial ^2 \phi}{\partial x^2},\frac{\partial ^2 \phi}{\partial x \partial y},\frac{\partial ^2 \phi}{\partial y^2}) = (0,0,0) $ & since this is a vector, we construct an equation of the form $ \mathcal{T}(\vec{v}) = \vec{w}$ & examine the case where a solution $ \vec{v}$ doesn't exist, i.e. a case where an infinite number of solutions exist.

    Using

    $$ d(\frac{\partial \phi}{\partial x}) = \frac{\partial}{\partial x} \frac{\partial \phi}{\partial x}dx + \frac{\partial }{\partial y} \frac{\partial \phi}{\partial x} dy = \frac{\partial ^2 \phi}{\partial x^2 }dx + \frac{\partial ^2 \phi}{\partial y \partial x} dy $$

    $$ d(\frac{\partial \phi}{\partial y}) = \frac{\partial}{\partial x} \frac{\partial \phi}{\partial y}dx + \frac{\partial }{\partial y} \frac{\partial \phi}{\partial y} dy = \frac{\partial ^2 \phi}{\partial x\partial y }dx + \frac{\partial ^2 \phi}{ \partial ^2 y} dy $$

    we arrive at

    $$ \mathcal{T}(\vec{v}) = \vec{w} \rightarrow \left[ \begin{matrix} A & B & C \\ dx & dy & 0 \\ 0 & dx & dy \end{matrix} \right] \left[ \begin{matrix} \phi_{xx} \\ \phi_{xy} \\ \phi_{yy} \end{matrix} \right] = \left[ \begin{matrix} -H \\ d( \frac{\partial \phi}{\partial x}) \\ d( \frac{\partial \phi}{\partial y}) \end{matrix} \right] $$

    so that $ \det{\mathcal{T}} = Ady^2 - Bdxdy + Cdx^2 = 0 \rightarrow A(\frac{dy}{dx})^2 - B(\frac{dy}{dx}) + C = 0$ implies $ \frac{dy}{dx} = \frac{B \pm \sqrt{B^2 - 4AC}}{2A}$

    These are the differential equations of the characteristics, splitting into three cases: two real equations, one real equation or two complex equations... Along the characteristics we then have to solve $H(x,y,\phi, \phi_x, \phi_y) = 0 $, & my guess is that the solution for this may not be the same solution for the general second order PDE, thus there will be discontinuities in the solution?

But then in this note characteristics were those curves along which the coefficient of the first term vanishes, so that a power series solution may be constructed. Here we apparently have two conflicting descriptions of what a characteristic is, definitions that don't seem to merge as equivalent.

Thirdly, in this note characteristics are shown to arise via a change of variables aimed at eliminating higher order derivatives, how does a change of variables relate to all of this? I imagine it's something to do with choosing coordinates so that the characteristics in the domain look nicer, but I have no concrete idea as to how it applies geometrically...

Finally, how does all of this, whatever the right way of looking at it actually is, relate to characteristics of first order PDE's exactly? On pages 5 - 6 a nice geometric interpretation of first order characteristics is given, but it looks like it has nothing to do with what has been described above.

What's going on?

$\endgroup$
7
+50
$\begingroup$

Warning: the following is by no means rigorous or even remotely constitute as proof of anything. This is just an intuition about characteristic lines to understand what purpose they really serve and how they relate to discontinuities.

TL;DR: Characteristic curves are curves where values propagate as in the transport equation. This is why derivative = $0$. Only characteristic curves can propagate discontinuities, as in the discontinuities don't survive the next timestep outside those curves. Therefore, looking for discontinuities (or persistent discontinuities) amount to looking for characteristic curves. Hence, defining characteristic curves as where the higher order terms are discontinuous.


The way I understand characteristic curves is that I think of them as some parametric curves where things don't change much. (As in some derivatives/partial derivatives/gradient = $0$)

The reason these things are useful is because these curves allow us to simplify PDE's into simpler differential equations that are easier to analyze. (Pro-tip: it's always about $\frac{dy}{dx}$, at least in 2d)

In your first example, you showed it when you said that you want $\vec{v} = (\frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial x \partial y}, \frac{\partial^2 \phi}{\partial y^2}) = \vec{0}$. This is intuitively the second order equivalent of saying that $\frac{du}{dt} = 0 $ in first order equations.

More on intuition:

Think back to the transport equation (i know... basics). What does the transport equation do along characteristic curves? It just propagates some value forward in time in a very clean and simply describable way. The same thing is true even with second order equations except, it might not be as simple as in the transport equation case.


Example:

Say $u$ is a function of $x$ and $t$ and $x$ is a function of $t$, i.e: $u(x(t), t)$

Now, suppose you wanted to see how u changed with time: i.e: $\frac{du}{dt}$

$\frac{du}{dt} = u_x \frac{dx}{dt} + u_t$ by chain rule

But if you look closely, this is just the transport equation in disguise:

recall: transport eq: $0 = cu_x + u_t$. So the transport equation has the peculiarity of $\frac{dx}{dt} = c$ and

$\frac{du}{dt} = 0$ <----- Our famous zero derivative

(helpful source: http://www.youtube.com/watch?v=tNP286WZw3o )

So, if we apply the "pro-tip" with x as y and t as x, we have that the characteristic curves are described by $\frac{dx}{dt} = c$, then $x = ct + x_0$ defines our characteristic curves given that $x_0$ is the initial value for $x$

Then, since $\frac{du}{dt} = 0$, u is constant along the characteristic lines.

And in reality, this is all characteristic curves are. Think of them as little rivers of simple behavior in an otherwise complex topological landscape filled with scary discontinuities and shocks and other horrible monsters that come out of PDE's.


Now, why did the first source you cited focus on discontinuities on the higher order terms? Turns out it's a nice trick. Basically, only characteristics can handle discontinuities because there's no chance for characteristic curves to do crazy stuff. (their behavior is in some way loosely constant-ish...)

Check out this source: http://www.cims.nyu.edu/~corona/hw/PDE%20NOTES%20Shatah%2009.pdf under 1.5.2 Characteristics as the carriers of discontinuity

Hopefully the source will be enlightening a bit because you'll see some of the talk of the whole vanishing stuff.

So, if you know that the discontinuities will only be carried through by characteristic curves, then you can "define" that characteristic curves are the curves such that the higher order terms are discontinuous.

If you have a hard time seeing this whole characteristic curves as carriers of discontinuity, think of Burger's equation. (I know, it's non-linear, but bear with me).

Look at where in Burger's equation you are forced to insert the shocks: where characteristic lines meet. Burger's equation, as non-linear as it looks, doesn't get any other discontinuities but those due to the characteristic lines.

This is why your first source goes for the discontinuities at the higher order terms. Any discontinuities that may occur, as the professor in the video says, will just vanish and not be there at the next timestep. Especially in cases like the heat equation with infinite propagation speed.

Hope this helps.

$\endgroup$
  • 2
    $\begingroup$ Fantastic answer, a bit too much for me on anything more than an intuitive level - but I got the intuition thanks to you. Around minute 31 of this video youtube.com/watch?v=l6ghiYfkwKc there is, what seems to be, an alternative explanation involving Cramer's rule & dividing by zero on the matrix in my post, which is basically the one step missing from that whole explanation. I think this may remove the trickery of the higher order terms you've mentioned, or hopefully provide a clearer link (once I fully understand it all), thanks $\endgroup$ – bolbteppa Nov 11 '13 at 6:34

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.