Okay so I am super confused on what the method of characteristics is and what it means geometrically. So my first question is if anyone could kindly explain what characteristic lines are, why its considered a "change of coordinates" and just a little geometrical meaning behind it.

Now, after that consider the wave equation $$u_{tt} = c^2 u_{xx}$$

If its okay with everyone I'd like just reference a pdf because my latex speed isn't that great and its 4:15 am. I have finals in a couple of days. here is one I really like and could follow it easily: http://www.maths.uq.edu.au/~asj/PDES/set3.pdf

Now once you go to the example section, they have two examples.

  • The first one is no initial speed but an initial displacement.
  • The second one is a "hammer" problem where a "blow" slows down the initial speed.

Both solutions are given in terms of "cases". I understand them perfectly HOWEVER I do not understand the graphs behind it. Basically I want to time them for t = 0, 1, 2. This is given for the first example but not for the second (not that I get the first example picture).

Secondly, is there an analytical solution to D'Alembert's solution? I mean if that question shows up on the final (and i have a feeling a variation of it will) then can I leave my answer as cases or is there a way to find an explicit formula for it?

So to recap I am just looking for intuition behind method of characteristics and how it applies to the wave equation. Bonus points (or my eternal appreciation) to someone who can type it up from scratch going through every point and explaining it but thats me being selfish and asking for too much :P

Thank you!

edit: page 5 of this pdf http://www.math.ubc.ca/~seymour/400wave5.pdf has the sketches for the other example (the hammer blow example) but i just dont understand how they are coming up with those sketches.

  • $\begingroup$ Note this somewhat related question: math.stackexchange.com/questions/88443/…. $\endgroup$ – joriki Dec 6 '11 at 11:32
  • $\begingroup$ There's quite a bit of geometric interpretation in the Wikipedia article I pointed out in response to your other question. It would make your question a lot easier to answer if you tell us what part of that you don't understand. $\endgroup$ – joriki Dec 6 '11 at 11:35
  • 2
    $\begingroup$ The intuition behind the method of characteristics has been written by many authors over the years. You can start with Wikipedia, Courant-Hilbert Methods of Mathematics Physics vol.2 ch.2 p.62 et seq, or F. John's Partial Differential Equations. (Just to check: do you understand the method of characteristics for first order equations? You should do so before attempting higher order equations). $\endgroup$ – Willie Wong Dec 6 '11 at 13:12
  • $\begingroup$ No I dont, @WillieWong $\endgroup$ – Tyler Hilton Dec 6 '11 at 17:21
  • $\begingroup$ the links are broken. $\endgroup$ – timur Aug 12 '12 at 15:09

With reference to http://en.wikipedia.org/wiki/D%27Alembert%27s_formula :

The characteristics of this PDE are $x\pm ct=\mathrm{const}$

Let $\mu=x+ct$ , $\eta=x-ct$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial\mu}\dfrac{\partial\mu}{\partial x}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial x}=\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\mu}\left(\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\mu}{\partial x}+\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial x}=\dfrac{\partial^2u}{\partial\mu^2}+\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}=\dfrac{\partial^2u}{\partial\mu^2}+2\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial\mu}\dfrac{\partial\mu}{\partial t}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial t}=c\dfrac{\partial u}{\partial\mu}-c\dfrac{\partial u}{\partial\eta}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(c\dfrac{\partial u}{\partial\mu}-c\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\mu}\left(c\dfrac{\partial u}{\partial\mu}-c\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\mu}{\partial t}+\dfrac{\partial}{\partial\eta}\left(c\dfrac{\partial u}{\partial\mu}-c\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial t}=c^2\dfrac{\partial^2u}{\partial\mu^2}-c^2\dfrac{\partial^2u}{\partial\mu\eta}-c^2\dfrac{\partial^2u}{\partial\mu\eta}+c^2\dfrac{\partial^2u}{\partial\eta^2}=c^2\dfrac{\partial^2u}{\partial\mu^2}-2c^2\dfrac{\partial^2u}{\partial\mu\eta}+c^2\dfrac{\partial^2u}{\partial\eta^2}$

$\therefore c^2\dfrac{\partial^2u}{\partial\mu^2}-2c^2\dfrac{\partial^2u}{\partial\mu\eta}+c^2\dfrac{\partial^2u}{\partial\eta^2}=c^2\dfrac{\partial^2u}{\partial\mu^2}+2c^2\dfrac{\partial^2u}{\partial\mu\eta}+c^2\dfrac{\partial^2u}{\partial\eta^2}$





  • $\begingroup$ The correct partial derivatives can be found here. You can patch this up. $\endgroup$ – user26872 Jul 11 '12 at 2:54
  • $\begingroup$ @oen: Thanks to reminder. My answer has been corrected now. $\endgroup$ – doraemonpaul Jul 11 '12 at 8:34

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