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Characteristics - What's Going On?

  1. Aug 16, 2013 #1
    What's really going on as regards characterstics?

    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 [itex]B^2 - 4AC[/itex] classification of partial differential equations.

    For those who are interested, the method is that given

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

    we basically want [itex] \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) [/itex] & since this is a vector, we construct an equation of the form [itex] \mathcal{T}(\vec{v}) = \vec{w} [/itex] & examine the case where a solution [itex] \vec{v}[/itex] doesn't exist, i.e. a case where an infinite number of solutions exist.


    [tex] 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 [/tex]

    [tex] 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 [/tex]

    we arrive at

    [tex] \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] [/tex]

    so that [itex] \det{\mathcal{T}} = Ady^2 - Bdxdy + Cdx^2 = 0 \rightarrow A(\frac{dy}{dx})^2 - B(\frac{dy}{dx}) + C = 0[/itex] implies [itex] \frac{dy}{dx} = \frac{B \pm \sqrt{B^2 - 4AC}}{2A} [/itex]
    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 [itex] H(x,y,\phi, \phi_x, \phi_y) = 0 [/itex], & 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 (is that tiny discussion given an actual proof of the Cauchy-Kovalevsky theorem, just without ε-δ?'s???). 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?
  2. jcsd
  3. Aug 19, 2013 #2


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    The basic idea is we can factor a partial differential equation into ordinary differential equations. A usual example is
    $$\dfrac{\partial ^2}{\partial t^2}-c^2\dfrac{\partial^2}{\partial x^2}=\left( \dfrac{\partial }{\partial t}-c\dfrac{\partial}{\partial x}\right) \left( \dfrac{\partial }{\partial t}+c\dfrac{\partial}{\partial x}\right)=\dfrac{\partial }{\partial u}\dfrac{\partial }{\partial v} \\ \text{where} \\
    \dfrac{\partial }{\partial u}=\dfrac{\partial }{\partial t}-c\dfrac{\partial}{\partial x} \\
    \dfrac{\partial }{\partial v}=\dfrac{\partial }{\partial t}+c\dfrac{\partial}{\partial x} \\$$
    Thus we have reduced the solution of a partial differential equation to the solution of two ordinary differential equations.
    Last edited: Aug 19, 2013
  4. Aug 19, 2013 #3
    Thanks for the input, unfortunately I don't yet see how that explains the situation as regards characteristics. In one case characteristics are defined by the fact that the highest partials are all simultaneously zero, in the next they are defined by the fact that (only one of? or all of) the coefficients are zero. If both conditions hold then it looks like you'd get two sets of completely different characteristics, one associated to the coefficients, the other to the partials... How a change of variables affects this is beyond me.

    Also, I don't really understand why it is that the characteristics being parallel to the solution curve is a bad thing? In the picture I linked to above for first order PDE's, surely the fact you know your initial curve lies along the characteristic curve gives you the information you get otherwise thinking of normal vectors - no? Further, how does this picture tie into 2nd order PDE's? It just doesn't make sense to me because even the basic definitions seem to contradict or go against one another...

    Finally, I don't have any intuitive feel for why curves partitioning the domain lead to such drastically different solutions, i.e. why the characteristics cause such drastically different solutions :frown:
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