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Homework Help: Hyperbolic PDE with only one characteristic

  1. Jul 21, 2015 #1
    Hello all,

    1. The problem statement, all variables and given/known data
    $$x{u_{xy}} - y{u_{yy}} = 0$$ Assume $$x,y \in {\rm{Reals}}$$

    2. Relevant equations
    I have been able to solve this using different methods, but my classmates and I are trying to figure out if there is a way to do this using the methods from the course's text. The problem is out of Kreyszig's Advanced Engineering Mathematics. It introduces the concept of characteristics to solve PDEs.

    The method puts a PDE in the form:
    $$A{u_{xx}} + 2B{u_{xy}} + C{u_{yy}} = F\left( {x,y,u,{u_x},{u_y}} \right)$$
    Then if
    $$AC - {B^2}\left\{ {\matrix{
    { < 0} & \Rightarrow & {{\rm{Hyperbolic}}} \cr
    { = 0} & { \Rightarrow {\rm{ }}} & {{\rm{Parabolic}}} \cr
    { > 0} & { \Rightarrow {\rm{ }}} & {{\rm{Elliptic}}} \cr

    } } \right.$$
    Then solve the characteristic equation $$A{\left( {y'} \right)^2} - 2By' + C = 0$$ where $$y' = {{dy} \over {dx}}$$ to find constants ##\phi \left( {x,y} \right)## and ##\psi \left( {x,y} \right)##.

    Knowing the type of equation allows one to make a substitution of variables. The equation of interest is hyperbolic, so $$v = \phi \left( {x,y} \right)$$ and $$w = \psi \left( {x,y} \right)$$.
    The normal form of such a PDE is $${u_{vw}} = F$$
    With this background, please give input on this problem.

    3. The attempt at a solution
    $$x{u_{xy}} - y{u_{yy}} = 0$$
    & \Rightarrow A = 0,B = {x \over 2},C = - y \cr
    & \Rightarrow AC - {B^2} = - {{{x^2}} \over 4} \cr} $$
    If ##x = 0## the PDE is parabolic. My professor said to ignore this case, as it does not produce meaningful solutions. (This may be the source of the error).

    If ##x \ne 0##, the solution is hyperbolic.
    $$A{\left( {y'} \right)^2} - 2By' + C = - {{{x^2}} \over 2}y' - y = 0$$
    $$ \Rightarrow y = {c \over x},c \buildrel \wedge \over = {\rm{const}}{\rm{.}}$$
    $$ \Rightarrow c = xy$$
    $$\phi = xy,\psi = ?$$
    Here is the rub! There exists only one, non-repeating characteristic which is not consistent with the hyperbolic change of variables! Therefore,
    $$v = xy,w = ?$$
    Using the internet, the author chose ##v = xy,w = x##, which he introduces as the change of variables for the parabolic case, yet in his solution, he defines the equation as hyperbolic.

    From here, we cannot find a way to arrive at the solution using the method described in the text. We have substituted ##x \leftrightarrow y##, and were able to arrive at a solution, but this is not a method discussed in the text.

    Is this bad text writing, or are we missing something? If we are, I imagine it has to do with the assertion that the parabolic case is not meaningful.

    Thank you for any help you can provide to sate our curiosity,
    ##\Psi \left( {x,t} \right)##

    EDIT: Small fixes and formatting.
    Last edited: Jul 21, 2015
  2. jcsd
  3. Jul 21, 2015 #2


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    Science Advisor

    The fact that the equation 'becomes parabolic at x= 0 is, in a sense, responsible. The difficulty, and solution, is that we can write the equation in terms of [itex]u_y[/itex], not just u. If we let [itex]U= u_y[/itex] the equation becomes [itex]xU_x- yU_y= 0[/itex], [itex]xU_x= yU_y[/itex] which gives characteristic [itex]\frac{dy}{y}= \frac{dx}{x}[/itex] and so [itex]y= Cx[/itex] or [itex]\frac{y}{x}= C[/itex]. The two characteristics are [itex]\phi(x,y)= xy[/itex] and [itex]\psi(x,y)= \frac{y}{x}[/itex].
  4. Jul 21, 2015 #3
    That's a much more elegant method than my solution! Thank you for your input. Just to clarify, to get a characteristic for the original equation and get to the author's ##w=x##, one would differentiate both sides by ##y##, and define a new constant, correct?

    So it would appear, that using the exact method of detailed in the section of the problem will not yield any solution? The book is not very well edited to begin with, so this would not surprise me.

    Thank you for your time,
    ##\Psi \left(x,t \right)##
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