- #1
psiofxandt
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Hello all,
$$x{u_{xy}} - y{u_{yy}} = 0$$ Assume $$x,y \in {\rm{Reals}}$$
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.
$$x{u_{xy}} - y{u_{yy}} = 0$$
$$\eqalign{
& \Rightarrow A = 0,B = {x \over 2},C = - y \cr
& \Rightarrow AC - {B^2} = - {{{x^2}} \over 4} \cr} $$
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.
Homework Statement
$$x{u_{xy}} - y{u_{yy}} = 0$$ Assume $$x,y \in {\rm{Reals}}$$
Homework 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.
The Attempt at a Solution
$$x{u_{xy}} - y{u_{yy}} = 0$$
$$\eqalign{
& \Rightarrow A = 0,B = {x \over 2},C = - y \cr
& \Rightarrow AC - {B^2} = - {{{x^2}} \over 4} \cr} $$
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: