Transforming 2nd order PDE into canonical form
Dear MHF members,
first of all glad to see the form back again.
I have the following problem $yz_{xx}+z_{yy}=0$.
When I check the discriminant, I see that $\Delta=0^2-4y=-4y$.
Hence, the following cases occur.
(a) For $y<0$, the equation is of hyperbolic type.
(b) For $y=0$, the equation is of parabolic type.
(c) For $y>0$, the equation is of elliptic type.
I have no problem with the cases (b) and (c).
The equation is already in its canonical form in (b),
and in (c) I can transform the equation in this case
into its canonical form $z_{\xi\xi}+z_{\eta\eta}+\frac{1}{3\eta}z_{\eta}=0 $
by the change of variables $\xi=\frac{2}{3}x$ and $\eta=y^{3/2}$.
But I really stuck at the part (a). I don't know why
I cannot transform the equation into its canonical form.
There must to be a trick in this case, which I miss.
Thank you.
bkarpuz
Re: Transforming 2nd order PDE into canonical form
Quote:
Originally Posted by
bkarpuz 
I have the following problem $yz_{xx}+z_{yy}=0$.
When I check the discriminant, I see that $\Delta=0^2-4y=-4y$.
Hence, the following cases occur.
(a) For $y<0$, the equation is of hyperbolic type.
...
But I really stuck at the part (a). I don't know why
I cannot transform the equation into its canonical form.
There must to be a trick in this case, which I miss.
Okay, I got the solution. While obtaining the characteristics I, in general, used
$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm i\frac{1}{\sqry{y}}$ but here it makes problem.
Thus I took the reciprocal equation $\frac{\mathrm{d}y}{\mathrm{d}x}=\pm i\sqrt{y}$,
which gives $\xi=\frac{3}{2}x+(-y)^{3/2}$ and $\eta=\frac{3}{2}x-(-y)^{3/2}$.
Thus, the equation transforms into $z_{\xi\eta}+\frac{1}{6(\xi-\eta)}(z_{\xi}-z_{\eta})=0$.
Have a good day.
bkarpuz
