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**