Canonical Transformation of Parabolic PDEs

1. Nov 21, 2007

maverick280857

Hi again

I am studying PDEs and came across a solved problem in my textbook, which describes the transformation of a parabolic second order PDE to canonical form. I want to know how to find the second canonical substitution when one has been computed from the characteristic equation.

(PS--This is not a homework problem.)

For instance, suppose the given equation is

$$y^{2}u_{xx} - 2xyu_{xy} + x^{2}u_{yy} = \frac{y^2}{x}u_{x} + \frac{x^2}{y}u_{y}$$

The solution is as follows:

Compare it with the 'standard' semi-linear second order PDE:

$$a(x,y)u_{xx} + 2b(x,y)u_{xy} + c(x,y)u_{yy} = \phi(x,y,u,u_{x},u_{y})$$

to get $a(x,y) = y^{2}$, $b(x,y) = -xy$, $c(x,y) = x^{2}$. Since $b^{2}-ac = 0$, the equation is parabolic. Considering level curves

$$\zeta(x,y) = c_{1}$$
$$\eta(x,y) = c_{2}$$

corresponding to the new independent variables $(\zeta,\eta)$, the characteristic equation is

$$a\left(\frac{dy}{dx}\right)^{2} - 2b\left(\frac{dy}{dx}\right) + c = 0$$

It has a double root $y^{2}+x^{2} = c_{1}$. Thus

$$\zeta(x,y) = x^{2} + y^{2}$$

But this determines only one of the canonical variables. The only condition on $\eta$ is that

$$\frac{\partial(\zeta,\eta)}{\partial(x,y)} \neq 0$$

which means that $\eta$ should not be explicitly dependent on $\zeta$ or conversely.

Here, it seems "natural" to take $\eta(x,y) = x^{2}-y^{2}$. But how does one find a $\eta$ in the general case?

Thanks.

Last edited: Nov 21, 2007
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