Canonical Transformation of Parabolic PDEs

[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

[tex]
y^{2}u_{xx} - 2xyu_{xy} + x^{2}u_{yy} = \frac{y^2}{x}u_{x} + \frac{x^2}{y}u_{y}
[/tex]

The solution is as follows:

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

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

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

[tex]\zeta(x,y) = c_{1}[/tex]
[tex]\eta(x,y) = c_{2}[/tex]

corresponding to the new independent variables [itex](\zeta,\eta)[/itex], the characteristic equation is

[tex]a\left(\frac{dy}{dx}\right)^{2} - 2b\left(\frac{dy}{dx}\right) + c = 0[/tex]

It has a double root [itex]y^{2}+x^{2} = c_{1}[/itex]. Thus

[tex]\zeta(x,y) = x^{2} + y^{2}[/tex]

But this determines only one of the canonical variables. The only condition on [itex]\eta[/itex] is that

[tex]\frac{\partial(\zeta,\eta)}{\partial(x,y)} \neq 0[/tex]

which means that [itex]\eta[/itex] should not be explicitly dependent on [itex]\zeta[/itex] or conversely.

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

Thanks.

 

[gtoguy97]

In order to find the second canonical substitution in the general case, you can use the method of characteristics. The idea is to look at the curves that are defined by the equation a(x,y)u_{xx} + 2b(x,y)u_{xy} + c(x,y)u_{yy} = 0. These curves are known as the characteristic curves and they help determine the form of the canonical substitution. Let us apply this method to the equation given in the question:y^{2}u_{xx} - 2xyu_{xy} + x^{2}u_{yy} = \frac{y^2}{x}u_{x} + \frac{x^2}{y}u_{y}The characteristic equation is:a\left(\frac{dy}{dx}\right)^{2} - 2b\left(\frac{dy}{dx}\right) + c = 0where a = y^2, b = -xy, c = x^2. This equation has a double root y^2 + x^2 = c_1. Therefore, one possible substitution is: \zeta = x^2 + y^2To find the second substitution, we need to find a function \eta such that \frac{\partial(\zeta,\eta)}{\partial(x,y)} \neq 0. This means that \eta should not be explicitly dependent on \zeta or vice versa. A possible choice for \eta is:\eta = xyTherefore, the two canonical substitutions are: \zeta = x^2 + y^2\eta = xy

 

[tacman]

To find the second canonical variable \eta, we can use the fact that the characteristic equation gives us a relation between the two variables \zeta and \eta. In this case, the characteristic equation is y^{2}+x^{2} = c_{1}, which can also be written as \zeta = c_{1} - \eta. This means that \eta can be expressed in terms of \zeta as \eta = c_{1} - \zeta. However, we still need to ensure that \eta is not explicitly dependent on \zeta, which can be done by choosing a suitable constant c_{1}. In this case, we can choose c_{1} = 0, which gives us \eta = -\zeta. Therefore, the second canonical variable is \eta = -x^{2}-y^{2}.

In general, to find the second canonical variable, we can use the characteristic equation to express one variable in terms of the other and then choose a suitable constant to ensure that the second variable is not explicitly dependent on the first. This process may require some trial and error, but it is based on the fundamental idea that the characteristic equation relates the two canonical variables.