Answer to Question #215395 in Differential Equations for Asmita

Question

Reduce the PDE uxx+xuyy
into a canonical form in the region
x<0.

Answer 1

Comparing the given DE with

$aU_{xx} + 2bU_{xy} + cU_{yy}$

We have that

$a = 1, b=0, c = x$

But

$\dfrac{dy}{dx} = \dfrac{b \pm \sqrt{b^2-ac}}{a}$

In this case,

$\dfrac{dy}{dx} = \pm \sqrt{-x}$

This value will be a real number since we are considering the region x < 0

$dy = \pm (-x)^\frac{1}{2}dx \\ y = \pm \dfrac{ 2(-x)^\frac{3}{2}}{3} + c$

Set

$\xi = 3y -2 (-x)^\frac{3}{2} ....... (*)$

And

$\eta = 3y + 2 (-x)^\frac{3}{2}...... (**)$

$\xi_x = -3(-x)^{\frac{1}{2}} \\ \eta_x = 3(-x)^{\frac{1}{2}}$

$\xi_y = 3 \\ \eta_y = 3$

$\xi_{xx} =- \frac{3}{2}(-x)^{-\frac{1}{2}} \\ \eta_{xx}= \frac{3}{2}(-x)^{-\frac{1}{2}}$

$\xi_{yy}= \eta_{yy} = 0$

So we have that

$U_x = U_{\xi} \xi_x + U_{\eta} \eta_{x} = -3(-x)^{\frac{1}{2}} U_{\xi} + 3(-x)^{\frac{1}{2}} U_{\eta}$

$U_{xx} = -\frac{3}{2}(-x)^{-\frac{1}{2}} U_{\xi}-9x U _{\xi \xi} + \frac{3}{2}(-x)^{-\frac{1}{2}} U_{\eta}-9x U _{\eta \eta}$

$U_y = U_{\xi} \xi_y + U_{\eta} \eta_{y} = 3 U_{\xi} + 3 U_{\eta}$

$U_{yy} = 9U_{\xi \xi} + 3 U_{\eta \eta}$

From (*) and (**), we have that

$\eta - \xi = 4 (-x)^{\frac{3}{2}} \\ (\dfrac{4}{\eta - \xi})^ \frac{1}{3} = (-x)^{-\frac{1}{2}} .... (***)$

Substituting the value of $U_{xx}$ and $U_{yy}$ into the equation, we have

$U_{xx} + xU_{yy} = (U_{\eta} - U_{\xi}) \frac{3}{2}(-x)^{-\frac{1}{2}}$

And substituting (***), we have

$U_{xx}+xU_{yy} = \dfrac{3(U_{\eta} - U_{\xi})}{[2(\eta - \xi)]^ \frac{1}{3}}$