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Given $$x^2u_{xx} + 2xyu_{xy} - 3y^2u_{yy} - 2xu_x +4yu_y +16x^4u = 0$$ I have to find the canonical form. The determinant is the following $$ D = 16x^2y^2$$ I considered cases when $x$ and $y$ can be $0$. When calculating for the remaining case (where both $x$ and $y$ is not $0$ and the type is hyperbolic), I am not getting its canonical form. I am getting this $$(-3y^2 - 6x^3y + 9x^6)u_{\xi\xi} + (-3y^2 + \frac{2y}{x} + \frac{1}{x^2})u_{\eta\eta} + (-6y^2 - 6x^2 + \frac {2y(1-3x^4)}x)u_{\eta\xi} + (\frac{-4}{x}+4y)u_\eta + 4yu_\xi + 16x^4u = 0$$ But the thing is, that this doesn't meet the requirements for canonical form of hyperbolic type. Any help will be appreciated.

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  • $\begingroup$ $x^2 \partial_x^2+2xy\partial_x\partial_y -3y^2\partial_y^2 = (x\partial_x-y\partial_y)(x\partial_x+3y\partial_y)$ $\endgroup$ – Cesareo Mar 28 '19 at 8:49
  • $\begingroup$ What does this give me? $\endgroup$ – Vahe Karamyan Mar 28 '19 at 8:54
  • $\begingroup$ It gives you the characteristic curves. $\endgroup$ – Cesareo Mar 28 '19 at 10:03
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Let $$a=x^2,\quad b=xy,\quad c=-3y^2$$ Characteristic equations: $$y'=\frac{b\pm\sqrt{b^2-ac}}{a}$$ or $$y'=\frac{3y}{x}, \quad y'=-\frac{y}{x}$$ Solutions: $$\frac{y}{x^3}=c_1,\quad xy=c_2$$ Change variables $$\xi=\frac{y}{x^3},\quad \eta=xy$$ gives canonical form $$u_{\xi\eta}-\frac{4u_\xi \xi^2+\xi\eta u_\eta+4\eta u}{4\eta\xi^2}=0$$

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