Given the 2nd order linear PDE
\begin{align} x^2u_{xx}-2xtu_{xt}+t^2u_{tt}+xu_x+tu_t &=0, & &x>0,\ t \in \mathbb{R} \tag 1 \end{align}
The principal part of the 2nd order equation is: $$x^2u_{xx}-2xtu_{xt}+t^2u_{tt}$$
1.Where $$a=x^2,\ b=-xt,\ c=t^2$$ So, the discriminant is $$D=b^2-ac=(-xt)^2-x^2\cdot t^2=x^2t^2-x^2t^2=0$$
That means it has a parabolic form.
- How does this equation look in canonical form?
We have to find the characteristics $$\frac{dt}{dx}=\frac{b\pm \sqrt{D}}{a}\Leftrightarrow \frac{dt}{dx}=\frac{-xt}{x^2}=-\frac{t}{x}$$
and the solution is $$t=\frac{c}{x}\Leftrightarrow c=x \cdot t$$
We consider the transformation to be \begin{align} \xi&=xt\\ \eta&=x \end{align}
So
$$J=\begin{vmatrix} \xi_x & \xi_t \\ \eta_x & \eta_t \end{vmatrix}=\begin{vmatrix} t & x \\ 1 & 0 \end{vmatrix}=x \neq 0$$
We know that $x>0$ by assumption.
Then the transformation is smooth because $x \neq 0$
We also know that $$u=u(\xi,\eta)=u(\xi(x,y),\eta(x,y))$$
Thus,
- $u_x=tu_{\xi}+u_{\eta}$
- $u_t=xu_{\xi}$
- $u_{x x}= t^2u_{\xi \xi}+t u_{\xi \eta}+tu_{\xi \eta}+u_{\eta \eta}$
- $u_{tt}=x^2u_{\xi \xi}$
- $u_{xt}=xtu_{\xi \xi}+xu_{\xi \eta}+u_{\xi}$
If we plug in all the above to the $(1)$ gives $$x^2u_{\eta \eta}+u_{\eta}=0$$.
Is it possible something like that?
Could anyone verify the second-order partial derivative, please?