Variable substitution in second order PDE

Question

Consider the the PDE

$$A(x, y)\partial_{xx} u + B(x, y)\partial_{xy}u + C(x, y)\partial_{yy}u=h(x, y) $$

Now I want to make a variable substitution $\xi=f(x, y), \eta=g(x,y)$, so I can get $u$ as a function of $\xi$ and $\eta$. So

$$\partial_{x}u=\partial_{x}\xi\partial_{\xi}u + \partial_{x}\eta\partial_{\eta}u$$

Then $$\partial_{xx}u=\partial_{x}\left(\partial_{x}\xi\partial_{\xi}u + \partial_{x}\eta\partial_{\eta}u\right)\overset{?}{=}\partial_{xx}\xi\partial_{\xi}u + \partial_{x}\xi\partial_{x\xi}u + \partial_{xx}\eta\partial_{\eta}u + \partial_{x}\eta\partial_{x\eta}u$$

I am supposed to find terms like $\partial_{\xi\xi} u$, but how? The question-mark over the equality is that I am unsure if and how the chain rule should be applied.

Answer 1

The equality $\partial_{xx}u=\partial_{x}\left(\partial_{x}\xi\partial_{\xi}u + \partial_{x}\eta\partial_{\eta}u\right)\overset{?}{=}\partial_{xx}\xi\partial_{\xi}u + \partial_{x}\xi\partial_{x\xi}u + \partial_{xx}\eta\partial_{\eta}u + \partial_{x}\eta\partial_{x\eta}u$ is wrong, because $\partial_x\xi$ depends on $x$ and $\partial_\xi u$ too.

I suppose you need only linear change of variable (when $A,B,C$ are constant). In this case it's easier.

Let $f(x,y)=ax+by$ and $g(x,y)=cx+dy$. In this case $\partial_x \xi=a$ and $\partial_x \eta=c$, so:

$\partial_{x}u=\partial_{x}\xi\partial_{\xi}u + \partial_{x}\eta\partial_{\eta}u=a\partial_{\xi}u+c\partial_{\eta}u$

Then as you write:

$\partial_{xx}u=\partial_x(a\partial_{\xi}u+c\partial_{\eta}u)=a^2\partial_{\xi \xi} u+ 2ac\partial_{\xi \eta} u+ c^2\partial_{\eta \eta}u$

$\bf{Edit:}$You can apply chain rule again to $v=\partial_{\xi}u$, you get:

$\partial_{x}v=\partial_{x}\xi\partial_{\xi}v + \partial_{x}\eta\partial_{\eta}v=\partial_{x}\xi\partial_{\xi \xi}u + \partial_{x}\eta\partial_{\eta \xi}u$

The same with $\partial_{\eta x}u$

Answer 2

Let $\begin{cases}\xi=f(x,y)\\\eta=g(x,y)\end{cases}$ ,

Then $\partial_xu=\partial_\xi u\partial_x\xi+\partial_\eta u\partial_x\eta=\partial_xf(x,y)\partial_\xi u+\partial_xg(x,y)\partial_\eta u$

$\partial_{xx}u=\partial_x(\partial_xf(x,y)\partial_\xi u+\partial_xg(x,y)\partial_\eta u)=\partial_xf(x,y)\partial_x(\partial_\xi u)+\partial_{xx}f(x,y)\partial_\xi u+\partial_xg(x,y)\partial_x(\partial_\eta u)+\partial_{xx}g(x,y)\partial_\eta u=\partial_xf(x,y)(\partial_\xi(\partial_\xi u)\partial_x\xi+\partial_\eta(\partial_\xi u)\partial_x\eta)+\partial_{xx}f(x,y)\partial_\xi u+\partial_xg(x,y)(\partial_\xi(\partial_\eta u)\partial_x\xi+\partial_\eta(\partial_\eta u)\partial_x\eta)+\partial_{xx}g(x,y)\partial_\eta u=\partial_xf(x,y)(\partial_xf(x,y)\partial_{\xi\xi}u+\partial_xg(x,y)\partial_{\xi\eta}u)+\partial_{xx}f(x,y)\partial_\xi u+\partial_xg(x,y)(\partial_xf(x,y)\partial_{\xi\eta}u+\partial_xg(x,y)\partial_{\eta\eta}u)+\partial_{xx}g(x,y)\partial_\eta u=(\partial_xf(x,y))^2\partial_{\xi\xi}u+2\partial_xf(x,y)\partial_xg(x,y)\partial_{\xi\eta}u+(\partial_xg(x,y))^2\partial_{\eta\eta}u+\partial_{xx}f(x,y)\partial_\xi u+\partial_{xx}g(x,y)\partial_\eta u$

$\partial_yu=\partial_\xi u\partial_y\xi+\partial_\eta u\partial_y\eta=\partial_yf(x,y)\partial_\xi u+\partial_yg(x,y)\partial_\eta u$

$\partial_{xy}u=\partial_x(\partial_yf(x,y)\partial_\xi u+\partial_yg(x,y)\partial_\eta u)=\partial_yf(x,y)\partial_x(\partial_\xi u)+\partial_{xy}f(x,y)\partial_\xi u+\partial_yg(x,y)\partial_x(\partial_\eta u)+\partial_{xy}g(x,y)\partial_\eta u=\partial_yf(x,y)(\partial_\xi(\partial_\xi u)\partial_x\xi+\partial_\eta(\partial_\xi u)\partial_x\eta)+\partial_{xy}f(x,y)\partial_\xi u+\partial_yg(x,y)(\partial_\xi(\partial_\eta u)\partial_x\xi+\partial_\eta(\partial_\eta u)\partial_x\eta)+\partial_{xy}g(x,y)\partial_\eta u=\partial_yf(x,y)(\partial_xf(x,y)\partial_{\xi\xi}u+\partial_xg(x,y)\partial_{\xi\eta}u)+\partial_{xy}f(x,y)\partial_\xi u+\partial_yg(x,y)(\partial_xf(x,y)\partial_{\xi\eta}u+\partial_xg(x,y)\partial_{\eta\eta}u)+\partial_{xy}g(x,y)\partial_\eta u=\partial_xf(x,y)\partial_yf(x,y)\partial_{\xi\xi}u+(\partial_yf(x,y)\partial_xg(x,y)+\partial_xf(x,y)\partial_yg(x,y))\partial_{\xi\eta}u+\partial_xg(x,y)\partial_yg(x,y)\partial_{\eta\eta}u+\partial_{xy}f(x,y)\partial_\xi u+\partial_{xy}g(x,y)\partial_\eta u$

$\partial_{yy}u=\partial_y(\partial_yf(x,y)\partial_\xi u+\partial_yg(x,y)\partial_\eta u)=\partial_yf(x,y)\partial_y(\partial_\xi u)+\partial_{yy}f(x,y)\partial_\xi u+\partial_yg(x,y)\partial_y(\partial_\eta u)+\partial_{yy}g(x,y)\partial_\eta u=\partial_yf(x,y)(\partial_\xi(\partial_\xi u)\partial_y\xi+\partial_\eta(\partial_\xi u)\partial_y\eta)+\partial_{yy}f(x,y)\partial_\xi u+\partial_yg(x,y)(\partial_\xi(\partial_\eta u)\partial_y\xi+\partial_\eta(\partial_\eta u)\partial_y\eta)+\partial_{yy}g(x,y)\partial_\eta u=\partial_yf(x,y)(\partial_yf(x,y)\partial_{\xi\xi}u+\partial_yg(x,y)\partial_{\xi\eta}u)+\partial_{yy}f(x,y)\partial_\xi u+\partial_yg(x,y)(\partial_yf(x,y)\partial_{\xi\eta}u+\partial_yg(x,y)\partial_{\eta\eta}u)+\partial_{yy}g(x,y)\partial_\eta u=(\partial_yf(x,y))^2\partial_{\xi\xi}u+2\partial_yf(x,y)\partial_yg(x,y)\partial_{\xi\eta}u+(\partial_yg(x,y))^2\partial_{\eta\eta}u+\partial_{yy}f(x,y)\partial_\xi u+\partial_{yy}g(x,y)\partial_\eta u$

But still very difficult to completely eliminate $x$ and $y$ generally