Partial derivatives for multivariate functions

Question

If $u = u(x, y)$ and $\xi = x + ay, \eta = x + by$, find the values of $a$ and $b$ such that they transform the equation $$\frac{\partial^2u}{\partial x^2}+4\frac{\partial^2u}{\partial x \partial y}+3\frac{\partial^2u}{\partial y^2}=0$$ into $$\frac{\partial^2u}{\partial \xi \partial \eta}=0. $$

Answer 1

Hint: By the chain rule $$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}= \frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta}. $$ Similarly, $$ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial y}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial y}= a\frac{\partial u}{\partial \xi}+b\frac{\partial u}{\partial \eta}. $$ Then, taking another derivative, $$ \frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta}\right)= \frac{\partial^2 u}{\partial \xi^2}\frac{\partial \xi}{\partial x}+\frac{\partial^2 u}{\partial\eta\partial \xi}\frac{\partial \eta}{\partial x}+ \frac{\partial^2 u}{\partial \xi\partial\eta}\frac{\partial \xi}{\partial x}+\frac{\partial^2 u}{\partial\eta^2}\frac{\partial \eta}{\partial x} = \frac{\partial^2 u}{\partial \xi^2}+2\frac{\partial^2 u}{\partial\eta\partial \xi}+ \frac{\partial^2 u}{\partial\eta^2}. $$ Similarly, you can do this for your other partial derivatives. Then, choose $a$ and $b$ so that all but the cross term cancels.

To make the last line make more sense, let $\frac{\partial u}{\partial \xi}=v(x,y)$ and $\frac{\partial u}{\partial \eta}=w(x,y)$. Then $$ \frac{\partial v}{\partial x}=\frac{\partial v}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial v}{\partial \eta}\frac{\partial \eta}{\partial x}= \frac{\partial v}{\partial \xi}+\frac{\partial v}{\partial \eta}=\frac{\partial^2 u}{\partial \xi^2}+\frac{\partial^2 u}{\partial\eta\partial\xi}. $$ The other partial can be computed similarly.