# Chain Rule & Partial derivatives

Assuming we have a function $$u(x,t)$$ and using the change of variables $$\xi=x-t$$ and $$\eta=x+t$$ we define a new function $$v(\xi,\eta)=u(x,t)$$.

I am attempting to take compute the partial derivatives $$\frac{\partial^2u}{\partial t^2}$$ and $$\frac{\partial^2u}{\partial x^2}$$. For the first order partial derivatives I have found that: $$$$\frac{\partial u}{\partial t}=\frac{\partial v}{\partial \xi}\frac{\partial \xi}{\partial t}+\frac{\partial v}{\partial \eta}\frac{\partial \eta}{\partial t}=-\frac{\partial v}{\partial \xi}+\frac{\partial}{\partial \eta}$$$$ Similarly, for differentiation w.r.t $$x$$ I have: $$$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial \xi}+\frac{\partial v}{\partial \eta}$$$$ Now for the second order derivatives I have used: $$$$\frac{\partial^2 u}{\partial^2 t}=\frac{\partial}{\partial t}\left( -\frac{\partial v}{\partial \xi}+\frac{\partial}{\partial \eta} \right)=-\left( \frac{\partial^2 v}{\partial \xi^2}\frac{\partial \xi}{\partial t} +\frac{\partial^2 v}{\partial \xi \partial\eta}\frac{\partial \eta}{\partial t}\right)+\left( \frac{\partial^2 v}{\partial \eta^2}\frac{\partial \eta}{\partial t} +\frac{\partial^2 v}{\partial \eta \partial\xi}\frac{\partial \xi}{\partial t}\right)=\frac{\partial^2v}{\partial \xi^2}-2\frac{\partial^2 v}{\partial \eta\partial\xi}+\frac{\partial^2 v}{\partial \eta^2}$$$$ Is my attempt correct or I have missed something?

• – amd May 6 at 20:12