PDE - Change of coordinate axes.

I'm confused how these second partials are calculated. Maybe someone can me remind me. I think it's from Calc3 stuff.

$$\Large \xi = x\cos\alpha+y\sin\alpha$$$$\Large \eta = -x\sin\alpha+y\cos\alpha$$

$$\Large u(x,y)\to u\left[\xi(x,y),\eta(x,y)\right]$$

First partials:$$\Large u_x = u_{\xi}\xi_x+u_{\eta}\eta_x$$$$\Large u_y = u_{\xi}\xi_y+u_{\eta}\eta_y$$

Which simplify to:$$\Large u_x = \cos\alpha u_{\xi}-\sin\alpha u_{\eta}$$$$\Large u_y = \sin\alpha u_{\xi}+\cos\alpha u_{\eta}$$

Question

PDE - Change of coordinate axes.

I'm confused how these second partials are calculated. Maybe someone can me remind me. I think it's from Calc3 stuff.

$$\Large \xi = x\cos\alpha+y\sin\alpha$$$$\Large \eta = -x\sin\alpha+y\cos\alpha$$

$$\Large u(x,y)\to u\left[\xi(x,y),\eta(x,y)\right]$$

First partials:$$\Large u_x = u_{\xi}\xi_x+u_{\eta}\eta_x$$$$\Large u_y = u_{\xi}\xi_y+u_{\eta}\eta_y$$

Which simplify to:$$\Large u_x = \cos\alpha u_{\xi}-\sin\alpha u_{\eta}$$$$\Large u_y = \sin\alpha u_{\xi}+\cos\alpha u_{\eta}$$

zepdrix · Answer

Teacher wrote that the second partials are giving us ...$$\Large u_{xx}= \cos^2\alpha (u_{\xi\xi})\color{royalblue}{-2\sin \alpha \cos \alpha (u_{\xi \eta})}+\sin^2\alpha(u_{\eta \eta})$$

zepdrix · Answer

I guess I'm especially confused on that middle term.. Maybe I'm just not remembering how to chain rule these functions of multiple variables.

zepdrix · Answer

$$\Large u_{yy}=\sin^2\alpha (u_{\xi \xi})+2\sin \alpha \cos \alpha (u_{\xi \eta})+\cos^2\alpha (u_{\eta \eta})$$

zepdrix · Answer

Phew! Luckily it wasn't on the test. :O