# Rotation by 45 degrees

• Jan 19th 2011, 02:53 PM
dwsmith
Rotation by 45 degrees
I don't need help with the problem just with how do I do a rotation of the axes by $\pm\frac{\pi}{4}$.

$u_{xx}-u_{yy}=0$

$u_{xy}=0$

Thanks.
• Jan 19th 2011, 02:58 PM
dwsmith
Would it just be making alpha $\pm\frac{\pi}{4}$ of

$\displaystyle u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) \mbox{?}$
• Jan 20th 2011, 12:51 PM
dwsmith
I am going to transform $u_{xx}-u_{yy}=0$ into $u_{xy}=0$ via the specified rotations.

$\omega_{\xi\xi}(\cos^2{\alpha}-\sin^2{\alpha})-4\omega_{\xi\eta}\cos{\alpha}\sin{\alpha}+\omega_{ \eta\eta}(\sin^2{\alpha}-\cos^2{\alpha})=0$

$\displaystyle\alpha=\pm\frac{\pi}{4}$

$\pm 2\omega_{\xi\eta}=0\Rightarrow\omega_{\xi\eta}=0\R ightarrow u_{xy}=0$

Since $u_{xx}-u_{yy}=0$ can be transformed into $u_{xy}=0$, does it suffice to show the one transformation?

Or if I want to show that $u_{xy}=0$ can be transformed into $u_{xx}-u_{yy}=0$, must I demonstrate it as well?
• Jan 22nd 2011, 12:33 PM
dwsmith
Since both these equations can be transformed into each other, should they have the same general solution?

For $u_{xx}-u_{yy}=0$, I obtained $u(x,y)=[c_1\exp{(x\sqrt{\lambda})}+c_2\exp{(-x\sqrt{\lambda})}]\cdot[c_3\exp{(y\sqrt{\lambda)}+c_4\exp{(-y\sqrt{\lambda})}]$, and if $c_1=c_2 \ \ \mbox{and} \ \ c_3=c_4$, then $u(x,y)=k\cosh{(x\sqrt{\lambda})}\cdot\cosh{(y\sqrt {\lambda})}$.

However, for $u_{xy}=0$, I obtained $u(x,y)=f(x)+g(y)$.

Thanks.