Rotation by 45 degrees

**dwsmith** · January 19th 2011, 01:53 PM

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.

**dwsmith** · January 19th 2011, 01:58 PM

Would it just be making alpha $\pm\frac{\pi}{4}$ of

$\displaystyle u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) \mbox{?}$

**dwsmith** · January 20th 2011, 11:51 AM

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?

**dwsmith** · January 22nd 2011, 11:33 AM

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.

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