I don't need help with the problem just with how do I do a rotation of the axes by $\displaystyle \pm\frac{\pi}{4}$.
$\displaystyle u_{xx}-u_{yy}=0$
$\displaystyle u_{xy}=0$
Thanks.
I am going to transform $\displaystyle u_{xx}-u_{yy}=0$ into $\displaystyle u_{xy}=0$ via the specified rotations.
$\displaystyle \omega_{\xi\xi}(\cos^2{\alpha}-\sin^2{\alpha})-4\omega_{\xi\eta}\cos{\alpha}\sin{\alpha}+\omega_{ \eta\eta}(\sin^2{\alpha}-\cos^2{\alpha})=0$
$\displaystyle \displaystyle\alpha=\pm\frac{\pi}{4}$
$\displaystyle \pm 2\omega_{\xi\eta}=0\Rightarrow\omega_{\xi\eta}=0\R ightarrow u_{xy}=0$
Since $\displaystyle u_{xx}-u_{yy}=0$ can be transformed into $\displaystyle u_{xy}=0$, does it suffice to show the one transformation?
Or if I want to show that $\displaystyle u_{xy}=0$ can be transformed into $\displaystyle u_{xx}-u_{yy}=0$, must I demonstrate it as well?
Since both these equations can be transformed into each other, should they have the same general solution?
For $\displaystyle u_{xx}-u_{yy}=0$, I obtained $\displaystyle u(x,y)=[c_1\exp{(x\sqrt{\lambda})}+c_2\exp{(-x\sqrt{\lambda})}]\cdot[c_3\exp{(y\sqrt{\lambda)}+c_4\exp{(-y\sqrt{\lambda})}]$, and if $\displaystyle c_1=c_2 \ \ \mbox{and} \ \ c_3=c_4$, then $\displaystyle u(x,y)=k\cosh{(x\sqrt{\lambda})}\cdot\cosh{(y\sqrt {\lambda})}$.
However, for $\displaystyle u_{xy}=0$, I obtained $\displaystyle u(x,y)=f(x)+g(y)$.
Thanks.