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.

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- Jan 19th 2011, 01:53 PMdwsmithRotation by 45 degrees
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. - Jan 19th 2011, 01:58 PMdwsmith
Would it just be making alpha $\displaystyle \pm\frac{\pi}{4}$ of

$\displaystyle \displaystyle u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) \mbox{?}$ - Jan 20th 2011, 11:51 AMdwsmith
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? - Jan 22nd 2011, 11:33 AMdwsmith
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.