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Math Help - Rotation by 45 degrees

  1. #1
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    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.
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  2. #2
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    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{?}
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  3. #3
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    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?
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  4. #4
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    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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