# Second Order

• Jan 11th 2011, 08:55 PM
dwsmith
Second Order
$3u_{xx}+4u_{yy}-u=0$

$a=3, \ \ b=0, \ \ c=4$

$ac-b^2=3*4-0=12>0$

$\mbox{Elliptic} \ \ \omega_{\xi\xi}+\omega_{\eta\eta}+\gamma\omega=0$

$u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) =\omega(\xi,\eta)$

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

$u_{yy}=(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos {\alpha})^2=\omega_{\xi\xi}\sin^2{\alpha}+2\omega_ {\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{\eta\eta }\cos^2{\alpha}$

$\omega_{\xi\xi}(3\cos^2{\alpha}+4\sin^2{\alpha})+\ omega_{\eta\eta}(3\sin^2{\alpha}+4\cos{\alpha})+2\ sin{\alpha}\cos{\alpha} \ \omega_{\xi\eta}+\omega=0$

Since I am trying to eliminate $\omega_{\xi\eta}$, does it matter what alpha I pick?

$2\sin{\alpha}\cos{\alpha}=0\Rightarrow \alpha=\frac{\pi}{2}+\pi k \ \mbox{or} \ \pi k \ \ \ k\in\mathbb{Z}$
• Jan 13th 2011, 01:44 PM
dwsmith
$\cos{\alpha}=1 \ \ \ \sin{\alpha}=0$

$\displaystyle 3\omega_{\xi\xi}+4\omega_{\eta\eta}-\omega=0\Rightarrow\omega_{\xi\xi}+\frac{4}{3}\ome ga_{\eta\eta}-\frac{1}{3}\omega=0$

$\displaystyle u_{xx}+\frac{4}{3}u_{yy}-\frac{1}{3}u=0$

$u=\exp{(\beta x)}\omega(x,y)$

$\exp{(\beta x)}[\omega_{xx}+\omega_{yy}+\omega_{x}2\beta+\omega(\b eta^2-1)]=0$

$2\beta=0\Rightarrow \beta=0$

$\omega_{xx}+\omega_{yy}-\omega=0$

$x=\mu\xi \ \ \ y=\nu\eta$

$\displaystyle\frac{1}{\mu^2}\omega_{\xi\xi}+\frac{ 1}{\nu^2}\omega_{\eta\eta}-\omega=0$

$\displaystyle\frac{1}{\mu^2}=\frac{1}{\nu^2}=-1\Rightarrow\mu=\pm i \ \ \ \nu=\pm i$

$\omega_{\xi\xi}+\omega_{\eta\eta}-\omega=0$

However, the solution is $u_{xx}+u_{yy}-u=0$

Why is it changed to u from omega?