Second Order Elliptic

• Jan 13th 2011, 04:22 PM
dwsmith
Second Order Elliptic
$u_{xx}+u_{yy}+3u_x-4u_y+25u=0$

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

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

$u_x=\omega_{\xi}\cos{\alpha}-\omega{\eta}\sin{\alpha}$

$u_y=\omega_{\xi}\sin{\alpha}+\omega{\eta}\cos{\alp ha}$

$\displaystyle\omega_{\xi\xi}(\cos^2{\alpha}+\sin^2 {\alpha})+\omega_{\eta\eta}(\cos^2{\alpha}+\sin^2{ \alpha})+\omega_{\eta\xi}(2\sin{\alpha}\cos{\alpha }-2\sin{\alpha}\cos{\alpha})+\omega_{\xi}(3\cos{\alp ha}-4\sin{\alpha})+\omega_{\eta}(-4\cos{\alpha}-3\sin{\alpha})+25\omega=0$

$\omega_{\xi\xi}+\omega_{\eta\eta}+\omega_{\xi}(3\c os{\alpha}-4\sin{\alpha})+\omega_{\eta}(-4\cos{\alpha}-3\sin{\alpha})+25\omega=0$

Since $\omega_{\xi\eta}$ is already gone, should I be working to eliminate $\omega_{\xi} \ \mbox{and} \ \omega_{\eta}\mbox{?}$
• Jan 14th 2011, 04:12 AM
Jester
Yes, that's the idea but let me say that it's not always possible :-)
• Jan 14th 2011, 12:39 PM
dwsmith
So continuing....

$\displaystyle\tan{\alpha}=\frac{3}{4} \ \ \cos{\alpha}=\frac{4}{5} \ \ \sin{\alpha}=\frac{3}{5}$

$\omega_{\xi\xi}+\omega_{\eta\eta}-5\omega_{\eta}+25\omega=0$

$u_{xx}+u_{yy}-5u_y+25u=0$

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

$u_{xx}=\exp{(\beta x)}[\beta^2\omega+2\beta\omega_x+\omega_{xx}], \ \ u_{yy}=\exp{(\beta x)\omega_{yy}, \ \ u_y=\exp{(\beta x)}\omega_y$

$\exp{(\beta x)}[\omega_{xx}+\omega_{yy}+2\beta\omega_x-5\omega_y+(\beta^2+25)\omega]=0$

$\beta=0$

Doing this didn't eliminate $\omega_y$

$\omega_{xx}+\omega_{yy}-5\omega_y+25\omega=0$

The answer is $u_{xx}+u_{yy}+u=0$

How do I solve this equation correctly?
• Jan 15th 2011, 05:34 AM
Jester
Try $u = e^{\beta y} \omega(x,y)$
• Jan 15th 2011, 12:40 PM
dwsmith
Quote:

Originally Posted by Danny
Try $u = e^{\beta y} \omega(x,y)$

$\exp{(\beta y)}[\omega_{xx}+\omega_{yy}+\omega_y(2\beta-5)+(\beta^2+\beta+25)\omega=0$

$\displaystyle\beta =\frac{5}{2}$

$\displaystyle\omega_{xx}+\omega_{yy}+\frac{135}{4} \omega=0$

$\displaystyle\frac{1}{\mu^2}\omega_{xx}+\frac{1}{\ nu^2}\omega_{yy}+\frac{135}{4}\omega=0$

$\displaystyle\frac{1}{\mu^2}=\frac{1}{\nu^2}=\frac {135}{4}\Rightarrow \mu=\nu=\pm\frac{2}{3\sqrt{15}}$

How from this am I able to say $u_{xx}+u_{yy}+u=0\mbox{?}$

I know I am supposed to conclude that once I solve for mu and nu, but I have no clue why.
• Jan 15th 2011, 04:02 PM
Jester
Let's start over. First let

$u = e^{a x + b y} v(x,y)$

and choose $a$ and $b$ so the first order terms vanish. Then scale $x$ and $y$ so that you can cancel the coefficient on the $v$ term.
• Jan 15th 2011, 04:03 PM
dwsmith
Quote:

Originally Posted by Danny
Let's start over. First let

$u = e^{a x + b y} v(x,y)$

and choose $a$ and $b$ so the first order terms vanish. Then scale $x$ and $y$ so that you can cancel the coefficient on the $v$ term.

Why?

You last suggestion worked fine.
• Jan 15th 2011, 04:22 PM
Jester
Why. Because (1) You have three different transformations when you can use two and (2), you seem a little fixated on using a rotation of coordinates. I personally won't rely on this so much on this (it really only works for constant coefficient PDEs.)
• Jan 15th 2011, 04:27 PM
dwsmith
Quote:

Originally Posted by Danny
Why. Because (1) You have three different transformations when you can use two and (2), you seem a little fixated on using a rotation of coordinates. I personally won't rely on this so much on this (it really only works for constant coefficient PDEs.)

I have to rely on what the book presents, because I am not going to be able to guess another form. Also, I haven't made it to variable coefficients yet.