# Thread: Second Order Elliptic

1. ## 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 $\omega_{\xi} \ \mbox{and} \ \omega_{\eta}\mbox{?}$

2. Yes, that's the idea but let me say that it's not always possible :-)

3. 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?

4. Try $u = e^{\beta y} \omega(x,y)$

5. 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.

6. 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.

7. 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.

8. 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.)

9. 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.