u_x-u_y+u=1

Printable View

January 8th 2011, 07:40 PM
dwsmith

u_x-u_y+u=1

$u_x-u_y+u=1$

Auxiliary condition $u(x,0)=\sin{x}$

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

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

$u_y=\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\al pha}$

$\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}-(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\alpha })+\omega=1$

$\omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=1$

$\sin{\alpha}+\cos{\alpha}=0$

$\displaystyle\tan{\alpha}=-1\Rightarrow\alpha=\frac{3\pi}{4}+\pi k \ k\in\mathbb{Z}$

$\displaystyle\cos{\alpha}=\frac{-\sqrt{2}}{2} \ \mbox{and} \ \sin{\alpha}=\frac{\sqrt{2}}{2}$

$\displaystyle -\sqrt{2}\omega_{\xi}+\omega=1\Rightarrow\omega_{\x i}-\frac{1}{\sqrt{2}}\omega=-\frac{1}{\sqrt{2}}$

$x=A\xi+B\eta$

$y=C\xi+D\eta$

$\omega_{\xi}=Au_x+Cu_y=u_x-u_y$

$A=1 \ \mbox{and} \ B=-1$

$x=\xi+B\eta$

$y=-\xi+D\eta$

$u_x-u_y+u=1\Rightarrow\omega_{\xi}+\omega=1$

$P(\xi)=1 \ \ Q(\xi)=1$

$\exp{\left(\int P(\xi)d\xi\right)}=\exp{\left(\int d\xi\right)}=\exp{\xi}$

$\displaystyle\int\frac{\partial}{\partial\xi}\left[\exp{\left(\int P(\xi)d\xi\right)}\omega\right]d\xi=\int\exp{\left(\int P(\xi)d\xi\right)}Q(\xi)d\xi$

$\displaystyle\int\frac{\partial}{\partial\xi}\left[\exp{(\xi)}\omega\right]d\xi=\int\exp{(\xi)}d\xi\Rightarrow\exp{(\xi)}\ome ga=\exp{(\xi)}+g(\eta)$

$\omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)$

How do I solve for the auxiliary equation?
January 8th 2011, 08:37 PM
snowtea

I think you made a typo, should be

$\omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=1$

but it seems to correct itself later.

Quote:

Originally Posted by dwsmith

Auxiliary condition $u(x,0)=\sin{x}$

$\omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)$

How do I solve for the auxiliary equation?

You can use the inverse rotation to get your solution in terms of x and y, and then solve using the auxiliary condition.
Or you can transform the auxiliary condition into your new coordinate system.
January 8th 2011, 08:38 PM
dwsmith

Can you show me how that is done? Yes, I made a typo.
January 8th 2011, 08:51 PM
snowtea

$y=0$ is the line $-\xi \tan\alpha = \eta$ .

Taking the partial w.r.t x at y=0 for the constraint gives:
$\cos(x) = u_x = \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}$
when $-\xi \tan\alpha = \eta$

Take the partials. Substitute to get rid of $\xi$ , and solve for $g(\eta)$ .
January 8th 2011, 08:52 PM
dwsmith

Quote:

Originally Posted by snowtea

$y=0$ is the line $\xi \tan\alpha = \eta$ .

Where did you get this equation?
January 8th 2011, 08:56 PM
snowtea

$0 = y = \xi\sin{\alpha}+\eta\cos{\alpha}$

implies $-\xi\sin{\alpha} = \eta\cos{\alpha}$
$-\xi\tan{\alpha} = \eta$

actually I made a mistake :p

Let me fix that.
January 8th 2011, 08:58 PM
dwsmith

Quote:

Originally Posted by snowtea

$\cos(x) = u_x = \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}$

Can you explain this part, too?
January 8th 2011, 09:05 PM
snowtea

Sorry about the poor explanations :(

$u(x,0)=\sin(x)$ so $u_x(x,0)=\cos(x)$
This is the same thing as the pair of constraints:
$ y=0 $
$ u_x = \cos(x) $

Transforming each equation gives:
$ -\xi \tan\alpha = \eta $
$ \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha} = \cos(\xi\cos{\alpha}-\eta\sin{\alpha}) $

Last part inside the cosine is because:
$x = \xi\cos{\alpha}-\eta\sin{\alpha}$
January 8th 2011, 09:17 PM
dwsmith

Quote:

Originally Posted by snowtea

Sorry about the poor explanations :(

$u(x,0)=\sin(x)$ so $u_x(x,0)=\cos(x)$
This is the same thing as the pair of constraints:
$ y=0 $
$ u_x = \cos(x) $

Transforming each equation gives:
$ -\xi \tan\alpha = \eta $
$ \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha} = \cos(\xi\cos{\alpha}-\eta\sin{\alpha}) $

Last part inside the cosine is because:
$x = \xi\cos{\alpha}-\eta\sin{\alpha}$

I am still not sure how to solve for $\eta$ though. :(
January 9th 2011, 08:18 AM
snowtea

Quote:

Originally Posted by dwsmith

I am still not sure how to solve for $\eta$ though. :(

You are trying to solve for $g(\eta)$ .

If you just simplify with the constraints (using what you found for $\omega$ ), you should get something of the form:
$Ag'(\eta) + Bg(\eta) = e^{C\eta}cos(D\eta)$ where $A,B,C,D$ are known constants (expressed in terms of $\alpha$ ).
Remeber $\alpha$ is just a constant, and so is $sin(\alpha)$ and $cos(\alpha)$ .
January 9th 2011, 11:33 AM
dwsmith

Quote:

Originally Posted by snowtea

$Ag'(\eta) + Bg(\eta) = e^{C\eta}cos(D\eta)$

I don't know how you obtained this equation though.

I already found A and C which are 1 and -1, respectively.
January 9th 2011, 12:34 PM
dwsmith

Snowtea,

unfortunately, I have no idea how to solve the problem using your method but I was able to obtain the solution.

$\displaystyle \begin{vmatrix}A&B\\C&D\end{vmatrix}\neq 0$

B and D are arbitrary and since $y=0\Rightarrow y=0=-\xi+D\eta$ , let D = 0.

$y=0=-\xi$

Now, let B = 1.

$x=\xi+\eta\Rightarrow\eta=x-\xi=x+y$

$\omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)\Rightarrow u(x,y)=1+\exp{(y)}g(x+y)$

$u(x,0)=1+g(x)=\sin{x}\Rightarrow g(x)=\sin{(x)}-1$

$u(x,y)=1+\exp{(y)}[\sin{(x+y)}-1]$
January 9th 2011, 01:58 PM
snowtea

Quote:

Originally Posted by dwsmith

I don't know how you obtained this equation though.

I already found A and C which are 1 and -1, respectively.

A,B,C,D were actually new constants (just to indicate the form of the DE for g(eta)). They are different from the ones you used above. I forgot that you used them.
Sorry for the confusion.

Glad you got the right solution though.