1. ## 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?

2. 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.

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.

3. Can you show me how that is done? Yes, I made a typo.

4. $y=0$ 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 $g(\eta)$.

5. Originally Posted by snowtea
$y=0$ is the line $\xi \tan\alpha = \eta$.
Where did you get this equation?

6. $0 = y = \xi\sin{\alpha}+\eta\cos{\alpha}$

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

Let me fix that.

7. Originally Posted by snowtea
$\cos(x) = u_x = \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}$
Can you explain this part, too?

8. 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:
$x = \xi\cos{\alpha}-\eta\sin{\alpha}$

9. Originally Posted by snowtea

$u(x,0)=\sin(x)$ 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.

10. 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 $cos(\alpha)$.

11. 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.

12. 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.

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

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

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

$u(x,y)=1+\exp{(y)}[\sin{(x+y)}-1]$

13. 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.