Finding Auxiliary Conditions for u(x,y) and Solving for ω(ξ,η)

[Dustinsfl]

2.2.2 $$3u_x+4u_y-2u=1\Rightarrow \omega_{\xi}+k\omega=\varphi(\xi,\eta)$$

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

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

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

$$3(\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha})+4(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\alpha})-2w=1$$

$$\omega_{\xi}(3\cos{\alpha}+4\sin{\alpha})+\omega_{\eta}(4\cos{\alpha}-3\sin{\alpha})-2\omega=1$$

$$\displaystyle 4\cos{\alpha}-3\sin{\alpha}=0\Rightarrow \tan{\alpha}=\frac{4}{3}$$

We have a 3,4,5 right triangle.

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

Substitution:

$$\displaystyle \omega_{\xi}\left(3\frac{3}{5}+4\frac{4}{5}\right)+\omega_{\eta}(0)-2\omega=1$$

$$\displaystyle \omega_{\xi}\left(\frac{9+16}{5}\right)-2\omega=1\Rightarrow\omega_{\xi}-\frac{2}{5}\omega=1$$

I understand everything posted above; however, I don't understanding anything below.

Let

2.2.8 $$x=A\xi+B\eta \ \mbox{and} \ y=C\xi+D\eta$$

A,B,C,D are constants to be determined, and set $$u(x,y)=\omega(\xi,\eta)$$ 2.2.9

Then, from $$\displaystyle\frac{\partial\omega}{\partial\xi}=u_x\frac{\partial x}{\partial\xi}+u_y\frac{\partial y}{\partial\xi}$$ and 2.2.8, 2.2.9, we see that with the choice A = 3, C = 4, Equation 2.2.2 becomes $$\omega_{\xi}-2\omega=1$$(nope don't see it).

 

[Fredrik]

2.2.2 $$3u_x+4u_y-2u=1\Rightarrow \omega_{\xi}+k\omega=\varphi(\xi,\eta)$$

...

Let

2.2.8 $$x=A\xi+B\eta \ \mbox{and} \ y=C\xi+D\eta$$

A,B,C,D are constants to be determined, and set $$u(x,y)=\omega(\xi,\eta)$$ 2.2.9

Then, from $$\displaystyle\frac{\partial\omega}{\partial\xi}=u_x\frac{\partial x}{\partial\xi}+u_y\frac{\partial y}{\partial\xi}$$ and 2.2.8, 2.2.9, we see that with the choice A = 3, C = 4, Equation 2.2.2 becomes $$\omega_{\xi}-2\omega=1$$(nope don't see it).

I didn't read any of the stuff that I didn't include in the quote, but it doesn't look like we need it. Do you realize that $\partial x/\partial \xi=A$ and so on? So

$$\omega_\xi=\frac{\partial\omega}{\partial\xi}=Au_x+Cu_y=3u_x+4u_y$$

and you're almost done.

 

[Dustinsfl]

Yes, I understand. Thank you.

 

[Dustinsfl]

I have a follow-up question now though.

The choice for B and D is arbitrary, except that $$AD - BC\neq 0$$.

Looking ahead to the effort to satisfy the auxiliary condition $$u(x,0)=u_0(x)$$, we shall choose B and D so that the line $$\xi=0$$ is the line on which the auxiliary data is prescribed, namely y = 0. This requires D = 0, and, since B is arbitrary we make the convenient choice B = 1.

Can you explain this?