Linear Equations in two independent variables

**dwsmith** · January 8th 2011, 10:27 AM

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

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

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

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

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

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

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

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. Everything here is understood now.

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 $\omega_{\xi}-2\omega=1$

**dwsmith** · January 8th 2011, 04:28 PM

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 someone explain this and why?

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