u_x-3u_y=0

**dwsmith** · January 10th 2011, 09:09 PM

$u_x-3u_y=0$

$u(x,x)=x^2$

$\omega_{\xi}(\cos{\alpha}-3\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+3\cos{\alpha})=0$

$\displaystyle\cos{\alpha}=\frac{1}{\sqrt{10}} \ \ \ \sin{\alpha}=\frac{-3}{\sqrt{10}}$

$\displaystyle\sqrt{10}\omega_{\xi}=0\Rightarrow\om ega_{\xi}=0$

$x=\xi+B\eta \ \ \ y=-3\xi+D\eta$

$\displaystyle\int\omega_{\xi}d\xi=\int 0d\xi\Rightarrow\omega(\xi,\eta)=g(\eta)$

I am lost with initial condition and what to set D and B equal too.

**PaulRS** · January 11th 2011, 02:12 AM

You may choose any numbers you like for B and D as long as the change of variables is invertible. - because these do not affect your original equation.

Say you could choose $B = \frac{1}{3}$ and $0$ .

- it looks nice that way.

Once you have undone your change of variables, set $x = y$ to use your initial condition.

Why do you use the $\cos$ and $\sin$ for coefficients?

**dwsmith** · January 11th 2011, 07:09 AM

Originally Posted by PaulRS

Why do you use the $\cos$ and $\sin$ for coefficients?

That is how my book has taught me. Here is what is in the book.

$\xi=x\cos{\alpha}+y\sin{\alpha}, \ \ \ x=\xi\cos{\alpha}-\eta\sin{\alpha}, \ \ \ \eta=-x\sin{\alpha}+y\cos{\alpha}, \ \ \ y=\xi\sin{\alpha}+\eta\cos{\alpha}$

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

**dwsmith** · January 11th 2011, 10:00 AM

$x=\xi+\frac{1}{3}\eta\Rightarrow\eta=3x-3\xi \ \ \ y=-3\xi=x$

$x=3x+y$

$\omega(\xi,\eta)=u(x,y)=g(3x+y)$

$u(x,x)=g(3x+x)=g(4x)=x^2$

How do I incorporate $g(4x)=x^2$ into the general solution $u(x,y)=g(3x+y)\mbox{?}$