# Thread:

1. ## u_x-3u_y=0

$u_x-3u_y=0$

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

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

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

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

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

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

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

2. 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?

3. 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.

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

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

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

$x=3x+y$

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

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

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

5. Well, note that then: $g(x) = \left(\frac{x}{4}\right)^2$ - from the equation you have there.

Now to get $u(x,y)$, plug $3x+y$ into $g$.

6. Originally Posted by PaulRS
Well, note that then: $g(x) = \left(\frac{x}{4}\right)^2$ - from the equation you have there.

Now to get $u(x,y)$, plug $3x+y$ into $g$.
Wouldn't it be $\displaystyle\left(\frac{x}{2}\right)^2\mbox{?}$

7. We want $g(u)$ , but to calculate this we let $u = 4\cdot x$ and then $g(u) = g(4\cdot x) = x^2 = ...\text{in terms of } u$
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