1. ## au_x+bu_y+cu=0

$au_x+bu_y+cu=0$

$a,b,c\in\mathbb{C} \ \ \mbox{and} \ \ b\ne 0$

a,b,c aren't necessarily Complex numbers.

$u(x,0)=u_0(x)$

$u_0(x)$ is differentiable.

$\omega_{\xi}(a\cos{\alpha}+b\sin{\alpha})+\omega_{ \eta}(b\cos{\alpha}-a\sin{\alpha})+c\omega=0$

$\displaystyle\tan{\alpha}=\frac{b}{a}$

Right Triangle:

Legs a and b

Hypotenuse $d=\sqrt{a^2+b^2}$

$\displaystyle\cos{\alpha}=\frac{a}{d} \ \ \sin{\alpha}=\frac{b}{d}$

$\displaystyle\frac{a^2+b^2}{d}\omega_{\xi}+c\omega = 0$

$\displaystyle\omega_{\xi}+\frac{c}{d}\omega=0$

$x=A\xi+B\eta \ \ \ y=C\xi+D\eta$

$au_x+bu_y=\omega_{\xi}=Au_x+Cu_y\Rightarrow \ \ A=a \ \ C=b$

Let $D=0 \ \ B=1$

$\displaystyle x=a\xi+\eta \ \ y=b\xi\Rightarrow \eta=x-\frac{ay}{b} \ \ \xi=\frac{y}{b}$

$\omega_{\xi}+c\omega=0$

$\exp{(c\xi)}\omega=g(\eta)\Rightarrow \omega(\xi,\eta)=\exp{(-c\xi)}g(\eta)$

$\displaystyle\omega(\xi,\eta)=u(x,y)=\exp{\left(\f rac{-cy}{b}\right)}g\left(x-\frac{ay}{b}\right)$

$\displaystyle u(x,0)=g(x)=u_0(x)$

$\displaystyle u(x,y)=\exp{\left(\frac{-cy}{b}\right)}u_0\left(x-\frac{ay}{b}\right)$

Is this correct?

2. An easy way of determining whether your answer is correct is

(1) Does it satisfy the IC?

(2) Does it satisfy the PDE when you substitute the solution in?

3. I wasn't sure how to differentiate $\displaystyle u(x,y)=\exp{\left(\frac{-cy}{b}\right)}u_0\left(x-\frac{ay}{b}\right)$ due to the u_0 piece.

4. $u_0$ is an arbitrary function of its argument so

$\dfrac{\partial}{\partial x} u_0\left(x - \dfrac{a}{b} y\right) = u_0'\left(x - \dfrac{a}{b} y\right)$

$\dfrac{\partial}{\partial y} u_0\left(x - \dfrac{a}{b} y\right) = u_0'\left(x - \dfrac{a}{b} y\right)\left(-\dfrac{a}{b}\right)$.