au_x+bu_y+cu=0

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January 10th 2011, 01:35 PM
dwsmith

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?
January 10th 2011, 01:52 PM
Danny

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?
January 10th 2011, 01:56 PM
dwsmith

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.
January 10th 2011, 02:00 PM
Danny

$u_0$ is an arbitrary function of its argument so

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

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