au_x+bu_y+cu=0

**dwsmith** · January 10th 2011, 01:35 PM

$au_x+bu_y+cu=0$

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

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?

**Danny** · January 10th 2011, 01:52 PM

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?

**dwsmith** · January 10th 2011, 01:56 PM

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.

**Danny** · January 10th 2011, 02:00 PM

$u_0$ is an arbitrary function of its argument so

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

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

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