1. ## Linear PDE u_x+u_y=0

Folks,

trying 2 methods, neither working out for me

1)Method of Characteristics

$\displaystyle \displaystyle \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0$ given u(x,0)=x
$\displaystyle \frac{dy}{dx}=1 \implies c=y-x$, $\displaystyle \frac{du}{dx}=0 \implies u=f(c)$

Using IC's, $\displaystyle x=f(-x)$, let$\displaystyle t=-x \implies x=-t=f(t) \therefore u(x,y)=y-x$

but the correct is $\displaystyle u(x,y)=-y+x$

2) Parameterise x and y with s with the given IC's

$\displaystyle x(s); x(0)=x, y(s);y(0)=0$

$\displaystyle \frac{d}{ds} u(x(s),y(s))= \frac{dx}{ds}\frac{\partial u}{\partial x}+\frac{dy}{ds}\frac{\partial u}{\partial y} = 0$

$\displaystyle \frac{dx}{ds}=1 \implies x=s+A, \frac{dy}{ds}=1 \implies y=s+B, \frac{du}{ds}=0 \implies u=f(c)$

Not sure what is next, when I use the IC's, it doesnt look right.

THanks

2. I am not familiar with the method in part one, but for the 2nd one you have

$\displaystyle x=s+a \quad y=s+b \quad u=c$

Now using the intial conditions gives

$\displaystyle x_0=0+a \quad 0=0+b \quad u(x_0,0)=x_0$

This gives

$\displaystyle x=s+x_0 \quad y=s \implies x_0=x-s=x-y$

This gives

$\displaystyle u(x,y)=x-y$

3. You have a sign error in part (i). From what you have

$\displaystyle u = f(c) = f(y-x)$ so when $\displaystyle y = 0$ then $\displaystyle u = x$ so $\displaystyle f(-x) = x$ or $\displaystyle f(t) = -t$

this is everything you have.

Thus $\displaystyle u = \color{red}{-}\color$$\displaystyle \;(y-x) = x - y$ (note the red negative sign)

4. Perfect,
For (1), I used x=f(t) which threw me off.

Thanks guys