1. ## Linear PDE u_x+u_y=0

Folks,

trying 2 methods, neither working out for me

1)Method of Characteristics

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

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

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

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

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

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

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

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

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

Now using the intial conditions gives

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

This gives

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

This gives

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

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

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

this is everything you have.

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

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

Thanks guys