Linear PDE by characteristics I am studying the characteristics method from these notes I found online http://web.stanford.edu/class/math220a/handouts/firstorder.pdf I can't seem to get my solutions to work out though, even in the simplest cases.
For example, I want to solve $-\frac{1}{2}u_x + u_y = c$, with $u|_{d\Omega} = 0$, for the unit circle.  According to page 7, I get
$\frac{\partial x}{\partial s} = -\frac{1}{2}, \frac{\partial y}{\partial s} = 1, \frac{\partial z}{\partial s} = c$, with initial conditions $x(r,0) = \cos(r) , y(r,0) = \sin(r) , z(r,0) = 0$.  
Of course, we can easily solve these, since none of them involve $s$!  We get $x = -\frac{1}{2}s + \cos(r)$,
$y= s + \sin(r)$,
$z = cs$.
Ok.. so then our solution should be $z$.  If you solve for $s$ from $y$, say you get $z = c(y-\sin(r))$.  If I solve the system of $x$ and $y$ for $r$, I should get $2x+y = \cos(r) + \sin(r)$, from which $r = \sin^{-1}(\frac{2x+y}{\sqrt2})-\frac{\pi}{4}$.  According to this then, my solution is 
$$ z=c\left(y-\sin\left(\sin^{-1}\left(\frac{2x+y}{\sqrt2}\right)-\frac{\pi}{4}\right) \right)$$.  
Something seems wrong here. Why is this $0$ on the boundary? For example, if I go a little further, using trig to solve for the angles, and using the formula for sine of a sum, I apparently get
$$c\left(y- \frac{2x+y}{2} -\frac{\sqrt2}{2}\sqrt{\sqrt{2} - (2x+y)^2}\right)$$
And this is just for a circle!  What I really want is more general domains.  Can you help at all?
 A: There are no global solutions to your problem (unless $c=0$, in which case the solution is $u \equiv 0$).
This is clear if you realize that what the PDE is requiring is that
$$
\frac{d}{dt}\big( u(a-\tfrac12 t,b+t) \bigr) = c
$$
for any point $(a,b)$ and any $t$; in other words, $u$ is growing (or decreasing) at a constant rate along all lines with direction vector $(-\tfrac12,1)$.
So if $(a,b)$ is a point on the unit circle, so that $u(a,b)=0$ according to the boundary condition, then along the line $L$ passing through $(a,b)$ with direction $(-\tfrac12,1)$ you have
$$
u(a-\tfrac12 t,b+t) = ct
.
$$
Now, unless $L$ happened to be tangent to the unit circle (which is an exceptional case), it will intersect the unit circle at some other point $(a',b')$ (for $t=t'$, say), where you then will have $u(a',b')=ct' \neq 0$, contradicting the desired boundary condition that $u=0$ on the unit circle.
A: for any function $g$ this equation has a solution 
$$u(x,y)=-2 c x + g(2 x + y)$$
$$g(x)= \frac{2}{5} \left(2 x\pm \sqrt{5-x^2}\right) c$$
satisfies $u[x,\mp\sqrt{1-x^2}]=0$ but not for entire unit circle (note the $\pm,\mp$).
If you read the pdf in your link, you can find that the boundary curve should never be tangent with characteristic lines, which creates the problem.
