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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?

MarsOneRover
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2 Answers2

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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.

Hans Lundmark
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  • I understand your argument. But... I found an explicit solution above according to the procedure. Why doesn't it work? – MarsOneRover Nov 02 '15 at 08:53
  • There's a computational mistake: you write $2x+y=\cos(r)+\sin(r)=\sqrt2 \sin(r+\pi/4)$, but it should be $2\cos(r)+\sin(r)=\sqrt5 \sin(x+\arctan 2)$. If you correct that, you get something which should satisfy the PDE, but it can't satisfy the boundary conditions everywhere. (I haven't thought this through carefully, but maybe the boundary conditions will be OK on one half of the circle, but not the other half, or something like that. And that probably has something to do with $a=\arcsin b$ not being the only solution to the equation $b=\sin a$.) – Hans Lundmark Nov 02 '15 at 14:53
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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.

Michael Medvinsky
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