I'm struggling with a PDE homework question (see below)
Solve the equation: $au_x + bu_y + u = 0$
With Cauchy data $u(s) = e^{−s}$ along the straight lines $(x(s), y(s)) = (cs, ds)$.
The solution breaks down for a particular combination of a, b, c and d. Describe the combination geometrically and explain it in terms of characteristic curves. Note a,b,c,d are constants and u = u(x,y)
So what I've done so far is....
$au_x+bu_y+u=0$, which I want to transform into $\frac{\partial}{\partial{s}}u(x(s),t(s))=F(u, x(s),t(s))$
So then, by the chain rule: $\frac{\partial}{\partial{s}}\bar{u} = \frac{\partial{u}}{\partial{x}}\frac{\partial{x}}{\partial{s}}+\frac{\partial{u}}{\partial{y}}\frac{\partial{y}}{\partial{s}}$ , then setting $\frac{\partial{x}}{\partial{s}}=a$ and $\frac{\partial{y}}{\partial{s}}=b$ (I'm using $\bar{u}$ to denote $u$ with it's change of co-ordinates)
Then I have: $\bar{u}_s=a\bar{u}_x+b\bar{u}_y=\bar{u} \Rightarrow \bar{u}_s=\bar{u}$, which is an ODE
The solns are along a line so $u(x_s,t_s)=u(x_0,0)$
Then solving and substituting:
$\frac{\partial{x}}{\partial{s}}=a \Rightarrow x=as+constant$
$\frac{\partial{y}}{\partial{s}}=b \Rightarrow y=bs+constant$
$\frac{\partial\bar{u}}{\partial{s}}=\bar{u}\Rightarrow \bar{u}=f_1e^{-s}+constant$
Then (this is where I am not 100%) I can use the data to find that:
$\bar{u}=e^{-1}, x=cs, y=ds\Rightarrow a=c, b=d$
Now, to me this would mean that the solution could exist anywhere? The one combination I could see where this wouldn't work would be $a=b=c=d=0$ but I feel like this is incorrect? I also have no idea how this has any geometric significance?
Thanks for any help!