Solving 1st order PDE (Finding where solution is invalid)

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!

• Big picture: if two characteristics intersect, then you wind up with more than one place to draw from the Cauchy data (I would say "initial condition" but this isn't really an initial condition) in order to compute the value of the solution. – Ian Aug 8 '16 at 12:10
• For example, this would happen in a simpler problem: consider $u_t+xu_x=0,u(0,x)=f(x)$. At $x=0$ for any $t>0$ there will be a characteristic extending backward in time to the left and a characteristic extending backward in time to the right. The characteristic equation then tries to tell you that $u(t,0)=f(y)=f(-y)$ for some $y>0$ that depends on $t$. (This $y$ can be explicitly calculated but it really doesn't matter what it is.) If $f$ is even then this is all well and good but if it is not then this is a contradiction. – Ian Aug 8 '16 at 12:10
• So what you're trying to do in your problem is get the characteristics to intersect and then check that the characteristic equation gives you a contradiction. – Ian Aug 8 '16 at 12:10
• Sorry, I made a small error, my example should be $u_t+(-x)u_x=0$. – Ian Aug 8 '16 at 12:17