I'm reading page 55 of a differential equation text. §3.5 Domain of definition gives an example
As a simple example which illustrates this behaviour you can consider the equation $$ x∂_xz + y∂_yz = −z^2 $$ with prescribed data of $z(x, y) = α ∈\Bbb R$ for on $\{(x, y) : x + y = 1, x ∈ [0, 1]\}$.
My working:
Introduce a parameter $s$ for the data, say $γ(s) = (s,1-s,\alpha)$, for $0 ≤ s ≤ 1$, and then solve the characteristic equations with this as data at $t = 0$: \begin{align} x&=s \exp (t)\\ y&=(1-s) \exp (t)\\ z&=\frac{1}{\frac{1}{\alpha }+t}\quad(\alpha\ne0)\end{align} From the first two equations $t=\log(x+y)$, plugging in the third equation, $$z=\frac{1}{\frac{1}{\alpha }+\log(x+y)}$$ which is singular on $x+y=\exp(-1/\alpha)$.
The characteristic projections are rays through the origin in the first quadrant, so
- for $α=0$: the solution is $z=0$, defined on $\{(x,y):x>0, y>0\}$
- for $α>0$: domain of definition is $\{(x,y):x>0, y>0 ,\text{ and }x+y>\exp(-1/α)\}$
- for $α<0$: domain of definition is $\{(x,y):x>0, y>0 ,\text{ and }x+y<\exp(-1/α)\}$
Could you check if my solution contains any error? Thanks!