Dear all,

in the 8th lecture notes in the example b) we end up at

$$u(x,t)=\frac{1}{2}(sin(x+tt)+sin(x-t))$$ for $x>t$

and $$u(x,t)=\frac {1}{2}(sin(x+t)-sin(x-t))$$ for 0<x<t. But isn't the latter $u$ valid on $-t<x<t$, and can't we even say that

$$u(x,t)=\frac{1}{2}(-sin(x+t)-sin(x-t))$$ for $x<-t$?