I have been trying to figure out how to apply the method of characteristics to the non-homogeneous traffic equation. $$ u_t + u_x = sin(x), \\ u(x, 0) = e^{-x^2} $$ I thought that forming the related system of ODEs and solving them would be sufficient: \begin{align} \frac{dx}{dt} &= 1, \\ \frac{du}{dt} &= \sin(x), \end{align} Solving for $x$ then for $u$, I get that $u(x, t) = \sin(x)t + e^{-(x-t)^2}$, which does not satisfy the original PDE.
I thought it could be due to the fact that the two ODEs are coupled, but not sure what the basis for such an argument would be.
What am I missing?
Thank you for your time!
The initial curve is given $$x(r,0)=r $$ $$ t(r,0)=0$$ $$z(r,0)=e^{-r^2}$$ and the characteristic ODEs are $$ \frac{dx}{ds}(r,s)=1$$ $$\frac{dt}{ds}(r,s)=1$$ $$\frac{dz}{ds}(r,s)= \sin x $$ we find $$ x(r,s)=s+\omega_1(r)$$ $$t(r,s)=s+\omega_2(r)$$ $$\frac{dz}{ds}(r,s)=\sin x$$ Then $$ x(r,s)=s+r$$ $$t(r,s)=s$$ $$ \frac{dz}{ds}(r,s)=\sin(r+s)$$ therefore $z(r,s)=\int^s_0\sin(y+r)dy+\omega_3(r)$
we have $\omega_3(r)=z(r,0)=e{-r^2}$
then
$z(r,s)=\int^s_0\sin(y+r)dy+e^{-r^2}$
on the other hand
$$r(x,t)=x-t$$ $$s(x,t)=t$$ So $$u(x,t)=e^{-(x-t)^2}+\int^t_0\sin(y+x-t)dy$$