I am trying to solve this nonlinear PDE using method of characteristics
$$u_t + uu_x = -u$$ $$u(x,0) = u_0(x)$$
We get two ODEs: $du/dt = -u$ and $dx/dt = u$
The solution is
$$u(x,t) = u_0(x_0)e^{-t}$$
$$x(t) = -u_0(x_0)e^{-t} + K$$
Finding $K$ using the initial conditions $x(0) = x_0$ we get
$$x(t) = -u_0(x)e^{-t} + x_0 + \dfrac{u_0(x_0)}{x_0}$$
Now to get the explicit solution, we need to solve for $x_0$ in the last equation above and substitute it into $u(x,t) = u_0(x_0)e^{-t}$. This is easy if for example $u_0(x) = x$. But what if $u_0(x) = x^2$? Substituting into $x(t)$ expression and solving for $x_0$ results in two possible solutions (since it is a quadratic formula). How do I deal with this situation?
$$u_t+u\,u_x=-u$$ Charpit-Lagrange Characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{-u}$$ A first characteristic equation comes from solving $\frac{dx}{u}=\frac{du}{-u}$ : $$u+x=c_1$$ A second characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{-u}$ : $$u\,e^{t}=c_2$$ The general solution of the PDE on the form of implicit equation $c_1=F(c_2)$ is : $$\boxed{u+x=F(u\,e^{t})}$$ $F$ is an arbitrary function ( To be determined according to some boundary and/or initial condition).
Note that $F$ isn't necessarily one-to-one. This function can be multivalued as well depending on the kind of bondary and/or initial condition.
Initial condition : $u(x,0)=u_0(x)$
$$u_0(x)+x=F\big(u_0(x)\big)$$ Let $X=u_0(x)$ and the inverse function $x=u_0^{-1}(X)$ $$X+u_0^{-1}(X)=F(X)$$ The function $F(X)$ is determined : $\quad F(X)=X+u_0^{-1}(X)$
We put it into the above general solution where $X=u\,e^{t}$ $$u+x=u\,e^{t}+u_0^{-1}(u\,e^{t})$$ Thus the particular solution $u(x,t)$ satisfying both the PDE and the specified condition is : $$\boxed{(1-e^{t})u=-x+u_0^{-1}\big(u\,e^{t}\big)}$$ This is the answer expressed on the form of implicit equation. The explicit form is possible if this equation is solvable for $u$ which isn't always the case.
Case of $u_0(x)=x^2$ :
$X=x^2$ and the inverse function $x=u_0^{-1}(X)=\pm\sqrt{X}$
$$(1-e^{t})u=-x\pm\sqrt{u\,e^{t}}$$ This equation is solvable for explicitly $u(x,t)$ .This is a quadratic equation wrt $\sqrt{u}$ .