I know the motivation to define weak solution to a PDE, for example if we consider transport equation
$$ u_{t}+(f(u))_{x}=0, ~~ (x,t) \in \mathbb{R} \times (0, \infty) $$ $$ u(x,0)=u_{0}(x), ~~ x \in \mathbb{R}. $$ Then even if $f$ and $u_{0}$ are smooth, we can get solution which is not even continuous. So, we relaxed the definition of solution to define weak solution. But what is the motivation to define measure valued solutions ?
I read the answer, but it does not fully answer my question.
I don't fully understand what is your problem. Nevertheless I post the solving for the general solution. Hoping that this will be helpful in some way.
$$u_t+(f(u))_x=0$$ $$u_t+f'(u)u_x=0$$ $f'(u)$ means $\frac{df(u)}{du}$
The Charpit-Lagrange system of characteristic ODEs is :}{1} $$\frac{dt}{1}=\frac{dx}{f'(u)}=\frac{du}{0}$$ A first characteristic equation comes from $du=0$ $$u=c_1$$ A second characteristic equation comes from $dt=\frac{dx}{f'(c_1)}\quad\implies\quad x=tf'(c_1)+c_2$
$$x-tf'(u)=c_2$$
The general solution on implicit form $c_1=\Phi(c_2)$ is : $$u=\Phi\big(x-tf'(u)\big)$$ $\Phi$ is an arbitrary function (continuous or not) to be determined according to the specified condition.
$u(x,0)=u_0(x)=\Phi\big(x-0f'(u)\big)=\Phi\big(x\big)$
The function $\Phi(X)$ is determined : $$\Phi(X)=u_0(X)$$ We put it into the above general solution where $X=x-tf'(u)$ : $$\boxed{u=u_0\left(x-t\frac{df(u)}{du}\right)}$$ As expected the result is on the form of implicit equation. It is doubtful that one can go further for explicit solution without knowing what kind of functions $f(u)$ and $u_0(x)$ are.