Given the 1st order linear PDE $$\begin{align} u_t+u_x &=x, & &x>0,\ t>0 \tag 1 \\ u(x,0) &= \sin x, & &x\geq 0 \tag 2\\ u(0,t)&=t, & &t\geq0 \tag 3 \end{align}$$
And we want to find if the solution is classical or weak.
Solution
The method of characteristics gives $$\frac{dx}{1}=\frac{dt}{1}=\frac{du}{x}$$
Edited
Νow from the first equality it follows that $$\frac{dt}{1}=\frac{dx}{1} \implies x=t+c_1 \implies c_1=x-t$$ and from the second $$\frac{dt}{1}=\frac{du}{x}=\frac{du}{t+c_1} \implies u=c_2+\frac{(t+c_1)^2}{2} \implies c_2=u-\frac{x^2}{2} $$
The general solution of the PDE expressed in the form of implicit equation $$c_2=G(c_1)$$ So $$u- \frac{x^2}{2}=G(x-t) \implies u=G(x-t)+\frac{x^2}{2} \tag4$$
We want to determine $G$ so conditions $(2)$ and $(3)$ are satisfied.
Now, applying the initial values, we get \begin{align} u(x,0) &= \sin x-\frac{x^2}{2} = G(x), & &x \geq 0 \\ u(0,t) &= t = G(-t), & &t \geq 0 \end{align}
It is sufficient then to say that a solution $u(x,t)$ of the given Boundary Value Problem, is the function defined as such :
$$u(x,t) = G(x-t)+\frac{x^2}{2} \quad \text{where} \quad \begin{cases} G(x) =sinx-\frac{x^2}{2}, \ x>t>0 \\ G(-t) = t ,\ 0<x<t \end{cases}$$
Letting $x := x-t$ and $t := t-x$ in the case of the boundary values. Then :
$$G(x-t) = sinx-\frac{x^2}{2} \quad \text{and} \quad G(x-t) = t-x\quad$$
For $x<t \implies x-t<0$,so $(4)$ gives $$u(x,t)=\frac{x^2}{2}+(t-x),\ 0<x<t$$
For $x>t \implies x-t>0$, so $(4)$ gives
$$u(x,t)=\frac{x^2}{2}-\frac{(x-t)^2}{2}+sin(x-t) ,\ x>t>0 $$
Therefore the solution of the system, if we do the identity at the second branch, is $$u(x,t)=\begin{cases} \frac{x^2}{2}+(t-x) ,\quad 0<x<t \quad \\ xt-\frac{t^2}{2}+sin(x-t),\quad x>t>0 \quad \end{cases}$$
Now we have to check if the solution is classical or weak.
$$ \lim_{x\to\ t^{-}} u= \lim_{x\to\ t^{+}} u= \frac{t^2}{2}$$
Thus $u$ is a continuous function $\forall x>0,\ t>0$.