I want to get the solution of $$\frac{\partial u(x,t)}{\partial t}=v(x,t)\cdot \nabla u(x,t)$$ with $u(x,0)=f(x)$
The problem I encountered using method of characteristics is that: I cannot write the characteristics curve explicitly.
What I get is $$u(x,t)=f(x+\int_0^t v(x,s)ds)$$ but it definitely should not be $v(x,s)$ in the integral. It should be independent of $x$, but I do not know how to express.
Let $x \in \mathbf{R}^d$ and $t \in [0, T]$. The characteristic equation is : $$\left\{\begin{aligned}X'(s) &= -v(X(s), s)\\X(t) &= x.\end{aligned}\right.$$ By Cauchy-Lipschitz (if say $v$ is global lipschitz in $x$, uniformly in $t$) this equation has a unique solution $X(s; x, t)$. And the characteristic curve should be the curve : $$\{(X(s;x,t), s) \colon s \in [0, T]\}.$$ The solution (if say $f$ is continuously differentiable) is given as you said by : $$u(x, t) = f(X(0 ; x, t)).$$ The solution depends on $X$ which you get by solving the characteristic equation first, it does not have a closed form like $x + \int_0^t v(x, -)$.
Take for example $v$ to be constant. Then the characteristic equation is solved as $X(s;x,t) = x + (t-s)v$ and you get $u(x, t) = f(x + tv)$. Your formula works here.
But take another example, say $v(x, t) = -x$. Then the characteristic equation is solved as $X(s;x,t) = x\mathrm{e}^{s-t}$, which yields $u(x, t) = f(x\mathrm{e}^{-t})$ which is not of the form you gave.