Uniqueness of solution of Kolmogorov (deterministic) equation pde Consider the following PDE (Kolmogorov equation?):
\begin{equation}
\frac{\partial}{\partial t}u(x,t)=\mu(x,t)\frac{\partial}{\partial x}u(x,t) + \frac{1}{2}\sigma^2(x,t)\frac{\partial^2}{\partial x^{2}}u(x,t),
\end{equation}
With initial conditions $u(x,0) = \phi(x)$.
Now consider the one (deterministic Kolmogorov equation?) with $\sigma = 0$ that is a linear first-order PDE.
For this equation, when for example $\mu$ and $\phi$ are $C^1$ and Lipschitz and they are only functions of the space component $x$, we can find a solution $u(x,t)$ in the following way:
"there exists a unique continuous function $X = (X^x_s)_{x \in \mathbb R^d, s \in [0, t]} \colon [0, t] \times \mathbb R^d \to \mathbb R^d$ which satisfies 
for all $s \in [0, t]$, $x \in \mathbb R^d$ that
\begin{equation}
  X^x_s = x + \int_0^s \mu(X^x_r) dr,
\end{equation}
and the function $u(x,t) \colon = \phi(X^x_t)$ is a solution of the PDE above."
Is this solution unique? Why?
I am sorry if I made some mistake, is my first question.
 A: When $\sigma=0$ and $\mu = \mu(x)$, your equation 
$$ \partial _t = \mu  \partial _x u$$
can be solved by the method of characteristics. To be clear the above equation should be written as 
$$ \partial_t u = \mu_1 \partial_{x_1} u + \cdots + \mu_d \partial_{x_d} u,$$
where $x = (x_1, \cdots, x_d)$ and $\mu = (\mu_1, \cdots, \mu_d)$. Moving all terms to one side, the equation is equivalent to 
$$ (\partial_t u , \partial_{x_1} u , \cdots, \partial_{x_d} u) \cdot (-1, \mu_1, \cdots, \mu_d) = 0.$$
In particular, $u$ satisfies the equation if and only if $u$ is constant along any integral curve of the vector fields 
$$Z = (-1, \mu_1, \cdots, \mu_d)$$
in $\mathbb R\times \mathbb R^d$. 
Now it is clear how to get existence and uniqueness at once: for any $x\in \mathbb R^n$, let $X^x(t)$ be a solution to the ODE: 
$$ \begin{cases} Y' = \mu (Y), & \\
Y(0) = x.\end{cases}$$
I am assuming that $\mu$ is globally Lipschitz (i.e. with a uniform Lipschitz constant). Thus $Y_x(t) $ is defined for all $t\in \mathbb R$, for all $x$ and for each fixed $t$, $\Phi_t (x):=Y_x(t)$ is a homeomorphism. 
Then the above ODE is the same as your integral equation
$$ X^x(t) = x + \int_0^t \mu (X^x(s)) ds.$$
With this $X^x$, it is clear that $Y(s) = (t-s, X^x(s))$ is an integral curve of the vector fields $Z$ and $Y(t) = (0,X^x(t))$. 
Thus 
\begin{align}
u(x, t) &= u(Y(0))\\
&= u(Y(t))  \ \ \ \ (u \text{ is constant along the integral curve of } Z)\\
&= u(X^x(t), 0) \\
&= \phi (X^x(t))
\end{align}
This shows that any solutions to the equation has to be of the form $\phi (X^x(t))$. This in particular shows that the solution must be unique (Indeed you find an explicit form using $\phi, \mu$). 
