# Uniqueness of solution of Kolmogorov (deterministic) equation pde

Consider the following PDE (Kolmogorov equation?):

$$$$\frac{\partial}{\partial t}u(x,t)=\mu(x,t)\frac{\partial}{\partial x}u(x,t) + \frac{1}{2}\sigma^2(x,t)\frac{\partial^2}{\partial x^{2}}u(x,t),$$$$ With initial conditions $$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 $$$$X^x_s = x + \int_0^s \mu(X^x_r) dr,$$$$ and the function $$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.

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 $$\phi, \mu$$).