My understanding of Lecture #33, 34: The Characteristic Function for a Diffusion:

As an alternative to directly computing the characteristic function of a random variable $X_t$ in a stochastic process $\{X_t\}_{t \in [0,T]}$, we can solve a (boundary?) value problem, whose PDE has parameters are given by the dynamics of the stochastic process, and then conclude by Feynman-Kac that it is the characteristic function of said random variable.

An example is Arithmetic Brownian Motion:

If we solve

$$ \frac{\partial f}{\partial t} + \mu \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 \frac{\partial^2 f}{\partial x^2} = 0, x \in \mathbb R, t \in [0,T]$$ $$f(T,x) = e^{i \theta x} \tag{1}$$

then we get a function $f(t,x)$ s.t. $f(0,x)$ is the characteristic function of $X_t$ where $$dX_t = \sigma dW_t + \mu dt$$

So what does this mean for the (boundary?) value problem

$$ \frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 x^2\frac{\partial^2 f}{\partial x^2} = 0, x \in \mathbb R, t \in [0,T]$$ $$f(T,x) = e^{i \theta x} \tag{2}$$

?

My guess is that solution of $(2)$, $f(t,x)$, will be s.t. $f(0,x)$ is the characteristic function of $X_t$ where $$dX_t = \sigma X_t dW_t + \mu X_t dt$$ i.e. $X_t$ is Geometric Brownian Motion and hence is lognormal, the distribution of which doesn't have a characteristic function.

So $(2)$ has no solution then?

I'm looking for an answer like

'We don't expect $(2)$ to have a solution. This can be proven through (some PDE things).'

or

'While we don't expect $(2)$ to have a solution, it actually does because (some PDE things), but then (some PDE things).'

So the (some PDE things) may or may not prove lognormal distribution doesn't have a characteristic function, but I'm looking more for consistency e.g. because

- Lognormal doesn't have a characteristic function
- For any random variable, its characteristic function is supposed to be able to be computed by solving a PDE
- of Feynman-Kac

said PDE must have no solution.