My understanding of [Lecture #33, 34: The Characteristic Function for a Diffusion](http://stat.math.uregina.ca/%7Ekozdron/Teaching/Regina/441Fall14/Notes/L33-34-Nov24.pdf):




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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 $$dX\_t = \sigma dW\_t + \mu dt$$




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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](https://en.wikipedia.org/wiki/Geometric_Brownian_motion) and [hence is](https://en.wikipedia.org/wiki/Geometric_Brownian_motion#Properties) lognormal, the distribution of which doesn't have a [characteristic function](https://en.wikipedia.org/wiki/Log-normal_distribution#Characteristic_function_and_moment_generating_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. PDE must have no solution because


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



Expected properties for a PDE whose solution is supposed to be something that doesn't exist
1. Lognormal does have a characteristic function
2. It has no closed form
3. The value problem solution doesn't have an Ansatz directly related to its terminal condition as with the value problems for ABM and OU
4. The value problem solution rather has a series solution.
5. Such series solution gives the characteristic function of lognormal for $t=0$.


