TITLE: How to solve the differential equation $\dfrac{\partial f}{\partial t}(x,t) = c_1 f(x,t)+c_2f(x-1,t)-c_2 f(x+1,t)$?
QUESTION [5 upvotes]: I am trying to solve the differential equation$$\frac{\partial f}{\partial t}(x,t) = -ix^3f(x,t)+Cf(x-1,t)-C f(x+1,t)$$ where $C$ is a purely imaginary constant. I am not sure how to go about this as normally when solving differential equations the first coordinate is the same.
I can't seem to think of any strategy to employ.

REPLY [2 votes]: This is not an answer, since it's probably not what you're looking for, but I just wanted to point out that you can construct solutions which are distributions (generalized functions).  For simplicity, consider the equation with $f_t=0$ and set $C=i/c$:
$$
f(x+1)-f(x-1)=cx^3f(x).
$$
As an example, let us seek a solution of the form $$f=\sum_{k=-\infty}^\infty f_k\delta_{k},$$ where $\delta_k(x)=\delta(x-k)$ is the Dirac delta, and $f_k\in\mathbb C$.  (Note that this is a well defined distribution, but possibly not a tempered distribution if $f_k$ grows too quickly.)  Substituting this into the equation and equating coefficients of $\delta_k$ to zero gives a recursive system for the $f_k$'s:
$$
f_{k+1}-f_{k-1}=ck^3f_k,\qquad k\in\mathbb Z.
$$
To understand this system, let us write some equations down:
\begin{align}
k&=0:&&f_1=f_{-1},\\
k&=1:&&f_2=f_0+cf_1,\\
k&=-1:&&f_{-2}=f_0+cf_{-1},
\end{align}
where the last equation follows from rearranging.  Inductively, we see that $f_{-k}=f_k$ for each $k\neq 0$, with $f_0,f_1\in\mathbb C$ arbitrary constants which determine the rest.  Specifically, we have shown there exists a weak solution of the form
$$
f=f_0\delta_0+\sum_{k=1}^\infty f_k(\delta_k+\delta_{-k}),
$$
where $f_k,k=1,2,3,\dots$ solves a second order recursion relation:
$$
f_{k+1}=f_{k-1}+ck^3f_k,\qquad k=1,2,3,\dots,
$$
and $f_0,f_1\in\mathbb C$ are arbitrary initial conditions.
There is little hope of solving this recursion relation in closed form, but because this equation is linear, we can reduce its order by the usual (Riccati) substitution
$$
f_k=\Pi_{m=0}^k g_m,\qquad k=0,1,2,3,\dots,
$$
since then $f_{k+1}=g_{k+1}f_k$ and $f_{k-1}=f_k/g_{k}$.  This gives a nonlinear recursion equation of first order for $g_k$:
$$
g_{k+1}=1/g_k+ck^3,\qquad k=1,2,\dots.
$$
This can be used to show that $g_k\sim c(k-1)^3$ for large $k$, such that $f_k$ grows exponentially as $k\to\infty$.  In particular, $f(x)$ cannot be a tempered distribution unless $f_0=f_1=0$.  (The book by Bender and Orszag gives more details on such asymptotic analysis for large $k$, called the method of dominant balance.)