How to solve the differential equation $\dfrac{\partial f}{\partial t}(x,t) = c_1 f(x,t)+c_2f(x-1,t)-c_2 f(x+1,t)$? 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.
 A: EDIT: the following is an answer to the originally posed question in which the coefficients c1,c2,c3 'appeared' to be constant - at least i read them that way.  The OP has since changed the question with specific forms for c1,c2,c3.
I am going to proceed in a very formal manner.  Consider writing your equation as
$${{f}_{t}}\left( x,t \right)=bf\left( x,t \right)+cf\left( x-1,t \right)-af\left( x+1,t \right)$$
Let  $$f\left( x,t \right)=\int\limits_{0}^{\infty }{{{y}^{x-1}}F\left( y,t \right)dy}$$ where F is the inverse Mellin transform of f.  substituting we have
$$\int\limits_{0}^{\infty }{{{y}^{x-1}}\left( {{F}_{t}}\left( y,t \right)-bF\left( y,t \right)-\frac{c}{y}F\left( y,t \right)+ayF\left( y,t \right) \right)dy}=0$$
hence
$${{F}_{t}}\left( y,t \right)=\left( b+\frac{c}{y}-ay \right)F\left( y,t \right)$$
which upon solving yields
$$F\left( y,t \right)={{e}^{\left( b+\frac{c}{y}-ay \right)t}}+G\left( y \right)$$
Now take the Mellin transform to obtain
$$f\left( x,t \right)=g\left( x \right)+{{e}^{bt}}\int\limits_{0}^{\infty }{{{y}^{x-1}}{{e}^{\left( \frac{c}{y}-ay \right)t}}dy}$$
The form of 1/y -y in the exponential smacks of a Bessel function.  And indeed it is! Look up Gradshtyn 3.471.9 where
   $$\int\limits_{0}^{\infty }{{{x}^{v-1}}{{e}^{-\frac{c}{x}-ax}}dx}=2{{\left( \frac{c}{a} \right)}^{v/2}}{{K}_{v}}\left( 2\sqrt{ca} \right)$$ for $$\operatorname{Re}\left( a,c \right)>0$$
Hence in this case assume $\operatorname{Re}\left( c \right)<0$, $\operatorname{Re}\left( a \right)>0$ and $t>0$, then
$$f\left( x,t \right)=g\left( x \right)+2{{e}^{bt}}{{\left( -\frac{c}{a} \right)}^{x/2}}{{K}_{x}}\left( 2t\sqrt{-ca} \right)$$
Here g must satisfy the linear difference equation
$$bg\left( x \right)+cg\left( x-1 \right)-ag\left( x+1 \right)=0$$
For this equation, take a slightly different approach.  Let
$$g\left( x \right)={{\beta }^{x}}$$
Then
$${{\beta }^{x-1}}\left( b\beta +c-a{{\beta }^{2}} \right)=0$$
And so
$${{\beta }_{\pm }}=\frac{b\pm \sqrt{{{b}^{2}}+4ac}}{2a}$$
The general solution therefore should be 
$$f\left( x,t \right)={{C}_{1}}\beta _{+}^{x}+{{C}_{2}}\beta _{-}^{x}+2{{C}_{3}}{{e}^{bt}}{{\left( -\frac{c}{a} \right)}^{x/2}}{{K}_{x}}\left( 2t\sqrt{-ca} \right)$$
for arbitrary constants C. You may confirm that this is indeed a solution by using the well known properties of the bessel function (it is a tedious calculation). 
A: The factor $-ix^3$ (not present in the title of the question) changes everything, since Fourier or some other transform may no longer be used in $x$-space. But you can still separate variables:
Look for solutions of the form $f(x,t):=g(x)e^{\lambda t}$, $\lambda$ a complex parameter. In this way you obtain the following functional equation for $g$:
$$(\lambda +ix^3) g(x)=C\bigl(g(x-1)-g(x+1)\bigr)\ .$$
The outlook here is very grim!
A: 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.)
