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I have

$$a(x,t+2) - a(x,t) = -\cos{\theta} [a(x+1,t+1) + a(x-1,t+1)]$$ Setting $t' = -t\cos(\theta)$ fetches under the continuum approximation $$2 \frac{\partial a(x,t')}{\partial t'} = a(x-1,t') - a(x+1,t')$$ which resembles Bessel functions's recurrence relations.

I used RSolve but it is of no help.

In[1]:= RSolve[a[x, t + 2] - a[x, t] - a[x - 1, t + 1] + a[x + 1, t + 1] == 0, a[x, t], {x, t}]
Out[1]= RSolve[-a[-1 + x, 1 + t] - a[x, t] + a[x, 2 + t] + a[1 + x, 1 + t] == 0, a[x, t], {x, t}]

Also, I tried using partial differential equation version which wasn't helpful either.

In[6]:= RSolve[2 D[a[x, t], t] - a[x - 1, t + 1] + a[x + 1, t + 1] == 0, a[x, t], {x, t}]
Out[6]= RSolve[-a[-1 + x, 1 + t] + a[1 + x, 1 + t] + 2*Derivative[0, 1][a][x, t] == 0, a[x, t], {x, t}]

This is one of the cases I have to verify and I have several such equations for which I don't know the solutions.

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  • $\begingroup$ Your first attempt, RSolve[a[x, t + 2] - a[x, t] - a[x - 1, t + 1] + a[x + 1, t + 1] == 0, a[x, t], {x, t}], works just fine (with result a[x, t] -> C[1][-t + x] + (-1)^t C[2][t + x]). Is this what you wanted? $\endgroup$ – AccidentalFourierTransform Jul 6 at 20:09
  • $\begingroup$ ... also, what are the initial conditions? note that the trivial solution $a(t,x)\equiv0$ solves your equations, but I guess it does not satisfy your initial conditions? $\endgroup$ – AccidentalFourierTransform Jul 6 at 20:15
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Without boundary/initial conditions, the problem is incomplete. For example, being homogeneous, we always have the trivial solution $a(t,x)\equiv0$. A more general solution is, for example, (letting $\alpha:=\cos\theta$ and $n\in\mathbb Z$), $$ a(t,x)=f_1(t-x)+(-1)^{nt}f_2(t-x) $$ where $f_i$ is any pair of periodic functions $f_i(r)\equiv f_i(r+2)$:

a[x, t + 2] - a[x, t] + α (-a[x - 1, t + 1] + a[x + 1, t + 1]) == 0 /. a -> (f1[#1 - #2] + (-1)^(#1 n) f2[#1 - #2] &)
FullSimplify[% /. {f1[-2 + r_] -> f1[r], f2[-2 + r_] -> f2[r]}, n ∈ Integers]
(* True *)

There are many more possible solutions. It is impossible to fix one uniquely without further data.

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