About the equation $\frac{\partial }{\partial t}x(n,t)=x(n+1,t)$

Question

For the equation $\frac{\partial }{\partial t}x(n,t)=x(n+1,t)$,

1. What is the general solution?

2. How to represent $x(n,t+1)$ using $\{x(n+k,t)\}_{k\in \mathbb{Z}}$?

This equation seems easy, but I have no idea to solve that. Besides, the second question is the discretization about $t$.

Thanks

Answer 1

Quoting the beginning of a review, on page 268 in the May 1950 issue of Bulletin of the American Mathematical Society, of An essay toward a unified theory of special functions by C. Truesdell:

It is very difficult to draw the line between mathematical physics and applied mathematics but this book shows that there does exist a definite and important difference between them. In mathematical physics many special functions such as the Legendre, Hermite, or Lauguerre polynomials, Bessel or hypergeometric functions are used as tools to solve particular problems. As a consequence many properties of these fucntions and connections between them have been established. The author, as an applied mathematician, has posed the question of finding a unified approach to these different special functions so that from it most of the known properties could be found directly. The monograph under review gives an answer to this question.

The author found that many of the special fuunctions, not only those previously mentioned but also such as the generalized Riemann zeta function, the incomplete gamma function, or the Poisson-Charlier polynomials, can be transformed into solutions of the equation $$ \tag{1} \frac{ \partial F(z,\alpha)} {\partial z} = F(z,\alpha+1). $$

Answer 2

The equation $\frac\partial{\partial t} x(n,t)=x(n+1,t)$ says that the function $t\mapsto x(n+1,t)$ is the derivative of the function $t\mapsto x(n,t)$, nothing more, nothing less. So for any integer $k>0$, $x(n+k,t)=\frac{\partial^k}{\partial t^k}x(n,t)$ is the $k$-th derivative, and $x(n-k,\cdot)$ is a $k$-th antiderivative of $x(n,\cdot)$.

That describes the general solution of the equation---note $x(n,\cdot)$ can be any smooth function. Without more information, the value of $x(n,t+1)$ cannot be determined from the values of $x(n+k,t)$. If, though, $x(n,\cdot)$ is analytic and its Taylor series has radius of convergence larger than 1, then of course $$x(n,t+1) = \sum_{k=0}^\infty \frac{x(n+k,t)}{k!}.$$

As elementary as this is, it is curious that the similar-looking system $$\beta_n'(t)= n\beta_n(t)-n\beta_{n+1}(t)$$ appears to be significant in this work on zero-free regions of the Riemann zeta function.