Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?

Question

Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?

(Asked by bcross at math.iuiui.edu on the Q&A board at JMM.)

Answer 1

Yes, there exist nonlinear solutions.

Multiplying by $e^{x+1}$ and setting $g(x):=e^x f(x)$ transforms the question into finding a solution to $g(x+1)=eg'(x)$ not of the form $e^x(ax+b)$.

Start with any $C^\infty$ function on $\mathbb{R}$ whose Taylor series centered at $0$ and $1$ are identically $0$, but which is nonzero somewhere inside $(0,1)$. Restrict it to $[0,1]$. Let $g(x)$ on $[0,1]$ be this. Using $g(x+1):=eg'(x)$ for $x \in [0,1]$ extends $g(x)$ to a $C^\infty$ function $g(x)$ on $[0,2]$, which can then be extended to $[0,3]$, and so on. In the other direction, use $g(x) := \int_0^x e^{-1} g(t+1) dt$ to define $g(x)$ for $x \in [-1,0]$, and then for $x \in [-2,-1]$, and so on. These piece together to give a $C^\infty$ function $g(x)$ on all of $\mathbb{R}$. The corresponding $f(x)$ satisfies $f(0)=0$ and $f(1)=0$ but is not identically $0$, so it is not linear.

Answer 2

This is an elaboration of Qiaochu Yuan's prior comment: there are complex solutions (in fact, infinitely many) to $e^t-1 = t$, and then $e^{tx}$ is a solution.

One root, the only real root, is $t=0$ which is actually a double root. Thus we have a two-term solution for this value of $t$, which is the familiar $y=ax+b$.

The other roots for $t$ are complex and so appear as conjugate pairs.

Answer 3

Theorem 1 in [Sugiyama, Shohei. On the existence and uniqueness theorems of difference-differential equations. Kōdai Math. Sem. Rep. 12 1960 179--190. MR0121552] (which you can probably get from here) gives an existence and uniqueness theorem which provides non-linear solutions on finite intervals.

Answer 4

After this question popped up again, it seemed to me to scream out for a use of the Fourier Transform (FT). I have decided to post this as an answer, since this approach is transparent and provides a systematic method to obtain all (distributional) solutions.

A simple formal manipulation (to be justified below) shows that the FT $g$ (with independent variable $z$ since, for reasons which will become clear below, we are regarding it as an entire function defined on the complex plane) of a solution satisfies the equation $$(e^{-iz}+1+iz)g(z)=0.$$

In classical terms, this only provides the zero solution but in the sense of distributions it has many non-trivial ones, in fact, suitable combinations of $\delta$-functions with singularities at the zeros of the function in brackets. The resulting solution for a simple pole, i.e. the inverse FT of the corresponding delta function is easily seen to be exactly one of the solutions given in the previous answers. This can be used to give a precise description of all possible solutions of the original equation.

In order to motivate this approach, we begin with a historical aside on the FT for distributions: The problem of extending the FT to the latter setting was motivated by applications in physics and was studied in detail by many authors at the middle of the last century. Laurent Schwartz famously considered the case of tempered distributions in his seminal monograph but similar constructions can be carried out in many other situations. The basic starting point for such an extension is an initial setting where the FT establishes an isomorphism between two specific l.c. spaces of test functions (that is, functions with good smoothness and/or growth properties). One then uses transposition (in the sense of duality theory for l.c.´s) to translate this into an isomorphism between corresponding dual spaces. The latter consist of (generalised) distributions. Schwartz used the (symmetric) case of the smooth functions of rapid decrease to extend the FT to tempered distributions but there are many non symmetric cases which are of interest, as is the case here.

We require the concept and simple properties of the FT applied to ARBITRARY distributions on the line and so used the fact that it is an isomorphism between the smooth functions on the line with compact support and a suitable space of entire functions of exponential growth. (This is the Paley-Wiener-Schwartz theorem, details of which can be found in Strichartz´ monograph "A Guide to Distribution Theory and Fourier Transforms", where it is Th. 7.2.1 on p. 113). Dualising, we get a version of the FT which works for ALL distributions--it takes its values in a space of analytic functionals (a superspace of the dual space of the Fréchet space of entire functions). This can be described explicitly, but for our purposes it suffices to know that it contains the delta functions on the complex plane, together with suitable combinations. With this apparatus, the above formal computations can be tidied up to give a rigorous treatment which identifies all possible solutions of the above equation.

Answer 5

Notation: the endpoints of the intervals are separated by ";" (instead of ",") -- e.g. $\ [0;1].$

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Given a function $\ C^{\infty}$-function $\ f:\mathbb R\to\mathbb R\ $ such that $$ \forall_{x\in\mathbb R}\quad f(x+1)-f(x)\ =\ f'(x) $$ all derivatives are also solutions: $$ \forall_{n\in\mathbb N}\,\forall_{x\in\mathbb R}\quad f^{(n)}(x+1)-f^{(n)}(x)\ =\ f^{(n+1)}(x) $$

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Given a function $\ C^{\infty}$-function $\ f_0:[0;1]\to\mathbb R\ $ such that $$ f_0(1)\ =\ f_0(0)\ + f_0'(0) $$ and additionally (thank you, @PietroMayer -- please, see Pietro's comment under this "Answer"):

$$ \forall_{n\in\mathbb N}\quad f_0^{(n)}(1)\ =\ f_0^{(n)}(0)\ + f_0^{(n+1)}(0) $$

(there is a continuum of such different functions $\ f_0)$, we have (recursively) functions $\ f_n:[n;\,n+1]\to\mathbb R\ $ defined as follows:

$$ \forall_{n\in\mathbb R}\,\forall_{x\in[n;\,n+1]}\quad f_n(x)\ :=\ f_{n-1}(x-1)+f_{n-1}'(x-1) $$

(we work here with the one-sided derivatives at the integer points) hence

$$ \forall_{n\in\mathbb R}\,\forall_{x\in[n;\,n+1]}\quad f'_n(x)\ :=\ f'_{n-1}(x-1)+f_{n-1}''(x-1) $$

Then, by induction, $$ f_n(n)\,\ =\ f_{n-1}(n-1)+f_{n-1}'(n-1)\ =\,\ f_{n-1}(n) $$ and $$ f'_n(n)\,\ =\ f'_{n-1}(n-1)+f_{n-1}''(n-1)\ =\,\ f'_{n-1}(n), $$ and the same for the higher derivatives. Thus, there is a unique $C^{(\infty)}$-function $\ f:[0;\infty)\to\mathbb R\ $ that extends all $\ f_n,\ $ and it satisfies the OT's differential equation.

Clearly, a similar construction exists for arbitrary half-line $\ [a;\infty)\to\mathbb R;\ $ we may even simply define $\ g_a:[a;\infty)\to\mathbb R\ $ by $\ g(x):=f_0(x-a),\ $ where $\ f_0\ $ would be as above, etc.

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It's easy to see that the extension of zero function $\ \theta:[0;1]\to\mathbb R $ over $\ [0;\infty),\ $ that safisfies the OT's equation, is the zero function.

Also, any such extension $\ \Theta\ $ of $\ \theta,\ $ over any interval $\ (a;1]\ $ for $\ a<0,\ $ must satisfy $$ \forall_{x\in[-1;0]}\quad \Theta(x)+\Theta'(x)\ =\ 0 $$ hence the extension would be of the form $\ C\cdot\exp(-x)\ $ hence $\ C=0.\ $ Thus the extension is the zero function. And step by step, any such extension in the negative direction must be the zero function. In particular, there is exactly one extension of $\ \theta\ $ onto $\ \mathbb R,\ $ namely the zero function.

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Now, it follows easily, that any OT's extension in the negative direction, if it exists, is unique. However, I have not addressed the question of the existence of the extensions in the negative direction.