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

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

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

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Shouldn't that be iupui instead of iuiui? – Jonas Meyer Jan 14 2010 at 7:55
There's a thread about this somewhere on the Art of Problem Solving forums, and I remember a tricky solution involving an exponential a^x satisfying a - 1 = log a, but I'm forgetting some important detail. – Qiaochu Yuan Jan 14 2010 at 13:44
If you look for exponential solutions, you get that equation. However, $a-1 \gt= ln a$ with equality at $a=1$, which doesn't make for an interesting $a^x$.  – Douglas Zare Jan 14 2010 at 14:09
What I remember someone doing in the thread I referred to is that they proved the existence of a complex value of a and did something like taking the real part.  Again, I think I'm forgetting an important detail. – Qiaochu Yuan Jan 14 2010 at 15:05
check also mathoverflow.net/questions/40275/… and mathoverflow.net/questions/39853/… – Pietro Majer Nov 15 2011 at 14:47

## 3 Answers

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.

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And if you don't want complex solutions to the original problem, just take real and imaginary parts... they are also solutions. – Gerald Edgar Jan 14 2010 at 15:13

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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.

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Perhaps I'm confused, but if g is all nonnegative on [0,1], it seems to me that you are getting a discontinuity at 0.  – Douglas Zare Jan 14 2010 at 9:08
Douglas Zare, how is that?  The limit from the left at 0 is clearly 0=g(0), whether or not g is nonnegative on [0,1] – Jonas Meyer Jan 14 2010 at 9:36
The limit from the left is the average value of g on $[0,1]$ divided by $e$, no? – Douglas Zare Jan 14 2010 at 9:39
No. For z a a very small positive number, g(-z) is $- e^{-1} \int_{1-z}^1 g(u) du$, which approaches zero as z goes to 0. (I've just rewritten Bjorn's formula to not have any hidden negatives.) – David Speyer Jan 14 2010 at 14:27
Ok, now I understand. Thanks. I misread the integral, and incorrectly assumed that these functions would be 0 at -1.  – Douglas Zare Jan 14 2010 at 14:56

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.

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