Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be a smooth function such that
$$f'(x) = f(1+x)$$ How can we find the general form of $f$? I thought of some differential equations, but not sure how to use them here. Thanks.

  • $\begingroup$ This is an example of a "differential delay equation." There is some discussion of these at en.wikipedia.org/wiki/Delay_differential_equation $\endgroup$ – Gerry Myerson Sep 20 '12 at 13:13
  • $\begingroup$ @Gerry Myerson: thank you for the piece of information. I was wondering if there is some easy approach to find its general form. $\endgroup$ – user 1357113 Sep 20 '12 at 13:14
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    $\begingroup$ If you can write $f(x)$ as power series $f(x)=\sum_{n=0}^\infty a_n x^n$, this should give you a recursive formula for $a_n$. $\endgroup$ – celtschk Sep 20 '12 at 13:16
  • $\begingroup$ Also you got $f^{(n)}(x) = f(x+n)$. Proof by induction: $f^{(0)}(x)=f(x)$ is trivial. $f^{(n+1)}(x) = (f^{(n)}(x))' = (f(x+n))' = f'(x+n) = f(x+n+1)$ where the second step was the induction assumption and the last step the defining equation. If you insert the power series from the previous comment and set $x=0$, you get $f(n) = n! a_n$. $\endgroup$ – celtschk Sep 20 '12 at 13:40
  • $\begingroup$ OK, thinking again about it, the series does not give a recursive formula for $a_n$ (at least not immediately) because you get infinite sums. $\endgroup$ – celtschk Sep 20 '12 at 14:03

By repeated application of the equation you have there, it's clear that $f^{(n)}(x) = f(n + x)$, and hence that $f$ is smooth (i.e. has continuous derivatives of all orders). Furthermore, the equation you give is linear. This means both that $f = 0$ is a solution, and that any linear combination of solutions is a solution.

Suppose we know what $f$ is on $[0,1]$. We can then work out what $f$ is everywhere else: on $[1,2]$ we need to have $f^\prime$, on $[n,n+1]$ more generally we need to have $f^{(n)}$. So the values on positive reals are already determined. On the negatives, we analogously need to put antiderivatives of $f$. You may think that this isn't unique, because of course antiderivatives aren't unique, but the continuity of the function at the endpoints of the intervals is going to fix the constants of integration, so in fact there are no choices.

However, that hints at a problem we might encounter on the positive reals: if $f^\prime(0)\not=f(1)$, the stitched-together function will not be continuous at $1$. Likewise, if $f^{\prime\prime}(0)\not=f^\prime(1)$, the solution will run into difficulties at $2$.

Hence, if my analysis is correct, the solutions are in one-to-one correspondence with smooth functions $g:[0,1]\to \mathbb R$ satisfying the additional constraint that $g^{(n+1)}(0)=g^{(n)}(1)$. In particular, the latter constraint is easy to satisfy if $g$ is a smooth bump function with support a subset of $[0+\epsilon,1-\epsilon]$ for some $\epsilon$. All such functions (excluding the zero function) are non-analytic, so wouldn't show up in a power series analysis.

I haven't actually got a general solution, still, but I have managed to find an infinite family of linearly-independent bump-solutions, and a criterion for a general solution.

  • $\begingroup$ To obtain the general solution: Let $g$ be smooth on $[-\frac23,\frac23]$, then use a smooth partition $\phi_1(x)+\phi_2(x)=1$ of one where $\phi_1(x)=1$ for $0\le x\le \frac13$, $\phi_3(x)=1$ for $\frac23\le x\le 1$, then consider $\phi_1(x)g'(x)+\phi_2(x)g(1+x)$ on $[0,1]$ (and extend to $\mathbb R$ as you described). $\endgroup$ – Hagen von Eitzen Jul 7 '14 at 6:14

If you assume your solution is $ f(x)= {\rm e}^{\lambda x } $ and substituting back in the equation, you get,

$$ \lambda {\rm e}^{\lambda x } = {\rm e}^{\lambda (x+1) } \Rightarrow \lambda = {\rm e}^{\lambda } \Rightarrow \frac{1}{\lambda}{\rm e}^{\lambda } = 1 \,. $$

To find $\lambda$, we appeal to the Lambert W function which gives

$$ \lambda = -W(-1)\, $$

Then, we can write our solution as

$$ f(x) = c \,{\rm e}^{-W(-1) x } \,.$$

But note that, this is not a real function, since $ -W(-1)=0.3181315052- 1.337235701\,i $ is complex. If we consider the other values contributed from the other branches of the Lambert W function, we can write the general solution, as pointed out in the comments, as

$$ f(x) = \sum_{n= -\infty}^{\infty} c_n {\rm e}^{-W_n(-1) x}\,, $$ where $ c_n $ are constants.

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    $\begingroup$ This gives a solution of the equation, but not the general solution. $\endgroup$ – user12477 Sep 20 '12 at 13:54
  • $\begingroup$ Chris's sister asked for the general form of $f$. This is just one solution. More solutions can be found by multiplying the function with a constant: $f(x)=a\mathrm e^{-W(-1)x}$. However is this already the most general form? $\endgroup$ – celtschk Sep 20 '12 at 13:57
  • $\begingroup$ As I said it is a complex function. $\endgroup$ – Mhenni Benghorbal Sep 20 '12 at 14:01
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    $\begingroup$ @MhenniBenghorbal Also worth noting that equation $\lambda \mathrm{e}^{-\lambda} = 1$ has infinitely many solutions $-W_k(-1)$, for $k \in \mathbb{Z}$. All of them are complex. The general solution is then $$ y(x) = \sum_{k=-\infty}^{\infty} c_k \exp\left(-W_k(-1) x\right)$$ $\endgroup$ – Sasha Sep 20 '12 at 14:03
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    $\begingroup$ @Sasha: That is only the general solution if there are no solutions which are not of the form $\mathrm e^{\lambda x}$. For which there is no proof yet. $\endgroup$ – celtschk Sep 20 '12 at 14:07

First of all, let's turns into a delayed form. Let $g(x) = f(-x)$, then we have: $$ g'(x) = -g(x-1) $$ Therefore consider more general equation $y'(t) = -y(t-\tau)$, and initial condition $y(t) = \upsilon(t)$ for $0<t<\tau$. Then, for $\tau < t < 2 \tau$, $$ y(t) = \upsilon(\tau) - \int_\tau^t y(s-\tau) \mathrm{d}s = \upsilon(\tau)- \int_\tau^t \upsilon(s-\tau) \mathrm{d}s = \upsilon(\tau) - \int_0^{t-\tau} \upsilon(s) \mathrm{d}s $$ for $2 \tau < t <3 \tau$: $$ y(t) = y(2 \tau) - \int_{\tau}^{t-\tau} y(s) \mathrm{d} s = \upsilon(\tau)-\int_0^\tau \upsilon(s) \mathrm{d}s - (t-2 \tau) \upsilon(\tau) + \int_\tau^{t-\tau} \int_{0}^{s-\tau} \upsilon(u) \mathrm{d} u $$ and so on.

Here is an implementation of this idea in Mathematica, and comparison with built-in delayed ODE solver: enter image description here

  • $\begingroup$ I don't think this resolves the question of which initial conditions lead to full solutions. Obviously $v$ being a constant function other than zero does not. $\endgroup$ – Ben Millwood Sep 20 '12 at 13:58
  • $\begingroup$ @BenMillwood The full solution has the initial condition function unspecified, i.e. the solution space is not finite-dimensional, like in the non-delayed counterpart $y^\prime(x) = -y(x)$. $\endgroup$ – Sasha Sep 20 '12 at 14:01
  • $\begingroup$ right, but only some initial condition functions are viable, and I think that in order to claim a general solution you'd have to identify that set. $\endgroup$ – Ben Millwood Sep 20 '12 at 14:20

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