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Today, via Dan Piponi's answer on Quora, I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and

$$e^D \triangleq \sum_{i=0}^{\infty} \frac{D^i}{i!}.$$

I was curious as to whether the differential equation

$$\frac{df(t)}{dt} = e^Df(t) = f(t+1)$$

has any solutions apart from $f = 0$? I am not sure whether this question is even well-posed.

Rodrigo de Azevedo
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Thiagarajan
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  • @sos440 : Your comment helped. Thanks. But why did you delete it? – Thiagarajan Aug 22 '12 at 07:20
  • Sorry if I'm making a mistake, but where it's defined, don't we have $$ e^D f(x) = \sum \frac{f^{(n)}(x)}{n!}1^n = f(1-x) ~~? $$ – Eugene Shvarts Aug 22 '12 at 07:22
  • @Thiagarajan, Becuase I missed the dot in $\dot{f}$. (My Chrome browser's MathJax performance has been deteriorated recently.) So my previous answer was utterly wrong. – Sangchul Lee Aug 22 '12 at 07:22
  • @ Eugene Shvarts The infinite sum $\sum \dfrac{f^n(x)}{n!}1^n$ is the Taylor series representation of $f(x + 1)$. – Thiagarajan Aug 22 '12 at 07:29
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    This type of differential equations is known as forward (advanced) differential equations. On the other, if you have $f'(t)=f(t-1)$, then it is known as delay differential equation. Off course, these are special cases of more general equations. – Mhenni Benghorbal Aug 22 '12 at 07:56
  • See also: https://math.stackexchange.com/questions/389847/when-fx1-fx-fx-what-are-the-solutions-for-fx for the step method. – Bcpicao Apr 01 '21 at 19:46
  • Terminology: The term "differential operator" describes a huge category of operators on functions and generalized functions. The correct term for the OP's D is "the derivative operator", specifically d/dx. – Dan Asimov Sep 05 '22 at 23:23

2 Answers2

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We first make a heuristic calculation.

The equation in question is $Df = e^{D}f$. Thus it suffices to solve that $D = e^D$. Treating $D$ as a complex number, it is not hard to see that this equation has a solution in $\Bbb{C}$. For example, we may exploit Lambert-W function to specify such a number. Now let $r$ satisfy $r = e^r$. Then the function $f(t) = e^{r t}$ satisfies the desired property.

Indeed, we prove that this is the case. We have

$$Df(t) = r e^{r t} = e^{r}e^{r t} = e^{r(t+1)} = f(t+1) = e^{D}f(t).$$

Therefore $f(t)$ satisfies the desired property.

If we confine ourselves to real-valued functions, taking real part or imaginary part will also give such non-trivial examples.

Martin Sleziak
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Sangchul Lee
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  • :) . the answer's delightful. Thanks. – Thiagarajan Aug 22 '12 at 07:48
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    Note also that linear combinations of solutions are solutions. – Robert Israel Aug 22 '12 at 08:00
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    @sos440 Just a question. What does it mean(formally) to solve for $D = e^D$ ? The reason I ask is that the operators $D$ and $e^D$ are known quantities. So, I am not sure what is being solved for. Maybe I am misinterpreting something? – Thiagarajan Aug 22 '12 at 08:05
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    This is called the characteristic method. Our ansatz shows that functions of the form $f(x) = e^{rx}$ is a strong candidate for the equation. Plugging this to the equation actually yields the condition $r = e^r$, called the characteristic equation. Now it remains to show that the function $f(x) = e^{rx}$ corresponding to a zero of the characteristic equation is indeed a solution. Usually the characteristic equation is obtained by treating $D$ just as a variable. – Sangchul Lee Aug 22 '12 at 08:25
  • @sos440 All right. Thanks. – Thiagarajan Aug 22 '12 at 08:33
  • I'm late but to add something. Given that $$De^{zx} = ze^{zx} $$ To shift $a$ units, you need to solve the equation $z = e^{za}$, with solutions $z=W(-a)/-a$, where $W$ is the Lambert W function. So the function: $$e^{W(-a)/-a}x$$ is the solution of: $$Df(x)=e^{aD}f(x) $$ for $a \in \Bbb C$ – Daniel Muñoz Nov 24 '22 at 12:04
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Let us elaborate on the differential operator $ {\rm e}^{aD} $. It is defined by the power series $ \sum_{k=0}^{\infty}\frac{D^k}{k!} $. We would like to know its effect on the function $f(x)$. First let's see its effect on the function $x^m$. $$ {\rm e}^{ aD } x^m = \sum_{k=0}^{\infty} \frac{a^kD^k}{k!} x^m = \sum_{k=0}^{\infty}\frac{\Gamma(m+1)}{\Gamma(m-k+1)k!} a^k x^{m-k} = \sum_{k=0}^{\infty}\frac{m!}{(m-k)!k!} a^k x^{m-k}$$ $$= \sum_{k=0}^{\infty}{m \choose k} a^k x^{m-k} = \sum_{k=0}^{m}{m \choose k} a^k x^{m-k} = (x+a)^m\,, $$ by binomial theorem. That means the differential operator works as a shift operator that shift the argument by $a$. Now, we can use this result to see the effect on a function $f(x)$. If the function $f(x)$ has a power series at $x=0$, then $$ {\rm e}^{aD} \sum_{k=0}^{\infty} f_k x^k = \sum_{k=0}^{\infty} f_k {\rm e}^{aD} x^k = \sum_{k=0}^{\infty} f_k (x+a)^k = f(x+a)\,.$$ So, the general effect of the differential operator ${\rm e}^{aD}$ on the function $f(x)$ is nothing but shifts the argument by $a$. In your case $a=1$.

No pie
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Mhenni Benghorbal
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    Why do you have to use the gamma function if only integer power of $D$ are ever taken? Why doesn't a simple factorial work? – Zach Effman Jan 23 '15 at 02:42