Exponential of the differential operator

I am not sure whether this question is even well-posed. But today I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and

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

So I was curious as to whether the differential equation

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

has any solutions apart from $f = 0$?

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@sos440 : Your comment helped. Thanks. But why did you delete it? –  Thiagarajan Aug 22 '12 at 7:20
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 7:22
@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. –  sos440 Aug 22 '12 at 7:22
@ 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 7:29
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 7:56

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 zero 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.

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:) . the answer's delightful. Thanks. –  Thiagarajan Aug 22 '12 at 7:48
Note also that linear combinations of solutions are solutions. –  Robert Israel Aug 22 '12 at 8:00
@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 8:05
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. –  sos440 Aug 22 '12 at 8:25
@sos440 All right. Thanks. –  Thiagarajan Aug 22 '12 at 8:33

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 $x^m$. $${\rm e}^{ aD } x^m = \sum_{k=0}^{\infty} \frac{a^kD^k}{k!} x^m = \sum_{k=0}^{\infty}\frac{\Gamma(m+1)}{\Gamma(k-m+1)k!} a^k x^{m-k} = \sum_{k=0}^{\infty}\frac{m!}{\Gamma(k-m)!k!} a^k x^{m-k}$$ $$= \sum_{k=0}^{\infty}{m \choose k} a^k x^{m-k} = \sum_{k=0}^{m}{m \choose k} a^k x^{m-k} = (x+a)^m\,,$$ 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 $a=1$.

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