Functional equation with integral: $ f(x) = \int_0^x f(s) \, \mathrm ds $

Question

I was hoping for some help with this integral relation. I would prefer hints to answers. I'm trying to find any solution to this relation:

$$ f(x) = \int_0^x f(s) \, \mathrm ds $$

I feel like it should be some variation of the exponential function that is $0$ for $x \le 0$, but I'm not sure.

Any help would be appreciated!

Try taking the derivative of both sides and remember FToC. You're going to be sad to know that the only analytic solution is $y = 0$. — Kaynex, Apr 19 '17 at 16:27
@kaynex how is that the only solution? the fundamental theorem of calculus yields $$\frac{df}{dx} = f(x)$$ — Chinny84, Apr 19 '17 at 16:38
That's what I arrived at, leading me to believe that $e^x$ is a solution, but plugging into integral does not confirm that. I get $e^x = e^x - 1$. Am I doing something wrong? — Enrico Borba, Apr 19 '17 at 16:40
So the function can't change since the initial rate of change is 0 is the intuition I guess. @Kaynex I am sad indeed. — Enrico Borba, Apr 19 '17 at 16:47

score 2 · Accepted Answer · answered Apr 19 '17 at 18:12

Take the derivative of both sides

$$\frac{d}{dx}f(x)=\frac{d}{dx}\int_{0}^{x}f(s)\space ds$$

Note that $\int_{0}^{x} f(s) ds=F(x)-F(0)$. The derivative of $F(x)$ is $f(x)$ and the derivative of $F(0)=0$. Hence

$$f'(x)=f(x)$$

There are two possiblities.

Once is $f(x)=0$, another is $f(x)=ce^x$

But if we substitute $f(x)=ce^x$ into the first equation the left hand side is

$$\frac{d}{dx}ce^x=ce^x$$

And the right hand side is

$$\int_{0}^{x}ce^s\space ds=ce^x-c$$

If set both side to equal eachother we get

$$0=-c$$

Hence $f(x)=0$ is the only solution.

I think the procedure is fine but you need to make some assumption on $f$ — user8469759, Oct 08 '21 at 22:34

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