Solving a Simple Integral Equation I'm looking to solve the integral equation
$$f(x)=a\int_x^{x+b}f(t)\,dt,$$
for positive $a,b$. However, I've gotten stuck. My current approach is to apply a Laplace Transform to obtain (denoting $F=\mathscr{L}\{f\}$ as the Laplace transform of $f$)
$$F(s)=\frac{f(0)}{s}+\frac{a}{s}\left(\mathscr{L}\{f(x+b)\}-F(s)\right).$$
At this point I am unsure how to handle the Laplace transform of $f(x+b)$ since $b$ is positive. If I can express it in terms of $F$, I can solve the resulting equation for $F$ and invert the transform. Any tips on how to solve this equation would be appreciated!
As a side note, the solution has the form $f\sim e^{Bx}$ for constant $B$, so I feel that directly solving this equation should be possible.
 A: Differentiating, we get the equation
$$
f'(x)=a(f(x+b)-f(x))\tag1
$$
Taking the Fourier Transform, we get
$$
\begin{align}
\int_{-\infty}^\infty f'(x)e^{-2\pi i\gamma x}\,\mathrm{d}x&=a\int_{-\infty}^\infty(f(x+b)-f(x))e^{-2\pi i\gamma x}\,\mathrm{d}x\tag{2a}\\
2\pi i\gamma \hat{f}(\gamma)&=a\!\left(e^{2\pi ib\gamma}-1\right)\hat{f}(\gamma)\tag{2b}
\end{align}
$$
Either $\hat{f}=0$ or $2\pi i\gamma=a\!\left(e^{2\pi ib\gamma}-1\right)$. The latter has the solutions
$$\newcommand{\W}{\operatorname{W}}
\begin{align}
2\pi i\gamma&=a\!\left(e^{2\pi ib\gamma}-1\right)\tag{3a}\\
2\pi ib\gamma&=ab\!\left(e^{2\pi ib\gamma}-1\right)\tag{3b}\\
2\pi ib\gamma+ab&=abe^{2\pi ib\gamma}\tag{3c}\\
2\pi ib\gamma+ab&=abe^{-ab}e^{2\pi ib\gamma+ab}\tag{3d}\\
2\pi ib\gamma+ab&=-\W\left(-abe^{-ab}\right)\tag{3e}\\
2\pi ib\gamma&=-ab-\W\left(-abe^{-ab}\right)\tag{3f}\\
\gamma&=\frac i{2\pi b}\left(ab+\W\left(-abe^{-ab}\right)\right)\tag{3g}\\
\end{align}
$$
Explanation:
$\text{(3a)}$: cancelling $\hat{f}(\gamma)\ne0$ from $\text{(2b)}$
$\text{(3b)}$: multiply both sides by $b$
$\text{(3c)}$: add $ab$ to both sides
$\text{(3d)}$: multiply the right side by $e^{-ab}e^{ab}=1$
$\text{(3e)}$: multiply both sides by $-e^{-(2\pi ib\gamma+ab)}$, getting
$\phantom{\text{(3e):}}$ $-(2\pi ib\gamma+ab)e^{-(2\pi ib\gamma+ab)}=-abe^{-ab}$
$\phantom{\text{(3e):}}$ then solve for $2\pi ib\gamma+ab$ by applying $-\W$ to both sides (Lambert W)
$\text{(3f)}$: subtract $ab$ from both sides
$\text{(3g)}$: divide both sides by $2\pi ib$
So $\hat{f}=0$ or $\hat{f}$ is supported at $\gamma=\frac i{2\pi b}\left(ab+\W\left(-abe^{-ab}\right)\right)$. That is,
$$
f(x)=ce^{-\frac1b\left(ab+\W\left(-abe^{-ab}\right)\right)x}\tag4
$$
If $ab\le0$ or $ab=1$, then $\W\left(-abe^{-ab}\right)=-ab$, and $(4)$ says that $f(x)=c$.
If $ab\gt0$ and $ab\ne1$, then there are two branches for $\W\left(-abe^{-ab}\right)$; one gives $f(x)=c$ and the other gives exponential growth for $0\lt ab\lt1$ and exponential decay for $ab\gt1$.
A: $$f(x)=a\int_x^{x+b}f(t)\,dt \tag{1}$$
Two steps:
A) First of all, let us look for solutions of (1) of the form $f(x)=Ke^{Bx}$ ($K$ arbitrary constant), we get:
$$Ke^{Bx}=\frac{a}{B}[Ke^{Bt}]_{t=x}^{t=x+b}$$
$$e^{Bx}=\frac{a}{B}(e^{B(x+b)}-e^{Bx})$$
$$e^{Bx}=\frac{a}{B}(e^{Bb}-1)e^{Bx} \ \ \text{for all} \ x.$$
Otherwise said, $B$ must fulfill the following equation:
$$\frac{a}{B}(e^{Bb}-1)=1 \iff B=a(e^{Bb}-1) \iff $$
$$Bb=ab(e^{Bb}-1)\tag{*}$$
which has the solution:
$$B=- \ \frac{1}{b}(ab+W(-e^{-ab})) \tag{**}$$
where $W$ denotes Lambert function defined [here] (https://en.wikipedia.org/wiki/Lambert_W_function)
(**) is obtained from (*) by using the following result about $W$ function that can be found in this article:
"The general solution of equation
$$x=\alpha+\beta e^{\gamma x} \ \ \text{is} \ \ x=\alpha - \frac{1}{\gamma}W(-\beta \gamma e^{\alpha \gamma})"$$
(take $\alpha=-ab, \beta=\gamma=1$)
Remark: A special case occurs when $ab=1$: indeed, as $W(-e^{-1})=-1$, (**) gives $B=0$. Therefore constant functions $f(x)=Ke^0$ can be solutions.

B) Now, let us transform directly equation (1) into a convolution:
$$f=a \mathbb{1}_{[-b,0]} \star f \tag{2}$$
where $\mathbb{1}_{[-b,0]}$ is the characteristic function of interval $[-b,0]$, also called "rectangular" function  on the interval.
Let us use on (2) the differentiation property of a convolution:
$$(\phi \star \psi)'=\phi' \star \psi$$
giving
$$f'= a(\delta_{-b}-\delta_0) \star f$$
(indeed, differentiation at the 'jumps' of the Rect function give Dirac $\delta$s in $-b$ and $0$).
$$f'= a(\delta_{-b} \star f-\delta_0 \star f)$$
Finally, we get the following differential-difference equation for $f$:
$$f'(x)= a(f(x+b)-f(x)) \tag{3}$$
(convolution by a Dirac = shift)
Up to you for applying Laplace Transform to (3)...
Remarks:
a) Take into account the constraint $f(0)=\int_0^b f(u)du$.
b) It could be interesting to briefly explain how you have obtained the Laplace Transform of the initial equation.
c) An interesting particular case in (3) is when $a=\frac{1}{b}$ giving:
$$f'(x)= \dfrac{f(x+b)-f(x)}{b} \tag{3}$$
which is verified by any affine function with equation
$$f(x)=mx+n$$
But constraint (1) gives
$$f(0)=\int_0^b f(t)dt \ \ \implies \ \ n=a(\tfrac12 mb^2+nb)$$
which (taking into account relationship $ab=1$) forces $m=0$. Therefore, only constant functions "survive" ; in this way, we find back the functions we had already considered in the remark of part A)... nothing new under the sun...
d) Equation (3) is reached in a much more direct manner in robjon's answer. The merit of my approach is to browse different useful analysis techniques.
