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.
Best Answer
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 $ab\gt1$.