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Solve $f(x)=\int_{x-1}^{x+1} f(t) \text{d}t$. We know that $f(x)=0$ is a solution, are there any other solutions?

I suppose I can start with $f(x)=\frac{dg(x)}{dx}$, and then:

$$\begin{align*} \frac{dg(x)}{dx}&=\int_{x-1}^{x+1} \frac{dg(t)}{dt} dt \\ \frac{dg(x)}{dx}&=g(x+1)-g(x-1) \\ \end{align*}$$

But then I'm stuck.

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    $\begingroup$ Where did you encounter this problem? So-called 'delay-differential' equations are notoriously difficult and tend not to have clean analytical solutions. $\endgroup$ – Steven Stadnicki Jan 31 '14 at 16:58
  • $\begingroup$ Also, you have 'past' and 'future' values in the ODE which complicates matters even more... $\endgroup$ – copper.hat Jan 31 '14 at 16:59
  • $\begingroup$ I was trying to model a random walk where a frog lies on a one-dimensional rope and jumps to a uniform random position in the [x-1,x+1] neighborhood on each iteration. $\endgroup$ – Ricbit Jan 31 '14 at 17:05
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    $\begingroup$ If you're trying to model the random walk, you should have a normalizing factor of $\frac12$ in front of the integral, which makes the problem rather different as now all linear functions are solutions. $\endgroup$ – Greg Martin Jan 31 '14 at 18:51
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    $\begingroup$ @Ricbit I have edited my post significantly. It actually now includes the solution given in the Math Overflow post you made. $\endgroup$ – Cameron Williams Feb 1 '14 at 0:11
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Better yet you can differentiate both sides since $f$ will be smooth (convince yourself of this). Then what you have is (by FToC)

$$f'(x) = f(x+1)-f(x-1)$$

We want to solve this equation. As per user Doubt, note that $f(x)=a$ is a solution to the ODE but not to the integral equation (unless $a=0$).

I will then assume that $f$ is nonconstant and is a true function. Further, let's assume the function has a Fourier transform. If we take the Fourier transform, where

$$\mathcal{F}f(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}f(x)dx,$$

we get that

$$ik\mathcal{F}f(k)=e^{ik}\mathcal{F}f(k)-e^{-ik}\mathcal{F}f(k).$$

This has to hold true for all $k$. So we can rewrite this as

$$(ik-2i\sin(k))\mathcal{F}f(k) = 0.$$

Since $k=2\sin k$ only for $k=0,\pm1.89549$, we see that $\mathcal{F}f(k) = 0$ almost everywhere and so $f(x) = 0$ almost everywhere.


Let us suppose then that $f$ is distributional in nature, particularly a tempered distribution. We can make sense of distributional derivatives so the above differential still holds but some of the later steps will be different.

Since $f$ is a tempered distribution, it has a Fourier transform. Let $\phi\in\mathcal{S}(\mathbb{R})$, i.e. a Schwartz function, then

$$\begin{eqnarray}\langle f',\phi\rangle &=& \langle f_1-f_{-1},\phi\rangle\\ &=&\langle f_1,\phi\rangle-\langle f_{-1},\phi\rangle\\ &=&\langle f,\phi_{-1}\rangle-\langle f,\phi_1\rangle\end{eqnarray}$$

where $f_a(x) = f(a+x)$. We also have that $\langle f',\phi\rangle = -\langle f,\phi'\rangle$ so that

$$-\langle f,\phi'\rangle = \langle f,\phi_{-1}-\phi_1\rangle.$$

Equivalently, $\langle f,\phi'-\phi_1+\phi_{-1}\rangle = 0$. Taking a Fourier transform, we see that

$$\langle \mathcal{F}^{-1}f,\mathcal{F}(\phi'-\phi_1+\phi_{-1})\rangle = 0.$$

Then we see that

$$\langle \mathcal{F}^{-1}f,-ik\mathcal{F}\phi-e^{-ik}\mathcal{F}\phi+e^{ik}\mathcal{F}\phi\rangle_k = 0,$$

where $\langle\cdot,\cdot\rangle_k$ indicates that the distributional integral is over $k$. Then by linearity,

$$\langle (-ik-e^{-ik}+e^{ik})\mathcal{F}^{-1}f,\mathcal{F}\phi\rangle_k = 0.$$

Since the Fourier transform is an automorphism of $\mathcal{S}(\mathbb{R})$, $\mathcal{F}\phi$ is just another Schwartz function. Thus

$$\langle (-ik+2i\sin k)\mathcal{F}^{-1}f,\psi\rangle_k = 0,$$

where $\psi$ is an arbitrary element of $\mathcal{S}(\mathbb{R})$. Thus we see that $(2\sin k-k)\mathcal{F}^{-1}f(k) = 0$, in a distributional sense, just like above. To be extremely handwavy about it: If $2\sin k -k\neq 0$, then the (inverse) Fourier transform of $f$ must be zero. However, if $2\sin k = k$, then the (inverse) Fourier transform of $f$ can take on any value. This is exactly why we have the solution that was noted on your Math Overflow post. If you were to take the Fourier transform of the solution given by Noam, you would see that it is a Dirac delta centered at the point $k=2\sin k$.

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    $\begingroup$ "and so $f(x) = 0$ almost everywhere": and so, from the original equation, $f(x) = 0$ everywhere. $\endgroup$ – TonyK Jan 31 '14 at 17:23
  • $\begingroup$ Actually your argument works in Schwartz class $\mathscr S'(\mathscr R)$, where $$\langle f,\varphi\rangle=\langle {\mathcal F}f,{\mathcal F}^{-1}\varphi\rangle=0,$$ for all $\varphi\in{\mathscr S}(\mathbb R)$. $\endgroup$ – Yiorgos S. Smyrlis Jan 31 '14 at 17:25
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    $\begingroup$ @Ricbit I tried working it out and got something different. Formally manipulating the integral, we have $$\int_{x-1}^{x+1}\delta(\sin(2\pi t))dt = \frac{1}{2\pi}\int_{\sin(2\pi(x-1))}^{\sin(2\pi(x+1))}\delta(u)\frac{1}{\sqrt{1-u^2}}du,$$ where I've let $u=\sin(2\pi t)$. I think you get $\frac{1}{2\pi}$ from this. $\endgroup$ – Cameron Williams Jan 31 '14 at 17:57
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    $\begingroup$ Note that $f(x) = a \in\mathbb{R}$ satisfies $f^\prime(x) = f(x+1) - f(x-1)$, but not the original problem $f(x) = \int^{x+1}_{x-1}f(t)\,dt$. $\endgroup$ – Doubt Jan 31 '14 at 18:10
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    $\begingroup$ @Doubt Good point! $\endgroup$ – Cameron Williams Jan 31 '14 at 18:13
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A set of solutions is $$f(x) = a\sin(\omega x) + b\cos(\omega x)$$ for any $a,b\in\mathbb{R}$. Note that $$\int_{x-1}^{x+1}f(t)\,dt = \frac{2\sin\omega}{\omega}\left(a\sin(\omega x)+b\cos(\omega x)\right)= \frac{2\sin\omega}{\omega}f(x)$$ so $\omega$ must be chosen to satisfy $$\frac{2\sin\omega}{\omega} = 1.$$ Numerically, $\omega \approx \pm 1.89549$.

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  • $\begingroup$ This is similar to a solution proposed by Yiorgos below. $\endgroup$ – copper.hat Jan 31 '14 at 19:14
  • $\begingroup$ Nice! Is there a way to find this solution without guessing its form? $\endgroup$ – Ricbit Jan 31 '14 at 21:01
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An attempt for a solution is to try $$ f(x)=\mathrm{e}^{ax}, $$ and this implies that $a$ satisfies equation $\mathrm{e}^a-\mathrm{e}^{-a}=a$. Unfortunately, the only solution of this equation is $a=0$, and the corresponding exponential function is not a solution of our equation.

Clearly, we need to assume some smoothness for $f$. Let's assume that $f$ is continuous. Then, the right hand side is continuously differentiable, and thus we can differentiate, and obtain, as you have already done the equation $$ f'(x)=f(x+1)-f(x-1), \quad \text{for all}\,\, x. $$

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    $\begingroup$ ${x \over 2} = \sinh x$ only has $0$ as a solution (over the reals, at least). $\endgroup$ – copper.hat Jan 31 '14 at 16:56
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    $\begingroup$ You actually don't need to assume smoothness as long as you assume $f$ is measurable (otherwise why would you be integrating it?) because $f$ is then the integral of a measurable function, which is always continuous. Repeat inductively to see that $f$ is smooth. $\endgroup$ – Cameron Williams Jan 31 '14 at 16:56
  • $\begingroup$ Still, it a reasonable first shot to assume smoothness and then validate afterwards... $\endgroup$ – copper.hat Jan 31 '14 at 16:58
  • $\begingroup$ @copper.hat: Undoubtedly! $\endgroup$ – Yiorgos S. Smyrlis Jan 31 '14 at 16:59
  • $\begingroup$ $f(x) = e^{0x} = 1$ is not a solution even though $a=0$ fulfills the condition $e^a-e^{-a}=a$. $\endgroup$ – Juho Kokkala Jan 31 '14 at 17:31
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(Written up as a solution since it's a bit too long for a comment: )

Another approach that doesn't work: suppose $g$ is a polynomial of degree $d$. Then $\frac{dg}{dx}$ will be a polynomial of degree $d-1$, and so will $g(x+1)-g(x-1)$ (the leading term cancels). Since the equation is linear, we can assume $g$ a monic polynomial (i.e., with a leading term of $x^d$). But then the leading term of $\frac{dg}{dx}$ will be $dx^{d-1}$, whereas the leading term of $g(x+1)-g(x-1)$ will be $2dx^{d-1}$.

In fact, you should be able to extend this approach to show that the only analytic solution is the constant function, since $g'(x)$ is 'twice as large' as would be expected from a naive central-difference approximation everywhere.

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