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During my recent work I encountered a weird type of differential equation containing scalar factors. In order to understand them better I thought it would make sense to look at a simpler example like:

$$ \dot x(t) = x(2t)$$

How does one solve this equation? Does it even have a solution? It seems kinda related to Functional Differential Equations which I am not familiar with.


marked as duplicate by Martin R, Community Jun 7 '18 at 12:55

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    $\begingroup$ Do you need exact solutions? $\endgroup$ – user223391 Jun 4 '18 at 19:13
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    $\begingroup$ It's not just related, it is a functional differential equation. $\endgroup$ – Robert Israel Jun 4 '18 at 19:14
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    $\begingroup$ More precisely it's a functional differential equation with advanced argument. It is not clear whether there exist solutions except the trivial one, $x \equiv 0$. For example a solution that is analytic about $t=0$ and satisfies $x(0) = 1$ would have to satisfy $x^{(k)}(t) = 2^{k-1}x(2^kt)$ for all $t$ s.th. $2^k|t|$ is sufficiently small. Thus in particular $x^{(k)}(0) = 2^{k-1}$. But then $x(t)$ would have to be an exponential function, which is clearly not a solution. $\endgroup$ – Hans Engler Jun 4 '18 at 19:28
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    $\begingroup$ This question seems quite similar in some ways to MSE question 2802986 about $f(x) = f'(x^2)$ $\endgroup$ – Somos Jun 4 '18 at 23:17
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    $\begingroup$ I used the answer in the question provided by @IvanNeretin to obtain this series: $$ x = a_0 \sum_{n=0}^\infty \frac{(-1)^n}{2^{n(n+1)/2}} \exp -2^n t, $$ i checked it and it seems to satisfy indeed the equation, but I could not replicate it numerically... $\endgroup$ – rafa11111 Jun 5 '18 at 15:22

Disclaimer: It is not a complete answer, but the development of an idea. I encourage anyone who wants to edit and complement this answer. Furthermore, I have no experience with functional differential equations.

Supposing that our solution is an analytical function of $t$, we write it as a power series $$ x(t) = \sum_{n=0}^\infty a_n t^n. $$ Substituting it in the equation, we get $$ \sum_{n=1}^\infty n a_n t^{n-1} = \sum_{n=0}^\infty a_n 2^n t^n, $$ then, changing the index, $$ \sum_{n=0}^\infty (n+1) a_{n+1} t^{n} = \sum_{n=0}^\infty a_n 2^n t^n, $$ which lead us to the following recurrence relation: $$ a_{n+1} = \frac{2^n}{n+1} a_n. $$

As pointed by Alex R. and by martin cohen, there is no analytical solution to the equation other than the trivial, (as suggested by Hans Engler).

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    $\begingroup$ So $a_n= \frac{2^\binom{n}{2}}{n!}a_0$, and hence the series for $x(t)$ seems to diverge by the ratio test, so it looks like the trivial solution is the only one. $\endgroup$ – Alex R. Jun 4 '18 at 19:51
  • $\begingroup$ Thanks for providing this closed form expression for $a_n$! Can you really affirm the non-existence of non-trivial solutions in a broad sense, or only about eventual analytical solutions? $\endgroup$ – rafa11111 Jun 4 '18 at 20:00
  • $\begingroup$ so $\dot x(t)=x(mt)$ would have a converging solution for $m \le 1$, that is when the "present" depends from the "present/past" and not from the "future" $\endgroup$ – G Cab Jun 4 '18 at 20:07
  • $\begingroup$ @GCab why can’t things go vice versa? e.g. the displacement depends on the past velocity. $\endgroup$ – Szeto Jun 4 '18 at 22:48
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    $\begingroup$ Great! reminds me of the old mantra 'when in doubt what to do, try a power series'. I will try to apply Padé's to this series, maybe they generate a solution. $\endgroup$ – Hyperplane Jun 5 '18 at 9:05

Continuing rafa11111's answer, if $a_{n+1} = \frac{2^n}{n+1} a_n$ then $\dfrac{a_{n+1}}{a_n} = \dfrac{2^n}{n+1} $ so

$\begin{array}{} a_m &=& \Big(\prod\limits_{n=0}^{m-1} \dfrac{a_{n+1}}{a_n}\Big)\,a_0 &=& \Big(\prod\limits_{n=0}^{m-1} \dfrac{2^n}{n+1}\Big)\,a_0 &=& \dfrac{2^{m(m-1)/2}}{m!}\,a_0 &=& \dfrac{2^{T_{m-1}}}{m!}\,a_0 \end{array} $

Where $T_m$ is the $m$-th triangle number. This, as well as the obvious use of the ratio test, shows that the power series does not converge for any $x \ne 0$.


Following the suggestion by Ivan Neretin, I used the idea from this answer to a similar question: Solving functional differential equation $f'(x)=2f(2x)-f(x)$

I will solve the more generic equation $\dot{x} = x(at)$. As pointed by Hyperplane, I will use the following series: $$ x = \sum_{n=-\infty}^\infty c_n \exp (-a^n t). $$ Therefore, $$ \dot{x} = -\sum_{n=-\infty}^\infty c_n a^n \exp (-a^n t), $$ $$ x(at) = \sum_{n=-\infty}^\infty c_n \exp (-a^{n+1} t) = \sum_{n=-\infty}^\infty c_{n-1} \exp (-a^n t). $$ Then, plugging in the equation, we get $$ -\sum_{n=-\infty}^\infty c_n a^n \exp (-a^n t) = \sum_{n=-\infty}^\infty c_{n-1} \exp (-a^n t), $$ which requires $$ \frac{c_n}{c_{n-1}} = -\frac{1}{a^n}. $$ Solving this recurrence relation with an "initial condition" $c_0 = A$, we get $$ c_m = A \prod_{n=0}^m \frac{c_n}{c_{n-1}} = A \prod_{n=0}^m -\frac{1}{a^n} = \frac{(-1)^m}{a^{m(m+1)/2}} A. $$ (See that $m(m+1)/2$ is the m-th triangular number). Therefore, our solution is $$ x = A \sum_{n=-\infty}^\infty \frac{(-1)^n}{a^{n(n+1)/2}} \exp (-a^n t), $$ which one can easily check to satisfy the equation. Since it is an alternating series, we can apply the Leibiniz criterion. The ratio between two consecutive terms is $a^{n+1} \exp(a^n(a-1)t)>1 \ \forall a>1$. Therefore, the series converges for $a>1$.

To obtain the value of $x(0)$ one must split the series: $$ x(0) = A \sum_{n=0}^\infty \frac{(-1)^n}{a^{n(n+1)/2}} + A \sum_{n=-1}^{-\infty} \frac{(-1)^n}{a^{n(n+1)/2}} = A \sum_{n=0}^\infty \frac{(-1)^n}{a^{n(n+1)/2}} + A \sum_{n=0}^{-\infty} \frac{(-1)^{n-1}}{a^{(n-1)n/2}} = $$ $$ A \sum_{n=0}^\infty \frac{(-1)^n}{a^{n(n+1)/2}} + A \sum_{n=0}^{\infty} \frac{(-1)^{-n-1}}{a^{-n(-n-1)/2}} = A \sum_{n=0}^\infty \frac{(-1)^n}{a^{n(n+1)/2}} - A \sum_{n=0}^{\infty} \frac{(-1)^{n}}{a^{n(n+1)/2}} = 0. $$

It is interesting to see the behavior of the derivatives of this function. We know that $\dot{x} = x(at)$. Differentiating it, we have $$ \ddot{x} = \frac{d}{dt} x(at) = \frac{d(at)}{dt} \frac{d}{d(at)} x(at) = a \dot{x}(at) = a x(a^2 t), $$ $$ \dddot{x} = \frac{d}{dt} a x(a^2 t) = a \frac{d(a^2t)}{dt} \frac{d}{d(a^2t)} x(a^2t) = a^3 \dot{x}(a^2t) = a^3 x(a^3 t), $$ or, in general, $$ x^{(n)} = a^{n(n-1)/2} x(a^n t), $$ which can be readly proved by induction. Therefore, it is easy to see that $x^{(n)}(0) = 0$ for all $n$, i.e., showing that the solution is non-analytical at $t=0$. Here is the graph of the solution for $a=2$ and $A=1$. enter image description here

And here is a graph comparing $\dot{x}(t)$ to $x(2t)$. This graph is rougher on purpose to allow an easy distinction between both curves. For a thinner resolution, the matching is perfect.

enter image description here

I also tried to solve the equation numerically using a simple finite differences scheme. Defining the domain as $t \in (0,T)$ and discretizing it in $N$ equally spaced points, the equation for the point $i$ is $$ x_{i+1} - x_i - hx_{[ai]} = 0, $$ in which $h$ is the step size, and $[ai]$ indicates that the result of $a\cdot i$ must be rounded to the closer integer. This method has a serious deficiency because the solution only will have mathematical meaning in the interval $t\in (0,T/a^2)$. That's because the finite differences equation can be applied only in the interval $t\in (0,T/a)$, as $x_{[ai]}$ would be undefined for $t>T/a$. Therefore, the equation for $t>T/a$ will not represent the original equation and will be only a "placeholder". However, the spurious solution in the region $T/a<t<T$ will cause an effect in the remaining part of the domain, specifically in the region $T/a^2<t<T/a$. Therefore, the solution only holds for $0<t<T/a^2$ and is spurious for $T/a^2<t<T$.

For numerical purposes, the matrix $A_{ij}$ is defined as $A_{ij} = A^1_{ij} + A^2_{ij}$, being $$ A^1_{ij} = \begin{cases} 1, & i+1=j \\ -1, & i=j \\ \end{cases}, \ \ \ A^2_{ij}=\begin{cases} -h, & [ai]=j \\ 0, & \mathrm{otherwise} \\ \end{cases}. $$

As the last line of $A^1_{ij}$ is undefined, it can be used to providing the "boundary" condition. It is important to notice that all solutions are equal up to a multiplicative constant. Therefore, the solution can be obtained by $$ x_i = (A_{ij})^{-1} B_j, $$ in which $B_j = 0$ for all $j$, except for the position in which the condition was imposed. Here is a numerical solution obtained for $a=2$, $h = 4\times 10^{-2}$ and $T=160$: enter image description here

Both solutions were 'normalized' to allow a better comparison. There is a good agreement among the solutions. It is interesting to see that the numerical solution does not satisfy $x(0) = 0$, only approximately. It is also important that the domain be sufficiently large in order that the most relevant oscillations are contained in the region $0<t<T/a^2$. The size of the domain is more important to the accuracy of the solution than the step size.

For $1\leq a\leq 1$ the power series solution, $$ x = A\sum_{n=0}^\infty \frac{a^{n(n-1)/2}}{n!} t^n, $$ converges. For $a=0$ the solution degenerates to $x=kt$, for $a=1$, to $x = k \exp t$ and for $a=-1$ to $x=k\sin (t+\pi/4)$ (I found this one quite surprising!). Solutions for intermediates values of $a$ are shown in the graph:enter image description here

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    $\begingroup$ 2 Remarks: First, we can extend the solution from $[0,\infty)$ to $(-\infty,+\infty)$ by simply letting $x(t)=0$ for $t<0$. Since the original solution is flat at the origin as you checked, the combined one is $C^\infty$-smooth. Secondly, this idea only supplies a non-trivial solution for $x(0)=0$, so the question about the case $x(0)=1$ remains open. Anyway it seems like you had quite your fun with the problem :) $\endgroup$ – Hyperplane Jun 6 '18 at 8:03
  • $\begingroup$ @Hyperplane Indeed, the solution can be expanded that way, still satisfying the equation. I was not aware of the initial condition. Checking the numerical scheme now, I noticed that it provides nice solutions for $x(0) \neq 0$. I thought that those solutions were spurious, but $\dot{x}$ seems to match with $x(2t)$. Therefore, I think that there are solutions for $x(0)\neq 0$, but I don't know how to approach it analytically. If you find this relevant, I can improve the answer with numerical results for $x(0)\neq 0$. $\endgroup$ – rafa11111 Jun 6 '18 at 12:51

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