The problem is as follows:

Let $f: \mathbb{Q} \to \mathbb{Q}$ and consider the differential equation $f' = f$, with the standard definition of differentiation. Do there exist any nontrivial solutions?

(Note that of course $f \equiv 0$ is a solution - by "nontrivial solutions", I mean anything else).


Differentiation and continuity are much weaker concepts on the rationals. For example, $H(x-\sqrt{2}) : \mathbb{Q} \to \mathbb{Q}$ is continuous and everywhere differentiable, where $H$ is the Heaviside step function.

If there exists a nontrivial solution $f_0$, then there are uncountably many solutions. For example, $H(x-\alpha)f_0$ is also a solution for any irrational $\alpha$ (which already gives uncountably many solutions), and thus any* linear combination $k_0 f_0 + \sum_{\alpha \in A} H(x-\alpha)k_\alpha f_0$ (with $A \subset \mathbb{R}\setminus\mathbb{Q}$) is also a solution, by linearity of the DE.

We can answer in the negative if there is a way to show that any such solution could be extended to a solution to $f' = f$ on $\mathbb{R}$, because those solutions are simply $ke^x$, which takes irrational values over the rationals unless $k = 0$. Unfortunately, the solutions $f : \mathbb{Q} \to \mathbb{Q}$ for the a differential equation $\mathcal{L}y = 0$ are not, in general, a subset of the real solutions. e.g. $y' = 0$ has solution $H(x-\sqrt{2})$ but every solution on $\mathbb{R}$ must be constant.

If the answer is "yes", maybe we'd hope to be able to construct a solution via some iterative method, but since Cauchy sequences are not in general convergent, we'd need some sort of machinery to guarantee rational limits.

*you can either insert the word "finite" here, or stipulate that the $k_\alpha$ are such that the quantity $\sum_{\alpha<q} k_\alpha$ is finite and rational for all $q \in \mathbb{Q}$, but the point is that we can construct a bunch of "different looking" solutions.

  • $\begingroup$ I have no experience in this field, however I would imagine that if $F$ is the family of solutions to a differential equation in functions $\mathbb R\to \mathbb R$, then the family $G$ of solutions $\mathbb Q\to \mathbb Q$ would be a subset of $F$ (when extended to the reals). $\endgroup$
    – David P
    Aug 22, 2016 at 3:29
  • $\begingroup$ I don't really have any experience in this field either. However, from discussions and thinking about this problem, I've found that often (possibly intuitive) statements of this kind are false. For example, consider the differential equation $f' = 0$; then the example in the question of $H(x-\sqrt{2})$ is a solution $\mathbb{Q} \to \mathbb{Q}$, but solutions on $\mathbb{R}$ must be constant! $\endgroup$ Aug 22, 2016 at 3:32
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    $\begingroup$ @fleablood A function can be continuous on $\mathbb{Q}$ without being continuous (or rather, lifting to anything continuous) on $\mathbb{R}$ . . . Also, note that $ke^x$ is never a map from $\mathbb{Q}$ to $\mathbb{Q}$ if $k\not=0$. (I made the same mistake in my now-deleted answer; look at the codomain!) $\endgroup$ Aug 22, 2016 at 3:40
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    $\begingroup$ @fleablood But it's not a solution (to the problem as posed). If it were, it would indeed be nontrivial. By "trivial solution", I'm referring exclusively to $f \equiv 0$. $\endgroup$ Aug 22, 2016 at 3:51
  • 2
    $\begingroup$ From either answer, though neither explicitly states it, it can be found out that every ordinary differential equation $f^{(n)}(x)=g(x,f(x),f'(x),\ldots,f^{(n-1)}(x))$ has a solution where $g:\mathbb Q^{n+1}\rightarrow\mathbb R$. I guess differential equations aren't very restrictive when taken over $\mathbb Q$. $\endgroup$ Aug 22, 2016 at 5:19

2 Answers 2


More generally, for any function $g:\mathbb Q\rightarrow\mathbb Q$ and any point $(x_0,y_0)\in\mathbb Q^2$, there exists $f:\mathbb Q\rightarrow\mathbb Q$ such that $f'(x_0)=y_0$ and $f'(x)=g(f(x))$.

Choose an enumeration $x_0,x_1,\ldots$ of $\mathbb Q$ starting with $x_0$. Let $Q_n=\{x_0,\ldots,x_n\}$, so $\mathbb Q=\bigcup_n Q_n$. We will inductively construct continuous functions $a_n,b_n:\mathbb Q\rightarrow\mathbb Q$ with the properties

  1. $a_{n-1}(x)\leq a_n(x)\leq b_n(x)\leq b_{n-1}(x)$
  2. If $x\in Q_n$ then $a_n(x)=b_n(x)$ and $a_n'(x)=b_n'(x)=g(a_n(x))$.
  3. If $x\in\mathbb Q\setminus Q_n$ then $a_n(x)<b_n(x)$.

We'll use the parabolic functions $c(s,t)$ and $d(s,t)$ defined by $$ c(s,t)(x)=t+g(t)(x-s)-(x-s)^2, $$ $$ d(s,t)(x)=t+g(t)(x-s)+(x-s)^2. $$ Note that $c(s,t)(x)<d(s,t)(x)$ for $x\neq s$ and both functions pass through $(s,t)$ with derivative $g(t)$. We can take $a_0=c(x_0,y_0)$ and $b_0=d(x_0,y_0)$.

Suppose $n>0$ and $a_{n-1},b_{n-1}$ are constructed. Then $a_{n-1}(x_n)<b_{n-1}(x_n)$, so choose $y_n$ strictly between these. Choose an open interval $I$ containing $x_n$ such that $c(x_n,y_n)>a_{n-1}$ and $d(x_n,y_n)<b_{n-1}$ on $I$. Shrink $I$ so that its closure doesn't intersect $Q_{n-1}$. Let $J$ be an open interval containing $x_n$ whose closure is inside $I$. We define $a_n$ to equal $a_{n-1}$ outside $I$, $c(x_n,y_n)$ inside $J$, and interpolate linearly between $I$ and $J$ so that the result is still continuous. Define $b_n$ similarly.

Since $\bigcup_n Q_n=\mathbb Q$, both $a_n$ and $b_n$ converge pointwise to a function $f$ as $n\rightarrow\infty$. For any $x\in\mathbb Q$ we have $x=x_n$ for some $n$, and $a_n\leq f\leq b_n$, so property 2 and the squeeze theorem imply that $f$ satisfies the required equation.


There are non-trivial solutions $f:\mathbb Q\rightarrow\mathbb Q$ to any differential equation of the form $f'(q)=g(q,f(q))$. Somewhat more strongly, we may find a solution $f$ such that for every $q\in \mathbb Q$ there is some $\delta>0$ such that $\left|\frac{f(q')-f(q)}{q'-q}-g(q,f(q))\right|<(q'-q)^2$ for every $q'$ with $|q'-q|<\delta$.

In this spirit, define the following subsets of $\mathbb Q^2$ $$S(x,y,\delta)=\left\{(x',y'):\left|y'-y-g(x,y)(x'-x)\right|<\left|x'-x\right|^3\text{ or }|x'-x|>\delta\right\}\cup \{(x,y)\}.$$ When $|x'-x|\leq \delta$, this is a region bounded by two parabolas tangent at $(x,y)$ with slope $m$ at that point, which is related to the condition we are requiring on $f$. We will define the function by defining it at particular points and choosing a suitable open set $S$ in which we place every further point. This will suffice to ensure differentiability.

In particular, let $\{p_n\}_{n=1}^{\infty}$ be an enumeration of the rationals. We will construct a sequence $\{q_n\}_{n=1}^{\infty}$ of rationals such that $f(p_n)=q_n$ defines a suitable function. We will, during the construction, use an auxiliary sequence $U_n$ of open subsets of $\mathbb Q^2$, letting $U_0=\mathbb Q^2$. At each step in the construction, we will demand the following of $U_{n}$ for all $n$:

  • Property 1: $U_{n}=U_{n-1}\cap S(p_n,q_n,\delta)$ for some $\delta>0$

  • Property 2: $U_{n}\setminus \{(p_1,q_1),(p_2,q_2),\ldots,(p_n,q_n)\}$ is open.

  • Property 3: For all rational $p\in\mathbb Q$ there exists $q\in \mathbb Q$ such that $(p,q)\in U_{n}$.

  • Property 4: For all $n'\leq n$ we have $(p_{n'},q_{n'})\in U_{n}$.

Given that the graph of the function being a subset of $S(p,f(p),\delta)$ for every $p$ implies that $f$ satisfies the differential equation, the only business we have is to show that such a triple of sequences exists.

To do so, suppose we are given the first $n-1$ terms of the sequences $\{q_n\}$ and $\{U_n\}$ along with the whole sequence $p_n$ and need to find a suitable $q_n$ and $U_n$ to extend the sequence. By property $3$ of $U_{n-1}$ there exists some $q$ such that $(p_n,q)\in U_{n-1}$. Set $q_n$ to any such $q$ and let $m=|g(p_n,q_n)|+1$. Now, using property 2 of $U_{n-1}$ choose some $\delta\in (0,1)$ such that $(p_n-\delta,p_n+\delta)\times (q_n-m\delta,q_n+m\delta)\subseteq U_{n-1}$. One can see that $$S(p_n,q_n,\delta)\cap(p_n-\delta,p_n+\delta)\times \mathbb R\subseteq(p_n-\delta,p_n+\delta)\times (q_n-m\delta,q_n+m\delta)\subseteq U_{n-1}.$$ Set $U_n=U_{n-1}\cap S(p_n,q_n,\delta)$. Now we check that we have satisfied the conditions:

  • Property 1: Trivial, from definition of $U_n$.

  • Property 2: Write $$U_n\setminus \{(p_1,q_1),\ldots,(p_n,q_n)\}=\left(U_{n-1}\setminus\{(p_1,q_1),\ldots,(p_{n-1},q_{n-1})\}\right)\cap \left(S(p_n,q_n,\delta)\setminus \{(p_n,q_n)\}\right).$$ Thus the given set is the intersection of two open sets and thus open.

  • Property 3: Suppose that $|p-p_n|<\delta$. Then there is a $q$ such that $(p,q)$ is in $S(p_n,q_n,\delta)\cap (p_n-\delta,p_n+\delta)\times \mathbb R\subseteq U_n$. If $|p-p_n|\geq \delta$, then $(p,q)\in U_n$ exactly when $(p,q)\in U_{n-1}$, so the theorem is satisfied.

  • Property 4: Due to property $1$, if there is a $q$ such that $(p_{n'},q)\in U_n$ then $q=q_n$. By property $3$, there is such a $q$.

This shows that we may extend the sequences given any finite prefix of it. It is then easy to check that the resulting function $f(p_n)=q_n$ satisfies the hypotheses.

  • $\begingroup$ This is insane. Get a Ph.d. $\endgroup$
    – Arbuja
    Sep 22, 2016 at 1:19

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