I was wondering about the following:

Let's say there is a differential equation whose solution is $f$

And $f$ also satisfies a functional equation.

Can anyone construct an (non-trivial) example where knowing the functional equation gives some sort of advantage in solving the differential equation or visa-versa? And if the functional equation does not help can you please give your reasoning on why so?

My attempt

For example, take $$\frac {\mathrm d f}{\mathrm d x} = f$$ And the functional equation is of the form: $$ A f(x+y) = f(x)f(y)$$ where, $A$ is a constant. I can't see any sort manipulation where knowing the functional equation has given be an advantage in solving the differential equation.

  • $\begingroup$ Usually this goes the other way round; if you know that the solutions to a functional equation have some regularity, then you can extract from it a differential equation. The latter is, typically, much easier. $\endgroup$ Jul 20, 2021 at 10:26

2 Answers 2


You can transform functional equations into differential equations and into difference equations. For example, if $f$ satisfies $$f(x+y)=f(x)f(y)$$ then it must satisfy $f(x+1)-f(x)=f(x)(f(1)-1)$ or $$\Delta f=Cf.$$ This difference equation is trivial and has the solution $f=f(1)^x$. Likewise, the function $f$ in

$$f(x+y)=yf^2(x)+(y^2+1)f(x)$$ must solve (by putting $y=dx$)

$$\frac{f(x+dx)-f(x)}{dx}=f'(x)=f^2(x).$$ Or it must solve $$\Delta f =f^2+f.$$ Just make sure to match the appropriate constants from the functional equation to the difference or differential equation. A differential equation might not always be derivable from a functional equation, but a difference equation is guaranteed. Most of the time you will need series solutions.

  • $\begingroup$ All true, BUT you might not know a priori that the solutions to the functional equations are differentiable. In that case, you cannot differentiate and you have to solve the functional equation as is, which is much harder. This happens, for example, for the equation $f(x+y)=f(x)+f(y)$. If $f$ is differentiable, then it is easy to conclude that $f'$ must be constant, and so $f(x)=Ax+B$ for constants $A, B$. But there are non-differentiable solutions to that functional equation. en.wikipedia.org/wiki/Cauchy%27s_functional_equation $\endgroup$ Jul 20, 2021 at 10:31
  • $\begingroup$ @Giuseppe Negro Of course, but I prefer the practical approach of just assuming it’s differentiable and then ending up with a contradiction or some nonsense, at which point I try the difference equation approach. The most general approach would be an integral equation approach combined with a certain sum, but I have a case of lazy in me too great for demonstrating that. $\endgroup$ Jul 20, 2021 at 13:18
  • 1
    $\begingroup$ Can we do the other way around? Transforming a differential equation into a function equation sounds much more exciting! $\endgroup$
    – High GPA
    May 26, 2022 at 1:00
  • 1
    $\begingroup$ @HighGPA I'll get back to you if I figure something out. $\endgroup$ Jun 16, 2022 at 15:01

I think your example is suboptimal. After all, your differential equation is, to a certain extent, trivial. After all, it is ordinal, homogeneous, linear, whole (first)-order, non-stochastic, non-integro, and undelayed DE. It doesn't get much easier than that.

  • But these functional equations can help you not to leave integration constants at random but to determine them concretely. E.g. given are $A \cdot f'\left( x \right) + f\left( x \right) = \cos\left( x \right) + \sin\left( x \right)$ and $-f\left( x + \pi \right) = f\left( x \right)$. We would consider $f\left( x \right) = c \cdot e^{-x} + \sin\left( x \right)$ as a solution to the ODE but with the functional equation we know that $f$ has to be periodic wich is not the case for $c \cdot e^{-x} + \sin\left( x \right) \wedge x \ne 0$. But with $f\left( x + \pi \right) = f\left( x \right)$ we get that $c = 0$, so $f\left( x \right) = \sin\left( x \right)$.

  • But the whole thing is not only useful for differential equations. Suppose given the equations $f\left( x + a \right) - f\left( x \right) = \left( x + 1 \right)^{x + 1} \cdot a$ and $f\left( x + a \right) - f\left( x \right) = \frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} \cdot a$. Solving the functional equation is really hard (there is no closed form solution for it), but solving the RDDE (Retarded Differential Difference Equation) is easy. Substituting $f\left( x \right) = e^{\lambda \cdot x}$ so $e^{\lambda \cdot \left( x + a \right)} - e^{\lambda \cdot x} = \lambda \cdot e^{\lambda \cdot x} \cdot a$ so we get $e^{\lambda \cdot a} - 1 = \lambda \cdot a \Leftrightarrow e^{\lambda \cdot a} = \lambda \cdot a + 1 \Leftrightarrow \frac{1}{\lambda \cdot a + 1} \cdot e^{\lambda \cdot a} = 1 \Leftrightarrow -\left( \lambda \cdot a + 1 \right) \cdot \frac{1}{e} \cdot e^{-\lambda \cdot a} = -\frac{1}{e} \Leftrightarrow -\left( \lambda \cdot a + 1 \right) \cdot \frac{1}{e} \cdot e^{-\lambda \cdot a} = -\frac{1}{e} \Leftrightarrow \left( -\lambda \cdot a - 1 \right) \cdot e^{-\lambda \cdot a - 1} = -\frac{1}{e} \Leftrightarrow -\lambda \cdot a - 1 = W_{k}\left( -\frac{1}{e} \right) \Leftrightarrow \lambda = \frac{-W_{k}\left( -\frac{1}{e} \right) - 1}{a}$ aka the solution is $f\left( x \right) = \text{constant} + \sum\limits_{k = -\infty}^{\infty}\left[ e^{\Re\left( \lambda_{k} \right) \cdot x} \cdot \left( c_{2 \cdot k} \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + c_{2 \cdot k + 1} \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right) \right) \right]$. Referring this to our functional equation, we see that the equation is always true for $a = r \cdot \varepsilon$ where $r \in \mathbb{R} \setminus \left\{ 0 \right\} \wedge \varepsilon^{2} = 0 \ne \varepsilon$ and $f\left( x \right) = \int \left( x + 1 \right)^{\left( x + 1 \right)}\, \operatorname{d}x$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .