Functional equation + differential equation = way of finding solution?

Question

Question

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.

Answer 1

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.

Answer 2

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 $f\left( x \right) = \int \left( x + 1 \right)^{\left( x + 1 \right)}\, \operatorname{d}x$.