Why This Question?

I have often asked myself what a solution to the Ordinary Differential Equation (ODE) would be and when I recently saw this ODE again, tried it again and failed again, I wanted to ask how this ODE was solved could become?

The latest reason I'm asking this question has to do with Fractional Calculus, for which I'm asking myself a similar but a lot more complex question.

Clarification Of The Question

To make the equation clearer:

Let's say that $\operatorname{D_{x}^{v}}\left[ f \right]$ is the $v$th derivative with respect to the variable $x$ of the function $f$ and that the "Power Tower" $a^{b^{c^{d^{\cdot^{\cdot^{\cdot}}}}}}$ is $a^{\left( b^{\left( c^{\left( d^{\cdot^{\cdot^{\cdot}}} \right)} \right)} \right)}$.

This gives us the following ODE: $$ \begin{align*} \left( \operatorname{D_{x}^{0}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ y\left( x \right) \right] \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \end{align*} $$

The whole thing goes up to the $n$th derivative of $y$ where $n \in \mathbb{N}$: $$ \begin{align*} \left( \operatorname{D_{x}^{0}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ y\left( x \right) \right] \right)^{\cdot^{\cdot^{\cdot^{\left( \operatorname{D_{x}^{n - 1}}\left[ y\left( x \right) \right] \right)^{\operatorname{D_{x}^{n - 1}}\left[ y\left( x \right) \right]}}}}}}} &= f\left( x \right)\\ \end{align*} $$

Now let us rewrite left-hand side to $P_{n}\left( y\left( x \right) \right) \equiv \left( \operatorname{D_{x}^{0}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ y\left( x \right) \right] \right)^{\cdot^{\cdot^{\cdot^{\left( \operatorname{D_{x}^{n - 1}}\left[ y\left( x \right) \right] \right)^{\operatorname{D_{x}^{n - 1}}\left[ y\left( x \right) \right]}}}}}}}$ were $P_{n}$ is an operator defined by this equation.

With that, we can write the ODE from the question title as follows: $$ \begin{align*} \lim\limits_{n \to \infty}\left[ P_{n}\left( y\left( x \right) \right) \right] = f\left( x \right) \tag{1}\\ \end{align*} $$

Now the question again: How to solve the ODE $\lim\limits_{n \to \infty}\left[ P_{n}\left( y\left( x \right) \right) \right] = f\left( x \right)$?

My Best Attempts At Finding A Solution

Try And Error / "Guessing"

Trying $y\left( x \right) = c \cdot \exp\left( x \right)$

$y\left( x \right) = c \cdot \exp\left( x \right)$ where $c$ is some constant gives us: $$ \begin{align*} \left( \operatorname{D_{x}^{0}}\left[ c \cdot \exp\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ c \cdot \exp\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ c \cdot \exp\left( x \right) \right] \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \left( \operatorname{D_{x}^{0}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ y\left( x \right) \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ y\left( x \right) \right] \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \left( c \cdot \exp\left( x \right) \right)^{\left( c \cdot \exp\left( x \right) \right)^{\left( c \cdot \exp\left( x \right) \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \left( c \cdot \exp\left( x \right) \right) \uparrow\uparrow \infty &= f\left( x \right) \tag{2.1}\\ \end{align*} $$ where $\uparrow \uparrow$ comes from the Knuth's Up-Arrow Notation.

A plot of $f$ for $c = 1$:

plot by desmos

That would give us a ($y\left( x \right) = c \cdot \exp\left( x \right)$) solution to the special case $\lim\limits_{n \to \infty}\left[ P_{n}\left( y\left( x \right) \right) \right] = \left( c \cdot \exp\left( x \right) \right) \uparrow\uparrow \infty$.

Trying $y\left( x \right) = c \cdot \sin\left( x + n \cdot \frac{\pi}{2} \right)$

$y\left( x \right) = c \cdot \sin\left( x + n \cdot \frac{\pi}{2} \right)$ where $c$ and $n$ with $n \in \mathbb{Z}$ are some constants gives us some simple observtions: $$ \begin{align*} c \cdot \sin\left( x + 0 \cdot \frac{\pi}{2} \right) &= c \cdot \sin\left( x \right)\\ c \cdot \sin\left( x + 1 \cdot \frac{\pi}{2} \right) &= c \cdot \cos\left( x \right)\\ c \cdot \sin\left( x + 2 \cdot \frac{\pi}{2} \right) &= -c \cdot \sin\left( x \right)\\ c \cdot \sin\left( x + 3 \cdot \frac{\pi}{2} \right) &= -c \cdot \cos\left( x \right)\\ c \cdot \sin\left( x + 4 \cdot k \cdot n \cdot \frac{\pi}{2} \right) &= c \cdot \sin\left( x + n \cdot \frac{\pi}{2} \right)\\ \end{align*} $$ where $k$ is some intiger ($k \in \mathbb{Z}$) constant.

This allows us to: $$ \begin{align*} \left( \operatorname{D_{x}^{0}}\left[ c \cdot \sin\left( x + n \cdot \frac{\pi}{2} \right) \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ c \cdot \sin\left( x + n \cdot \frac{\pi}{2} \right) \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ c \cdot \sin\left( x + n \cdot \frac{\pi}{2} \right) \right] \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \left( c \cdot \sin\left( x + \left( n + 0 \right) \cdot \frac{\pi}{2} \right) \right)^{\left( c \cdot \sin\left( x + \left( n + 1 \right) \cdot \frac{\pi}{2} \right) \right)^{\left( c \cdot \sin\left( x + \left( n + 2 \right) \cdot \frac{\pi}{2} \right) \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right) \tag{2.2}\\ \end{align*} $$

A plot of $f$ for $c = 1$ and $n = 0$:

Not displayed because the plot is probably too complex. Someone is welcome to add a plot of this.

Trying $y\left( x \right) = \frac{1}{x}$

This gives us: $$ \begin{align*} \left( \operatorname{D_{x}^{0}}\left[ \frac{1}{x} \right] \right)^{\left( \operatorname{D_{x}^{1}}\left[ \frac{1}{x} \right] \right)^{\left( \operatorname{D_{x}^{2}}\left[ \frac{1}{x} \right] \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \left( \frac{1}{x} \right)^{\left( -\frac{1}{x^{2}} \right)^{\left( \frac{1}{x^{3}} \right)^{\cdot^{\cdot^{\cdot}}}}} &= f\left( x \right)\\ \sqrt[\ddots]{\frac{1}{x}} &= f\left( x \right)\\ \end{align*} $$

A Plot of $f$:

plot by desmos


Since the ODE contains a sort of infinite "Power Tower" and $0^{0}$ is undefined, $\operatorname{D_{x}^{n}}\left[ y\left( x \right) \right] \ne 0,\, \text{for}\, n \in \mathbb{N}$ must hold 'cause some $\operatorname{D_{x}^{z}}\left[ y\left( x \right) \right] = 0$ where $z$ is some Number in $\mathbb{N}$ implies $\cdot^{\operatorname{D_{x}^{z}}\left[ y\left( x \right) \right]^{\operatorname{D_{x}^{z + 1}}\left[ y\left( x \right) \right]^{\cdot}}} = \cdot^{0^{0^{0^{\cdot}}}}$. This means that $y$ must be derivable infinitely often in at least one interval without ever yielding $0$ in this interval, which already rules out a large number of functions. As I write this I realize that it is difficult to find a legible but correct and accurate notation for this. I would also be open to anyone suggesting a better notation.

The simplest case of functions for which this is the case are functions which have a derivative of order $n$, which gives the function or another derivative of the function itself or in formuls: $\operatorname{D_{x}^{v}}\left[ y\left( x \right) \right] = \operatorname{D_{x}^{w}}\left[ y\left( x \right) \right],\, \text{for}\, \mathbb{N} \ni w \ne v \in \mathbb{N}$

I would solve this ODE like this (for this I'll assume $v > w$):$^{\left[ 1 \right]}$ $$ \begin{align*} \operatorname{D_{x}^{v}}\left[ y\left( x \right) \right] &= \operatorname{D_{x}^{w}}\left[ y\left( x \right) \right]\\ \operatorname{D_{x}^{v - w}}\left[ y\left( x \right) \right] &= y\left( x \right)\\ \operatorname{D_{x}^{v - w}}\left[ e^{\lambda \cdot x} \right] &= e^{\lambda \cdot x}\\ \lambda^{v - w} \cdot e^{\lambda \cdot x} &= e^{\lambda \cdot x}\\ \lambda^{v - w} &= 1\\ \lambda &= 1^{\frac{1}{v - w}}\\ \lambda_{k} &= \left( \exp\left( 2 \cdot k \cdot \pi \cdot i \right) \right)^{\frac{1}{v - w}}\\ \lambda_{k} &= \exp\left( \frac{2 \cdot k \cdot \pi}{v - w} \cdot i \right)\\ \\ \lambda_{k} &= \operatorname{cis}\left( \frac{2 \cdot k \cdot \pi}{v - w} \right)\\ \lambda_{k} &= \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) + \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot i\\ \end{align*} $$ where $\operatorname{cis}$ is the [CiS-Function][8]. That means $\lambda^{v - w} + 1 = \prod\limits_{k = 1}^{\left| \mathbb{S} \right|}\left( \lambda - \operatorname{cis}\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \right)$ where $\mathbb{S}$ is the set of all roots of the equation and $\left| \mathbb{S} \right|$ is the [Cardinality][9] of $\mathbb{S}$. This gives the multiplicity of the all roots $= 1$. With $y_{h}\left( x \right) = Q_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + P_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right)$ we'll get: $$ \begin{align*} y_{h}\left( x \right) &= Q_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + P_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right)\\ y_{h}\left( x \right) &= \sum\limits_{k = 1}^{\left| \mathbb{S} \right|}\left[ c_{2 \cdot k} \cdot \exp\left( \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \cdot \cos\left( \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) + c_{2 \cdot k + 1} \cdot \exp\left( \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \cdot \sin\left( \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \right] \tag{3.1.1}\\ \end{align*} $$ where $\mathbb{S} = v - w$. But there are also the solutions given by: $$ \begin{align*} \operatorname{D_{x}^{v}}\left[ y\left( x \right) \right] &= \operatorname{D_{x}^{w}}\left[ y\left( x \right) \right]\\ \operatorname{D_{x}^{v - 1}}\left[ y\left( x \right) \right] &= \operatorname{D_{x}^{w - 1}}\left[ y\left( x \right) \right] + c_{-1}\\ \operatorname{D_{x}^{v - 2}}\left[ y\left( x \right) \right] &= \operatorname{D_{x}^{w - 1}}\left[ y\left( x \right) \right] + c_{-2} + c_{-1} \cdot x\\ &\cdots\\ \operatorname{D_{x}^{v - w}}\left[ y\left( x \right) \right] &= y\left( x \right) + \sum\limits_{k = 1}^{w}\left[ c_{-k} \cdot x^{k} \right] \tag{3.1.2}\\ \end{align*} $$

So $y\left( x \right) = \sum\limits_{k = 1}^{w}\left[ c_{-k} \cdot x^{k} \right] + \sum\limits_{k = 1}^{\left| v \right|}\left[ c_{2 \cdot k} \cdot \exp\left( \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \cdot \cos\left( \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) + c_{2 \cdot k + 1} \cdot \exp\left( \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \cdot \sin\left( \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \right]$ aka $y\left( x \right) = \sum\limits_{k = 1}^{w}\left[ c_{-k} \cdot x^{k} \right] + \sum\limits_{k = 1}^{v - w}\left[ c_{2 \cdot k} \cdot \exp\left( \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \cdot \cos\left( \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) + c_{2 \cdot k + 1} \cdot \exp\left( \cos\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \cdot \sin\left( \sin\left( \frac{2 \cdot k \cdot \pi}{v - w} \right) \cdot x \right) \right] \tag{3.1.3}$.

Which would be one possible form that $y$ could take to be defined and thus be defined for a given $f$.

Another way to satisfy the condition is that we solve the inequality $\operatorname{D_{x}^{z}}\left[ y\left( x \right) \right] \ne 0$. But we have already solved that for $z = w$ (a similar equation), see Formula $\left( 3.1.2 \right)$, with: $$y\left( x \right) \ne \sum\limits_{k = 1}^{z}\left[ c_{-k} \cdot x^{k} \right] \tag{3.1.4}$$ But that's obvious, 'since every finite polynomial is no solution.

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    $\begingroup$ You literally don't need any of those parentheses in the title, other than the $(x)$ part. $a^{b^c}$ means $a^{(b^c)},$ and $(y(x))$ can just be written $y(x).$ $\endgroup$ Apr 12, 2023 at 1:35

1 Answer 1


Some Observations + Full Exact Solution (of Inverse Function)

Begin by defining: $$U_n=[D^ny]^{[D^{n+1}y]^{\&c.}}$$ so that $$\ln U_n =U_{n+1}\ln D^ny \iff U_n^{\frac{1}{U_{n+1}}}=D^ny.$$ This is the first quality. Now, by your question, if $U_0=f(x),$ then $$y=f(x)^{1/U_1}$$where $$U_{1}=\frac{\ln f}{\ln Dy},$$ so, assuming one wants an exact answer (and not iterating $y_{k+1}=[Dy_k]^{(\ln f)^{-2}}$), notice that $$y=[Dy]^{(\ln f)^2}\iff Dy=y^{\frac{1}{(\ln f)^2}}\equiv Dy=y^{\alpha(x)}.$$ This means that the differential equation we wish to solve is of the form $Dy=y^{\alpha(x)}$. So, differentiating $$\frac{D^2y}{Dy}=\alpha' \ln y+\alpha\frac{Dy}{y},$$ and again, $$\boxed{\boxed{D\left\{\frac{D^2y}{\alpha' Dy}-\frac{\alpha Dy}{\alpha' y}\right\}=\frac{Dy}{y}}}.$$ This is the nonlinear equality we must solve for $\alpha=\frac{1}{(\ln f)^2}$. A much easier approach to this is obtained by putting $y=e^{u}$ in $$Dy=y^{\alpha}$$ to get $$u'e^u=e^{\alpha u}\iff \ln u'=(\alpha-1)u\\ \iff {{{u''=\alpha'uu'+(\alpha-1)u'^2}}}.$$ This is an effective reduction of your power tower to a much simpler, much nicer nonlinear differential equation. However, solving it will involve nonlinear iteration schemes, or very janky power series, or something of the like, unless we consider the inverse function of $u$. Notice that $$u_x=\frac{1}{x_u}$$ and $$u_{xx}=-\frac{x_{uu}}{x^3_u},$$ and subbing this into the ODE in $u$, (I took into account $\alpha_x=\frac{\alpha_u}{x_u}$)

$$-\frac{x''}{x'^3}=\alpha'_uu\frac{1}{x'^2}+(\alpha-1)\frac{1}{x'^2}\\ \iff -x''=[\alpha+u\alpha'_u-1]x',\\ {\boxed{\boxed{\therefore x(u)=C_0\int e^{\int [1-\alpha-u\alpha'_u ]du}du+C_1}}}.$$ Admittedly this is not a good solution under 4 conditions, that is

    1. The initial conditions are ambiguous, extraneous or difficult to obtain,
    1. $\frac{1}{(\ln f)^2}$ is not easily obtainable as a function of $u=\ln y$,
    1. $x(u)$ is difficult to invert.
    1. The iteration equivalent to your original expression doesn't converge, or doesn't converge to a stable solution equivalent to that obtained here.

Otherwise, this is the complete solution.

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    $\begingroup$ That's a really good observation. That might lead to a solution, thanks. If nobody writes a better answer, you get the bounty. $\endgroup$ Apr 18, 2023 at 0:29
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    $\begingroup$ @KevinDietrich Happy to help :) $\endgroup$ Apr 18, 2023 at 0:34
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    $\begingroup$ @KevinDietrich I have an exact solution for you. $\endgroup$ Apr 18, 2023 at 17:12
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    $\begingroup$ That looks pretty good. Could you explain your notation a bit more to me? What is $u_{x}$, $x_{u}$ and $\alpha_{u}$? If you tell me I'll give you the bounty and accept the answer. If there is no other answer you will also get the bounty... $\endgroup$ Apr 20, 2023 at 5:41
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    $\begingroup$ @KevinDietrich $u_x$, $x_u$, and $\alpha_u$ are, respectively, $\frac{du}{dx}$, $\frac{dx}{du}$, and $\frac{d\alpha}{du}$. $\endgroup$ Apr 21, 2023 at 15:30

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