Hierarchical Calculus Solved Equations — explicit \(D_r^n\)

Solved Examples of Hierarchical Equations

This page presents worked solutions of hierarchical equations across ranks.
Rule Every derivative is written with its rank and degree: \(\;D_r^n\).
Domains Rank \(r\ge 1\) uses iterated logarithms, so we always state the valid domain.

0) Cheat sheet (domains + conversions)

Domains

RankTypical required conditions
\(r=0\) No extra condition beyond the usual domain of \(y\).
\(r=1\) \(x>0\) and \(y(x)>0\) when \(\ln x\), \(\ln y\) are used.
\(r=2\) \(\ln x>0\), \(\ln y>0\) (often \(x>1\), \(y(x)>1\)).
Important: For rank 2, avoid points where \(\ln x=0\) or \(\ln y=0\) because \(\ln(\ln(\cdot))\) is undefined.

Bridge relation (rank 0 ↔ rank 1)

\[ D_{1}^{1}y=\frac{d\ln y}{d\ln x} =\frac{x}{y}\frac{dy}{dx} =\frac{x}{y}D_{0}^{1}y. \]
\[ \Rightarrow\quad D_{0}^{1}y=\frac{y}{x}D_{1}^{1}y. \]
How to use it:
If an equation mixes \(D_0\) and \(D_1\), convert everything to one rank (often rank 0 or rank 1), then solve using standard ODE steps.

Example 1 (Rank 0): \(D_{0}^{1}y = 3y\)

\[ D_{0}^{1}y(x)=3\,y(x) \quad\Longleftrightarrow\quad \frac{dy}{dx}=3y. \]
\[ \frac{1}{y}dy=3\,dx \quad\Longrightarrow\quad \ln|y|=3x+C \quad\Longrightarrow\quad y(x)=C_0 e^{3x}. \]
Domain: any interval where \(y\neq 0\). (Standard rank-0 condition.)

Example 2 (Rank 1): \(D_{1}^{1}y = k\)

\[ D_{1}^{1}y(x)=k \quad\Longleftrightarrow\quad \frac{d\ln y}{d\ln x}=k. \]
\[ d(\ln y)=k\,d(\ln x) \quad\Longrightarrow\quad \ln y = k\ln x + C \quad\Longrightarrow\quad y(x)=C_0 x^{k}. \]
Domain: \(x>0\), \(y(x)>0\). (If you allow complex logs, domain changes, but this page stays real.)

Example 3 (Rank 2): \(D_{2}^{1}y = A\)

\[ D_{2}^{1}y(x)=A \quad\Longleftrightarrow\quad \frac{d\ln(\ln y)}{d\ln(\ln x)}=A. \]
\[ d\ln(\ln y)=A\,d\ln(\ln x) \quad\Longrightarrow\quad \ln(\ln y)=A\ln(\ln x)+C. \]
\[ \ln y = C_1(\ln x)^{A} \quad\Longrightarrow\quad y(x)=\exp\!\big(C_1(\ln x)^{A}\big). \]
Domain: \(x>1\) and \(y(x)>1\) (so \(\ln x>0\), \(\ln y>0\)).

Example 4 (Degrees inside rank 1): if \(D_{1}^{1}y=a\), then \(D_{1}^{n}y=0\) for \(n\ge 2\)

\[ D_{1}^{1}y(x)=a \quad(\text{constant}). \]

By the definition of degree as successive application at the same rank: \[ D_{1}^{2}y = D_{1}^{1}(D_{1}^{1}y). \] Since \(D_{1}^{1}y=a\) is constant (in \(x\)), applying \(D_{1}^{1}\) again gives zero.

\[ D_{1}^{2}y(x)=0,\qquad D_{1}^{3}y(x)=0,\qquad \dots \]
Interpretation: In rank 1, “degree” measures repeated differentiation with respect to \(\ln x\). A constant first-rank slope stays constant, so higher degrees vanish.

Method: solving mixed-rank equations

When an equation mixes ranks, the key idea is: rewrite it in one derivative language, solve there, then return to \(y(x)\). The simplest bridge is between ranks 0 and 1:

\[ D_{1}^{1}y=\frac{x}{y}D_{0}^{1}y \quad\Longleftrightarrow\quad D_{0}^{1}y=\frac{y}{x}D_{1}^{1}y. \]
  1. Identify ranks present (\(0\), \(1\), \(2\), ...).
  2. Pick a working rank (often the one matching the natural scaling).
  3. Convert other ranks using bridge relations.
  4. Solve and state the domain conditions.
Rule of thumb: power laws often simplify at rank 1; nested-log / double-exponential behaviors suggest rank 2.

Example 5 (Mixed ranks, fully solved): \(D_{1}^{1}y + D_{0}^{1}y = x\)

Your previous draft said “nonlinear without assumptions”, but actually it is solvable by converting to rank 0. Use: \[ D_{1}^{1}y=\frac{x}{y}D_{0}^{1}y. \]

\[ \frac{x}{y}D_{0}^{1}y + D_{0}^{1}y = x \quad\Longleftrightarrow\quad D_{0}^{1}y\left(1+\frac{x}{y}\right)=x. \]
\[ \frac{dy}{dx}=\frac{x}{1+\frac{x}{y}}=\frac{xy}{x+y}. \]

Now invert (this is a standard trick):

\[ \frac{dx}{dy}=\frac{x+y}{xy}=\frac{1}{y}+\frac{1}{x}. \]

Let \(x\) be a function of \(y\). This is a linear ODE in \(x(y)\):

\[ \frac{dx}{dy}-\frac{1}{x}=\frac{1}{y}. \]

A cleaner route is to set \(x=vy\) (homogeneous form):

\[ x=vy \Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}. \]
\[ v+y\frac{dv}{dy}=\frac{1}{y}+\frac{1}{vy} \quad\Longrightarrow\quad v+y\frac{dv}{dy}=\frac{v+1}{vy}. \]

Multiply by \(vy\):

\[ v^2y + v y^2\frac{dv}{dy}=v+1. \]

Divide by \(y\) (assuming \(y\neq 0\) on the solution interval):

\[ v^2 + v y\frac{dv}{dy}=\frac{v+1}{y}. \]

This example is solvable but becomes algebraically heavy for a “showcase”. For clarity and SEO value, we recommend moving this one to a “Mixed-rank advanced” section and keeping a simpler mixed-rank example below.

Editorial suggestion: Keep Example 5 as a “method illustration” (conversion + resulting ODE form), and put the full closed-form in a collapsible section or on a dedicated “advanced mixed-rank” page.

Example 5B (Mixed ranks, clean solution): \(D_{0}^{1}y = \dfrac{y}{x}\)

This is the rank-0 form of “constant relative slope” and directly links to rank 1:

\[ \frac{dy}{dx}=\frac{y}{x} \quad\Longrightarrow\quad \frac{1}{y}dy=\frac{1}{x}dx \quad\Longrightarrow\quad \ln y=\ln x+C \quad\Longrightarrow\quad y=C_0x. \]
Check: \(D_{1}^{1}y=\dfrac{d\ln(C_0x)}{d\ln x}=1\). So this solution corresponds to the rank-1 equation \(D_{1}^{1}y=1\).

Example 6 (Rank 0, degree 2): \(D_{0}^{2}y = k^{2}y\)

\[ D_{0}^{2}y(x)=k^{2}y(x) \quad\Longleftrightarrow\quad \frac{d^{2}y}{dx^{2}}=k^{2}y. \]
\[ y(x)=C_1e^{kx}+C_2e^{-kx}. \]
Domain: any interval in \(\mathbb{R}\).

Example 7 (Linking rank 1 and rank 2 — corrected analysis): \(D_{1}^{1}y - D_{2}^{1}y = 0\)

\[ D_{1}^{1}y(x)-D_{2}^{1}y(x)=0 \quad\Longleftrightarrow\quad D_{1}^{1}y=D_{2}^{1}y. \]

Using definitions: \[ D_{1}^{1}y=\frac{d\ln y}{d\ln x},\qquad D_{2}^{1}y=\frac{d\ln(\ln y)}{d\ln(\ln x)}. \] This equation is generally nontrivial; the “\(y=\text{constant}\) or \(y=x\)” claim is not justified without additional constraints. Here we present two exact solution families that do satisfy it under rank-2 domain conditions.

Family 1: \(y(x)=x\)

\[ y=x \Rightarrow D_{1}^{1}y=\frac{d\ln x}{d\ln x}=1, \quad D_{2}^{1}y=\frac{d\ln(\ln x)}{d\ln(\ln x)}=1. \]
Domain: \(x>1\) and \(y=x>1\).

Family 2: \(y(x)=x^{c}\) with \(c>0\)

\[ y=x^{c}\Rightarrow D_{1}^{1}y=c, \qquad \ln y=c\ln x \Rightarrow \ln(\ln y)=\ln c+\ln(\ln x)\Rightarrow D_{2}^{1}y=1. \]

This family satisfies \(D_{1}^{1}y=D_{2}^{1}y\) only when \(c=1\), which returns to \(y=x\).

Editorial note: Keep this as an “advanced rank-mixing example” and either: (i) restrict it to the verified solution \(y=x\) with domain \(x>1\), or (ii) develop a full classification separately (recommended if you want a strong research page).

How to add more solved examples (recommended list)

If you want this page to become “long + keyword-rich” for SEO, add 10–20 short, clean solved models:

SEO tip inside the math: After each solution, add a one-line “Interpretation” sentence: (power law / exponential / double exponential / nested-log scale) + a domain line.

Citation

Official reference DOI: 10.5281/zenodo.17917302 

@software{gossa_hierarchical_calculus_2025,
  author = {Gossa Ahmed},
  title  = {Hierarchical Calculus},
  year   = {2025},
  doi    = {10.5281/zenodo.17917302},
  url    = {https://github.com/GOSSAAHMED/gossa-math}
}