An actual attempt to solve $f^{(n)}(x)=f(x)^n$

Question

Earlier today I saw this cool differential equation question which was closed because the asker didn't show any effort in solving it: $$f^{(n)}(x)=f(x)^n$$ But I still wanted to know the answer, you know? It looks like a super cool equation. But I had no clue where to start, since it didn't look separable or anything, so I just looked through cases for very low $n$. $$f(x)=f'(x)$$ Oh, that's easy. Just any function of the form $f(x)=Ae^{x}$. So, how hard can $n=2$ be? $$f''(x)=f(x)^2$$ .... Uh oh. My first thought was perhaps a monomial of some sort. Something of the form $Ax^n$. I plugged in and got $$A(n)(n-1)x^{n-2}=A^2x^{2n}$$ so clearly, $n$ has to satisfy both $n-2=2n$ and $(n)(n-1)=A$. $n=-2$, which means $A=-6$. But is that the only solution? $6x^{-2}$? Seems oddly isolated. So, I looked to Mathematica for help finding a general solution.

... oh my god. What the hell is the "Weierstrass elliptic function"???

It gave me this nightmarish form: $$\sqrt[3]{6}\,\wp\!\left(\frac{x+c_1}{\sqrt[3]{6}},0,c_2\right)$$ Looking up the function on Wikipedia, I was able to rewrite that as $$\sqrt[3]{6}\left(\frac{6^{\frac{2}{3}}}{(x+c_1)^2}+\sum_{k=1}^{\infty} \frac{1}{\left(\frac{x+c_1}{\sqrt[3]{6}}-c_2k\right)^2}-\frac{1}{c_2^2k^2}+\sum_{k=1}^{\infty} \frac{1}{\left(\frac{x+c_1}{\sqrt[3]{6}}+c_2k\right)^2}-\frac{1}{c_2^2k^2}\right)$$ That spooked me. So, I tried looking up $f^{(3)}(x)=f(x)^3$, and it gave... nothing. No searches coming up on ApproachZero either. Maybe it's an issue with odd $n$s? But nope, $n=4$ gives the same problem. So, I tried myself, first to prove that the Weierstrass elliptic function actually satisfies this, and if I could modify it to somehow fit with $f'''(x)=f(x)^3$ and beyond.

Can I even prove that the Weierstrass function actually works? I took $c_1=0$, because it was fairly unnecessary, and renamed $c_2=C$. $$\frac{6}{x^2}+\sum_{k=1}^{\infty} \frac{\sqrt[3]{6}}{\left(\frac{x}{\sqrt[3]{6}}-Ck\right)^2}-\frac{\sqrt[3]{6}}{C^2k^2}+\sum_{k=1}^{\infty} \frac{\sqrt[3]{6}}{\left(\frac{x}{\sqrt[3]{6}}+Ck\right)^2}-\frac{\sqrt[3]{6}}{C^2k^2}$$ Uh oh. So, I took the derivative twice. And that gave me $$\frac{36}{x^4}+\sum_{k=1}^{\infty} \frac{36}{\left(\frac{x}{\sqrt[3]{6}}-Ck\right)^4}+\sum_{k=1}^{\infty} \frac{36}{\left(\frac{x}{\sqrt[3]{6}}+Ck\right)^4}$$ Off to a fine-ish start. But then I remembered I had to square $f(x)$. How do you deal with the square of an infinite sum?? And I gave up.

@Gonçalo's comment showing me that there was an easy particular solution I missed for $n=2$ remotivated me to look for general solutions. So I tried... $$A(k)(k-1)\cdots (k-n+1)x^{k-n}=A^nx^{kn}$$ which implies $k-n=kn \Rightarrow k=\frac{n}{1-n}$ and

$$A=\left((\frac{n}{1-n})(\frac{n}{1-n}-1)\cdots (\frac{n}{1-n}-n+1)\right)^{\frac{1}{n-1}}$$

That's fine for even $n$, but the expression under $\frac{1}{n-1}$ is negative for odd $n$, which screws things up a bit. But, there's still definitely a real solution for odd $n$, as shown by the numerical approximations. So what's up with those? And can we get a general solution for even ones?

Update on question 3: I tried to solve this for $n=-2$. So, we have... $$f(x)^2\iint f(x)\ dx = 1$$ I tried differentiating both sides, which gave me $$\int f(x)\ dx = -2f(x)^{-3}f'(x)$$ and then differentiated again, which gave $$f(x)=6f(x)^{-4}f'(x)^2-2f(x)^{-3}f''(x)$$ or $$f(x)^5=6f'(x)^2-2f(x)f''(x)$$ Eh... I'm not feeling it. What if we tried plugging in another solution of the form $Ax^k$? $$A\int \left(\int x^k \ dx\right) \ dx=\frac{1}{A^2}x^{2k}$$ $$\frac{A}{(k+1)(k+2)}x^{k+2}=\frac{1}{A^2}x^{2k}$$ So $k+2=2k$, which implies $k=2$. That means $$\frac{A}{12}=\frac{1}{A^2} \Rightarrow A=\sqrt[3]{12}$$ I'm pretty convinced at this point that $\sqrt[3]{12}x^2$ is not the only solution, and also that $n=-3$ will be a little more complex. Any ideas about it?

1. Has this equation been studied? Where can I find references on it?

2. Is there even a power series solution to $n=3$, let alone beyond? (And, if not, could you please show that the solution for $n=2$ is correct?)

3. Bonus question: is this even solvable for negative $n$, which would give an equation of the form $$\underbrace{\int \cdots \int}_{\text{$k$ integrals}}f(x)\ dx = \frac{1}{f(x)^k}$$ The solution for $k=1$ ($n=-1$) can be written as $$\int f(x)\ dx = \frac{1}{f(x)} \Rightarrow g(x)g'(x)=1 \Rightarrow g(x)=\pm \sqrt{2x+C} \Rightarrow f(x)=g'(x)=\pm\frac{1}{\sqrt{2x+C}}$$ which seemingly has no relation to the solution for $n=1$. What about $n=-2$ ($k=2$)? What if we use fractional derivatives? etc.

Answer 1

For the case $n=2$, the differential equation is $$f''(x)=f(x)^2.$$ To solve this first multiply both sides by $f'(x)$ to get $$f'(x)f''(x)=f(x)^2f'(x).$$ Integrating bout sides by $x$, we get $$\int f'(x)f''(x)dx=\int f(x)^2f'(x)dx$$ $$\frac{1}{2}f(x)'^2=\frac{1}{3}f(x)^3+C_1.$$ Moving all the $f(x)$ terms to one side gives $$ \pm(\frac{2}{3}f(x)^{3}+C_1)^{-\frac{1}{2}}f'(x)=1.$$ Integrating again with respect to $x$ yields a non-elementary from of the solution for $n=2$. Wolfram Alpha says that $f(x)$ is the inverse of $$C_2 \pm \frac{x \sqrt{3 + \frac{2 x^3}{C_1}} \; {}_2F_1\left( \frac{1}{3}, \frac{1}{2}; \frac{4}{3}; -\frac{2 x^3}{3 C_1} \right)}{\sqrt{3 C_1 + 2 x^3}}$$ where ${}_2 F_1$ is the "hypergeometric function".

Answer 2

Here is way to seemingly solve $y^{(3)}=y^3$ even though expansions like for $\wp(z;0,a)$ at $z=0$ can have non-natural powers, can be extended a generalization of $y^{(n)}=y^n$. If one were to substitute $y=\sum\limits_{j=0}^\infty a_jx^j$, the $y^3$ term would give a product of 3 sums, so instead a series solution to $3y^{(3)}y’-yy^{(4)}=0 \iff \frac d{dx}\ln(y^{(3)})=\frac d{dx}\ln(y^3)\iff y^{(3)}=y^3 $ will be found:

$$3y^{(3)}y’-yy^{(4)}=0\iff3\sum_{k=0}^\infty a_kk(k-1)(k-2)x^{k-3}\sum_{m=0}^\infty a_mmx^{m-1}-\sum_{k=0}^\infty a_k x^k\sum_{m=0}^\infty a_m m(m-1)(m-2)(m-3)=0\\\sum_{k=0}^\infty\sum_{m=0}^\infty a_ka_mm(3 (k-2) (k-1) k - (m-3) (m-2) (m-1))x^{m+k-4}=0\\\implies a_n= \sum_{k=0}^\infty\sum_{m=0}^\infty a_ka_mm(3 (k-2) (k-1) k - (m-3) (m-2) (m-1))\delta_{n,m+k}=0 $$

Therefore, using a recurrence relation for the coefficients:

$$3y^{(3)}y’-yy^{(4)}=0 \implies y=\sum_{j=0}^\infty a_j x^j=a_0+a_1x+a_2x^2+a_3x^3+\frac{3a_1a_3}{4a_0}x^4+\frac{3a_3(a_1^2+a_0a_2)}{10a^2}x^5+\frac{a_3(a_1^3+6a_0a_1a_2+3a_0^2a_3}{20a^3}x^6+\dots\\ a_j=\sum_{m=1}^{j-1}\frac{(j-4)!}{a_0j!}(3 (-2 + j-m) (-1 +j-m) (j-m) - (-3 + m) (-2 + m) (-1 + m)) m a_{j-m} a_m $$ as implied here. The $n=2$ case is:

$$2y’’y’-yy^{(3)}=0\implies y=\sum_{j=0}^\infty a_jx^j=a_0+a_1x+a_2x^2+\frac{2a_1a_2}{3a_0}x^3+\frac{a_1^2a_2+2a_0a_2^2}{6a_0^2}x^4+\frac{a_1a_2^2}{3a_0^2}x^5+\frac{a_1^2a_2^2+a_0a_2^3}{9a_0^3}x^6+\dots\\a_j=\frac{(j-3)!}{a_0j!}\sum_{m=1}^{j-1}(2(j-m-1)(j-m)-(m-2)(m-1))ma_{j-m}a_m$$

as implied here. Notice the graphs near $x=0$ are near $y=0$

Answer 3

As a supplement to @d ds's answer, for the $n=2$ case, the wiestrauss elliptic p form can be derived as follows, with more information here in section 1.3: https://arxiv.org/pdf/1706.07371

If we multiply both sides by $2y'$, we can simplify using the chain rule to get $$2y'y'' = 2y^2y'\iff\left(y'^2\right)'=\frac23\left(y^3\right)'\iff y'^2 = \frac{2}{3}y^3+C$$

This equation is separable, and we aim to turn it into wiestrauss p form, where $$\wp'^2=4\wp^3-g_2\wp-g_3$$ $$\wp^{-1}(z; g_2, g_3) = \int_{\infty}^z\frac{\text{d}t}{\sqrt{4t^3-g_2t-g_3}}= \int\frac{\text{d}t}{\sqrt{4t^3-g_2t-g_3}} \text{(up to}+C\text{)}$$

We substitute $u(x) = \frac{y(x)}{\sqrt[3]6}$, which has the differential $\text{d}y=\text{d}u\sqrt[3]6$, yielding $$y'^2 = \frac{2}{3}y^3+C\iff\left(\frac{\text{d}y}{\text{d}x}\right)^2 = 4\left(\frac y{\sqrt[3]{6}}\right)^3+C \tag{*}$$ $$ \left(\frac{\text{d}u\cdot \sqrt[3]6}{\text{d}x}\right)^2 = 4u^3+C \Longleftrightarrow \sqrt[3]6\, u' = \pm\sqrt{4u^3+C }\Longleftrightarrow \frac{\sqrt[3]6\, u'}{\sqrt{4u^3+C}} = \pm 1$$

We now integrate both sides, getting \begin{align*} \int\frac{\sqrt[3]6\, u'}{\sqrt{4u^3+C}}\text{ d}x &= \int\pm 1\text{ d}x\\ \sqrt[3]6\,\wp^{-1}(u; 0, -C) &= D\pm x\\ u &= \wp\left(\frac{D\pm x}{\sqrt[3]6}; 0, -C\right) \end{align*} We know that the Weierstrass elliptic is even, so we rid ourselves of the $\pm$ since both solutions will be the same. Collecting constants and solving for $y(x)$, our final answer is $$\boxed{y(x) = \sqrt[3]6\, \wp\left(\frac{D + x}{\sqrt[3]6}; 0, C\right)}$$

Care should be taken with equations of similar form when doing the manipulation $(*)$ as taking the cube root messes with branch cuts.