$x\in\mathbb{C}$ and $f,g,h:\mathbb{C}\to\mathbb{C}$
$f,g,h$ are Meromorphic functions.
$$f(x)^3+g(x)^3+h(x)^3=1 \tag{1}$$ $$f'(x)=g(x)^2-h(x)^2 \tag{2}$$ $$g'(x)=h(x)^2-f(x)^2 \tag{3}$$
Initial conditions: $f(0)=m$, $g(0)=n$
We have 3 unknown functions, and we also have 3 different equations . I would like to find out $f(x),g(x),h(x),$
If we derivative Equation $(1)$ ,we can get
$$3f(x)^2f'(x)+3g(x)^2g'(x)+3h(x)^2h'(x)=0 $$
$$h'(x)=f(x)^2-g(x)^2 \tag{4}$$
It can be shown easily from sum of Equation (2), (3) , (4) that
$$f'(x)+g'(x)+h'(x)=0 $$
If we integrate it,
$$f(x)+g(x)+h(x)=k$$
and put $x=0$
$$f(0)+g(0)+h(0)=k$$
$k=m+n+(1-(m^3+n^3))^\frac{1}{3}$
I would like to find differential equation for $f(x),g(x),h(x) $
Using Equation $(2)$ and apply derivative both sides
$$f''(x)=2g'(x)g(x)-2h'(x)h(x)$$
$$f''(x)=2g(x)(h(x)^2-f(x)^2)-2h(x)(f(x)^2-g(x)^2)$$
$$f''(x)=2(g(x)h(x)-f(x)^2)(g(x)+h(x))$$ $$f''(x)=2(k-f(x))(g(x)h(x)-f(x)^2) \tag {5}$$
If we use Equation (2) $$f'(x)=(g(x)-h(x))(g(x)+h(x)) $$ $$f'(x)=(g(x)-h(x))(k-f(x)) $$
$$g(x)-h(x)=\frac{f'(x)}{k-f(x)} $$
$$g(x)+h(x)=k-f(x)$$
$$2g(x)=\frac{f'(x)}{k-f(x)} +k-f(x)$$
$$g(x)=\frac{f'(x)+(k-f(x))^2}{2(k-f(x))} $$ $$h(x)=\frac{(k-f(x))^2-f'(x)}{2(k-f(x))} $$
If we put $g(x),h(x)$ in Equation (5)
$$f''(x)=2(k-f(x))(\frac{(k-f(x))^4-f'(x)^2}{4(k-f(x))^2}-f(x)^2) \tag {6}$$
$$2f''(x) (k-f(x)) +f'(x)^2 =(k-f(x))^4-4f(x)^2(k-f(x))^2 \tag {7}$$
Let's assume that the solution of the differential equation is :
$$f'(x)^2=(f(x)^2-k^2)^2+c(f(x)-k) \tag{8}$$ Where c is another constant
The Proof of my assumption above:
$$2 f'(x)f''(x)=4(f(x)^2-k^2)f(x)f'(x)+cf'(x)$$
$$2 f''(x)=4(f(x)^2-k^2)f(x)+c$$
Multiply both side by $ (k-f(x))$
$$2 f''(x)(k-f(x))=4(f(x)^2-k^2)(k-f(x))f(x)+c(k-f(x)) \tag{9}$$
If we add Equation $(8)$ and $(9)$,
$$2 f''(x)(k-f(x))+f'(x)^2=4(f(x)^2-k^2)(k-f(x))f(x)+(f(x)^2-k^2)^2$$ $$2 f''(x)(k-f(x))+f'(x)^2=(f(x)^2-k^2)[4(k-f(x))f(x)+(f(x)^2-k^2)]$$
$$2 f''(x)(k-f(x))+f'(x)^2=(f(x)^2-k^2)(k-f(x))[4f(x)-f(x)-k)]$$
$$2 f''(x)(k-f(x))+f'(x)^2=(f(x)^2-k^2)(k-f(x))(3f(x)-k)$$
$$2 f''(x)(k-f(x))+f'(x)^2=(k-f(x))^4-4f(x)^2(k-f(x))^2$$
It proves that my assumption is true in Equation $(8)$
Solution for $f(x)$ can be written as
$$f'(x)^2=(f(x)^2-k^2)^2+c_1(f(x)-k) $$ $$ x=\int_{f(0)}^{f(x)} \frac{du}{\sqrt{(u^2-k^2)^2+c_1(u-k)}}$$
It is an elliptic function and we apply the same procedure for $g(x)$ and $h(x)$ same result we will get
$$ x=\int_{g(0)}^{g(x)} \frac{du}{\sqrt{(u^2-k^2)^2+c_2(u-k)}}$$
$$ x=\int_{h(0)}^{h(x)} \frac{du}{\sqrt{(u^2-k^2)^2+c_3(u-k)}}$$
If we derivative all $f(x)$ $g(x)$ and $h(x)$
$$\frac{f'(x)}{(\sqrt{f(x)^2-k^2)^2+c_1(f(x)-k)}}=\frac{g'(x)}{\sqrt{(g(x)^2-k^2)^2+c_2(g(x)-k)}}=\frac{h'(x)}{\sqrt{(h(x)^2-k^2)^2+c_3(h(x)-k)}}=1$$ Where $c_1,c_2,c_3$ are constants
My Question:
- Are $f(x)$ $g(x)$ and $h(x)$ doubly periodic functions?
- What are their period formulas if they are doubly periodic functions ?
- Can We express the $f(x)$ $g(x)$ and $h(x)$ as the Weierstrass Elliptic function?
It is very nice result but I have not expected to have such result .Please write comments about your analyzes why the result is elliptic integral for cubic sum equation? I believe my method can be extended for $$f(x)^n+g(x)^n+h(x)^n=1 $$ $$f'(x)=g(x)^{n-1}-h(x)^{n-1}$$ $$g'(x)=h(x)^{n-1}-f(x)^{n-1}$$ where n is positive integer $n>3$. Please share if you find some elliptic integrals for $n>3$
Thanks a lot for answers and comments
EDIT: I have proved that $c_1=c_2=c_3$
Proof:
$f(0)=m$, $g(0)=n$ , $$f(0)+g(0)+h(0)=k$$ $$k=m+n+(1-(m^3+n^3))^\frac{1}{3}$$
$$h(0)=k-(m+n)$$
Let's define $$h(0)=k-(m+n)=r$$
$$\frac{f'(x)}{(\sqrt{f(x)^2-k^2)^2+c_1(f(x)-k)}}=1$$
$$f'(x)^2=(f(x)^2-k^2)^2+c_1(f(x)-k) $$
$$f'(x)^2-(f(x)^2-k^2)^2=c_1(f(x)-k) $$
$x=0$
$$f'(0)^2-(f(0)^2-k^2)^2=c_1(f(0)-k) $$
$$[f'(0)+(f(0)^2-k^2)][f'(0)-(f(0)^2-k^2)]=c_1(f(0)-k) $$
$$[\frac{f'(0)}{(f(0)-k)}+f(0)+k][f'(0)-f(0)^2+k^2]=c_1 $$ $$[\frac{n^2-r^2}{(-(n+r)}+m+m+n+r][n^2-r^2-m^2+(m+n+r)^2]=c_1$$
$$[-n+r+m+m+n+r][n^2-r^2-m^2+m^2+n^2+r^2+2mn+2mr+2rn]=c_1$$ $$2[m+r][2n^2+2mn+2mr+2rn]=c_1$$ $$c_1=4(m+r)(n+m)(n+r)$$
$c_1$ is symmetric , It means if we do the same actions for $g(x),h(x)$,
we will find same $c_1=c_2=c_3=4(m+r)(n+m)(n+r)$
If we write as k value, $$c_1=c_2=c_3=4(m+r)(n+m)(n+r)=c$$
$$c=4(m+k-m-n)(n+m)(n+k-n-m)$$ $$c=4(k-n)(n+m)(k-m)$$
$$\frac{f'(x)}{(\sqrt{f(x)^2-k^2)^2+c(f(x)-k)}}=\frac{g'(x)}{\sqrt{(g(x)^2-k^2)^2+c(g(x)-k)}}=\frac{h'(x)}{\sqrt{(h(x)^2-k^2)^2+c(h(x)-k)}}=1$$
$$ x=\int_{f(0)}^{f(x)} \frac{du}{\sqrt{(u^2-k^2)^2+c(u-k)}}$$ $$ x=\int_{g(0)}^{g(x)} \frac{du}{\sqrt{(u^2-k^2)^2+c(u-k)}}$$
$$ x=\int_{h(0)}^{h(x)} \frac{du}{\sqrt{(u^2-k^2)^2+c(u-k)}}$$
$$ x=\int_{m}^{f(x)} \frac{du}{\sqrt{(u^2-(m+n+r)^2)^2+4(m+n)(m+r)(r+n)(u-(m+n+r))}}=\int_{n}^{g(x)} \frac{du}{\sqrt{(u^2-(m+n+r)^2)^2+4(m+n)(m+r)(r+n)(u-(m+n+r))}}=\int_{r}^{h(x)} \frac{du}{\sqrt{(u^2-(m+n+r)^2)^2+4(m+n)(m+r)(r+n)(u-(m+n+r))}}$$
EDIT: I have found an easy way how to produce differential equations
$$f(x)^p+g(x)^p+h(x)^p=1 $$ $$f'(x)=g(x)^{p-1}-h(x)^{p-1}$$ $$g'(x)=h(x)^{p-1}-f(x)^{p-1}$$ and we can find other relations as I showed above that $$h'(x)=f(x)^{p-1}-g(x)^{p-1}$$ $$f'(x)+g'(x)+h'(x)=0 $$
$$f(x)+g(x)+h(x)=k $$ where p is positive integer . $k$ is a constant.
$p=2$
$$f(x)^2+g(x)^2+h(x)^2=1 $$ $$f(x)^2+(g(x)+h(x))^2-2g(x)h(x)=1 $$ $$f(x)^2+(k-f(x))^2-2g(x)h(x)=1 $$
$$2g(x)h(x)=1-f(x)^2-k^2+2kf(x)-f(x)^2=1-k^2+2kf(x)-2f(x)^2 $$ $$g(x)h(x)=\frac{1-k^2}{2}+kf(x)-f(x)^2$$
$$f'(x)^2=g(x)-h(x)$$ $$f'(x)^2=(g(x)+h(x))^2-4g(x)h(x)$$
$$f'(x)^2=(k-f(x))^2-2(1-k^2)+4kf(x)-4f(x)^2$$ $$f'(x)^2=k^2-2kf(x)+f(x)^2+2(1-k^2)+4kf(x)-4f(x)^2$$ $$f'(x)^2=-3f(x)^2+2kf(x)+2-k^2$$
If both side derivate, $$f''(x)+3f(x)=k$$
General solution for $p=2$
$f(0)=m$ ,$g(0)=n$ , $h(0)=r$ $$m^2+n^2+r^2=1$$ $$m+n+r=k$$
$$f(x)=\frac{m+n+r}{3}+\frac{2m-(n+r)}{3}\cos( \sqrt{3}x)+\frac{(n-r)}{\sqrt{3}}\sin( \sqrt{3}x)$$ $$g(x)=\frac{m+n+r}{3}+\frac{2n-(m+r)}{3}\cos( \sqrt{3}x)+\frac{(r-m)}{\sqrt{3}}\sin( \sqrt{3}x)$$ $$h(x)=\frac{m+n+r}{3}+\frac{2r-(m+n)}{3}\cos( \sqrt{3}x)+\frac{(m-n)}{\sqrt{3}}\sin( \sqrt{3}x)$$
$p=3$
$$f(x)^3+g(x)^3+h(x)^3=1 $$ $$f(x)^3+(g(x)+h(x))^3-3g(x)h(x)(g(x)+h(x))=1 $$ $$f(x)^3+(k-f(x))^3-3g(x)h(x)(k-f(x))=1 $$
$$k^3-3k^2f(x)+3kf(x)^2-3g(x)h(x)(k-f(x))=1$$ $$-3kf(x)(k-f(x))+k^3-1=3g(x)h(x)(k-f(x))$$ $$g(x)h(x)=-kf(x)+\frac{k^3-1}{3(k-f(x))}$$
$$f''(x)=2g'(x)g(x)-2h'(x)h(x)$$
$$f''(x)=2g(x)(h(x)^2-f(x)^2)-2h(x)(f(x)^2-g(x)^2)$$
$$f''(x)=2(g(x)h(x)-f(x)^2)(g(x)+h(x))$$ $$f''(x)=2(-kf(x)+\frac{k^3-1}{3(k-f(x))}-f(x)^2)(k-f(x))$$
$$f''(x)=2(-kf(x)+\frac{k^3-1}{3(k-f(x))}-f(x)^2)(k-f(x))$$ $$f''(x)=2(-kf(x)(k-f(x))+\frac{k^3-1}{3}-f(x)^2(k-f(x)))$$
$$f''(x)=2f(x)^3-2k^2f(x)+\frac{2(k^3-1)}{3}$$
if we put $k=m+n+r$ as shown above we can get the same differential Equation that I found above
Let's check out $p=4$
$$f(x)^4+g(x)^4+h(x)^4=1 $$ $$f(x)^4+(g(x)+h(x))^4-4g(x)h(x)(g(x)^2+h(x)^2)-6g(x)^2h(x)^2=1 $$ $$f(x)^4+(k-f(x))^4-4g(x)h(x)((k-f(x))^2-2g(x)h(x))-6g(x)^2h(x)^2=1 $$ $$f(x)^4+(k-f(x))^4-4g(x)h(x)(k-f(x))^2+2g(x)^2h(x)^2=1 \tag{12} $$
$$f'(x)=g(x)^3-h(x)^3$$
$$f''(x)=3g'(x)g(x)^2-3h'(x)h(x)^2$$
$$f''(x)=3g(x)^2(h(x)^3-f(x)^3)-3h(x)^2(f(x)^3-g(x)^3)$$
$$f''(x)=3g(x)^2h(x)^2(h(x)+g(x))-3f(x)^3(h(x)^2+g(x)^2)$$
$$f''(x)=3g(x)^2h(x)^2(k-f(x))-3f(x)^3((h(x)+g(x))^2-2g(x)h(x))$$
$$f''(x)=3g(x)^2h(x)^2(k-f(x))-3f(x)^3((k-f(x))^2-2g(x)h(x)) $$ $$f''(x)= -3f(x)^3((k-f(x))^2+3g(x)^2h(x)^2(k-f(x))+6g(x)h(x)f(x)^3 \tag{13} $$
I just need to combine Equation (12) and (13) and cancel $g(x)h(x)$ but I noticed that the result looks ugly . It is not nice like $p=2$ and $p=3$. Maybe I need to check $f'''(x)=a_0+a_1f(x)+....a_nf(x)^n$ for polynomial result for differential equation. Is there any idea to organize better way? I have not found general formula yet for when $p$ is any integer. Maybe someone can see a way for general case too. Thanks a lot for helps and advice.