Hyperelliptic function addition formula

Question

$$x= \int_{0}^{f(x)} \frac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}\sqrt{1-l^2u^2}}$$

$$f(0)=0$$

If we apply derivative operation for both sides, we get:

$$f'(x)=\sqrt{(1-f^2(x))(1-k^2f^2(x))(1-l^2f^2(x))}$$ My Questions:

1. What is closed form addition formula of f(x)?
2. Is $f(x)$ a periodic function or doubly periodic function or not periodic function?
3. What is the period if it is a periodic function ( or what are the periods if it is a doubly periodic function)?

My attempt to solve the questions:

Let's define a new function $\phi(x)=\frac{\sin(ax)}{a}$: and it is single periodic function.

$$x= \int_{0}^{\phi(x)} \frac{du}{\sqrt{1-a^2u^2}}$$ $$\phi(0)=0$$ I realized that we can write addition formula of $\phi(x)$ as:

$$\phi(x+y)=\phi(x)\sqrt{(1-a^2\phi^2(y))}+\phi(y)\sqrt{(1-a^2\phi^2(x))}$$

$$\phi(x+y)=\phi(x)\phi'(y)+\phi'(x)\phi(y)$$ Doubly periodic Jacobi elliptic function sn(x) is defined

$$x= \int_{0}^{sn(x)} \frac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}}$$

$$sn(0)=0$$

Addition formula of sn(x):

$$sn(x+y)=\frac{sn(x)\sqrt{(1-sn^2(y))(1-k^2sn^2(y))}+sn(y)\sqrt{(1-sn^2(x))(1-k^2sn^2(x))}}{1-k^2sn^2(x)sn^2(y)}$$

$$sn(x+y)=\frac{sn(x)sn'(y)+sn'(x)sn(y)}{1-k^2sn^2(x)sn^2(y)}$$

Let's define a new function g(x):

$$x= \int_{0}^{g(x)} \frac{du}{\sqrt{1-a^2u^2}\sqrt{1-b^2u^2}}$$ $$g(0)=0$$ I realized that we can write addition formula of g(x) as:

$$g(x+y)=\frac{g(x)\sqrt{(1-a^2g^2(y))(1-b^2g^2(y))}+g(y)\sqrt{(1-a^2g^2(x))(1-b^2g^2(x))}}{1-a^2b^2g^2(x)g^2(y)}$$

$$g(x+y)=\frac{g(x)g'(y)+g'(x)g(y)}{1-a^2b^2g^2(x)g^2(y)}$$

Let's define a new function h(x): $$x= \int_{0}^{h(x)} \frac{du}{\sqrt{1-a^2u^2}\sqrt{1-b^2u^2}\sqrt{1-c^2u^2}}$$ If we follow the same rule for h(x) as we applied above, so my estimation for addition formula of h(x) is:

$$h(0)=0$$

$h(x+y)=\frac{h(x)\sqrt{(1-a^2h^2(y))(1-b^2h^2(y))(1-c^2h^2(y))}+h(y)\sqrt{(1-a^2h^2(x))(1-b^2h^2(x))(1-c^2h^2(x))}}{1-(a^2b^2+a^2c^2+b^2c^2)h^2(x)h^2(y)+ a^2b^2c^2P(h(x),h(y))}$

$$h(x+y)=\frac{h(x)h'(y)+h'(x)h(y)}{1-(a^2b^2+a^2c^2+b^2c^2)h^2(x)h^2(y)+ a^2b^2c^2P(h(x),h(y))}$$

Where $P(h(x),h(y))$ is symmetric function

To find $P(x,y)$:

If $a=b=c$ then

$$x= \int_{0}^{h(x)} \frac{du}{(1-a^2u^2)\sqrt{1-a^2u^2}}$$

$$x= \frac{h(x)}{\sqrt{1-a^2h^2(x)}}$$

$$h(x)= \frac{x}{\sqrt{1+a^2x^2}}$$

$$h'(x)= \frac{1}{(1+a^2x^2)\sqrt{1+a^2x^2}}$$

$$h(x+y)=\frac{h(x)h'(y)+h'(x)h(y)}{1-3a^4h^2(x)h^2(y)+ a^6P(h(x),h(y))}$$

I have not found a result from this direction yet.

Do you see a solution way for addition formula for h(x) from other methods?

Thanks a lot for helps

EDIT: We may use algebraic transform on some hyper elliptic function and can transfer into Weierstrass ℘ elliptic function.

$$h'(x)=\sqrt{(1-a^2h^2(x))(1-b^2h^2(x))(1-c^2h^2(x))}$$

If we apply the algebraic transform to the hyper elliptic function above, $h(x)=-\dfrac{1}{\sqrt{g(x)}}$

We will get : $$\dfrac{g'(x)}{2g(x)\sqrt{g(x)}}=\dfrac{ \sqrt{(g(x)-a^2)(g(x)-b^2)(g(x)-c^2)}}{g(x)\sqrt{g(x)}}$$

$$g'(x)^2=4(g(x)-a^2)(g(x)-b^2)(g(x)-c^2)$$

$$g'(x)^2=4g^3(x)-4(a^2+b^2+c^2)g^2(x)+4(a^2b^2+a^2c^2+b^2c^2)g(x)-4a^2b^2c^2 \tag{1}$$

Then We can consider on the Weierstrass $\wp$ elliptic function $\wp(z, g_2, g_3)$ with the invariants $g_2\in\mathbb{R}$ and $g_3\in\mathbb{R}$:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3$$

$$\wp(x)=g(x)+\beta$$ , where $\beta$ is a constant

$$g'(x)^2 = 4(g(x)+\beta)^3 - g_2 (g(x)+\beta) - g_3$$ $$g'(x)^2 = 4g^3(x)+12\beta g^2(x) - (g_2-12\beta^2)g(x) - (g_3+g_2\beta-4 \beta^3)$$

If we equal in Equation 1 :

We can find $\beta,g_2,g_3$ as $a,b,c$

$$\beta=-\dfrac{(a^2+b^2+c^2)}{3}$$

$$\wp(x+y)+\wp(x)+\wp(y) =\frac{1}{4} \left(\frac{\wp'(x)-\wp'(y)}{\wp(x)-\wp(y)}\right)^2 \tag{3}$$

We know addition formula of Weierstrass $\wp$ elliptic function.Thus we can find h(x) addition formula as:

$$h(x)=-\dfrac{1}{\sqrt{\wp(x)-\beta}}$$ $$\wp(x)=\dfrac{1}{h(x)^2}+\beta$$

If we use Equation 3 , we can write addition formula of $h(x)$ as:

$$\dfrac{1}{h(x+y)^2}+\dfrac{1}{h(x)^2}+\dfrac{1}{h(y)^2}+3\beta = \left(\frac{\frac{h'(y)}{h(y)^3}-\frac{h'(x)}{h(x)^3}}{\frac{1}{h(x)^2}-\frac{1}{h(y)^2}}\right)^2 \tag{4}$$

$$\dfrac{1}{h(x+y)^2}+\dfrac{1}{h(x)^2}+\dfrac{1}{h(y)^2}+3\beta = \left(\frac{\frac{\sqrt{(1-a^2h^2(y))(1-b^2h^2(y))(1-c^2h^2(y))}}{h(y)^3}-\frac{\sqrt{(1-a^2h^2(x))(1-b^2h^2(x))(1-c^2h^2(x))}}{h(x)^3}}{\frac{1}{h(x)^2}-\frac{1}{h(y)^2}}\right)^2 \tag{5}$$

Answer 1

$\color{brown}{\textbf{Primary transformations.}}$

Taking in account the possible substitution in the form of $$u_1= \dfrac um,\quad\text{where}\quad m=\max(1,k,l),$$ the given equation WLOG can be presented in the form of $$mx = I\left(mx,k_1,l_1\right),$$

or $$x = I\left(x,k_1,l_1\right),\tag1$$

where $$I(x,k,l) = \int\limits_0^{f(x)}\dfrac{\mathrm du}{\sqrt{(1-u^2)(1-k^2u^2)(1-l^2u^2)}},\quad f(-x)=f(x),\tag2$$

$$l_1=\dfrac{\min(1,k,l)}m,\quad k_1=\dfrac{\operatorname{med}(1,k,l)}m,\quad l_1\le k_1 \le 1,\quad m\ge1\tag3.$$

$\color{brown}{\textbf{Case $l=k=1.$}}$

$$I(x,1,1) = \int\limits_0^{f(x)}(1-u^2)^{-\frac32}\mathrm du = -\dfrac12\int\limits_0^{f(x)}\left(\dfrac1{u^2}-1\right)^{-\frac32} \,\mathrm d\left(\dfrac1{u^2}-1\right)\\ =\left(\dfrac1{u^2}-1\right)^{-\frac12}\bigg|_0^{f(x)} =\left(\dfrac1{f^2(x)}-1\right)^{-\frac12}.$$

Applying $(1),$ should $$\dfrac1{x^2}+1=\dfrac1{f^2(x)},$$ $$\color{green}{\mathbf{f(x) = \dfrac x{\sqrt{1+x^2}}.}}\tag4$$

Plot of $f(x)$ is shown below.

Since $f(x)$ has aperiodic closed form $(4)$, then $f(x+y)$ also has aperiodic closed form.

$\color{brown}{\textbf{Case $l=k<1.$}}$

$$I(x,k,k) = \int\limits_0^{f(x)}\dfrac{\mathrm du}{(1-k^2u^2)\sqrt{1-u^2}} = \int\limits_0^{\arcsin f(x)}\dfrac{\mathrm dv}{1-k^2\sin^2v} = \int\limits_0^{\arcsin f(x)}\dfrac{\csc^2v\,\mathrm dv}{\csc^2v-k^2}\\ = -\int\limits_0^{\arcsin f(x)}\dfrac{\mathrm d\cot v}{\cot^2v+1-k^2} = \int\limits_{\frac{\sqrt{1-f^2(x)}}{f(x)}}^\infty\dfrac{\mathrm dw}{w^2+1-k^2}\\ = \dfrac1{\sqrt{1-k^2}}\arctan\dfrac w{\sqrt{1-k^2}} \left|\phantom{\LARGE{\Bigg|}}\hspace{-6mu}\right._{\frac{\sqrt{1-f^2(x)}}{f(x)}}^\infty = \dfrac1{\sqrt{1-k^2}}\cot^{-1} \dfrac1{\sqrt{1-k^2}}\sqrt{\dfrac{1-f^2(x)}{f^2(x)}}.$$

Applying $(1),$ should $$(1-k_1^2)\cot^2\left(x\sqrt{1-k_1^2}\right) = \dfrac1{f^2(x)}-1,$$ $$f^2(x) = \dfrac1{1+(1-k_1^2)\cot^2\left(x\sqrt{1-k_1^2}\right)},$$ $$\color{green}{\mathbf{f(x) = \dfrac{\sin\left(x\sqrt{1-k_1^2}\right)}{\sqrt{1-k_1^2\cos^2\left(x\sqrt{1-k_1^2}\right)}}.}}\tag5$$

Plots of $f(x)$ for $k=\dfrac13,\dfrac12,\dfrac23$ are shown below.

Since $f(x)$ has periodic closed form $(5)$, then $f(x+y)$ has doubly periodic closed form.

$\color{brown}{\textbf{Case $l<k=1.$}}$

$$I(x,1,l) = \int\limits_0^{f(x)}\dfrac{\mathrm du}{(1-u^2)\sqrt{1-l^2u^2}} = l\int\limits_0^{lf(x)}\dfrac{\mathrm du}{(l^2-u^2)\sqrt{1-u^2}} = l\int\limits_0^{\arcsin(lf(x))}\dfrac{\mathrm dv}{l^2-\sin^2v}\\ = l\int\limits_0^{\arcsin(lf(x))} \dfrac{\csc^2v\,\mathrm dv}{l^2\csc^2v-1} = -\int\limits_0^{\arcsin(lf(x))} \dfrac{\mathrm d(l\cot v)}{l^2\cot^2v+l^2-1} = \int\limits_{\Large \frac{\sqrt{1-l^2f^2(x)}}{f(x)}}^\infty\dfrac{\mathrm dw}{w^2+l^2-1}\\ = \dfrac1{2\sqrt{1-l^2}}\ln\dfrac{w-\sqrt{1-l^2}}{w+\sqrt{1-l^2}} \left|\phantom{\LARGE{\Bigg|}}\hspace{-6mu}\right._{\Large \frac{\sqrt{1-l^2f^2(x)}}{f(x)}}^\infty = \dfrac1{2\sqrt{1-l^2}}\ln\dfrac{\sqrt{1-l^2f^2(x)}+f(x)\sqrt{1-l^2}}{\sqrt{1-l^2f^2(x)}-f(x)\sqrt{1-l^2}}.$$

Applying $(1),$ should $$e^{x\sqrt{1-l_1^2}} = \dfrac{\sqrt{1-l_1^2f^2(x)}+f(x)\sqrt{1-l_1^2}}{\sqrt{1-l_1^2f^2(x)}-f(x)\sqrt{1-l_1^2}},$$ $$\sqrt{1-l_1^2f^2(x)} = f(x)\coth\left(\dfrac x2\sqrt{1-l_1^2}\right),$$ $$1-l_1^2f^2(x) = f^2(x)\coth^2\left(\dfrac x2\sqrt{1-l_1^2}\right),$$ $$\color{green}{\mathbf{f(x) = \dfrac{\sinh\left(\dfrac x2\sqrt{1-l_1^2}\right)}{\sqrt{1+l_1^2+\sinh^2\left(\dfrac x2\sqrt{1-l_1^2}\right)}}.}} \tag6$$

Plots of $f(x)$ for $k=\dfrac13,\dfrac12,\dfrac23$ are shown below.

Since $f(x)$ has aperiodic closed form $(6)$, then $f(x+y)$ also has aperiodic closed form.

$\color{brown}{\textbf{Case $l^2 < k^2 < 1.$}}$

$$I(x,k.l) = -\dfrac12\int\limits_0^{f(x)}\dfrac{1}{\sqrt{\left(\dfrac1{u^2}-1\right)\left(\dfrac1{u^2}-k^2\right) \left(\dfrac1{u^2}-l^2\right)}} \,\mathrm d\left(\dfrac 1{u^2}\right)\\ = \dfrac12\int\limits_{\Large\frac1{f^2(x)}}^\infty \dfrac{\mathrm dv}{\sqrt{(v-1)(v-k^2)(v-l^2)}}.$$

Standard substitution

$$v=1+(1-l^2)\cot^2\varphi,\quad \mathrm dv=-2(1-l^2)\cot\varphi\csc^2\varphi\mathrm d\varphi,\quad \varphi = \arccos\sqrt{\dfrac{v-1}{v-l^2}},\tag7$$

presents the obtained integral in the form of $$I(x,k,l) = \int\limits_0^{\arccos\sqrt{\large\frac{1-f^2(x)}{1-l^2f^2(x)}}} \dfrac{(1-l^2)\cot\varphi\csc^2\varphi\mathrm d\varphi} {\sqrt{(1-l^2)\cot^2\varphi(1-k^2+(1-l^2)\cot^2\varphi)(1-l^2)\csc^2\varphi}}\\ = \int\limits_0^{\arccos\sqrt{\large\frac{1-f^2(x)}{1-l^2f^2(x)}}} \dfrac{\mathrm d\varphi}{\sqrt{(1-l^2)\cos^2\varphi+(1-k^2)\sin^2\varphi}}\\ = \dfrac1{\sqrt{1-l^2}}\int\limits_0^{\arcsin\left(\sqrt{\large\frac{1-l^2}{1-l^2f^2(x)}}f(x)\right)} \dfrac{\mathrm d\varphi}{\sqrt{1-\dfrac{k^2-l^2}{1-l^2}\sin^2\varphi}} = \dfrac1{\sqrt{1-l^2}}F\left(\arcsin\left(\sqrt{\frac{1-l^2}{1-l^2f^2(x)}}f(x)\right),\sqrt{\dfrac{k^2-l^2}{1-l^2}}\right),$$

$$I(x,k,l) = \dfrac1{\sqrt{1-l^2}}F\left(\arcsin\left(\sqrt{\frac{1-l^2}{1-l^2f^2(x)}}f(x)\right),\sqrt{\dfrac{k^2-l^2}{1-l^2}}\right),\tag8$$ where

$$F(\varphi,k) = \int\limits_0^\varphi\dfrac{\mathrm d\varphi}{\sqrt{1-k^2\sin^2\varphi}}$$ is the Elliptic Integral of the First Kind.

Formula $(8)$ was confirmed for $l_1=\dfrac13,\quad k_1=\dfrac12,\quad f(x)=0.9.$

Let $$F^{-1}(F(\varphi,k),k) = \varphi.$$

Applying $(1),$ should $$\frac{\sqrt{1-l^2}f(x)}{\sqrt{1-l_1^2f^2(x)}} = \sin F^{-1}\left(x\sqrt{1-l_1^2}, \sqrt{\dfrac{k_1^2-l_1^2}{1-l_1^2}}\right),$$ $$\color{green}{\mathbf{f(x) = \dfrac{\sin F^{-1}\left(x\sqrt{1-l_1^2}, \sqrt{\dfrac{k_1^2-l_1^2}{1-l_1^2}}\right)} {\sqrt{1-l_1^2\cos^2F^{-1}\left(x\sqrt{1-l_1^2}, \sqrt{\dfrac{k_1^2-l_1^2}{1-l_1^2}}\right)}}.}}\tag9$$ Since $F(\varphi,k)$ is not linear, then $f(x)$ is not periodic.

Plot of $f^{-1}(x)$ for $l=\dfrac13,k=\dfrac23$ is shown below.

$\color{brown}{\textbf{Results.}}$

1. Closed forms of $f(x)$ are defined by $(4-6),(9).$ Closed form of $f(x+y)$ can be obtained as $f(z),$ where $z=x+y.$ Since $f(x+0) = f(x),$ the other expressions for $f(x+y)$ should be more complex than $$f(x+y) = f\left(f^{-1}(x) + f^{-1}(y)\right),$$ where $$f^{-1}(z) = I(z, k,l).$$
2. $f(x)$ is periodic iff $l_1=k_1<1,$ with the period $\dfrac\pi{\sqrt{1-k_1^2}}.$

Answer 2

This is a long comment given in response to the comment by OP regarding algebraic addition theorems.

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Here are the key theorems taken from Theory of Elliptic Functions by Harris Hancock.

Theorem 1: A single valued analytic function with an algebraic addition theorem is either a rational function or a singly periodic function or a doubly periodic function.

Moreover in this case we can express $f(x+y) $ rationally in terms of $f(x), f'(x), f(y), f'(y) $.

Theorem 2: Any function with an algebraic addition theorem is the root of a polynomial equation with coefficients being functions described in theorem 1 above.

Your $h(x) $ is indeed an example for this theorem.

When you make the substitution $h(x) =-1/\sqrt{g(x)}$ you lose the single valued nature of the function and then instead of theorem 1, theorem 2 applies.

Answer 3

\begin{align*} w&=\sqrt{\frac{a^2 b^2 c^2 u^2 v^2 \left(u^2-v^2\right)^2+P^2}{4 a^2b^2c^2u^2 v^2 \left(u^2+v^2-a^2u^2v^2-b^2u^2v^2-c^2u^2v^2\right)+Q^2}}\\ \\ P&=u \sqrt{\left(1-a^2 v^2\right) \left(1-b^2 v^2\right) \left(1-c^2 v^2\right)}+v\sqrt{\left(1-a^2 u^2\right) \left(1-b^2 u^2\right) \left(1-c^2 u^2\right)}\\ Q&=1-a^2b^2 u^2v^2-a^2c^2u^2v^2-b^2c^2u^2v^2 \end{align*}

We assume that $0<a<b<c<1$.

We can transform hyperelliptic integral into compound function of elliptic integral by quadratic transformation.

\begin{align*} \int_{0}^{x}\frac{1}{\sqrt{\left(1-a^2t^2\right)\left(1-b^2t^2\right)\left(1-c^2t^2\right)}}\mathrm{d}t\quad\qquad\\ \\ = \frac{1}{\sqrt{c^2-a^2}}\int_{0}^{x\cdot\sqrt{\frac{c^2-a^2}{1-a^2x^2}}}\frac{1}{\sqrt{\left(1-t^2\right)\left(1-\frac{b^2-a^2}{c^2-a^2}t^2\right)}}\mathrm{d}t \end{align*}

Using the addition theorem of elliptic integral, we have

\begin{align*} U&=u\sqrt{\left(1-a^2u^2\right)\left(1-b^2v^2\right)\left(1-c^2v^2\right)}+v\sqrt{\left(1-a^2v^2\right)\left(1-b^2u^2\right)\left(1-c^2u^2\right)}\\ V&=1-a^2\left(u^2+v^2\right)+\left(a^2b^2+a^2c^2-b^2c^2\right)u^2v^2 \end{align*}

Hence

$$ h(u+v)=W=\dfrac{U}{\sqrt{a^2U^2+V^2\,}} $$

In addition, by changing the order of the letters ($a,b,c$), the equation still holds