Motions about a fixed point by hypergeometric functions: new non-complex analytical solutions and integration of the herpolhode

Abstract

The present study highlights the dynamics of a body moving about a fixed point and provides analytical closed form solutions. Firstly, for the symmetrical heavy body, that is the Lagrange–Poisson case, we compute the second (precession, $\psi$) and third (spin, $\varphi$) Euler angles in explicit and real form by means of multiple hypergeometric (Lauricella) functions. Secondly, releasing the weight assumption but adding the complication of the asymmetry, by means of elliptic integrals of third kind, we provide the precession angle $\psi$ completing the treatment of the Euler–Poinsot case. Thirdly, by integrating the relevant differential equation, we reach the finite polar equation of a special motion trajectory named the herpolhode. Finally, we keep the symmetry of the first problem, but without weight, and take into account a viscous dissipation. The use of motion first integrals—adopted for the first two problems—is no longer practicable in this situation; therefore, the Euler equations, faced directly, are driving to particular occurrences of Bessel functions of order $-\,1/2$.

Appendices

Appendix: Hypergeometric identities

We recall hereinafter an outline of the Lauricella hypergeometric functions. The first hypergeometric historical series appeared in the Wallis’s Arithmetica infinitorum (1656):

$$\begin{aligned} {}_2\mathrm {F}_{1}\left( \left. \begin{array}{c} a,b\\ c \end{array} \right| x\right) =1+\frac{a\, b}{1\cdot c}x+\frac{a\, (a+1)\, b \, (b+1)}{1\cdot 2\cdot c\cdot (c+1)}x^{2}+\cdots , \end{aligned}$$

for $|x|<1$ and real parameters $a,\,b,\,c.$ The product of n ascending factors:

$$\begin{aligned} (\lambda )_{n}=\lambda \left( \lambda +1\right) \cdots \left( \lambda +n-1\right) =\frac{\Gamma (\lambda +n)}{\Gamma (\lambda )}, \end{aligned}$$

called Pochhammer symbol (or truncated factorial), allows to represent $_{2}F_{1}$ as:

$$\begin{aligned} _2\mathrm {F}_{1}\left( \left. \begin{array}{c} a,b\\ c \end{array} \right| x\right) =\sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( b\right) _{n}}{\left( c\right) _{n}}\frac{x^{n}}{n!}. \end{aligned}$$

A meaningful contribution on various $_{2}F_{1}$ topics is ascribed to Euler in three papers (Euler 1738, 1769, 1792), but he does not seem (Dutka 1984) to have known the integral representation:

$$\begin{aligned} {}_2\mathrm {F}_{1}\left( \left. \begin{array}{c} a,b\\ c \end{array} \right| x\right) =\frac{\Gamma (c)}{\Gamma (a)\Gamma (c-a)}\,\int _{0}^{1} \frac{u^{a-1}(1-u)^{c-a-1}}{(1-xu)^{b}}\,\mathrm {d}u, \end{aligned}$$

really due to Legendre (1811), sect. 2. The above integral relationship is true if $c>a>0$ and for $\left| x\right| <1,$ even if this limitation can be discarded thanks to the analytic continuation.

Many functions have been introduced in nineteenth century for generalizing the hypergeometric functions to multiple variables. We recall the Appell $\mathrm{F}_1$ two-variable hypergeometric series, defined as:

$$\begin{aligned} \mathrm {F}_{1}\left( \left. \begin{array}{c} a;b_{1},b_{2} \\ c \end{array} \right| x_{1},x_{2}\right) =\sum _{m_{1}=0}^{\infty }\sum _{m_{2}=0}^{\infty } \frac{(a)_{m_{1}+m_{2}}(b_{1})_{m_{1}}(b_{2})_{m_{2}}}{(c)_{m_{1}+m_{2}}} \frac{x_{1}^{m_{1}}}{m_{1}!}\frac{x_{2}^{m_{2}}}{m_{2}!},\quad |x_{1}|<1,\,|x_{2}|<1. \end{aligned}$$

The analytic continuation of Appell’s function on $\mathbb {C}\setminus [1,\infty )\times \mathbb {C}\setminus [1,\infty )$ comes from its integral representation theorem: if $\mathrm {Re}\,(a)>0$, $\mathrm {Re}\,(c-a)>0$, then:

$$\begin{aligned} \mathrm {F}_{1}\left( \left. \begin{array}{c} a;b_{1},b_{2} \\ c \end{array} \right| x_{1},x_{2}\right) =\frac{\Gamma (c)}{\Gamma (a)\Gamma (c-a)} \int _{0}^{1}\frac{u^{a-1}\left( 1-u\right) ^{c-a-1}}{\left( 1-x_{1}\,u\right) ^{b_{1}}\left( 1-x_{2}\,u\right) ^{b_{2}}}\,\mathrm {d}u. \end{aligned}$$

(A.1)

The functions introduced and investigated by Lauricella (1893) and Saran (1954) are of our prevailing interest; and among them the hypergeometric function $F_{D}^{(n)}$ of $n\in \mathbb {N}^{+}$ variables (and $n+2$ parameters), see Saran (1954) and Lauricella (1893), defined as:

$$\begin{aligned} \mathrm {F}_{D}^{(n)}\left( \left. \begin{array}{c} a,b_{1},\ldots ,b_{n}\\ c \end{array} \right| x_{1},\ldots ,x_{n}\right) := \sum _{m_{1},\ldots ,m_{n}\in \mathbb {N}}\frac{(a)_{m_{1}+\cdots +m_{n}}(b_{1})_{m_{1}}\cdots (b_{n})_{m_{n}}}{(c)_{m_{1}+\cdots +m_{n}}m_{1}!\cdots m_{n}!}\,x_{1}^{m_{1}}\cdots x_{n}^{m_{m}} \end{aligned}$$

with the hypergeometric series usual convergence requirements $|x_{1}|<1,\ldots ,|x_{n}|<1$. If $\mathrm {Re}\,c>\mathrm {Re}\,a>0$ , the relevant Integral Representation Theorem provides:

$$\begin{aligned} \mathrm {F}_{D}^{(n)}\left( \left. \begin{array}{c} a,b_{1},\ldots ,b_{n}\\ c \end{array} \right| x_{1},\ldots ,x_{n}\right) =\frac{ \Gamma (c)}{\Gamma (a)\,\Gamma (c-a)}\,\int _{0}^{1}\,\frac{ u^{a-1}(1-u)^{c-a-1}}{(1-x_{1}u)^{b_{1}}\cdots (1-x_{n}u)^{b_{n}}}\,\mathrm {d }u \end{aligned}$$

(A.2)

allowing the analytic continuation to $\mathbb {C}^{n}$ deprived of the Cartesian n-dimensional product of the interval $]1,\infty [$ with itself.

Notice that the integral formula (A.2) allows an easy computer algebra implementation, being (A.2) a parametric integral, which is suitable for each possible numerical computation.

Some integrals evaluation

The reader can find hereinafter some details about the integrations used throughout the text.

1.1 Integral $I_1$

Let $a>b>y>c$ and $\alpha$ a real number. Consider

$$\begin{aligned} I_1(a,b,c;\alpha ;y)=\int _{c}^y\frac{1-\alpha u}{1-u^2}\frac{\mathrm{d}u}{\sqrt{(a-u)(b-u)(u-c)}}. \end{aligned}$$

To evaluate the integral we normalize the interval of integration, putting:

$$\begin{aligned} s=\frac{u-c}{y-c}; \end{aligned}$$

this leads to this expression for the given integral, where we remark that in the application of our interest we can limit to $0<c<1$:

$$\begin{aligned} I_1(a,b,c;\alpha ;y)= & {} \frac{\sqrt{y-c}}{(1-c^2) \sqrt{(a-c) (b-c)}}\\&\times \int _0^1\frac{1-c\alpha -\alpha (y-c) s}{\left( 1-\frac{y-c}{1-c}s\right) \left( 1+\frac{y-c}{1+c}s\right) \sqrt{s\left( 1+\frac{ y-c}{c-a}s\right) \left( 1+\frac{y-c}{c-b}s\right) }}\mathrm{d}s, \end{aligned}$$

which is therefore expressible, via the integral representation theorem, in terms of two Lauricella functions of four variables, that is:

$$\begin{aligned} I_1(a,b,c;\alpha ;y)=\frac{\sqrt{y-c}}{(1-c^2) \sqrt{(a-c) (b-c)}}\left( 2(1-c\alpha )X-\frac{2}{3}\alpha (y-c)Y\right) , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} X&=\mathrm {F}_{D}^{(4)}\left( \left. \begin{array}{c} \frac{1}{2};1,1,\frac{1}{2},\frac{1}{2} \\ \frac{3}{2} \end{array} \right| \frac{y-c}{1-c},-\frac{y-c}{1+c},-\frac{y-c}{c-a},-\frac{y-c}{c-b}\right) \\ Y&=\mathrm {F}_{D}^{(4)}\left( \left. \begin{array}{c} \frac{3}{2};1,1,\frac{1}{2},\frac{1}{2} \\ \frac{5}{2} \end{array} \right| \frac{y-c}{1-c},-\frac{y-c}{1+c},-\frac{y-c}{c-a},-\frac{y-c}{c-b}\right) . \end{aligned} \end{aligned}$$

Moreover integral $I_1$ could also have been evaluated in terms of elliptic integrals of third kind; in fact following (Byrd and Friedman 1971) entry 233.20 p. 74 we are led to

$$\begin{aligned} I_1(a,b,c;\alpha ;y)=\frac{1}{\sqrt{a-c}}\left( \frac{1-\alpha }{1-c} X_1+\frac{1+\alpha }{1+c}X_2\right) , \end{aligned}$$

(B.1)

where

$$\begin{aligned} \begin{aligned} X_1&=\Pi \left( \text {am}\left( F\left( \arcsin \left( \sqrt{\frac{y-c}{b-c}}\right) ,\sqrt{\frac{b-c}{a-c}}\right) ,\sqrt{\frac{b-c}{a-c}}\right) ,\frac{b-c}{1-c},\sqrt{\frac{b-c}{a-c}}\right) \\ X_2&=\Pi \left( \text {am}\left( F\left( \arcsin \left( \sqrt{\frac{y-c}{b-c}}\right) ,\sqrt{\frac{b-c}{a-c}}\right) ,\sqrt{\frac{b-c}{a-c}}\right) ,-\frac{b-c}{c+1}, \sqrt{\frac{b-c}{a-c}}\right) . \end{aligned} \end{aligned}$$

(B.2)

1.2 Integral $I_2$

With the same assumptions of integral $I_1$, we consider

$$\begin{aligned} I_2(a,b,c;\alpha ;y)=\int _{c}^y\frac{1-\alpha u}{1-u^2}\frac{u}{\sqrt{(a-u)(b-u)(u-c)}}\,\mathrm{d}u. \end{aligned}$$

Using the same change of variable used for $I_1$, we can also express $I_2$ in terms of Lauricella functions, namely:

$$\begin{aligned} I_2(a,b,c;\alpha ;y)= & {} \frac{\sqrt{y-c}}{(1-c^2) \sqrt{(a-c) (b-c)}}\\&\times \,\left( 2 c (1-c\alpha )X+\frac{2}{3} (1-2 c \alpha ) (y-c)Y-\frac{2}{5}\alpha \left( (c-y)^2\right) Z\right) , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} X&=\mathrm {F}_{D}^{(4)}\left( \left. \begin{array}{c} \frac{1}{2};1,1,\frac{1}{2},\frac{1}{2} \\ \frac{3}{2} \end{array} \right| \frac{y-c}{1-c},-\frac{y-c}{1+c},-\frac{y-c}{c-a},-\frac{y-c}{c-b}\right) \\ Y&=\mathrm {F}_{D}^{(4)}\left( \left. \begin{array}{c} \frac{3}{2};1,1,\frac{1}{2},\frac{1}{2} \\ \frac{5}{2} \end{array} \right| \frac{y-c}{1-c},-\frac{y-c}{1+c},-\frac{y-c}{c-a},-\frac{y-c}{c-b}\right) \\ Z&=\mathrm {F}_{D}^{(4)}\left( \left. \begin{array}{c} \frac{5}{2};1,1,\frac{1}{2},\frac{1}{2} \\ \frac{7}{2} \end{array} \right| \frac{y-c}{1-c},-\frac{y-c}{1+c},-\frac{y-c}{c-a},-\frac{y-c}{c-b}\right) . \end{aligned} \end{aligned}$$

Again following Byrd and Friedman (1971) entry 233.20 p. 74 we can express $I_2$ in terms of elliptic integral of third and first kind as

$$\begin{aligned} I_2(a,b,c;\alpha ;y)=\frac{1}{\sqrt{a-c}}\left( \frac{1-\alpha }{1-c} X_1-\frac{1+\alpha }{1+c}X_2+2\alpha X_3\right) , \end{aligned}$$

(B.3)

where $X_1$ and $X_2$ are the same as (B.2) while

$$\begin{aligned} X_3=F\left( \arcsin \left( \sqrt{\frac{y-c}{b-c}}\right) ,\sqrt{\frac{b-c}{a-c}}\right) . \end{aligned}$$

(B.4)

1.3 Integral $I_3$

For $c>1$ we consider

$$\begin{aligned} I_3(y,c)=\int _0^y\frac{\mathrm{sn}^2u}{1-c^2\mathrm{sn}^2u}\,\mathrm{d}u. \end{aligned}$$

The integral appears as entry 337.01 p. 201 of Byrd and Friedman (1971), namely

$$\begin{aligned} I_3(y,c)=\frac{1}{c^2}\left[ \Pi (\mathrm{am}(y),c^2,k)-y\right] . \end{aligned}$$

1.4 Integral $I_4$

For $c<1$ we consider:

$$\begin{aligned} I_4(y,c)=\int _0^y\frac{\mathrm{cn}^2u}{1-c^2\mathrm{sn}^2u}\,\mathrm{d}u. \end{aligned}$$

The integral appears as entry 338.01 p. 202 of Byrd and Friedman (1971), namely

$$\begin{aligned} I_4(y,c)=\frac{1}{c^2}\left[ (c^2-1)\Pi (\mathrm{am}(y),c^2,k)+y\right] . \end{aligned}$$

1.5 Integral $I_5$

If $c<a<y<b$ consider the integral

$$\begin{aligned} I_5(a,b,c;y)=\int _{\sqrt{b}}^y\frac{u}{\sqrt{(a-u^2)(u^2-b)(u^2-c)}}\,\mathrm{d}u. \end{aligned}$$

A natural step is a change of variable $u^2=x$ which gives the elliptic integral of first kind

$$\begin{aligned} I_5(a,b,c;y)=\frac{1}{2}\int _{b}^{y^2}\frac{\mathrm{d}x}{\sqrt{(a-x)(x-b)(x-c)}}, \end{aligned}$$

(B.5)

which is computed in Gradshteyn et al. (2007) entry 2.131.5 p. 250 and Byrd and Friedman (1971) entry 235.00 leading to:

$$\begin{aligned} I_5(a,b,c;y)=\frac{1}{\sqrt{a-c}}F(\varphi ,k), \end{aligned}$$

being

$$\begin{aligned} \varphi =\arcsin \sqrt{\frac{(a-c)(y^2-b)}{(a-b)(y^2-c)}},\quad k^2=\frac{a-b}{a-c}. \end{aligned}$$

1.5.1 Remark

This integral can, of course, be evaluated using the hypergeometric approach. In fact, if in (B.5) we introduce the change of variable $s=\frac{x-b}{y^2-b}$ and we obtain:

$$\begin{aligned} I_5(a,b,c;y)=\frac{1}{2}\sqrt{\frac{y^2-b}{(a-b) (b-c)}}\int _0^1\frac{s^{-\frac{1}{2}}}{ \sqrt{\left( 1-\frac{ y^2-b}{a-b}s\right) \left( 1+\frac{y^2-b}{b-c}s\right) }}\,\mathrm{d}s. \end{aligned}$$

Therefore we can use the relevant integral representation, which allows us to express $I_5$ in terms of the Appell two-variable hypergeometric function:

$$\begin{aligned} I_5(a,b,c;y)=\sqrt{\frac{y^2-b}{(a-b) (b-c)}}\;\mathrm {F}_{1}\left( \left. \begin{array}{c} \frac{1}{2};\frac{1}{2},\frac{1}{2} \\ \frac{3}{2} \end{array} \right| \frac{y^2-b}{a-b},-\frac{y^2-b}{b-c}\right) . \end{aligned}$$

1.6 Integral $I_6$

With the same parameters of $I_5$ we consider

$$\begin{aligned} I_6(a,b,c;y)=\int _{\sqrt{b}}^y\frac{\mathrm{d}u}{u\sqrt{(a-u^2)(u^2-b)(u^2-c)}}. \end{aligned}$$

Again, the change of variable $u^2=x$ leads to an elliptic integral, namely

$$\begin{aligned} I_6(a,b,c;y)=\frac{1}{2}\int _{b}^{y^2}\frac{\mathrm{d}x}{x\sqrt{(a-x)(x-b)(x-c)}}. \end{aligned}$$

(B.6)

This last integral is tabulated in Gradshteyn et al. (2007) entry 3.137.5 p. 259 and in Byrd and Friedman (1971) entry 235.17, leading to:

$$\begin{aligned} I_6(a,b,c;y)=\frac{1}{bc\sqrt{a-c}}\left[ b\,F(\varphi ,k) -(b-c)\Pi \left( \varphi ,\frac{c}{b}k^2,k\right) \right] , \end{aligned}$$

where $\varphi$ and k are the same introduced for the fifth integral.

1.6.1 Remark

This integral also can be evaluated using the hypergeometric approach, so using for (B.6) the same change of variable, used for the fifth integral, we obtain the expression:

$$\begin{aligned} I_6(a,b,c;y)=\frac{1}{2b}\sqrt{\frac{y^2-b}{(a-b) (b-c)}}\int _0^1\frac{s^{-\frac{1}{2}}}{\left( 1+\frac{y^2-b}{b}s \right) \sqrt{\left( 1-\frac{ y^2-b}{a-b}s\right) \left( 1+\frac{y^2-b}{b-c}s\right) }}\,\mathrm{d}s, \end{aligned}$$

which drives, using the integral representation, to express $I_6$ in terms of a Lauricella function of three variables:

$$\begin{aligned} I_6(a,b,c;y)=\frac{1}{b}\sqrt{\frac{y^2-b}{(a-b) (b-c)}}\;\mathrm {F}_{D}^{(3)}\left( \left. \begin{array}{c} \frac{1}{2};1,\frac{1}{2},\frac{1}{2} \\ \frac{3}{2} \end{array} \right| -\frac{y^2-b}{b},\frac{y^2-b}{a-b},-\frac{y^2-b}{b-c}\right) . \end{aligned}$$

1.7 Integral $I_7$

We go back to the parameters of the first and second integral $a>b>y>c$. Here, the integral is quite a simple one, but here we are concerned about its inversion. The integral, given in Gradshteyn et al. (2007) entry 3.131.3 p. 230 and Byrd and Friedman (1971) entry 233.00 is:

$$\begin{aligned} I_7(a,b,c;y)=\int _{c}^y\frac{\mathrm{d}u}{\sqrt{(a-u)(b-u)(u-c)}}=\frac{2}{\sqrt{a-c}}\,F\left( \arcsin \sqrt{\frac{y-c}{b-c}},\sqrt{\frac{b-c}{a-c}}\right) . \end{aligned}$$

The integral inversion, that is the solution with respect to y of the equation $I_7(a,b,c;y)=L$, is obtained by recalling the Jacobi amplitude $\mathrm{am}(u,k),$ which is the inverse of the elliptic integral of first kind $F(\mathrm{am}(u,k),u)=u$ and the Jacobi sinus amplitude $\mathrm{sn}(u,k)=\sin \mathrm{am}(u,k).$ The inversion formula is

$$\begin{aligned} I_7(a,b,c;y)=L\iff y=c+(b-c)\, \text {sn}^2\left( \frac{L\,\sqrt{a-c}}{2} ,\sqrt{\frac{b-c}{a-c}}\right) . \end{aligned}$$

Observe that the equation has a solution if L is such that

$$\begin{aligned} L\le \frac{2}{\sqrt{a-c}}\,\mathbf{K}\left( \sqrt{\frac{b-c}{a-c}}\right) . \end{aligned}$$

After having cited some books about Lauricella functions we deem to quote papers where their application is best shown as Mingari Scarpello and Ritelli (2011) and Kraniotis (2007).