Let α : $I\to {\mathbb{E}}_{1}^{2},\alpha (s)=(x(s),\phantom{\rule{thickmathspace}{0ex}}z(s))$ be a spacelike elastic Sturmian spiral in Lorentz-Minkowski plane which means that its curvature is given by the function κ = σ/*r*, where $r=\sqrt{{z}^{2}-{x}^{2}}$ and σ ∈ ℝ^{+}. Before making use of the intrinsic equation (5) let us integrate it with respect to *s*. This produces the equation

$${\dot{\kappa}}^{2}=\frac{1}{4}{\kappa}^{4}-\frac{\lambda}{2}{\kappa}^{2}+2E$$(19)in which *E* denotes the integration constant which can be interpreted as energy. For the special case of the Sturmian spirals it can be rewritten in the form

$${\dot{r}}^{2}=\frac{{\sigma}^{2}}{4}-\frac{\lambda}{2}{r}^{2}+\frac{2E}{{\sigma}^{2}}{r}^{4}.$$(20)As the tension λ is a physical property it can be assumed to be a positive constant. Therefore, there are only two cases to be considered, depending on the sign of *E*.

**Case 1**. Let us start by assuming that both *E* and λ are positive. We will write this fact formally as

$$E=\frac{{a}^{2}{c}^{2}}{8}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thickmathspace}{0ex}}\lambda =\frac{{a}^{2}+{c}^{2}}{2}$$with *a* and *c* being non-zero positive constants which comply with the condition *a* > *c*. Substituting the above expressions of *E* and λ in (20) we obtain

$${\dot{r}}^{2}=\frac{{a}^{2}{c}^{2}}{4{\sigma}^{2}}(\frac{{\sigma}^{2}}{{a}^{2}}-{r}^{2})(\frac{{\sigma}^{2}}{{c}^{2}}-{r}^{2}).$$(21)As a consequence, the solutions of the differential equation (21) are either

$$r(s)=\frac{\sigma}{a}\mathrm{s}\mathrm{n}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k)\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{1em}{0ex}}k=\frac{c}{a},\phantom{\rule{1em}{0ex}}r<\frac{\sigma}{a}$$(22)or

$$r(s)=\frac{\sigma}{c}\frac{1}{\mathrm{s}\mathrm{n}(\frac{as}{2},k)},\phantom{\rule{1em}{0ex}}k=\frac{c}{a},\phantom{\rule{1em}{0ex}}r>\frac{\sigma}{c}$$(23)where the first slot in the Jacobian sinus elliptic function sn ) is reserved for the argument, the second one for the so called elliptic modulus which is a real number between zero and one.

Now, when (22) holds we have

$$\kappa (s)=\frac{\sigma}{r(s)}=\frac{a}{\mathrm{s}\mathrm{n}(\frac{as}{2},k)},\phantom{\rule{thickmathspace}{0ex}}\dot{\kappa}({s}_{0})=0.$$(24)The equality on the right hand side is satisfied for *s*_{0} = 2*K*(*k*)/*a*, where *K*(*k*) is the complete elliptic integral of the first kind and κ(*s*_{0}) = *a*. Then using (24) in (13) we can conclude that (*c*_{1}, *c*_{2}) = ((*c*^{2}-*a*^{2}})/4,0) . Taking into account all above and Theorem 3.1 we end up with the parameterization

$$\begin{array}{ll}x(s)& =\frac{4a}{{a}^{2}-{c}^{2}}\frac{1}{\mathrm{s}\mathrm{n}(\frac{as}{2},k)},\phantom{\rule{1em}{0ex}}k=c/a\\ z(s)& =\frac{4a}{{a}^{2}-{c}^{2}}F(\mathrm{a}\mathrm{m}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k),\phantom{\rule{thickmathspace}{0ex}}k)-\frac{4a}{{a}^{2}-{c}^{2}}E(\mathrm{a}\mathrm{m}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k),\phantom{\rule{thickmathspace}{0ex}}k)\\ & \phantom{\rule{1em}{0ex}}-\frac{4a}{{a}^{2}-{c}^{2}}\frac{\mathrm{c}\mathrm{n}(\frac{as}{2},k)\mathrm{d}\mathrm{n}(\frac{as}{2},k)}{\mathrm{s}\mathrm{n}(\frac{as}{2},k)}-\frac{{a}^{2}+{c}^{2}}{{a}^{2}-{c}^{2}}s\end{array}$$(25)where (.,.) is the Jacobian amplitude function, cn and dn are the remaining Jacobian elliptic functions and *F*(.,.), *E*(.,.) denote the incomplete elliptic inegrals of the first, respectively second kind (for more details see [15]).

Now, let us switch to the solution presented in (23). Proceeding in the same manner we obtain

$$\kappa (s)=\frac{\sigma}{r}=c(\mathrm{s}\mathrm{n}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k))\phantom{\rule{1em}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{1em}{0ex}}\dot{\kappa}({s}_{0})=0$$(26)for *s*_{0} = 2(*K*(*k*))/*a* complemented by κ(*s*_{0}) = *c*. Entering with (26) in (13) we obtain that (*c*_{1}, *c*_{2}) = ((*a*^{2}}-*c*^{2})/4,0) . Substituting (26) in the formulas of Theorem 1, we find

$$\begin{array}{l}x(s)=\frac{4c}{{c}^{2}-{a}^{2}}\mathrm{s}\mathrm{n}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k)\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}k=c/a\\ z(s)=\frac{4a}{{c}^{2}-{a}^{2}}(F(\mathrm{a}\mathrm{m}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k),\phantom{\rule{thickmathspace}{0ex}}k)-E(\mathrm{a}\mathrm{m}(\frac{as}{2},\phantom{\rule{thickmathspace}{0ex}}k),\phantom{\rule{thickmathspace}{0ex}}k))-\frac{{a}^{2}+{c}^{2}}{{c}^{2}-{a}^{2}}s.\end{array}$$(27)**Case 2**. When *E* is negative, we can write respectively

$$E=-\frac{{a}^{2}{c}^{2}}{8}\phantom{\rule{1em}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{1em}{0ex}}\lambda =\frac{{a}^{2}-{c}^{2}}{2}.$$(28)Inserting *E* and λ in (20) we obtain the equation

$${\dot{r}}^{2}=\frac{{a}^{2}{c}^{2}}{4{\sigma}^{2}}\left(\frac{{\sigma}^{2}}{{a}^{2}}-{r}^{2}\right)\left(\frac{{\sigma}^{2}}{{c}^{2}}+{r}^{2}\right).$$(29)The solution of the above differential equation is

$$r(s)=\frac{\sigma}{a}\mathrm{c}\mathrm{n}(\frac{cs}{2k},\phantom{\rule{thickmathspace}{0ex}}k)\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}k=\frac{c}{\sqrt{{a}^{2}+{c}^{2}}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}r<\frac{\sigma}{a}$$(30)and therefore

$$\kappa (s)=\frac{\sigma}{r(s)}=\frac{a}{\mathrm{c}\mathrm{n}(\frac{cs}{2k},k)},\phantom{\rule{1em}{0ex}}\dot{\kappa}(0)=0\phantom{\rule{1em}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{1em}{0ex}}\kappa (0)=a.$$(31)All of this, along with (13), produces (*c*_{1}, *c*_{2}) = (-(*a*^{2} + *c*^{2})/4,0) . Relying on Theorem 3.1 we obtain finally the parameterization of the curve, i.e.,

$$\begin{array}{ll}x(s)& =\frac{4a}{{a}^{2}+{c}^{2}}\frac{1}{\mathrm{c}\mathrm{n}(\frac{cs}{2k},k)},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}k=\frac{c}{\sqrt{{a}^{2}+{c}^{2}}}\\ z(s)& =\frac{4k{a}^{2}}{c({a}^{2}+{c}^{2})}F(\mathrm{a}\mathrm{m}(\frac{cs}{2k},\phantom{\rule{thickmathspace}{0ex}}k),\phantom{\rule{thickmathspace}{0ex}}k)-\frac{4k}{c}E(\mathrm{a}\mathrm{m}(\frac{cs}{2k},\phantom{\rule{thickmathspace}{0ex}}k),\phantom{\rule{thickmathspace}{0ex}}k)\\ & \phantom{\rule{1em}{0ex}}-\frac{4k}{c}\frac{\mathrm{c}\mathrm{n}(\frac{cs}{2k},k)\mathrm{d}\mathrm{n}(\frac{cs}{2k},k)}{\mathrm{s}\mathrm{n}(\frac{cs}{2k},k)}-\frac{{a}^{2}-{c}^{2}}{{a}^{2}+{c}^{2}}s.\end{array}$$(32)
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