Integrals of Rational Functions of Jacobi Elliptic Functions

Note that this post will be continuously updated as results are required for evaluating other integrals.

We use results from the relationships among squares of Jacobi elliptic functions.

Also, we use the numbering scheme of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman.

We begin with the definition of the incomplete elliptic integral of the third kind
\begin{equation}
\Pi(\phi,\alpha^{2},k) = \int \frac{du}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}(u,k)}
\end{equation}

and
\begin{equation}
\int du = u
\end{equation}
which we note here as we drop the differential and dependent variables in the work below.

BF 337.01

We begin with
\begin{align}
\int \frac{\alpha^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} &= \int \frac{1 \,-\, 1 + \alpha^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \\
&= \int \frac{1}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \,-\, \int \frac{1 \,-\, \alpha^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} = \Pi \,-\, u
\end{align}
Rearranging yields
\begin{equation}
\int \frac{\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} = \frac{\Pi \,-\, u}{\alpha^{2}}
\end{equation}

BF 338.01

\begin{align}
\int \frac{\mathrm{cn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} &= \int \frac{1}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \,-\, \int \frac{\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \\
&= \Pi \,-\, \frac{\Pi \,-\, u}{\alpha^{2}} = \frac{1}{\alpha^{2}}\left[(\alpha^{2} \,-\, 1)\Pi + u \right]
\end{align}

BF 339.01

\begin{align}
\int \frac{\mathrm{dn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} &= \int \frac{1 \,-\, k^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \\
&= \Pi \,-\, \frac{k^{2}}{\alpha^{2}}(\Pi \,-\, u) = \frac{1}{\alpha^{2}}\left[(\alpha^{2} \,-\, k^{2})\Pi + k^{2}u \right]
\end{align}

Integrals of Jacobi Elliptic Functions Raised to the Power 4

We use combinations of the following to evaluate integrals in this post:

All results can be found in section 310 – 321 of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman. We use their equation numbering scheme.

BF 310.04

\begin{align}
\int \mathrm{sn}^{4} &= \int \mathrm{sn}^{2} (1 \,-\, \mathrm{cn}^{2}) = \int \mathrm{sn}^{2} \,-\, \int \mathrm{sn}^{2} \mathrm{cn}^{2} \\
&= \frac{1}{3k^{4}}\left[(k^{2} + 2)u \,-\, 2(k^{2} + 1)E + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{align}
We used GR 5.134.1 and BF 361.01.

BF 312.04

\begin{align}
\int \mathrm{cn}^{4} &= \int \mathrm{cn}^{2} (1 \,-\, \mathrm{sn}^{2}) = \int \mathrm{cn}^{2} \,-\, \int \mathrm{sn}^{2} \mathrm{cn}^{2} \\
&= \frac{1}{3k^{4}}[{k^{\prime}}^{2}(2 \,-\, 3k^{2})u + 2(2k^{2} \,-\, 1)E + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} ]
\end{align}
We used GR 5.134.2 and BF 361.01.

BF 314.04

\begin{align}
\int \mathrm{dn}^{4} &= \int \mathrm{dn}^{2} (1 \,-\, k^{2}\mathrm{sn}^{2}) = \int \mathrm{dn}^{2} \,-\, k^{2}\int \mathrm{sn}^{2} \mathrm{dn}^{2} \\
&= \frac{1}{3}[2(1 + {k^{\prime}}^{2})E \,-\, {k^{\prime}}^{2}u + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} ]
\end{align}
We used GR 5.134.3 and BF 361.02.

Continue reading “Integrals of Jacobi Elliptic Functions Raised to the Power 4”

Integrals of Pairs of Products of Squares of Jacobi Elliptic Functions

We use combinations of the following to evaluate integrals in this post:

The order of presentation here is done to avoid look ahead bias, thus avoiding circularity when referencing previous results.

All results can be found in section 361 of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman. We use their equation numbering scheme.

Note that this post will be continuously updated as results are required for evaluating other integrals.

BF 361.02

We begin with
\begin{equation}
\frac{\partial}{\partial u}\mathrm{cn} = -\,\mathrm{sn} \, \mathrm{dn}
\end{equation}
Squaring both sides, integrating by parts, and then making substitutions for \(\mathrm{cn}^{2}\), we have
\begin{align}
\int \mathrm{sn}^{2} \mathrm{dn}^{2} &= \int \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int \mathrm{cn}^{2} \mathrm{dn}^{2} – k^{2}\int \mathrm{cn}^{2} \mathrm{sn}^{2} \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int (1 – \,\mathrm{sn}^{2})\,\mathrm{dn}^{2} – \int (\mathrm{dn}^{2} – {k^{\prime}}^{2} ) \mathrm{sn}^{2} \\
&= \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{align}

BF 361.01

From BF 361.02 we have
\begin{equation}
\int \mathrm{sn}^{2} ({k^{\prime}}^{2} + k^{2}\mathrm{cn}^{2}) = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{sn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{4}}\left[(2 -\,k^{2})E \,-\,2{k^{\prime}}^{2}u \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}

BF 361.03

From BF 361.02 we have
\begin{equation}
\int (1 \,-\,\mathrm{cn}^{2}) \mathrm{dn}^{2} = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{dn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{2}}\left[(1+k^{2})E \,-\,{k^{\prime}}^{2}u + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}

Continue reading “Integrals of Pairs of Products of Squares of Jacobi Elliptic Functions”

Integrals of Gradshteyn and Ryzhik: 5.132, 5.133, 5.135 – Integrals of Jacobi Elliptic Functions

In this post we evaluate integrals of the 12 Jacobi elliptic functions. For most of these integrals, there are multiple expressions which are all equal upto a constant. We use a mixed strategy. Sometimes multiple results will be obtained from the integral directly using different substitutions. Other times a result will be shown to be the same as other results that appear in GR. This latter strategy usually consists of converting between log expressions and inverse trig and hyperbolic expressions.

Also, extensive use is made of derivatives and relationships between squares of the Jacobi elliptic functions. All equation numbers are references to the latter.

When there are multiple GR solutions, each is enclosed by a blue box.

For the sake of clarity, we drop the argument \(u\) and the modulus \(k\) from all expressions,
thus \(\mathrm{sn}(u,k) = \mathrm{sn}\). Additionally, we drop the \(du\) to avoid confusion with the function \(\mathrm{dn}\).

5.132.1

\begin{align}
\int \mathrm{ns} &= \int \frac{1}{\mathrm{sn}} = \int \frac{{k^{\prime}}^{2}\mathrm{sn}}{{k^{\prime}}^{2}\mathrm{sn}^{2}}
= {k^{\prime}}^{2} \int \frac{\mathrm{sn}}{\mathrm{dn}^{2} – \mathrm{cn}^{2}}
= {k^{\prime}}^{2} \int \frac{\mathrm{sn}}{\mathrm{cn}^{2}} \frac{1}{(\mathrm{dn}/\mathrm{cn})^{2} – 1} \\
&= \int \frac{dw}{w^{2}-1} = \frac{1}{2} \ln\left(\frac{1-w}{1+w}\right)
= \frac{1}{2} \ln\left(\frac{\mathrm{cn} – \mathrm{dn}}{\mathrm{cn} + \mathrm{dn}}\right)
\end{align}
We used equation 17 and the substitution \(w=\mathrm{dn}/\mathrm{cn} = \mathrm{dc}\).

To obtain the form of the solution in GR, we expand the logarithm and again use equation 17 (factored).
\begin{align}
\frac{1}{2} \ln\left(\frac{\mathrm{cn} – \mathrm{dn}}{\mathrm{cn} + \mathrm{dn}}\right) &=
\frac{1}{2} [\ln(\mathrm{cn} – \mathrm{dn}) – \ln(\mathrm{cn} + \mathrm{dn})] \\
&= \frac{1}{2} \ln(k^{2}-1) + \ln(\mathrm{sn})- \frac{1}{2} \ln(\mathrm{cn} + \mathrm{dn})
– \frac{1}{2} \ln(\mathrm{cn} + \mathrm{dn}) \\
&= \bbox[5px,border:2px solid blue] {\ln\left(\frac{\mathrm{sn}}{\mathrm{cn} + \mathrm{dn}}\right)} + \frac{1}{2} \ln(k^{2}-1)
\end{align}
the first term is the solution given in GR, while the second term is a constant of integration.

Using equation 17 again, we have
\begin{align}
\frac{1}{2} \ln\left(\frac{\mathrm{cn} – \mathrm{dn}}{\mathrm{cn} + \mathrm{dn}}\right) &=
\frac{1}{2} \ln\left(-\frac{(\mathrm{dn}\,-\mathrm{cn})^2}{\mathrm{dn}^{2}-\mathrm{cn}^{2}} \right)
= \frac{1}{2} \ln\left(-\frac{(\mathrm{dn}\,-\mathrm{cn})^2}{{k^{\prime}}^{2}\mathrm{sn}^{2}} \right) \\
&= \bbox[5px,border:2px solid blue] { \ln\left(\frac{\mathrm{dn}\,-\mathrm{cn}}{\mathrm{sn}} \right) } \,- \ln k^{\prime} + \frac{1}{2} \ln(-1)
\end{align}

Continue reading “Integrals of Gradshteyn and Ryzhik: 5.132, 5.133, 5.135 – Integrals of Jacobi Elliptic Functions”

Integrals of Gradshteyn and Ryzhik: 5.134 – Integrals of Squares of Jacobi Elliptic Functions – Revised

In this post, we make use of results from Relationships Between Squares of Jacobi Elliptic Functions, Derivatives of Jacobi Elliptic Functions, and Derivatives of Jacobi Elliptic Functions – Part 2. We will refer to equations in the first and third of these posts with i.e 12s and 12d2.

Note that only the first three integrals appear explicitly in Gradshteyn and Ryzhik. The other 9 appear implicitly as part of solutions of elliptic integrals. Thus we establish results here and label them for use in future posts. These 9 integrals can be found in Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman, sections 310 – 321.

Addendum – Previously, we used the integrals of products of pairs of Jacobi elliptic functions to evaluate some of the integrals in this post. This method was somewhat dubious as it was due to working backwards from known results. We have revised the work so that we only require knowledge of derivatives of single Jacobi elliptic functions and thus no need to “look ahead”. We now do this below and rearrange the order so that it is clear how all integrals were evaluated.

GR 5.134.3

\begin{equation}
\int \mathrm{dn}^{2}(u,k)\,du = \mathrm{E}(\mathrm{am}\,u,k) = \mathrm{E}(u)
\end{equation}
this is a definition of the incomplete elliptic integral of the second kind. \(\mathrm{am}\,u\) is the amplitude function.

GR 5.134.1

\begin{equation}
\int \mathrm{sn}^{2} = \int \frac{1}{k^2}\left[ 1 – \mathrm{dn}^{2}\right] = \frac{1}{k^2}\left[u -\mathrm{E}\right]
\end{equation}
We used GR 5.134.3.

GR 5.134.2

\begin{align}
\int \mathrm{cn}^{2} &= \int [1-\mathrm{sn}^{2}] = u-\frac{u}{k^2}+\frac{\mathrm{E}}{k^2} \\
&= \frac{1}{k^2}\left[\mathrm{E} \,-{k^{\prime}}^{2}u \right]
\end{align}
We used GR 5.134.3.

Continue reading “Integrals of Gradshteyn and Ryzhik: 5.134 – Integrals of Squares of Jacobi Elliptic Functions – Revised”

Derivatives of Jacobi Elliptic Functions – Part 2

Here we derive derivatives of combinations of Jacobi elliptic functions that will be useful in the evaluation of integrals of Jacobi elliptic functions which will be presented in future posts. We make use of results from Derivatives of Jacobi Elliptic Functions and Relationships Between Squares of Jacobi Elliptic Functions.

\begin{equation}
\frac{\partial}{\partial u} \mathrm{cn}^2 = -2\mathrm{cn}\,\mathrm{dn}\,\mathrm{sn}
\label{eq:dpjef-1}
\tag{1}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{dn}^2 = -2k^{2}\mathrm{cn}\,\mathrm{dn}\,\mathrm{sn}
\label{eq:dpjef-2}
\tag{2}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{sn}^2 = 2\mathrm{cn}\,\mathrm{dn}\,\mathrm{sn}
\label{eq:dpjef-3}
\tag{3}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{nc}^2 = \frac{\partial}{\partial u} \frac{1}{\mathrm{cn}^2}
= 2\,\mathrm{nc}\,\mathrm{sc}\,\mathrm{dc}
= 2\frac{\mathrm{dn}\,\mathrm{sn}}{\mathrm{cn}^3}
\label{eq:dpjef-4}
\tag{4}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{nd}^2 = \frac{\partial}{\partial u} \frac{1}{\mathrm{dn}^2}
= 2k^{2}\mathrm{nd}\,\mathrm{sd}\,\mathrm{cd}
= 2k^{2}\frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}^3}
\label{eq:dpjef-5}
\tag{5}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{ns}^2 = \frac{\partial}{\partial u} \frac{1}{\mathrm{sn}^2}
= -2\,\mathrm{ns}\,\mathrm{cs}\,\mathrm{ds}
= -2\frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}^3}
\label{eq:dpjef-6}
\tag{6}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{cn}\,\mathrm{dn} = -k^{2}\mathrm{cn}^{2}\mathrm{sn}\,-\mathrm{dn}^{2}\mathrm{sn}
\label{eq:dpjef-7}
\tag{7}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{cn}\,\mathrm{sn} = \mathrm{cn}^{2}\,\mathrm{dn} –
\mathrm{dn}\,\mathrm{sn}^{2}
\label{eq:dpjef-8}
\tag{8}
\end{equation}

\begin{equation}
\frac{\partial}{\partial u} \mathrm{dn}\,\mathrm{sn} = 2\mathrm{cn}\,\mathrm{dn}^{2}\,-\mathrm{cn}
\label{eq:dpjef-9}
\tag{9}
\end{equation}

Integrals of Gradshteyn and Ryzhik: 5.136 – 5.139 – Integrals of Combinations of Jacobi Elliptic Functions (Part 1)

5.136.1

\begin{equation}
\int \mathrm{sn}\,\mathrm{cn} = -\frac{1}{k^2}\mathrm{dn}
\end{equation}

5.136.2

\begin{equation}
\int \mathrm{sn}\,\mathrm{dn} = -\mathrm{cn}
\end{equation}

5.136.3

\begin{equation}
\int \mathrm{cn}\,\mathrm{dn} = \mathrm{sn}
\end{equation}
The three integrals above follow from the derivatives of the Jacobi elliptic functions.

5.137.1

\begin{equation}
\int \frac{\mathrm{sn}}{\mathrm{cn}^2} = \frac{1}{{k^{\prime}}^2} \frac{\mathrm{dn}}{\mathrm{cn}} = \frac{1}{{k^{\prime}}^2} \mathrm{dc}
\end{equation}

5.137.2

\begin{equation}
\int \frac{\mathrm{sn}}{\mathrm{dn}^2} = -\frac{1}{{k^{\prime}}^2} \frac{\mathrm{cn}}{\mathrm{dn}}
= -\frac{1}{{k^{\prime}}^2} \mathrm{cd}
\end{equation}

5.137.3

\begin{equation}
\int \frac{\mathrm{cn}}{\mathrm{sn}^2} = -\frac{\mathrm{dn}}{\mathrm{sn}} = -\mathrm{ds}
\end{equation}

5.137.4

\begin{equation}
\int \frac{\mathrm{cn}}{\mathrm{dn}^2} = \frac{\mathrm{sn}}{\mathrm{dn}} = \mathrm{sd}
\end{equation}

5.137.5

\begin{equation}
\int \frac{\mathrm{dn}}{\mathrm{sn}^2} = -\frac{\mathrm{cn}}{\mathrm{sn}} = -\mathrm{cs}
\end{equation}

5.137.6

\begin{equation}
\int \frac{\mathrm{dn}}{\mathrm{cn}^2} = \frac{\mathrm{sn}}{\mathrm{cn}} = \mathrm{sc}
\end{equation}
The six integrals above follow from the derivatives of the Jacobi elliptic functions.

5.138.1

\begin{align}
\int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} &= \int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} \frac{\mathrm{dn}}{\mathrm{dn}} = \int \frac{\mathrm{cn}}{\mathrm{dn}^2} \frac{1}{\mathrm{sd}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sd}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{dn}}\right)
\end{align}
We used the substitution \(w=\mathrm{sd}\).

5.138.2

\begin{align}
\int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} &= \int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}}{\mathrm{cn}^2} \frac{1}{\mathrm{dc}} \\
&= \frac{1}{{k^{\prime}}^2} \int \frac{1}{w} = \frac{1}{{k^{\prime}}^2} \ln(\mathrm{dc}) = \frac{1}{{k^{\prime}}^2} \ln\left(\frac{\mathrm{dn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution \(w=\mathrm{dc}\).

5.138.3

\begin{align}
\int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} &= \int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{sc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sc}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution \(w=\mathrm{sc}\).

5.139.1

\begin{equation}
\int \frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}} = \int \frac{1}{w} = \ln(\mathrm{sn})
\end{equation}

5.139.2

\begin{align}
\int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} &= \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{nc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{nc}) = \ln\left(\frac{1}{\mathrm{cn}}\right)
\end{align}
We used the substitution \(w=\mathrm{nc}\).

5.139.3

\begin{equation}
\int \frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}} = -\frac{1}{k^2} \int \frac{1}{w} = -\frac{1}{k^2} \ln(\mathrm{dn})
\end{equation}

Derivatives of Jacobi Elliptic Functions

Here we compute derivatives of 9 of the Jacobi elliptic functions with respect to the argument \(u\). Three were derived here, and are reproduced below.

For convenience, we drop the argument and modulus.

\begin{equation}
\frac{\partial \,\mathrm{sn}}{\partial u} = \mathrm{cn}\,\mathrm{dn}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{cn}}{\partial u} = -\mathrm{dn}\,\mathrm{sn}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{dn}}{\partial u} = -k^{2}\mathrm{cn}\,\mathrm{sn}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{ns}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{sn}} = -\frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}^2} = -\mathrm{cs}\,\mathrm{ds}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{nc}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{cn}} = \frac{\mathrm{dn}\,\mathrm{sn}}{\mathrm{cn}^2} = \mathrm{sc}\,\mathrm{dc}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{nd}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{dn}} = k^{2}\frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}^2} = k^{2}\mathrm{sd}\,\mathrm{cd}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{cs}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{sc}}
= \frac{\partial}{\partial u}\frac{\mathrm{cn}}{\mathrm{sn}} = -\frac{\mathrm{dn}}{\mathrm{sn}^2}
= -\mathrm{dn}\,\mathrm{ns}^2
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{sd}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{ds}}
= \frac{\partial}{\partial u}\frac{\mathrm{sn}}{\mathrm{dn}} = \frac{\mathrm{cn}}{\mathrm{dn}^2}
= \mathrm{cn}\,\mathrm{nd}^2
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{ds}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{sd}}
= \frac{\partial}{\partial u}\frac{\mathrm{dn}}{\mathrm{sn}} = -\frac{\mathrm{cn}}{\mathrm{sn}^2}
= -\mathrm{cn}\,\mathrm{ns}^2
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{sc}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{cs}}
= \frac{\partial}{\partial u}\frac{\mathrm{sn}}{\mathrm{cn}} = \frac{\mathrm{dn}}{\mathrm{cn}^2}
= \mathrm{nc}\,\mathrm{dc}
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{cd}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{dc}}
= \frac{\partial}{\partial u}\frac{\mathrm{cn}}{\mathrm{dn}} = -{k^{\prime}}^2\frac{\mathrm{sn}}{\mathrm{dn}^2}
= -{k^{\prime}}^2\mathrm{sn}\,\mathrm{nd}^2
\end{equation}

\begin{equation}
\frac{\partial \,\mathrm{dc}}{\partial u} = \frac{\partial}{\partial u}\frac{1}{\mathrm{cd}}
= \frac{\partial}{\partial u}\frac{\mathrm{dn}}{\mathrm{cn}} = {k^{\prime}}^2\frac{\mathrm{sn}}{\mathrm{cn}^2}
= {k^{\prime}}^2\mathrm{sn}\,\mathrm{nc}^2
\end{equation}

Relationships Between Squares of Jacobi Elliptic Functions

In this post, we derive a number of relationships between squares of Jacobi elliptic functions. The derivations are trivial, but the results will be required for subsequent posts involving derivatives and integrals of Jacobi elliptic functions. The numbering scheme will also be referenced in future posts.

We drop the arguments for convenience, thus \(\mathrm{cn}(u,k)=\mathrm{cn}\).

The 12 Jacobi Elliptic Functions

Beginning with \(\mathrm{cn}, \mathrm{dn}, \mathrm{sn}\) we define 9 other functions using Glaisher’s notation.
\begin{equation}
\mathrm{nc} = \frac{1}{\mathrm{cn}} \qquad \mathrm{nd} = \frac{1}{\mathrm{dn}} \qquad \mathrm{ns} = \frac{1}{\mathrm{sn}}
\end{equation}

\begin{equation}
\mathrm{cd} = \frac{\mathrm{cn}}{\mathrm{dn}} \qquad \mathrm{cs} = \frac{\mathrm{cn}}{\mathrm{sn}} \qquad \mathrm{ds}=\frac{\mathrm{dn}}{\mathrm{sn}}
\end{equation}

\begin{equation}
\mathrm{dc} = \frac{\mathrm{dn}}{\mathrm{cn}} \qquad \mathrm{sc} = \frac{\mathrm{sn}}{\mathrm{cn}} \qquad \mathrm{sd}=\frac{\mathrm{sn}}{\mathrm{dn}}
\end{equation}

Relationships Between Squares of Jacobi Elliptic Functions

We begin with 3 results from Theta Functions and Jacobi Elliptic Functions.

\begin{equation}
\mathrm{cn}^{2} + \mathrm{sn}^{2} = 1
\label{eq:pythag-1}
\tag{1}
\end{equation}

\begin{equation}
\mathrm{dn}^{2} + k^{2}\mathrm{sn}^{2} = 1
\label{eq:pythag-2}
\tag{2}
\end{equation}

\begin{equation}
k^{2} + {k^{\prime}}^{2} = 1
\label{eq:pythag-3}
\tag{3}
\end{equation}

Divide equation \eqref{eq:pythag-1} by \(\mathrm{cn}^{2}\) to obtain:
\begin{equation}
1 + \mathrm{sc}^{2} = \mathrm{nc}^{2}
\label{eq:pythag-4}
\tag{4}
\end{equation}

Divide equation \eqref{eq:pythag-1} by \(\mathrm{sn}^{2}\) to obtain:
\begin{equation}
\mathrm{cs}^{2} + 1 = \mathrm{ns}^{2}
\label{eq:pythag-5}
\tag{5}
\end{equation}

Divide equation \eqref{eq:pythag-2} by \(\mathrm{dn}^{2}\) to obtain:
\begin{equation}
\mathrm{cd}^{2} + \mathrm{sd}^{2} = \mathrm{nd}^{2}
\label{eq:pythag-6}
\tag{6}
\end{equation}

Divide equation \eqref{eq:pythag-2} by \(\mathrm{dn}^{2}\) to obtain:
\begin{equation}
1 + k^{2}\mathrm{sd}^{2} = \mathrm{nd}^{2}
\label{eq:pythag-7}
\tag{7}
\end{equation}

Divide equation \eqref{eq:pythag-2} by \(\mathrm{cn}^{2}\) to obtain:
\begin{equation}
\mathrm{dc}^{2} + k^{2}\mathrm{sc}^{2} = \mathrm{nc}^{2}
\label{eq:pythag-8}
\tag{8}
\end{equation}

Divide equation \eqref{eq:pythag-2} by \(\mathrm{sn}^{2}\) to obtain:
\begin{equation}
\mathrm{ds}^{2} + k^{2} = \mathrm{ns}^{2}
\label{eq:pythag-9}
\tag{9}
\end{equation}

Eliminate \(\mathrm{nd}^{2}\) from equations \eqref{eq:pythag-6} and \eqref{eq:pythag-7} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
\mathrm{cd}^{2} + {k^{\prime}}^{2}\mathrm{sd}^{2} = 1
\label{eq:pythag-10}
\tag{10}
\end{equation}

Eliminate \(\mathrm{sd}^{2}\) from equations \eqref{eq:pythag-6} and \eqref{eq:pythag-7} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
1 = k^{2}\mathrm{cd}^{2} + {k^{\prime}}^{2}\mathrm{nd}^{2}
\label{eq:pythag-11}
\tag{11}
\end{equation}

Eliminate \(\mathrm{nc}^{2}\) from equations \eqref{eq:pythag-4} and \eqref{eq:pythag-8} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
1 + {k^{\prime}}^{2}\mathrm{sc}^{2} = \mathrm{dc}^{2}
\label{eq:pythag-12}
\tag{12}
\end{equation}

Eliminate \(\mathrm{sc}^{2}\) from equations \eqref{eq:pythag-4} and \eqref{eq:pythag-8} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
\mathrm{dc}^{2} = k^{2} + {k^{\prime}}^{2}\mathrm{nc}^{2}
\label{eq:pythag-13}
\tag{13}
\end{equation}

Eliminate \(\mathrm{ns}^{2}\) from equations \eqref{eq:pythag-5} and \eqref{eq:pythag-9} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
\mathrm{cs}^{2} + {k^{\prime}}^{2} = \mathrm{ds}^{2}
\label{eq:pythag-14}
\tag{14}
\end{equation}

Multiply equations \eqref{eq:pythag-11} by \(\mathrm{ns}^{2}\) to obtain:
\begin{equation}
\mathrm{ds}^{2} = k^{2}\mathrm{cs}^{2} + {k^{\prime}}^{2}\mathrm{ns}^{2}
\label{eq:pythag-15}
\tag{15}
\end{equation}

Eliminate \(\mathrm{sn}^{2}\) from equations \eqref{eq:pythag-1} and \eqref{eq:pythag-2} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
\mathrm{dn}^{2} = k^{2}\mathrm{cn}^{2} + {k^{\prime}}^{2}
\label{eq:pythag-16}
\tag{16}
\end{equation}

Eliminate 1 from equations \eqref{eq:pythag-1} and \eqref{eq:pythag-2} and use equation \eqref{eq:pythag-3} to obtain:
\begin{equation}
\mathrm{dn}^{2} = \mathrm{cn}^{2} + {k^{\prime}}^{2}\mathrm{sn}^{2}
\label{eq:pythag-17}
\tag{17}
\end{equation}

References

1. Handbook of Elliptic Integrals for Engineers and Scientists – Byrd and Friedman

Theta Functions and Jacobi Elliptic Functions

While theta functions have applications in number theory and the heat equation, they are also used as helper functions in proving properties of Jacobi elliptic functions. The latter is what we will do in this post. This post will be followed by others to derive various properties of the Jacobi elliptic functions in order to evaluate integrals of them and eventually elliptic integrals.

We begin with the basic definitions of the theta functions. Note that notation for the theta functions is not standard.

Theta Functions

\(z\) = argument
\(\tau\) = lattice parameter, \(\mathfrak{I}(\tau) \gt 0\)
\(q = \mathrm{e}^{i\pi \tau}\) = nome, \(|q| \lt 1\)

a. Exponential forms
\begin{equation}
\theta_{1}(z|\tau) = \theta_{1}(z,q) = -i\sum_{n=-\infty}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z} \\
\theta_{2}(z|\tau) = \theta_{2}(z,q) = \sum_{n=-\infty}^{\infty} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z} \\
\theta_{3}(z|\tau) = \theta_{3}(z,q) = \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz} \\
\theta_{4}(z|\tau) = \theta_{4}(z,q) = \sum_{n=-\infty}^{\infty} (-1)^{n} q^{n^{2}} \mathrm{e}^{i2nz}
\end{equation}

b. Trigonometric forms
\begin{equation}
\theta_{1}(z|\tau) = \theta_{1}(z,q) = 2\sum_{n=0}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \sin[(2n+1)z] \\
\theta_{2}(z|\tau) = \theta_{2}(z,q) = 2\sum_{n=0}^{\infty} q^{(n+1/2)^{2}} \cos[(2n+1)z] \\
\theta_{3}(z|\tau) = \theta_{3}(z,q) = 1 + 2\sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\theta_{4}(z|\tau) = \theta_{4}(z,q) = 1 + 2\sum_{n=1}^{\infty} (-1)^{n} q^{n^{2}} \cos(2nz)
\end{equation}

Jacobi Elliptic Functions as Functions of Theta Functions

The modulus \(k\):
\begin{equation}
\sqrt{k} = \frac{\theta_{2}(0,q)}{\theta_{3}(0,q)}
\end{equation}
The complementary modulus \(k^{\prime}\):
\begin{equation}
\sqrt{k^{\prime}} = \frac{\theta_{4}(0,q)}{\theta_{3}(0,q)}
\end{equation}
3 of the 12 Jacobi elliptic functions:
\begin{align}
\mathrm{sn}(u,k) &= \frac{\theta_{3}(0,q)\theta_{1}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)} \\
\mathrm{cn}(u,k) &= \frac{\theta_{4}(0,q)\theta_{2}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)} \\
\mathrm{dn}(u,k) &= \frac{\theta_{4}(0,q)\theta_{3}(z,q)}{\theta_{3}(0,q)\theta_{4}(z,q)}
\end{align}
where \(z\) is the argument of the theta functions and \(u\) is the argument of the Jacobi elliptic functions. They are related by the following equation:
\begin{equation}
z=\frac{u}{\theta_{3}^{2}(0,q)}
\end{equation}

Proofs of Basic Properties of Jacobi Elliptic Functions

We assume all properties of theta functions as given. For proofs of these properties, see references 2 and 4 below.

1. First “Pythagorean” Identity
\begin{align}
k^{2} + {k^{\prime}}^{2} &=
\frac{\theta_{2}^{4}(0,q)}{\theta_{3}^{4}(0,q)} + \frac{\theta_{4}^{4}(0,q)}{\theta_{3}^{4}(0,q)}
= \frac{\theta_{2}^{4}(0,q) + \theta_{4}^{4}(0,q)}{\theta_{3}^{4}(0,q)} \\
&= \frac{\theta_{3}^{4}(0,q)}{\theta_{3}^{4}(0,q)} = 1
\end{align}
We used this identity in the last step.

2. Second “Pythagorean” Identity
\begin{align}
\mathrm{cn}^{2}(u,k) + \mathrm{sn}^{2}(u,k)
&= \frac{\theta_{4}^{2}(0,q)\theta_{2}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} +
\frac{\theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{\theta_{4}^{2}(0,q)\theta_{2}^{2}(z,q) + \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{[\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q) \,-\, \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)] + \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} = 1
\end{align}
We used this identity in the last step.

3. Third “Pythagorean” Identity
\begin{align}
\mathrm{dn}^{2}(u,k) + k^{2}\mathrm{sn}^{2}(u,k)
&= \frac{\theta_{4}^{2}(0,q)\theta_{3}^{2}(z,q)}{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)}
+ \frac{\theta_{2}^{4}(0,q)}{\theta_{3}^{4}(0,q)}\,
\frac{\theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{\theta_{4}^{2}(0,q)\theta_{3}^{2}(z,q) + \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{[\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q) \,-\, \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)]
\,+\, \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)} = 1
\end{align}
We used this identity in the last step.

Derivations of Derivatives of the 3 Basic Jacobi Elliptic Functions

1.
\begin{align}
\frac{\partial \,\mathrm{sn}(u,k)}{\partial u}
&= \frac{\theta_{3}(0,q)}{\theta_{2}(0,q)} \frac{\partial}{\partial z} \left(\frac{\theta_{1}(z,q)}{\theta_{4}(z,q)} \right) \frac{\partial z}{\partial u} \\
&= \frac{\theta_{3}(0,q)}{\theta_{2}(0,q)} \, \frac{\theta_{4}^{2}(0,q)\theta_{2}(z,q)\theta_{3}(z,q)}{\theta_{4}^{2}(z,q)} \, \frac{1}{\theta_{3}^{2}(0,q)} \\
&= \left[\frac{\theta_{4}(0,q)\theta_{2}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)}\right] \,
\left[\frac{\theta_{4}(0,q)\theta_{3}(z,q)}{\theta_{3}(0,q)\theta_{4}(z,q)}\right]
= \mathrm{cn}(u,k)\mathrm{dn}(u,k)
\end{align}
The derivative of the ratio of theta functions in the first line was obtained from Gradshteyn and Ryzhik 8.199(2).1.

2.
\begin{equation}
\frac{\partial}{\partial u} [\mathrm{cn}^{2}(u,k) + \mathrm{sn}^{2}(u,k) = 1]
\end{equation}
\begin{equation}
2\mathrm{cn}(u,k)\frac{\partial}{\partial u}\mathrm{cn}(u,k) + 2\mathrm{sn}(u,k)\frac{\partial}{\partial u}\mathrm{sn}(u,k) = 0
\end{equation}
\begin{equation}
\frac{\partial}{\partial u}\mathrm{cn}(u,k) = -\frac{\mathrm{sn}(u,k)}{\mathrm{cn}(u,k)}\frac{\partial}{\partial u}\mathrm{sn}(u,k) = -\mathrm{sn}(u,k)\mathrm{dn}(u,k)
\end{equation}

3.
\begin{equation}
\frac{\partial}{\partial u} [\mathrm{dn}^{2}(u,k) + k^{2}\mathrm{sn}^{2}(u,k) = 1]
\end{equation}
\begin{equation}
2\mathrm{dn}(u,k)\frac{\partial}{\partial u}\mathrm{dn}(u,k) + 2k^{2}\mathrm{sn}(u,k)\frac{\partial}{\partial u}\mathrm{sn}(u,k) = 0
\end{equation}
\begin{equation}
\frac{\partial}{\partial u}\mathrm{dn}(u,k) = -k^{2}\frac{\mathrm{sn}(u,k)}{\mathrm{dn}(u,k)}\frac{\partial}{\partial u}\mathrm{sn}(u,k) = -k^{2}\mathrm{sn}(u,k)\mathrm{cn}(u,k)
\end{equation}

References

1. NIST Digital Library of Mathematical Functions
2. A Course of Modern Analysis – Whittaker and Watson
3. Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 8th edition
4. Lectures on the Theory of Elliptic Functions – Hancock