Day: January 20, 2017

Derivatives of Jacobi elliptic am, sn, cn, dn

Posted on by arkajad

    Yesterday, in “The case of the swinging pendulum“, I started my writing about nonlinear mathematical pendulum, and how it can be solved using Jacobi elliptic functions. I did not finish. My idea was to follow the derivation in the book by Lawden. But then I slept over it and this morning I decided to do it differently, I believe in a better way. Or, at least, to do it the way that gives me a satisfaction of being “more elegant”. But in order to do it I had to return to pure math and derive some properties of the derivatives of our functions. So, here it is pure math. We will be using these properties later on.

    Differential equations satisfied by Jacobi elliptic functions am, sn, cn, dn.
    We will first deal with the case when the modulus k\leq 1.
    We start with the definition of the incomplete elliptic integral of the first kind defined in Eq. (1) of Jacobi amplitude- realism or cubism:

    (1)   \begin{equation*} F(\phi,m)=\int_0^\phi\frac{d\theta}{\sqrt{1-m\sin^2\theta}}.\end{equation*}

    The derivative of F with respect \phi is then simply the integrand function:

        \[F'(\phi,m)=\frac{1}{\sqrt{1-m\sin^2\phi}}.\]

    The amplitude function \mathrm{am}(u,m) is the inverse function of F(\phi,m). Therefore the derivative of \mathrm{am}(u,m) with respect to u is the the inverse of
    F'(\phi,m)

        \[\mathrm{am}'(u,m)=\frac{1}{F'(\phi,m)}=\sqrt{1-m\sin^2\phi}.\]

    But, from the definition of the function \mathrm{sn}, \sin(\phi)=\mathrm{sn}(u,m). Therefore

        \[\mathrm{am}'(u,m)=\frac{1}{F'(\phi,m)}=\sqrt{1-m\,\mathrm{sn}^2(u,m)}.\]

    Now, using the definition of the function \mathrm{dn} (Eq. (5) in Derivatives of Jacobi elliptic am, sn, cn, dn) we get

    (2)   \begin{equation*}\mathrm{am}'(u,m)=\mathrm{dn}(u,m).\end{equation*}

    We can now calculate the derivative \mathrm{sn}'(u,m). The function \mathrm{sn}(u,m) is a composed function: \mathrm{sn}(u,m)=\sin\mathrm{am}(u,m). Therefore

        \[ \mathrm{sn}'(u,m)=\frac{d}{du}\left(\sin\mathrm{am}(u,m)\right)=\sin'(\mathrm{am}(u,m))\,\mathrm{am}'(u,m)=\cos(\mathrm{am}(u,m)) \mathrm{dn}(u,m).\]

    From the definition of \mathrm{cn}(u,m) we know that \cos(\mathrm{am}(u,m)) =\mathrm{cn}(u,m), therefore using Eq. (2) we obtain

    (3)   \begin{equation*}\mathrm{sn}'(u,m)=\mathrm{cn}(u,m)\,\mathrm{dn}(u,m).\end{equation*}

    Much the same way we get

    (4)   \begin{equation*}\mathrm{cn}'(u,m)=-\mathrm{sn}(u,m)\,\mathrm{dn}(u,m).\end{equation*}

    To calculate \mathrm{dn}'(u,m) we use the definition

        \[ \mathrm{dn}'(u,m)=\frac{d}{du}\sqrt{1-m\,\mathrm{sn}^2(u,m)}= \frac{-2m\,\mathrm{sn}(u,m)\mathrm{sn}'(u,m)}{2\sqrt{1-m\mathrm{sn}^2(u,m)}}=\frac{-m\,\mathrm{sn}(u,m)\mathrm{cn}(u,m)\mathrm{dn}(u,m)}{\mathrm{dn}(u,m)},\]

    or

    (5)   \begin{equation*}\mathrm{dn}'(u,m)=-m\,\mathrm{sn}(u,m)\mathrm{cn}(u,m).\end{equation*}

    We will now derive differential equation satisfied by the function \mathrm{am}(u,m). We have

        \[(\mathrm{am}'(u,m))^2=1-m\,\mathrm{sn}^2(u,m).\]

    But \mathrm{sn}(u,m)=\sin\mathrm{am}(u,m), therefore

    (6)   \begin{equation*}\left(\mathrm{am}'(u,m)\right)^2=1-m\,\sin^2(\mathrm{am}(u,m)),\quad \mathrm{am}(0,m)=0.\end{equation*}

    This last equation will be just right for solving the pendulum problem,

    The formulas above were derived assuming m=k^2\leq 1. It can be now verified, in a quite straightforward way, that they are valid also for m>1.
    I am not posting the derivation here – they follow simply from the definitions and from the formulas above. The simple calculations can be downloaded form here: jacobi_cn_and_dn.pdf.