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22 Jacobian Elliptic FunctionsProperties

§22.13 Derivatives and Differential Equations

Contents

§22.13(i) Derivatives

Table 22.13.1: Derivatives of Jacobian elliptic functions with respect to variable.
ddz(snz)= cnzdnz ddz(dcz) = k2sczncz
ddz(cnz)= -snzdnz ddz(ncz) = sczdcz
ddz(dnz)= -k2snzcnz ddz(scz) = dczncz
ddz(cdz)= -k2sdzndz ddz(nsz) = -dszcsz
ddz(sdz)= cdzndz ddz(dsz) = -csznsz
ddz(ndz)= k2sdzcdz ddz(csz) = -nszdsz

Note that each derivative in Table 22.13.1 is a constant multiple of the product of the corresponding copolar functions. (The modulus k is suppressed throughout the table.)

For alternative, and symmetric, formulations of these results see Carlson (2004, 2006a).

§22.13(ii) First-Order Differential Equations

22.13.1 (ddzsn(z,k))2 =(1-sn2(z,k))(1-k2sn2(z,k)),
22.13.2 (ddzcn(z,k))2 =(1-cn2(z,k))(k2+k2cn2(z,k)),
22.13.3 (ddzdn(z,k))2 =(1-dn2(z,k))(dn2(z,k)-k2).
22.13.4 (ddzcd(z,k))2 =(1-cd2(z,k))(1-k2cd2(z,k)),
22.13.5 (ddzsd(z,k))2 =(1-k2sd2(z,k))(1+k2sd2(z,k)),
22.13.6 (ddznd(z,k))2 =(nd2(z,k)-1)(1-k2nd2(z,k)),
22.13.7 (ddzdc(z,k))2 =(dc2(z,k)-1)(dc2(z,k)-k2),
22.13.8 (ddznc(z,k))2 =(k2+k2nc2(z,k))(nc2(z,k)-1),
22.13.9 (ddzsc(z,k))2 =(1+sc2(z,k))(1+k2sc2(z,k)),
22.13.10 (ddzns(z,k))2 =(ns2(z,k)-k2)(ns2(z,k)-1),
22.13.11 (ddzds(z,k))2 =(ds2(z,k)-k2)(k2+ds2(z,k)),
22.13.12 (ddzcs(z,k))2 =(1+cs2(z,k))(k2+cs2(z,k)).

For alternative, and symmetric, formulations of these results see Carlson (2006a).

§22.13(iii) Second-Order Differential Equations

22.13.13 d2dz2sn(z,k) =-(1+k2)sn(z,k)+2k2sn3(z,k),
22.13.14 d2dz2cn(z,k) =-(k2-k2)cn(z,k)-2k2cn3(z,k),
22.13.15 d2dz2dn(z,k) =(1+k2)dn(z,k)-2dn3(z,k).
22.13.16 d2dz2cd(z,k) =-(1+k2)cd(z,k)+2k2cd3(z,k),
22.13.17 d2dz2sd(z,k) =(k2-k2)sd(z,k)-2k2k2sd3(z,k),
22.13.18 d2dz2nd(z,k) =(1+k2)nd(z,k)-2k2nd3(z,k),
22.13.19 d2dz2dc(z,k) =-(1+k2)dc(z,k)+2dc3(z,k),
22.13.20 d2dz2nc(z,k) =(k2-k2)nc(z,k)+2k2nc3(z,k),
22.13.21 d2dz2sc(z,k) =(1+k2)sc(z,k)+2k2sc3(z,k),
22.13.22 d2dz2ns(z,k) =-(1+k2)ns(z,k)+2ns3(z,k),
22.13.23 d2dz2ds(z,k) =(k2-k2)ds(z,k)+2ds3(z,k),
22.13.24 d2dz2cs(z,k) =(1+k2)cs(z,k)+2cs3(z,k).

For alternative, and symmetric, formulations of these results see Carlson (2006a).