构造第二类不完全椭圆积分反函数所满足的ODE
本帖最后由 葡萄糖 于 2019-2-16 15:53 编辑怎么构造第二类不完全椭圆积分\(E(u,k)\)反函数\(E^{-1}(u,k)\)所满足的常微分方程 本帖最后由 葡萄糖 于 2020-9-7 18:27 编辑
\begin{gather*}
F(u,k)=\int_0^u\dfrac{\mathrm{d}\,\theta}{\sqrt{1-k^2\sin^2\left(\theta\right)}}\\
\\
\dfrac{{\mathrm{d}}}{{\mathrm{d}}k}\left-k\,\!E(u,k)=-\dfrac{k\sin\left(u\right)\cos\left(u\right)}{\sqrt{\left^3}}\\
k(1-k^2)\dfrac{{\mathrm{d}^2F(u,k)}}{{\mathrm{d}}k^2}+(1-3k^2)\dfrac{{\mathrm{d}F(u,k)}}{{\mathrm{d}}k}-k\,\!F(u,k)=-\dfrac{k\sin\left(u\right)\cos\left(u\right)}{\sqrt{\left^3}}
\end{gather*}
\begin{gather*}
E(u,k)=\int_0^u\sqrt{1-k^2\sin^2\left(\theta\right)}\,\mathrm{d}\theta\\
\\
(1-k^2)\dfrac{{\mathrm{d}}}{{\mathrm{d}}k}\left+k\,\!E(u,k)=\dfrac{k\sin\left(u\right)\cos\left(u\right)}{\sqrt{1-k^2\sin^2\left(u\right)}}\\
k(1-k^2)\dfrac{{\mathrm{d}^2E(u,k)}}{{\mathrm{d}}k^2}+(1-k^2)\dfrac{{\mathrm{d}E(u,k)}}{{\mathrm{d}}k}+k\,\!E(u,k)=\dfrac{k\sin\left(u\right)\cos\left(u\right)}{\sqrt{1-k^2\sin^2\left(u\right)}}
\end{gather*}
https://dlmf.nist.gov/19.4#ii
\begin{align*}
F^{-1}(u,k)&=\operatorname{am}\left(u,k\right)\\
F\Big(\operatorname{am}\left(u,k\right),\,k\Big)&=\int_0^{\operatorname{am}\left(u,k\right)}\dfrac{\mathrm{d}\theta}{\sqrt{1-k^2\sin^2\theta}}=u\\
\dfrac{\mathrm{d}}{\mathrm{d}u}\operatorname{am}\left(u,k\right)&=\operatorname{dn}\left(u,k\right)=\sqrt{1-k^2\sin^2\big(\operatorname{am}\left(u,k\right)\big)}
\end{align*}
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