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# [讨论] 构造第二类不完全椭圆积分反函数所满足的ODE

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楼主| 发表于 2020-9-7 09:30:10 | 显示全部楼层
 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(1-k^2)\dfrac{{\mathrm{d}}}{{\mathrm{d}}k}E(u,k)\right]-k\,\!E(u,k)=-\dfrac{k\sin\left(u\right)\cos\left(u\right)}{\sqrt{\left[1-k^2\sin^2\left(u\right)\right]^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[1-k^2\sin^2\left(u\right)\right]^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\dfrac{{\mathrm{d}}}{{\mathrm{d}}k}E(u,k)\right]+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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