一种快速确定能量最优拦截预测命中点的数值优化方法与流程

技术特征：

1.一种快速确定能量最优拦截预测命中点的数值优化方法，其特征在于，具体步骤如下：

(1)、给定自变量X的试探值，并设定系数矩阵u和v；

(2)、由公式(22)和公式(24)计算偏导数矩阵FF′(X)；

(3)、由公式(25)计算新的自变量取值X_k+1；

(4)、判断是否收敛；如果ΔXX＝XX_k+1-XX_k小于误差限，计算结束；否则转至步骤(2)，重复进行步骤(2)-(4)；

其中，步骤(1)中，所述自变量X＝[t P f_M]^T，引入两组等式约束

$f_1=g+\sqrt{\frac{a^3}{\mathrm{\μ}}}\{\arccos \[1-\frac{r_1}{a}(1-f)\]+\frac{r_1{r}_2\stackrel{\·}{f}}{\sqrt{\mathrm{\μ}a}}\}-t$

$f_2=E_M\left(f_M\right)-e_{M}sin\[E_M\left(f_M\right)\]-\frac{2\mathrm{\π}}{T_M}(t-t_{pM})$

使拦截弹的速度修正量最小，描述为数学表达式，则目标函数为

J(X)＝(v_1x-v_0x)²+(v_1y-v_0y)²+(v_1z-v_0z)²；

至此，把目标存在轨道机动的拦截变轨最小速度修正问题，转化为一个含有等式约束的最优规划问题，该最优规划问题的表达式整理如下：

$\begin{array}{c}X={\left[\begin{array}{ccc}t& P& f_M\end{array}\right]}^T\\ J\left(X\right)={\left(v_{1x}-v_{0x}\right)}^2+{\left(v_{1y}-v_{0y}\right)}^2+{\left(v_{1z}-v_{0z}\right)}^2\\ f_1\left(t,P,f_M\right)=g+\sqrt{\frac{a^3}{\mathrm{\μ}}}\left\{\arccos \[1-\frac{r_1}{a}\left(1-f\right)\]+\frac{r_1{r}_2\stackrel{\·}{f}}{\sqrt{\mathrm{\μ}a}}\right\}-t\\ f_2\left(t,f_M\right)=E_M\left(f_M\right)-e_M\sin \[E_M\left(f_M\right)\]-\frac{2\mathrm{\π}}{T_M}\left(t-t_{pM}\right)\end{array}---\left(13\right)$

设带有Lagrange乘子的增广目标函数为

L＝J+λ₁f₁+λ₂f₂ (14)

由KKT条件，可得

$\frac{\∂L}{\∂t}=-{\mathrm{\λ}}_1-\frac{2\mathrm{\π}}{T_M}{\mathrm{\λ}}_2=0---\left(15\right)$

$\frac{\∂L}{\∂P}=\frac{\∂J}{\∂P}+{\mathrm{\λ}}_1\frac{\∂f_1}{\∂P}=0---\left(16\right)$

$\frac{\∂L}{\∂f_M}=\frac{\∂J}{\∂f_M}+{\mathrm{\λ}}_1\frac{\∂f_1}{\∂f_M}+{\mathrm{\λ}}_2\frac{\∂f_2}{\∂f_M}=0---\left(17\right)$

由公式(15)与(16)，解得

$\begin{array}{c}{\mathrm{\λ}}_1=-\frac{\∂J/\∂P}{\∂f_1/\∂P}\\ {\mathrm{\λ}}_2=-\frac{T_M}{2\mathrm{\π}}{\mathrm{\λ}}_1=\frac{T_M}{2\mathrm{\π}}\frac{\∂J/\∂P}{\∂f_1/\∂P}\end{array}---\left(18\right)$

将公式(18)代入公式(17)，得到等式

$f_3\left(P,f_M\right)=\frac{\∂J}{\∂f_M}-\frac{\∂J/\∂P}{\∂f_1/\∂P}\frac{\∂f_1}{\∂f_M}+\frac{T_M}{2\mathrm{\π}}\frac{\∂J/\∂P}{\∂f_1/\∂P}\frac{\∂f_2}{\∂f_M}---\left(19\right)$

所以，KKT条件最终转化为三个等式约束

$\begin{array}{c}{f}_1\left(t,P,f_M\right)=0\\ f_2(t,f_M)=0\\ f_3\left(P,f_M\right)=0\end{array}---\left(20\right)$

设F(X)＝[f₁ f₂ f₃]^T (21)

则方程组(21)的解就是拦截变轨最小速度修正问题的最优解；

步骤(2)中，所述公式(22)为方程组F(X)的Jacobi矩阵具体为：

$F^{\′}\left(X\right)=\left[\begin{array}{ccc}\frac{\∂f_1}{\∂t}& \frac{\∂f_1}{\∂P}& \frac{\∂f_1}{\∂f_M}\\ \frac{\∂f_2}{\∂t}& 0& \frac{\∂f_2}{\∂f_M}\\ 0& \frac{\∂f_3}{\∂P}& \frac{\∂f_3}{\∂f_M}\end{array}\right]---\left(22\right)$

其中，因为f₂中不显含P，f₃中不显含t，所以则自变量X的迭代公式可以写为：

X_K+1＝X_K-[F′(X_K)]^-1·F(X_K) (23)

为提高数值计算的稳定性，公式(22)可以转化为公式(24)，

所述公式(24)为：

其中，对角阵u和v为步骤(1)所述的系数矩阵；

所述公式(25)为：

2.如权利要求1所述的方法，其特征在于，所述公式(22)中，各项的解析表达式如下：

因为

$\begin{array}{c}\frac{\∂f_3}{\∂P}=\frac{{\∂}^{2}J}{\∂P\∂f_M}+\frac{\frac{{\∂}^{2}J}{\∂P^2}\·\frac{\∂f_1}{\∂P}-\frac{\∂J}{\∂P}\·\frac{{\∂}^2{f}_1}{\∂P^2}}{{\left(\frac{\∂f_1}{\∂P}\right)}^2}\·\left(\frac{T_M}{2\mathrm{\π}}\frac{\∂f_2}{\∂f_M}-\frac{\∂f_1}{\∂f_M}\right)-\frac{\∂J/\∂P}{\∂f_1/\∂P}\·\frac{{\∂}^2{f}_1}{\∂P\∂f_M}\\ \frac{\∂f_3}{\∂f_M}=\frac{{\∂}^{2}J}{\∂f_M^2}+\frac{\frac{{\∂}^{2}J}{\∂P\∂f_M}\·\frac{\∂f_1}{\∂P}-\frac{\∂J}{\∂P}\·\frac{{\∂}^2{f}_1}{\∂P\∂f_M}}{{\left(\frac{\∂f_1}{\∂P}\right)}^2}\·\left(\frac{T_M}{2\mathrm{\π}}\frac{\∂f_2}{\∂f_M}-\frac{\∂f_1}{\∂f_M}\right)+\frac{\∂J/\∂P}{\∂f_1/\∂P}\·\left(\frac{T_M}{2\mathrm{\π}}\frac{{\∂}^2{f}_2}{\∂f_M^2}-\frac{{\∂}^2{f}_1}{\∂f_M^2}\right)\end{array}---\left(26\right)$

并且，显然

$\begin{array}{c}\frac{\∂f_1}{\∂t}=-1\\ \frac{\∂f_2}{\∂t}=-\frac{2\mathrm{\π}}{T_M}\end{array}---\left(27\right)$

所以，只需获得所述J、f₁、f₂对P、f_M的一阶偏导，二阶纯偏导，以及二阶混合偏导，即可获得公式(22)的解析表达式；

将矢量投影在大地直角坐标系，并约定

$\left[\begin{array}{c}{h}_{2x}\\ h_{2y}\\ h_{2z}\end{array}\right]=\left[\begin{array}{c}cos(f_M+w)cos\mathrm{\Ω}-sin(f_M+w)\cos isin\mathrm{\Ω}\\ cos(f_M+w)sin\mathrm{\Ω}+sin(f_M+w)\cos icos\mathrm{\Ω}\\ sin(f_M+w)\sin i\end{array}\right]---\left(28\right)$

则

$v_1=\frac{r_2-{\mathrm{fr}}_1}{g}=\frac{1}{g}\{\left[\begin{array}{c}{h}_{2x}\\ h_{2y}\\ h_{2z}\end{array}\right]\·r_2-\left[\begin{array}{c}{r}_{1x}\\ r_{1y}\\ r_{1z}\end{array}\right]\·f\}=\left[\begin{array}{c}{v}_{1x}\\ v_{1y}\\ v_{1z}\end{array}\right]---\left(29\right)$

首先，给出J对P、f_M的一阶偏导，二阶纯偏导，以及二阶混合偏导，

设

$J_{Px1}=\frac{r_{2x}\left(f_M\right)}{g(P,f_M)}-r_{1x}\frac{f(P,f_M)}{g(P,f_M)}-v_{0x};J_{Px2}=\frac{r_{2x}\left(f_M\right)-r_{1x}\[2-f(P,f_M)\]}{2\·P\·g(P,f_M)};$

$J_{Py1}=\frac{r_{2y}\left(f_M\right)}{g(P,f_M)}-r_{1y}\frac{f(P,f_M)}{g(P,f_M)}-v_{0y};J_{Py2}=\frac{r_{2y}\left(f_M\right)-r_{1y}\[2-f(P,f_M)\]}{2\·P\·g(P,f_M)};$

$J_{Pz1}=\frac{r_{2z}\left(f_M\right)}{g(P,f_M)}-r_{1z}\frac{f(P,f_M)}{g(P,f_M)}-v_{0z};J_{Pz2}=\frac{r_{2z}\left(f_M\right)-r_{1z}\[2-f(P,f_M)\]}{2\·P\·g(P,f_M)};$

r_1h2＝r_1xh_2x+r_1yh_2y+r_1zh_2z；

$s_{r12h2}=\sqrt{r_1^2-{(r_{1x}{h}_{2x}+r_{1y}{h}_{2y}+r_{1z}{h}_{2z})}^2}$

则

$\frac{\∂J}{\∂P}=2\·(J_{Px1}\·J_{Px2}+J_{Py1}\·J_{Py2}+J_{Pz1}\·J_{Pz2})---\left(30\right)$

因为

$\frac{\∂r_2}{\∂f_M}=\frac{r_2\·e_M\·sin\left(f_M\right)}{1+e_M\·\cos \left(f_M\right)}$

$\left[\begin{array}{c}\frac{\∂h_{2x}}{\∂f_M}\\ \frac{\∂h_{2y}}{\∂f_M}\\ \frac{\∂h_{2z}}{\∂f_M}\end{array}\right]=\left[\begin{array}{c}-\sin \left(f_M+w\right)\cos \mathrm{\Ω}-\cos \left(f_M+w\right)\cos i\sin \mathrm{\Ω}\\ -\sin \left(f_M+w\right)\sin \mathrm{\Ω}+\cos \left(f_M+w\right)\cos i\cos \mathrm{\Ω}\\ \cos \left(f_M+w\right)\sin i\end{array}\right]$

$\frac{\∂g}{\∂f_M}=\frac{1}{\sqrt{\mathrm{\μ}P}}\·\[\frac{\∂r_2}{\∂f_M}\·s_{r12h2}-\frac{r_2\·r_{1ph2}\·r_{1h2}}{s_{r12h2}}\]$

$\frac{\∂f}{\∂f_M}=\frac{1}{P}\[\frac{\∂r_2}{\∂f_M}(cos\mathrm{\Δ}\mathrm{\ν}-1)+\frac{r_2\·r1ph2}{r_1}\]$

$\frac{\∂}{\∂f_M}\frac{f(P,f_M)}{g(P,f_M)}=\frac{\frac{\∂f}{\∂f_M}\·g-f\·\frac{\∂g}{\∂f_M}}{g^2}$

所以，设

$\frac{\∂}{\∂f_M}\frac{r_{2x}\left(f_M\right)}{g(P,f_M)}=\frac{(\frac{\∂r_2}{\∂f_M}\·h_{2x}+r_2\·\frac{\∂h_{2x}}{\∂f_M})g-r_2\·h_{2x}\·\frac{\∂g(P,f_M)}{\∂f_M}}{g^2(P,f_M)};J_{f_{M}x2}=\frac{\∂}{\∂f_M}\frac{r_{2x}\left(f_M\right)}{g(P,f_M)}-r_{1x}\·\frac{\∂}{\∂f_M}\frac{f(P,f_M)}{g(P,f_M)}$

$\frac{\∂}{\∂f_M}\frac{r_{2y}\left(f_M\right)}{g\left(P,f_M\right)}=\frac{\left(\frac{\∂r_2}{\∂f_M}\·h_{2y}+r_2\·\frac{\∂h_{2y}}{\∂f_M}\right)g-r_2\·h_{2y}\·\frac{\∂g\left(P,f_M\right)}{\∂f_M}}{g^2\left(P,f_M\right)};J_{f_{M}y2}=\frac{\∂}{\∂f_M}\frac{r_{2y}\left(f_M\right)}{g\left(P,f_M\right)}-r_{1y}\·\frac{\∂}{\∂f_M}\frac{f\left(P,f_M\right)}{g\left(P,f_M\right)}$

$\frac{\∂}{\∂f_M}\frac{r_{2z}\left(f_M\right)}{g\left(P,f_M\right)}=\frac{\left(\frac{\∂r_2}{\∂f_M}\·h_{2z}+r_2\·\frac{\∂h_{2z}}{\∂f_M}\right)g-r_2\·h_{2z}\·\frac{\∂g\left(P,f_M\right)}{\∂f_M}}{g^2\left(P,f_M\right)};J_{f_{M}z2}=\frac{\∂}{\∂f_M}\frac{r_{2z}\left(f_M\right)}{g\left(P,f_M\right)}-r_{1z}\·\frac{\∂}{\∂f_M}\frac{f\left(P,f_M\right)}{g\left(P,f_M\right)}$

则

$\frac{\∂J}{\∂f_M}=2\·(J_{Px1}\·J_{f_{M}x2}+J_{Py1}\·J_{f_{M}y2}+J_{Pz1}\·J_{f_{M}z2})---\left(31\right)$

$\frac{{\∂}^{2}J}{\∂P^2}=-\frac{1}{2P}\frac{\∂J}{\∂P}+2\·(J_{Px2}^2+J_{Py2}^2+J_{Pz2}^2)+\frac{1}{Pg}\frac{\∂f}{\∂P}\·(J_{Px1}\·r_{1x}+J_{Py1}\·r_{1y}+J_{Pz1}\·r_{1z})---\left(32\right)$

因为

$\frac{{\∂}^2}{\∂P\∂f_M}\frac{f\left(P,f_M\right)}{g\left(P,f_M\right)}=-\frac{1}{2P}\[\frac{2}{g^2}\frac{\∂}{\∂f_M}g\left(P,f_M\right)+\frac{\∂}{\∂f_M}\frac{f\left(P,f_M\right)}{g\left(P,f_M\right)}\]$

所以

$\frac{{\∂}^{2}J}{\∂P\∂f_M}=2\·\left\{\begin{array}{c}{J}_{Px2}\·J_{f_{M}x2}+J_{Px1}\·\[\frac{1}{2P}\·\frac{\∂}{\∂f_M}\frac{r_{2x}\left(f_M\right)}{g(P,f_M)}-r_{1x}\frac{{\∂}^2}{\∂P\∂f_M}\frac{f(P,f_M)}{g(P,f_M)}\]+\\ J_{Py2}\·J_{f_{M}y2}+J_{Py1}\·\[\frac{1}{2P}\·\frac{\∂}{\∂f_M}\frac{r_{2y}\left(f_M\right)}{g(P,f_M)}-r_{1y}\frac{{\∂}^2}{\∂P\∂f_M}\frac{f(P,f_M)}{g(P,f_M)}\]+\\ J_{Pz2}\·J_{f_{M}z2}+J_{Pz1}\·\[\frac{1}{2P}\·\frac{\∂}{\∂f_M}\frac{r_{2z}\left(f_M\right)}{g(P,f_M)}-r_{1z}\frac{{\∂}^2}{\∂P\∂f_M}\frac{f(P,f_M)}{g(P,f_M)}\]\end{array}\right\}---\left(33\right)$

因为

$\frac{{\∂}^2{r}_2}{\∂f_M^2}=\frac{2e^2-e^2{\cos }^2\left(f_M\right)+e\cos \left(f_M\right)}{{\[1+e\cos \left(f_M\right)\]}^2}{r}_2$

${\left[\begin{array}{ccc}\frac{{\∂}^2{h}_{2x}}{\∂f_M^2}& \frac{{\∂}^2{h}_{2y}}{\∂f_M^2}& \frac{{\∂}^2{h}_{2z}}{\∂f_M^2}\end{array}\right]}^T=-{\left[\begin{array}{ccc}{h}_{2x}& h_{2y}& h_{2z}\end{array}\right]}^T$

$\frac{{\∂}^{2}g}{\∂f_M^2}=\frac{1}{\sqrt{\mathrm{\μ}P}}\[\frac{{\∂}^2{r}_2}{\∂f_M^2}{s}_{r12h2}-2\·\frac{\∂r_2}{\∂f_M}\frac{r_{1ph2}\·r_{1h2}}{s_{r12h2}}-r2\·\frac{\left(r_{1ph2}^2-r_{1h2}^2\right)\·r_1^2+r_{1h2}^4}{s_{r12h2}^3}\]$

$\frac{{\∂}^{2}f}{\∂f_M^2}=\frac{1}{r_{1}P}\[2\·\frac{\∂r_2}{\∂f_M}\·r_{1ph2}-r_2\·r_{1h2}-\frac{{\∂}^2{r}_2}{\∂f_M^2}\·(r_1-r_{1h2})\]$

$\frac{{\∂}^2}{\∂f_M^2}\frac{f(P,f_M)}{g(P,f_M)}=\frac{\frac{{\∂}^{2}f}{\∂f_M^2}g-f\frac{{\∂}^{2}g}{\∂f_M^2}-2\frac{\∂f}{\∂f_M}\frac{\∂g}{\∂f_M}+\frac{2f}{g}{\left(\frac{\∂g}{\∂f_M}\right)}^2}{g^2}$

所以，设

$\frac{{\∂}^2}{\∂f_M^2}\frac{r_2\·h_{2x}}{g(P,f_M)}=\frac{(\frac{{\∂}^2{r}_2}{\∂f_M^2}\·h_{2x}+2\frac{\∂r_2}{\∂f_M}\frac{\∂h_{2x}}{\∂f_M}-r_2{h}_{2x})\·g-r_2\·h_{2x}\·\frac{{\∂}^{2}g}{\∂f_M^2}-\frac{\∂}{\∂f_M}\frac{r_2\·h_{2x}}{g(P,f_M)}\·2g\·\frac{\∂g}{\∂f_M}}{g^2(P,f_M)}$

$\frac{{\∂}^2}{\∂f_M^2}\frac{r_2\·h_{2y}}{g(P,f_M)}=\frac{(\frac{{\∂}^2{r}_2}{\∂f_M^2}\·h_{2y}+2\frac{\∂r_2}{\∂f_M}\frac{\∂h_{2y}}{\∂f_M}-r_2{h}_{2y})\·g-r_2\·h_{2y}\·\frac{{\∂}^{2}g}{\∂f_M^2}-\frac{\∂}{\∂f_M}\frac{r_2\·h_{2y}}{g(P,f_M)}\·2g\·\frac{\∂g}{\∂f_M}}{g^2(P,f_M)}$

$\frac{{\∂}^2}{\∂f_M^2}\frac{r_2\·h_{2z}}{g(P,f_M)}=\frac{(\frac{{\∂}^2{r}_2}{\∂f_M^2}\·h_{2z}+2\frac{\∂r_2}{\∂f_M}\frac{\∂h_{2z}}{\∂f_M}-r_2{h}_{2z})\·g-r_2\·h_{2z}\·\frac{{\∂}^{2}g}{\∂f_M^2}-\frac{\∂}{\∂f_M}\frac{r_2\·h_{2z}}{g(P,f_M)}\·2g\·\frac{\∂g}{\∂f_M}}{g^2(P,f_M)}$

则

其次，给出f₁对P、f_M的一阶偏导，二阶纯偏导，以及二阶混合偏导；因为

χ₀＝(2m-l²)P²+2klP-k²；$\frac{\∂}{\∂P}\frac{1}{aP}=\frac{2k}{m}\frac{1}{P^3}-\frac{2l}{m}\frac{1}{P^2}$

$\frac{\∂a}{\∂P}=\frac{mk\[\left(l^2-2m\right)P^2-k^2\]}{{\mathrm{\χ}}_0^2};\frac{{\∂}^{2}a}{\∂P^2}=2mk\frac{{\left(2m-l^2\right)}^2{P}^3+3\left(2m-l^2\right){\mathrm{Pk}}^2+2k^{3}l}{{\mathrm{\χ}}_0^3}$

并且，设

${\mathrm{\χ}}_1=1-\frac{r_1}{a}(1-f);{\mathrm{\χ}}_2=\sqrt{1-{\mathrm{\χ}}_1^2};{\mathrm{\χ}}_3=r_1{r}_{2}tan\frac{\mathrm{\Δ}\mathrm{\ν}}{2}\·\frac{1-cos\mathrm{\Δ}\mathrm{\ν}}{P};$

${\mathrm{\χ}}_4=r_1{r}_2\tan \frac{\mathrm{\Δ}\mathrm{\ν}}{2}\left(\frac{1-\cos \mathrm{\Δ}\mathrm{\ν}}{P}-\frac{1}{r_1}-\frac{1}{r_2}\right);\frac{{\∂}^2}{\∂P^2}\frac{1}{aP}=\frac{4l}{m}\frac{1}{P^3}-\frac{6k}{m}\frac{1}{P^4};$

${\mathrm{\φ}}_1=\arccos \[1-\frac{r_1}{a}(1-f(P,f_M\left)\right)\]+\frac{{\mathrm{\χ}}_4}{\sqrt{aP}};$

${\mathrm{\φ}}_2=\frac{r_1{r}_2(1-cos\mathrm{\Δ}\mathrm{\ν})}{{\mathrm{\χ}}_2}\·\frac{\∂}{\∂P}\frac{1}{aP}-\frac{{\mathrm{\χ}}_3}{P}\·\sqrt{\frac{1}{ap}}+{\mathrm{\χ}}_4\·\frac{\sqrt{aP}}{2}\·\frac{\∂}{\∂P}\frac{1}{aP};$

$\begin{array}{c}\frac{\∂{\mathrm{\φ}}_2}{\∂P}=k\·\frac{\frac{{\∂}^2}{\∂P^2}\frac{1}{aP}\·\[2-\frac{k}{aP}\]-{\left(\frac{\∂}{\∂P}\frac{1}{aP}\right)}^2\·\[aP-k\]}{{\[\frac{k}{aP}\]}^{\frac{1}{2}}\·{\[2-\frac{k}{aP}\]}^{\frac{3}{2}}}+\frac{\sqrt{aP}}{2}\·\frac{{\∂}^2}{\∂P^2}\frac{1}{aP}\\ +{\mathrm{\χ}}_3\·\left(\sqrt{\frac{1}{aP}}\frac{a\·\left(\frac{k}{m}\frac{1}{P}-\frac{l}{m}\right)-2}{P^2}+\frac{1}{2}\sqrt{\frac{a}{P}}\·\frac{\∂}{\∂P}\frac{1}{aP}\right)+{\mathrm{\χ}}_4\·\frac{1}{4}\frac{1}{\sqrt{aP}}\left(P\frac{\∂a}{\∂P}+a\right)\·\frac{\∂}{\∂P}\frac{1}{aP}\end{array}$

所以

设

χ₅₁＝χ₆₁·2P²+χ₆₂·2·P-2·r₂·χ₆·χ₆₃；

χ₆＝r₁-r_1h2；

${\mathrm{\χ}}_{61}=(r_{1h2}-r_2)\·\frac{\∂r_2}{\∂f_M}+r_2\·r_{1ph2};{\mathrm{\χ}}_{63}={\mathrm{\χ}}_6\·\frac{\∂r_2}{\∂f_M}-r_2\·r_{1ph2};$

${\mathrm{\χ}}_{64}=\frac{mk}{{\mathrm{\χ}}_0^2};{\mathrm{\χ}}_{65}={\mathrm{\χ}}_{63}-\frac{r_2\·{\mathrm{\χ}}_6}{a}\·\frac{\∂a}{\∂f_M};$

${\mathrm{\χ}}_7=\sqrt{\frac{r_1-r_{1h2}}{r_1+r_{1h2}}};\frac{\∂{\mathrm{\χ}}_7}{\∂f_M}=-\frac{r_1\·r_{1ph2}}{(r_1+r_{1h2})\·s_{r12h2}};$

${\mathrm{\χ}}_8=\frac{(r_1-r_{1h2})\·r_2}{P}-r_2-r_1;\frac{\∂{\mathrm{\χ}}_8}{\∂f_M}=\frac{{\mathrm{\χ}}_{63}}{P}-\frac{\∂r_2}{\∂f_M};\frac{\∂{\mathrm{\χ}}_8}{\∂P}=-\frac{(r_1-r_{1h2})\·r_2}{P^2};$

则，

$\begin{array}{c}\frac{\∂f_1}{\∂f_M}=\frac{\∂g}{\∂f_M}+\frac{3}{2}\sqrt{\frac{a}{\mathrm{\μ}}}\frac{\∂a}{\∂f_M}\arccos \[1-\frac{r_1}{a\left(P,f_M\right)}\left(1-f\left(P,f_M\right)\right)\]+\sqrt{\frac{a}{\mathrm{\μ}}}\frac{{\mathrm{\χ}}_{65}}{{\mathrm{P\χ}}_2}\\ +\frac{\frac{\∂a}{\∂f_M}\·{\mathrm{\χ}}_7\·{\mathrm{\χ}}_8+a\·\frac{\∂{\mathrm{\χ}}_7}{\∂f_M}{\mathrm{\χ}}_8+a\·{\mathrm{\χ}}_7\·\frac{\∂{\mathrm{\χ}}_8}{\∂f_M}}{\sqrt{\mathrm{\μ}P}}\end{array}---\left(36\right)$

$\frac{{\∂}^2{f}_1}{\∂P^2}=\frac{3}{4}\frac{g}{P^2}+\frac{3}{2}\[\frac{1}{2}\frac{1}{\sqrt{a\mathrm{\μ}}}{\left(\frac{\∂a}{\∂P}\right)}^2+\sqrt{\frac{a}{\mathrm{\μ}}}\frac{{\∂}^{2}a}{\∂P^2}\]\·{\mathrm{\φ}}_1+3\sqrt{\frac{a}{\mathrm{\μ}}}\frac{\∂a}{\∂P}\·{\mathrm{\φ}}_2+\sqrt{\frac{a^3}{\mathrm{\μ}}}\·\frac{\∂{\mathrm{\φ}}_2}{\∂P}---\left(37\right)$

因为

所以

$\begin{array}{c}\frac{{\∂}^2{f}_1}{\∂P\∂f_M}=-\frac{1}{2P}\frac{\∂}{\∂f_M}g\left(P,f_M\right)-\frac{1}{P}\frac{1}{{\mathrm{\χ}}_2}\sqrt{\frac{a}{\mathrm{\μ}}}\·{\mathrm{\χ}}_{63}\·\left(\frac{1}{P}+\frac{\∂{\mathrm{\χ}}_2}{\∂P}\frac{1}{{\mathrm{\χ}}_2}\right)\\ +\frac{1}{2a}\frac{\∂a}{\∂P}\·\left\{\frac{3}{2}\sqrt{\frac{a}{\mathrm{\μ}}}\frac{\∂a}{\∂f_M}\arccos \[1-\frac{r_1}{a\left(P,f_M\right)}\left(1-f\left(P,f_M\right)\right)\]+\sqrt{\frac{a}{\mathrm{\μ}}}\frac{{\mathrm{\χ}}_{65}}{{\mathrm{P\χ}}_2}\right\}\\ +\frac{3}{2}\sqrt{\frac{a}{\mathrm{\μ}}}\left\{\frac{{\∂}^{2}a}{\∂P\∂f_M}\arccos \[1-\frac{r_1}{a\left(P,f_M\right)}\left(1-f\left(P,f_M\right)\right)\]+\frac{\∂a}{\∂f_M}\frac{k}{{\mathrm{\χ}}_2}\·\frac{\∂}{\∂P}\frac{1}{aP}\right\}\\ -\sqrt{\frac{a}{\mathrm{\μ}}}{r}_2{\mathrm{\χ}}_6\[\left(\frac{\∂}{\∂P}\frac{1}{\∂P}\frac{\∂a}{\∂f_M}+\frac{1}{aP}\frac{{\∂}^{2}a}{\∂P\∂f_M}\right)\·\frac{1}{{\mathrm{\χ}}_2}-\frac{1}{{\mathrm{aP\χ}}_2^2}\frac{\∂a}{\∂f_M}\frac{\∂{\mathrm{\χ}}_2}{\∂P}\]\end{array}---\left(38\right)$

因为

${\mathrm{\χ}}_9=\frac{{\∂}^2{r}_2}{\∂f_M^2}{\mathrm{\χ}}_6-2\frac{\∂r_2}{\∂f_M}{r}_{1ph2}+r_2{r}_{1h2};{\mathrm{\χ}}_{10}=\frac{1}{P}\frac{{\mathrm{\χ}}_{65}}{{\mathrm{\χ}}_2};\frac{\∂{\mathrm{\χ}}_2}{\∂f_M}=\frac{{\mathrm{\χ}}_1}{{\mathrm{\χ}}_2}\·\frac{{\mathrm{\χ}}_{65}}{aP}$

$\frac{\∂{\mathrm{\χ}}_5}{\∂f_M}=\[2{\left(\frac{\∂r_2}{\∂f_M}\right)}^2+2r_2\frac{{\∂}^2{r}_2}{\∂f_M^2}\]\·s_{r12h2}^2-8r_2\frac{\∂r_2}{\∂f_M}{r}_{1h2}{r}_{1ph2}+2r_2^2(r_{1h2}^2-r_{1ph2}^2)$

$\begin{array}{c}\frac{\∂{\mathrm{\χ}}_{51}}{\∂f_M}=\[2r_{1ph2}\frac{\∂r_2}{\∂f_M}+\left(r_{1h2}-r_2\right)\·\frac{{\∂}^2{r}_2}{\∂f_M^2}-r_2\·r_{1h2}-{\left(\frac{\∂r_2}{\∂f_M}\right)}^2\]\·2P^2-2{\mathrm{\χ}}_{63}^2-2r_2{\mathrm{\χ}}_6{\mathrm{\χ}}_9\\ +\left\{\[2{\left(\frac{\∂r_2}{\∂f_M}\right)}^2+\left(r_1+2r_2\right)\·\frac{{\∂}^2{r}_2}{\∂f_M^2}\]\·{\mathrm{\χ}}_6-2\left(r_1+2r_2\right)\frac{\∂r_2}{\∂f_M}{r}_{1ph2}+\left(r_1{r}_2+r_2^2\right)r_{1h2}\right\}\·2P\end{array}$

$\frac{{\∂}^{2}a}{\∂f_M^2}=P\·(\frac{\∂{\mathrm{\χ}}_5}{\∂f_M}\·\frac{1}{{\mathrm{\χ}}_0}-{\mathrm{\χ}}_{64}\·\frac{\∂{\mathrm{\χ}}_{51}}{\∂f_M}-2\·\frac{{\mathrm{\χ}}_5{\mathrm{\χ}}_{51}}{{\mathrm{\χ}}_0^2}+2\frac{{\mathrm{\χ}}_{64}{\mathrm{\χ}}_{51}^2}{{\mathrm{\χ}}_0})$

$\frac{\∂{\mathrm{\χ}}_{10}}{\∂f_M}=\frac{1}{P}\·\frac{\{\frac{{\∂}^2{r}_2}{\∂f_M^2}\·{\mathrm{\χ}}_6-2\frac{\∂r_2}{\∂f_M}{r}_{1ph2}+r_2{r}_{1h2}-\[\frac{r_2}{a}\frac{{\∂}^{2}a}{\∂f_M^2}+\frac{1}{a}\frac{\∂a}{\∂f_M}\frac{\∂r_2}{\∂f_M}-\frac{r_2}{a^2}{\left(\frac{\∂a}{\∂f_M}\right)}^2\]\·{\mathrm{\χ}}_6+\frac{r_2}{a}\frac{\∂a}{\∂f_M}{r}_{1ph2}\}\·{\mathrm{\χ}}_2-{\mathrm{\χ}}_{65}\·\frac{\∂{\mathrm{\χ}}_2}{\∂f_M}}{{\mathrm{\χ}}_2^2}$

$\frac{{\∂}^2{\mathrm{\χ}}_7}{\∂f_M^2}=\frac{r_1{r}_{1h2}\·\[r_1+\left(r_{1x}{h}_{2x}+r_{1y}{h}_{2y}+r_{1z}{h}_{2z}\right)\]s_{r12h2}+r_1\·r_{1ph2}\·\left\{r_{1ph2}{s}_{r12h2}-\[r_1+\left(r_{1x}{h}_{2x}+r_{1y}{h}_{2y}+r_{1z}{h}_{2z}\right)\]\frac{r_{1h2}{r}_{1ph2}}{s_{r12h2}}\right\}}{{\[r_1+\left(r_{1x}{h}_{2x}+r_{1y}{h}_{2y}+r_{1z}{h}_{2z}\right)\]}^2{s}_{r12h2}^2}$

$\frac{{\∂}^2{\mathrm{\χ}}_8}{\∂f_M^2}=\frac{{\mathrm{\χ}}_9}{P}-\frac{{\∂}^2{r}_2}{\∂f_M^2}$

所以$\begin{array}{c}\frac{{\∂}^2{f}_1}{\∂f_M^2}=\frac{{\∂}^{2}g}{\∂f_M^2}+\frac{1}{2a}\frac{\∂a}{\∂f_M}\left\{\frac{3}{2}\sqrt{\frac{a}{\mathrm{\μ}}}\frac{\∂a}{\∂f_M}\arccos \[1-\frac{r_1}{a\left(P,f_M\right)}\left(1-f\left(P,f_M\right)\right)\]+\sqrt{\frac{a}{\mathrm{\μ}}}\frac{{\mathrm{\χ}}_{65}}{{\mathrm{P\χ}}_2}\right\}\\ +\sqrt{\frac{a}{\mathrm{\μ}}}\left\{\frac{3}{2}\frac{{\∂}^{2}a}{\∂f_M^2}\arccos \[1-\frac{r_1}{a\left(P,f_M\right)}\left(1-f\left(P,f_M\right)\right)\]-\frac{3}{2}\frac{\∂a}{\∂f_M}\frac{{\mathrm{\χ}}_{10}}{a}+\frac{\∂{\mathrm{\χ}}_{10}}{\∂f_M}\right\}\\ +\frac{\frac{{\∂}^{2}a}{\∂f_M^2}{\mathrm{\χ}}_7{\mathrm{\χ}}_8+a\frac{{\∂}^2{\mathrm{\χ}}_7}{\∂f_M^2}{\mathrm{\χ}}_8+{\mathrm{a\χ}}_7\frac{{\∂}^2{\mathrm{\χ}}_8}{\∂f_M^2}+2\frac{\∂a}{\∂f_M}\frac{\∂{\mathrm{\χ}}_7}{\∂f_M}{\mathrm{\χ}}_8+2\frac{\∂a}{\∂f_M}{\mathrm{\χ}}_7\frac{\∂{\mathrm{\χ}}_8}{\∂f_M}+2a\frac{\∂{\mathrm{\χ}}_7}{\∂f_M}\frac{\∂{\mathrm{\χ}}_8}{\∂f_M}}{\sqrt{\mathrm{\μ}P}}\end{array}---\left(39\right)$

最后，给出f₂对f_M的一阶偏导，二阶偏导；

因为

$\frac{\∂E}{\∂f_M}=\frac{2}{\sqrt{\frac{1+e_M}{1-e_M}}\[1+cos\left(f_M\right)\]+\sqrt{\frac{1-e_M}{1+e_M}}\[1-cos\left(f_M\right)\]}$

$\frac{{\∂}^{2}E}{\∂f_M^2}=\frac{2}{{\left\{\sqrt{\frac{1+e_M}{1-e_M}}\[1+\cos \left(f_M\right)\]+\sqrt{\frac{1-e_M}{1+e_M}}\[1-\cos \left(f_M\right)\]\right\}}^2}\sin \left(f_M\right)\·\left(\sqrt{\frac{1+e_M}{1-e_M}}-\sqrt{\frac{1-e_M}{1+e_M}}\right)$

所以

$\frac{\∂f_2}{\∂f_M}=\frac{\∂E}{\∂f_M}-e_{M}cos\left(E\right)\·\frac{\∂E}{\∂f_M}---\left(40\right)$

$\frac{{\∂}^2{f}_2}{\∂f_M^2}=\frac{{\∂}^{2}E}{\∂f_M^2}\[1-e_{M}cos\left(E\right)\]+e_{M}sin\left(E\right)\·{\left(\frac{\∂E}{\∂f_M}\right)}^2---\left(41\right)$

至此，由公式(30)至公式(41)，并结合公式(26)和公式(27)，获得公式(22)的解析表达式，算法完整执行。