Klassik mexanikada Liuvil dinamik tizimi aniq yechiladigan dinamik tizim boʻlib, unda kinetik energiya T va potensial energiya V umumlashgan koordinatalari q orqali quyidagicha ifodalanishi mumkin[ 1]
T
=
1
2
{
u
1
(
q
1
)
+
u
2
(
q
2
)
+
⋯
+
u
s
(
q
s
)
}
{
v
1
(
q
1
)
q
˙
1
2
+
v
2
(
q
2
)
q
˙
2
2
+
⋯
+
v
s
(
q
s
)
q
˙
s
2
}
{\displaystyle T={\frac {1}{2}}\left\{u_{1}(q_{1})+u_{2}(q_{2})+\cdots +u_{s}(q_{s})\right\}\left\{v_{1}(q_{1}){\dot {q}}_{1}^{2}+v_{2}(q_{2}){\dot {q}}_{2}^{2}+\cdots +v_{s}(q_{s}){\dot {q}}_{s}^{2}\right\}}
V
=
w
1
(
q
1
)
+
w
2
(
q
2
)
+
⋯
+
w
s
(
q
s
)
u
1
(
q
1
)
+
u
2
(
q
2
)
+
⋯
+
u
s
(
q
s
)
{\displaystyle V={\frac {w_{1}(q_{1})+w_{2}(q_{2})+\cdots +w_{s}(q_{s})}{u_{1}(q_{1})+u_{2}(q_{2})+\cdots +u_{s}(q_{s})}}}
Bu sistemaning yechimi ajraladigan integrallanuvchi tenglamalar toʻplamidan iborat
2
Y
d
t
=
d
φ
1
E
χ
1
−
ω
1
+
γ
1
=
d
φ
2
E
χ
2
−
ω
2
+
γ
2
=
⋯
=
d
φ
s
E
χ
s
−
ω
s
+
γ
s
{\displaystyle {\frac {\sqrt {2}}{Y}}\,dt={\frac {d\varphi _{1}}{\sqrt {E\chi _{1}-\omega _{1}+\gamma _{1}}}}={\frac {d\varphi _{2}}{\sqrt {E\chi _{2}-\omega _{2}+\gamma _{2}}}}=\cdots ={\frac {d\varphi _{s}}{\sqrt {E\chi _{s}-\omega _{s}+\gamma _{s}}}}}
Bu yerda E = T + V — saqlangan energiya va
γ
s
{\displaystyle \gamma _{s}}
q s dan φs ga oʻzgartirildi va u s va w s funksiyalari ularning oʻxshashlari χs va ωs bilan almashtirildi. Ushbu yechim Nyuton tortishish kuchi taʼsirida ikkita sobit yulduz atrofida kichik sayyoraning orbitasi kabi koʻplab ilovalarga ega. Liouville dinamik tizimi taniqli fransuz matematiki Jozef Liuvil nomi bilan atalgan bir nechta narsalardan biridir.
Bisentrik orbitalarga misol
tahrir
Klassik mexanikada Eylerning uch jismli muammosi zarrachaning har biri Nyuton tortishish kuchi yoki Kulon qonuni kabi teskari kvadrat kuch bilan zarrachani oʻziga tortadigan ikkita qoʻzgʻalmas markaz taʼsiri ostida tekislikdagi harakatini tasvirlaydi. Ikki markazli muammoga misollar orasida ikki sekin harakatlanuvchi yulduz atrofida harakatlanuvchi sayyora yoki ikkita musbat zaryadlangan yadroning elektr maydonida harakatlanuvchi elektron , masalan, vodorod molekulasi H2 ning birinchi ioni , yaʼni vodorod molekulyar ioni yoki H2 + kiradi. . Ikki diqqatga sazovor joyning kuchi teng boʻlishi shart emas; Shunday qilib, ikkita yulduzning massasi har xil yoki yadrolari ikki xil zaryadga ega boʻlishi mumkin.
Ruxsat etilgan tortishish markazlari x oʻqi boʻylab ± a da joylashgan boʻlsin. Harakatlanuvchi zarrachaning potentsial energiyasi bilan ifodalanadi
V
(
x
,
y
)
=
−
μ
1
(
x
−
a
)
2
+
y
2
−
μ
2
(
x
+
a
)
2
+
y
2
.
{\displaystyle V(x,y)={\frac {-\mu _{1}}{\sqrt {\left(x-a\right)^{2}+y^{2}}}}-{\frac {\mu _{2}}{\sqrt {\left(x+a\right)^{2}+y^{2}}}}.}
Ikki tortishish markazini ellipslar toʻplamining oʻchoqlari deb hisoblash mumkin. Agar biron bir markaz boʻlmasa, zarracha Kepler muammosining yechimi sifatida ushbu ellipslardan biri boʻylab harakatlanadi. Shuning uchun, Bonnet teoremasiga koʻra, bir xil ellipslar ikki markazli muammoning echimlari hisoblanadi.
Elliptik koordinatalar bilan tanishtirib,
x
=
a
cosh
ξ
cos
η
,
{\displaystyle x=a\cosh \xi \cos \eta ,}
y
=
a
sinh
ξ
sin
η
,
{\displaystyle y=a\sinh \xi \sin \eta ,}
potensial energiyani quyidagicha yozish mumkin
V
(
ξ
,
η
)
=
−
μ
1
a
(
cosh
ξ
−
cos
η
)
−
μ
2
a
(
cosh
ξ
+
cos
η
)
=
−
μ
1
(
cosh
ξ
+
cos
η
)
−
μ
2
(
cosh
ξ
−
cos
η
)
a
(
cosh
2
ξ
−
cos
2
η
)
,
{\displaystyle V(\xi ,\eta )={\frac {-\mu _{1}}{a\left(\cosh \xi -\cos \eta \right)}}-{\frac {\mu _{2}}{a\left(\cosh \xi +\cos \eta \right)}}={\frac {-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)}{a\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)}},}
va kinetik energiya sifatida
T
=
m
a
2
2
(
cosh
2
ξ
−
cos
2
η
)
(
ξ
˙
2
+
η
˙
2
)
.
{\displaystyle T={\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)\left({\dot {\xi }}^{2}+{\dot {\eta }}^{2}\right).}
Agar ξ va η mos ravishda φ1 va φ2 sifatida qabul qilinsa, bu Liouvil dinamik tizimidir; demak, Y funksiyasi quyidagiga teng:
Y
=
cosh
2
ξ
−
cos
2
η
{\displaystyle Y=\cosh ^{2}\xi -\cos ^{2}\eta }
va W funksiyasi quyidagiga teng:
W
=
−
μ
1
(
cosh
ξ
+
cos
η
)
−
μ
2
(
cosh
ξ
−
cos
η
)
{\displaystyle W=-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)}
Quyidagi Liouville dinamik tizimi uchun umumiy yechimdan foydalanib, bittasi olinadi
m
a
2
2
(
cosh
2
ξ
−
cos
2
η
)
2
ξ
˙
2
=
E
cosh
2
ξ
+
(
μ
1
+
μ
2
a
)
cosh
ξ
−
γ
{\displaystyle {\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)^{2}{\dot {\xi }}^{2}=E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }
m
a
2
2
(
cosh
2
ξ
−
cos
2
η
)
2
η
˙
2
=
−
E
cos
2
η
+
(
μ
1
−
μ
2
a
)
cos
η
+
γ
{\displaystyle {\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)^{2}{\dot {\eta }}^{2}=-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }
Formula boʻyicha u parametrini kiritish
d
u
=
d
ξ
E
cosh
2
ξ
+
(
μ
1
+
μ
2
a
)
cosh
ξ
−
γ
=
d
η
−
E
cos
2
η
+
(
μ
1
−
μ
2
a
)
cos
η
+
γ
,
{\displaystyle du={\frac {d\xi }{\sqrt {E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }}}={\frac {d\eta }{\sqrt {-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }}},}
parametrik yechimni beradi
u
=
∫
d
ξ
E
cosh
2
ξ
+
(
μ
1
+
μ
2
a
)
cosh
ξ
−
γ
=
∫
d
η
−
E
cos
2
η
+
(
μ
1
−
μ
2
a
)
cos
η
+
γ
.
{\displaystyle u=\int {\frac {d\xi }{\sqrt {E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }}}=\int {\frac {d\eta }{\sqrt {-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }}}.}
Bular elliptik integral boʻlgani uchun, ξ va η koordinatalarini u ning elliptik funksiyalari sifatida ifodalash mumkin.
Bisentrik muammo doimiy harakatga ega, yaʼni,
r
1
2
r
2
2
(
d
θ
1
d
t
)
(
d
θ
2
d
t
)
−
2
c
[
μ
1
cos
θ
1
+
μ
2
cos
θ
2
]
,
{\displaystyle r_{1}^{2}r_{2}^{2}\left({\frac {d\theta _{1}}{dt}}\right)\left({\frac {d\theta _{2}}{dt}}\right)-2c\left[\mu _{1}\cos \theta _{1}+\mu _{2}\cos \theta _{2}\right],}
undan muammoni oxirgi koʻpaytma usuli yordamida hal qilish mumkin.
v funksiyalarini bartaraf qilish uchun oʻzgaruvchilar ekvivalent toʻplamga oʻzgartiriladi
φ
r
=
∫
d
q
r
v
r
(
q
r
)
,
{\displaystyle \varphi _{r}=\int dq_{r}{\sqrt {v_{r}(q_{r})}},}
munosabatni berish
v
1
(
q
1
)
q
˙
1
2
+
v
2
(
q
2
)
q
˙
2
2
+
⋯
+
v
s
(
q
s
)
q
˙
s
2
=
φ
˙
1
2
+
φ
˙
2
2
+
⋯
+
φ
˙
s
2
=
F
,
{\displaystyle v_{1}(q_{1}){\dot {q}}_{1}^{2}+v_{2}(q_{2}){\dot {q}}_{2}^{2}+\cdots +v_{s}(q_{s}){\dot {q}}_{s}^{2}={\dot {\varphi }}_{1}^{2}+{\dot {\varphi }}_{2}^{2}+\cdots +{\dot {\varphi }}_{s}^{2}=F,}
yangi F oʻzgaruvchini belgilaydi. Yangi oʻzgaruvchilar yordamida u va w funksiyalarni χ va ω ekvivalent funksiyalar bilan ifodalash mumkin. χ funksiyalar yigʻindisini Y bilan belgilab,
Y
=
χ
1
(
φ
1
)
+
χ
2
(
φ
2
)
+
⋯
+
χ
s
(
φ
s
)
,
{\displaystyle Y=\chi _{1}(\varphi _{1})+\chi _{2}(\varphi _{2})+\cdots +\chi _{s}(\varphi _{s}),}
kinetik energiyani quyidagicha yozish mumkin
T
=
1
2
Y
F
.
{\displaystyle T={\frac {1}{2}}YF.}
Xuddi shunday, ō funksiyalarining yigʻindisini V bilan belgilab
W
=
ω
1
(
φ
1
)
+
ω
2
(
φ
2
)
+
⋯
+
ω
s
(
φ
s
)
,
{\displaystyle W=\omega _{1}(\varphi _{1})+\omega _{2}(\varphi _{2})+\cdots +\omega _{s}(\varphi _{s}),}
potentsial energiya V sifatida yozish mumkin
V
=
W
Y
.
{\displaystyle V={\frac {W}{Y}}.}
r chi oʻzgaruvchi uchun Lagrange tenglamasi
φ
r
{\displaystyle \varphi _{r}}
d
d
t
(
∂
T
∂
φ
˙
r
)
=
d
d
t
(
Y
φ
˙
r
)
=
1
2
F
∂
Y
∂
φ
r
−
∂
V
∂
φ
r
.
{\displaystyle {\frac {d}{dt}}\left({\frac {\partial T}{\partial {\dot {\varphi }}_{r}}}\right)={\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)={\frac {1}{2}}F{\frac {\partial Y}{\partial \varphi _{r}}}-{\frac {\partial V}{\partial \varphi _{r}}}.}
Ikkala tomonni koʻpaytirish
2
Y
φ
˙
r
{\displaystyle 2Y{\dot {\varphi }}_{r}}
T = YF munosabatini qayta tartibga solish va undan foydalanish tenglamani beradi.
2
Y
φ
˙
r
d
d
t
(
Y
φ
˙
r
)
=
2
T
φ
˙
r
∂
Y
∂
φ
r
−
2
Y
φ
˙
r
∂
V
∂
φ
r
=
2
φ
˙
r
∂
∂
φ
r
[
(
E
−
V
)
Y
]
,
{\displaystyle 2Y{\dot {\varphi }}_{r}{\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)=2T{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2Y{\dot {\varphi }}_{r}{\frac {\partial V}{\partial \varphi _{r}}}=2{\dot {\varphi }}_{r}{\frac {\partial }{\partial \varphi _{r}}}\left[(E-V)Y\right],}
deb yozilishi mumkin
d
d
t
(
Y
2
φ
˙
r
2
)
=
2
E
φ
˙
r
∂
Y
∂
φ
r
−
2
φ
˙
r
∂
W
∂
φ
r
=
2
E
φ
˙
r
d
χ
r
d
φ
r
−
2
φ
˙
r
d
ω
r
d
φ
r
,
{\displaystyle {\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2E{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2{\dot {\varphi }}_{r}{\frac {\partial W}{\partial \varphi _{r}}}=2E{\dot {\varphi }}_{r}{\frac {d\chi _{r}}{d\varphi _{r}}}-2{\dot {\varphi }}_{r}{\frac {d\omega _{r}}{d\varphi _{r}}},}
bu yerda E = T + V — (saqlangan) umumiy energiya. Bundan quyidagi kelib chiqadi:
d
d
t
(
Y
2
φ
˙
r
2
)
=
2
d
d
t
(
E
χ
r
−
ω
r
)
,
{\displaystyle {\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2{\frac {d}{dt}}\left(E\chi _{r}-\omega _{r}\right),}
hosil qilish uchun bir marta birlashtirilishi mumkin
1
2
Y
2
φ
˙
r
2
=
E
χ
r
−
ω
r
+
γ
r
,
{\displaystyle {\frac {1}{2}}Y^{2}{\dot {\varphi }}_{r}^{2}=E\chi _{r}-\omega _{r}+\gamma _{r},}
bu yerda
γ
r
{\displaystyle \gamma _{r}}
∑
r
=
1
s
γ
r
=
0.
{\displaystyle \sum _{r=1}^{s}\gamma _{r}=0.}
Inverting, kvadrat ildizni olish va oʻzgaruvchilarni ajratish ajraladigan integrallashuvchi tenglamalar toʻplamini beradi:
2
Y
d
t
=
d
φ
1
E
χ
1
−
ω
1
+
γ
1
=
d
φ
2
E
χ
2
−
ω
2
+
γ
2
=
⋯
=
d
φ
s
E
χ
s
−
ω
s
+
γ
s
.
{\displaystyle {\frac {\sqrt {2}}{Y}}dt={\frac {d\varphi _{1}}{\sqrt {E\chi _{1}-\omega _{1}+\gamma _{1}}}}={\frac {d\varphi _{2}}{\sqrt {E\chi _{2}-\omega _{2}+\gamma _{2}}}}=\cdots ={\frac {d\varphi _{s}}{\sqrt {E\chi _{s}-\omega _{s}+\gamma _{s}}}}.}
![{\displaystyle r_{1}^{2}r_{2}^{2}\left({\frac {d\theta _{1}}{dt}}\right)\left({\frac {d\theta _{2}}{dt}}\right)-2c\left[\mu _{1}\cos \theta _{1}+\mu _{2}\cos \theta _{2}\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b031f911bc9a8c9fd81800f52a5532e15fc921a)
![{\displaystyle 2Y{\dot {\varphi }}_{r}{\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)=2T{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2Y{\dot {\varphi }}_{r}{\frac {\partial V}{\partial \varphi _{r}}}=2{\dot {\varphi }}_{r}{\frac {\partial }{\partial \varphi _{r}}}\left[(E-V)Y\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1addebe0acd83dc477577658fa1c849a6171128d)
