w*****x
发帖数: 116 | 1 帮我看一下这个方程组， 是藕合的．不知道有高手吗／
谢谢了， | x******n
发帖数: 24 | 2 First you may use F=f/\rho^k, G=g/\rho^k, you can obtain the new equation as
dF/d\rho=(a/\rho+a\lambda/\rho^2)G
dG/d\rho=-(a/\rho+a\lambda/\rho^2)F
which you may get F^2+G^2=C, then, assume
F(\rho)=\sqrt{C}\sin\theta(\rho),
G(\rho)=\sqrt{C}\cos\theta(\rho)
substitute into above equations, you reduce to solve the ODE on \theta(\rho)
d\theta/d\rho=a/\rho+a\lambda/\rho^2
then you can get the solutions...
【在 w*****x 的大作中提到】
: 帮我看一下这个方程组， 是藕合的．不知道有高手吗／
: 谢谢了，
| w*****x
发帖数: 116 | 3 谢谢, 你的解答我看懂了,上面那个也可以用 X=f-g代换.
此外, 我计算对正电子作用,得到相似的方程组,
原来第二式中变为正号, 你的方法就不能用了.
df/d\rho-a*f/\rho=(a/\rho+a\lambda/\rho^2)g
dg/d\rho+a*g/\rho=-(a/\rho+a\lambda/\rho^2)f
而这里第二式的+a*g/\rho为正号.
请教中,谢谢. | x******n
发帖数: 24 | 4 For these eqs, same ideas may apply to, althouth it is unlikely to get a
nalytic solution:
first let F=f*\rho^{-a}, G=g*\rho^a, from eqs, you may get
\rho^{2a}(F^2)'+\rho^{-2a}(G^2)'=0 (*)
which may give you two eqs as:
(G^2)'+(\rho^{4a}F^2)'=4a\rho^{4a-1}F^2
(F^2)'+(\rho^{-4a}G^2)'=-4a\rho^{-4a-1}G^2
integrate these two eqs, the second one will give you
f^2+g^2
the first one will give you an integral ineq related to
\rho^{4a-1}F^2 after get rid of positive term G^2, by Gromell's ineq, y
【在 w*****x 的大作中提到】
: 谢谢, 你的解答我看懂了,上面那个也可以用 X=f-g代换.
: 此外, 我计算对正电子作用,得到相似的方程组,
: 原来第二式中变为正号, 你的方法就不能用了.
: df/d\rho-a*f/\rho=(a/\rho+a\lambda/\rho^2)g
: dg/d\rho+a*g/\rho=-(a/\rho+a\lambda/\rho^2)f
: 而这里第二式的+a*g/\rho为正号.
: 请教中,谢谢.
| x******n
发帖数: 24 | 5 For these eqs, same ideas may apply to, althouth it is unlikely to get
analytic solution:
first let F=f*\rho^{-a}, G=g*\rho^a, from eqs, you may get
\rho^{2a}(F^2)'+\rho^{-2a}(G^2)'=0 (*)
which may give you two eqs as:
(G^2)'+(\rho^{4a}F^2)'=4a\rho^{4a-1}F^2
(F^2)'+(\rho^{-4a}G^2)'=-4a\rho^{-4a-1}G^2
integrate these two eqs, and notice the definations of F and g, you will get
f^2+g^2 is bounded by two functions [f^2(1)+g^2(1)]\rho^{2a} and
[f^2(1)+g^2(1)]\rho^{-2a} (1)
Secondly, if you multipl
【在 x******n 的大作中提到】
: For these eqs, same ideas may apply to, althouth it is unlikely to get a
: nalytic solution:
: first let F=f*\rho^{-a}, G=g*\rho^a, from eqs, you may get
: \rho^{2a}(F^2)'+\rho^{-2a}(G^2)'=0 (*)
: which may give you two eqs as:
: (G^2)'+(\rho^{4a}F^2)'=4a\rho^{4a-1}F^2
: (F^2)'+(\rho^{-4a}G^2)'=-4a\rho^{-4a-1}G^2
: integrate these two eqs, the second one will give you
: f^2+g^2: the first one will give you an integral ineq related to
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