常微分方程
Posted eufisky
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了常微分方程相关的知识,希望对你有一定的参考价值。
利用首次积分法(First Integral)求解对称形式的常微分方程组:
[frac{{
m,d}x}{-x+y+z}=frac{{
m,d}y}{x-y+z}=frac{{
m,d}z}{x+y-z}]
[frac{{ m,d}x}{-x^2+y^2+z^2}=frac{{ m,d}y}{x^2-y^2+z^2}=frac{{ m,d}z}{x^2+y^2-z^2}]
egin{align*}
&&frac{{
m,d}x}{-x+y+z}&=frac{{
m,d}y}{x-y+z}\
&Rightarrow&frac{{
m,d}left(x-y
ight)}{-2left(x-y
ight)}&=frac{{
m,d}left(x+y+z
ight)}{x+y+z}\
&Rightarrow&frac{{
m,d}left(x-y
ight)}{left(x-y
ight)}&=-frac{2{
m,d}left(x+y+z
ight)}{left(x+y+z
ight)}\
&Rightarrow&lnleft|x-y
ight|&=-2lnleft|x+y+z
ight|+ln|C_1|\
&Rightarrow&left(x-y
ight)left(x+y+z
ight)^2&=C_1\
end{align*}
同理
In like manner
egin{align*}
&&frac{{
m,d}y}{x-y+z}&=frac{{
m,d}z}{x+y-z}\
&Rightarrow&frac{{
m,d}left(y-z
ight)}{-2left(y-z
ight)}&=frac{{
m,d}left(x+y+z
ight)}{x+y+z}\
&Rightarrow&frac{{
m,d}left(y-z
ight)}{left(y-z
ight)}&=-frac{2{
m,d}left(x+y+z
ight)}{left(x+y+z
ight)}\
&Rightarrow&lnleft|y-z
ight|&=-2lnleft|x+y+z
ight|+ln|C_2|\
&Rightarrow&left(y-z
ight)left(x+y+z
ight)^2&=C_2\
end{align*}
egin{align*}
left{ egin{aligned}
frac{ ext{d}x}{-x+y+z}&=frac{ ext{d}y}{x-y+z}\
frac{ ext{d}y}{x-y+z}&=frac{ ext{d}z}{x+y-z}\
end{aligned}
ight. Rightarrow left{ egin{aligned}
left( x-y
ight) left( x+y+z
ight) ^2&=C_1\
left( y-z
ight) left( x+y+z
ight) ^2&=C_2\
end{aligned}
ight.
end{align*}
egin{align*}
left{ egin{aligned}
frac{ ext{d}x}{-x+y+a}&=frac{ ext{d}y}{x-y+a}\
frac{ ext{d}y}{b-y+z}&=frac{ ext{d}z}{b+y-z}\
end{aligned}
ight. Rightarrow left{ egin{aligned}
x-y&=C_1e^{-frac{x+y}{z}}\
y-z&=C_2e^{-frac{y+z}{x}}\
end{aligned}
ight.
end{align*}
egin{align*}
left{ egin{aligned}
frac{ ext{d}x}{-x^2+y^2+a^2}&=frac{ ext{d}y}{x^2-y^2+a^2}\
frac{ ext{d}y}{b^2-y^2+z^2}&=frac{ ext{d}z}{b^2+y^2-z^2}\
end{aligned}
ight. Rightarrow left{ egin{aligned}
x-y&=C_1e^{-frac{left( x+y
ight) ^2}{2a^2}}\
y-z&=C_2e^{-frac{left( y+z
ight) ^2}{2b^2}}\
end{aligned}
ight.
end{align*}
另外:
egin{align*} &&frac{{
m,d}x}{-x^3+y^3+a^3}&=frac{{
m,d}y}{x^3-y^3+a^3}\ &Rightarrow&frac{{
m,d}left(x-y
ight)}{-2left(x^3-y^3
ight)}&=frac{{
m,d}left(x+y
ight)}{2a^3}\ &Rightarrow&frac{{
m,d}left(x-y
ight)}{{
m,d}left(x+y
ight)}&=frac{-2left(x^3-y^3
ight)}{2a^3}\ &Rightarrow&frac{{
m,d}left(x-y
ight)}{{
m,d}left(x+y
ight)}&=frac{-left(x-y
ight)left(x^2+xy+y^2
ight)}{a^2}\ &Rightarrow&frac{left(x-y
ight){
m,d}left(x-y
ight)}{{
m,d}left(x+y
ight)}&=frac{-left(x-y
ight)^2left(3left(x+y
ight)^2+left(x-y
ight)^2
ight)}{a^2}\ &Rightarrow&frac{{
m,d}left(left(x-y
ight)^2
ight)}{{
m,d}left(x+y
ight)}&=-frac{3left(x+y
ight)^2}{a^2}left(x-y
ight)^2-frac{1}{a^2}left(x-y
ight)^4\ &Rightarrow&frac{{
m,d}u}{{
m,d}v}&=-frac{3v^2}{a^2}u-frac{1}{a^2}u^2\ &Rightarrow&\ &Rightarrow&\ end{align*}
%http://kuing.orzweb.net/viewthread.php?tid=5752&tdsourcetag=s_pctim_aiomsg
[y‘‘‘=frac{3left(y‘‘
ight)^2+x!cdot!left(y‘
ight)^5}{y‘}]
egin{align*}
frac{{
m,d}x}{{
m,d}y}&=frac{1}{y‘}\
frac{{
m,d}^2x}{{
m,d}y^2}&=frac{{
m,d}}{{
m,d}y}left(frac{{
m,d}x}{{
m,d}y}
ight)
=frac{{
m,d}}{{
m,d}y}left(frac{1}{y‘}
ight)
=frac{{
m,d}}{{
m,d}x}left(frac{1}{y‘}
ight)frac{{
m,d}x}{{
m,d}y}\
&=-frac{y‘‘}{left(y‘
ight)^2}frac{{
m,d}x}{{
m,d}y}=-frac{y‘‘}{left(y‘
ight)^3}\
frac{{
m,d}^3x}{{
m,d}y^3}&=frac{{
m,d}}{{
m,d}y}left(frac{{
m,d}^2x}{{
m,d}y^2}
ight)
=frac{{
m,d}}{{
m,d}y}left(-frac{y‘‘}{left(y‘
ight)^3}
ight)
=frac{{
m,d}}{{
m,d}x}left(-frac{y‘‘}{left(y‘
ight)^3}
ight)frac{{
m,d}x}{{
m,d}y}\
&=-frac{y‘‘‘!cdot!left(y‘
ight)^3-3left(y‘
ight)^2y‘‘!cdot!y‘‘}{left(y‘
ight)^6}frac{{
m,d}x}{{
m,d}y}=-frac{y‘‘‘!cdot!left(y‘
ight)^3-3left(y‘
ight)^2left(y‘‘
ight)^2}{left(y‘
ight)^7}\
&=frac{3left(y‘
ight)^2left(y‘‘
ight)^2-y‘‘‘!cdot!left(y‘
ight)^3}{left(y‘
ight)^7}=frac{3left(y‘‘
ight)^2-y‘y‘‘‘}{left(y‘
ight)^5}
end{align*}
egin{align*}
&&frac{{
m,d}^3x}{{
m,d}y^3}&=frac{color{red}{3left(y‘‘
ight)^2-y‘y‘‘‘}}{left(y‘
ight)^5}=frac{color{red}{-x!cdot!left(y‘
ight)^5}}{left(y‘
ight)^5}=-x\
&Rightarrow&frac{{
m,d}^3x}{{
m,d}y^3}+x&=0\
&Rightarrow&x‘‘‘+x&=0\
end{align*}
egin{align*}
x‘‘‘+x&=0\
x&=C_1e^{-y}+C_2e^{frac{y}{2}}cosleft(frac{sqrt{3}}{2}y
ight)+C_3e^{frac{y}{2}}sinleft(frac{sqrt{3}}{2}y
ight)
end{align*}
更一般:
[
y ^ { prime } y ^ {prime prime prime } - 3 left( y ^ { prime prime }
ight) ^ { 2 } + a left( y ^ { prime }
ight) ^ { 2 } y ^ { prime prime } - b left( y ^ { prime }
ight) ^ { 4 } - c x cdot left( y ^ { prime }
ight) ^ { 5 } = 0
]
[xrightarrow[frac{{
m,d}^3x}{{
m,d}y^3}=frac{3left(y‘‘
ight)^2-y‘y‘‘‘}{left(y‘
ight)^5}]
{frac{{
m,d}x}{{
m,d}y}=frac{1}{y‘}quadfrac{{
m,d}^2x}{{
m,d}y^2}=-frac{y‘‘}{left(y‘
ight)^3}}]
[f ^ { prime } ( x ) y ^ { prime } y ^ { prime prime prime } - 3 f ^ { prime } ( x ) left( y ^ { prime prime } ight) ^ { 2 } + 3 f ^ { prime prime } ( x ) y ^ { prime } y ^ { prime prime } + a f ^ { prime } ( x ) left( y ^ { prime } ight) ^ { 2 } y ^ { prime prime } = a f ^ { prime prime } ( x ) left( y ^ { prime } ight) ^ { 3 } + b f ^ { prime } ( x ) left( y ^ { prime } ight) ^ { 4 } + f ^ { prime prime prime } ( x ) left( y ^ { prime } ight) ^ { 2 } + c f ( x ) left( y ^ { prime } ight) ^ { 5 }]
令$u(y)=f(x)$
[frac { mathrm { d } ^ { 3 } u ( y ) } { mathrm { d } y ^ { 3 } } + a frac { mathrm { d } ^ { 2 } u ( y ) } { mathrm { d } y ^ { 2 } } + b frac { mathrm { d } u ( y ) } { mathrm { d } y } + c u ( y ) = 0.]
end{spacing}
egin{align*}
min_xquad f(x)=sum_{i=1}^{n} f_i(x),\
s.t.quad g_i(x)leq 0, A_ix=b_i,xincap_{i=1}^nOmega_i.
end{align*}
egin{align*}
frac{d}{dt}left( egin{array}{c}
y_i\
lambda _i\
mu _i\
x_i\
end{array}
ight) &=left( egin{array}{c}
-y_i+x_i-
abla f_ileft( x_i
ight) -left(
abla g_ileft( x_i
ight)
ight) ^Tleft( lambda _i+g_ileft( x_i
ight)
ight) ^+-A_{i}^{T}left( mu _i+A_ix_i-b_i
ight) +u_i\
-lambda _i+left( lambda _i+g_ileft( x_i
ight)
ight) ^+\
A_ix_i-b_i\
x=P_{Omega _i}left( y_i
ight)\
end{array}
ight)
\
u_i &=mathcal{k}_Psum_{j=1}^n{a_{ij}left( x_j-x_i
ight)}+mathcal{k}_Iint_0^t{sum_{j=1}^n{a_{ij}left( x_jleft( s
ight) -x_ileft( s
ight)
ight) ds}}
end{align*}
以上是关于常微分方程的主要内容,如果未能解决你的问题,请参考以下文章