求解一个方程组（个人认为有点挑战性，目前我们同学和老师们全都不会做）

zsmiup

T = 1/8 (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l)
(\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) = (1/
  3) (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) Log[(\[Rho]l (3 \
- \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]
  
(* 第二方程可以简化为： (\[Rho]l-\[Rho]g)(6-\[Rho]g-\[Rho]l)=8/3*T*Log[(\[Rho]l \
(3-\[Rho]g))/(\[Rho]g (3-\[Rho]l))]， 因为T=1/8 (3-\[Rho]g)(3-\[Rho]l)(\
\[Rho]g+\[Rho]l)*)

T -> temperature , from 0, to 1 从0变化到1
\[Rho]l -> density of liquid
\[Rho]g -> density of gas
p -> pressure
P[\[Rho]_] := (8 \[Rho]*(T))/(3 - \[Rho]) - 3 \[Rho]^2
（* \[Rho]l，\[Rho]g 都满足上式，密度和压力方程*）

T = 1/8 (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l)
(\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) =
1/3 (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) Log[(\[Rho]l (3 \
- \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]
另有关于\[Rho]l，pg和温度的表达式。
目的，是数值法 (T = T0 + 0.01, T0 = 0,
   Tmax = 1) ，解出，在相同的温度T下的\[Rho]l和\[Rho]g。 并作 P - \[Rho]l 和 P - \
\[Rho]g （压力和密度图） 证明 这两曲线相切，切点为 气液共同点

我的思路是用 For loop ，来求解。
但是问题是这个方程组，太难解，Mathematica 解起来就没完了，我取T =
0.5 ， 就想算一个值，我等了20分钟都没有算完，所以想请教高手们怎么可以解下列方程组。
我试了， Solve， Reduce， FindRoot，Nsolve，我还分别解了这两个方程 【就是用\[Rho]g表示\[Rho]g】 \
，然后Plot 他们，想找到他们的交点。 但是第2个方程还是解不出来，想请教高手们出出主意
T = 1/8 (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l)
(\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) = (1/
  3) (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) Log[(\[Rho]l (3 \
- \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]
  
(* 第二方程可以简化为： (\[Rho]l-\[Rho]g)(6-\[Rho]g-\[Rho]l)=8/3*T*Log[(\[Rho]l \
(3-\[Rho]g))/(\[Rho]g (3-\[Rho]l))]， 因为T=1/8 (3-\[Rho]g)(3-\[Rho]l)(\
\[Rho]g+\[Rho]l)*)


  下面是我试着的解，谢谢

Test body
\!\(\*
ButtonBox["For",
BaseStyle->"Link",
ButtonData->"paclet:ref/For"]\)[start, test, incr, body]
executes start, then repeatedly evaluates body and incr until test \
fails to give \!\(\*
ButtonBox["True",
BaseStyle->"Link",
ButtonData->"paclet:ref/True"]\).




T = 1
Reduce[1/8 (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) ==
  T, {\[Rho]g, \[Rho]l}]


1

-3 + \[Rho]g !=
  0 && (\[Rho]l == (-9 +
     6 \[Rho]g - \[Rho]g^2 - (-1 + \[Rho]g) Sqrt[-15 +
       2 \[Rho]g + \[Rho]g^2])/(
    2 (-3 + \[Rho]g)) || \[Rho]l == (-9 +
     6 \[Rho]g - \[Rho]g^2 + (-1 + \[Rho]g) Sqrt[-15 +
       2 \[Rho]g + \[Rho]g^2])/(2 (-3 + \[Rho]g)))


T = 1
Reduce[(\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) ==
  1/3*T*Log[(\[Rho]l (3 - \[Rho]g))/(\[Rho]g (3 - \[Rho]l))], \
{\[Rho]g, \[Rho]l}]


1

$Aborted



Clear["Global"]
T = 0.01

Solve[{(3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) ==
   T, (\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) ==
   1/3*T*Log[(\[Rho]l (3 - \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]}, {\
\[Rho]l, \[Rho]g}]


Clear["\[Rho]l"] ，

NSolve[(\[Rho]l - 0.5) (6 - .5 - \[Rho]l) ==
  8/3*1/8 (3 - .5) (3 - \[Rho]l) (.5 + \[Rho]l)*
   Log[(\[Rho]l (3 - .5))/(0.5 (3 - \[Rho]l))], \[Rho]l] (* T=.5 *)

NSolve[(5.5` - \[Rho]l) (-0.5` + \[Rho]l) ==
  0.8333333333333333` (3 - \[Rho]l) (0.5` + \[Rho]l) (Log[
      5.` \[Rho]l] - Log[3 - \[Rho]l]), \[Rho]l]



NSolve[(5.5 - \[Rho]l) (-0.5 + \[Rho]l) ==
  0.833333 (3 - \[Rho]l) (0.5 + \[Rho]l) (-Log[3 - \[Rho]l] +
     Log[5. \[Rho]l]), \[Rho]l]


抱歉 帖子有点乱， 这个问题确实复杂，请有有兴趣的高手贴到mathematica上看看

TBE_Legend

1# zsmiup

zsmiup

wow 谢谢 高手

不过，我将您的代码 打进去了 却得到不同的结果

如下

1. ff[T_] := Module[{sol},

复制代码

sol = FindRoot[{(3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) == T,
       (\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) ==
      1/3*T*Log[(\[Rho]l (3 - \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]}, {{\
\[Rho]l, Random[]}, {\[Rho]g, Random[]}}, MaxIterations -> 100]]
Table[ff[x], {x, 0, 1, 0.1}][/code][/code]


During evaluation of In[35]:= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances. >>

During evaluation of In[35]:= FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>

During evaluation of In[35]:= FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>

During evaluation of In[35]:= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances. >>

During evaluation of In[35]:= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances. >>

During evaluation of In[35]:= General::stop: Further output of FindRoot::lstol will be suppressed during this calculation. >>

During evaluation of In[35]:= FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>

During evaluation of In[35]:= General::stop: Further output of FindRoot::cvmit will be suppressed during this calculation. >>

Out[36]= {{\[Rho]l -> 0.285354, \[Rho]g -> -0.285354}, {\[Rho]l ->
   0.00681528 + 2.04605*10^-15 I, \[Rho]g -> -0.00715281 -
    1.79037*10^-15 I}, {\[Rho]l -> -2.5948*10^-8 +
    1.20629*10^-8 I, \[Rho]g -> -2.59477*10^-8 +
    1.20634*10^-8 I}, {\[Rho]l -> -9.05927*10^-9, \[Rho]g -> \
-9.05932*10^-9}, {\[Rho]l ->
   0.0187468 - 0.0257193 I, \[Rho]g -> -0.0109141 +
    0.0154141 I}, {\[Rho]l ->
   0.0317468 - 0.000638149 I, \[Rho]g -> -0.0165133 +
    0.000337336 I}, {\[Rho]l -> -7.67425*10^-8 +
    1.23146*10^-10 I, \[Rho]g -> -7.67422*10^-8 +
    1.24441*10^-10 I}, {\[Rho]l -> -1.18052*10^-8, \[Rho]g -> \
-1.18052*10^-8}, {\[Rho]l -> -4.96428*10^-7, \[Rho]g -> \
-4.96425*10^-7}, {\[Rho]l -> -3.93091*10^-8, \[Rho]g -> \
-3.93093*10^-8}, {\[Rho]l ->
   0.11282 + 2.14641*10^-15 I, \[Rho]g -> -0.0525342 -
    4.15596*10^-16 I}}

chungyuandye

1. 
   
2. ff[T_] :=
   
3.   Module[{sol},
   
4.    sol = FindRoot[{(3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) ==
   
5.        T (\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l), (3 - \[Rho]g) \
   
6. (3 - \[Rho]l) (\[Rho]g + \[Rho]l) ==
   
7.        1/3*T*Log[(\[Rho]l (3 - \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]}, \
   
8. {{\[Rho]l, 1/10, 9/10}, {\[Rho]g, 1/10, 9/10}}, MaxIterations -> 100]];
   
9. 
   
10. ff[#] & /@ Range[0, 1, 1/10]
    
11. 
    
12. TableForm[{N@#, \[Rho]l, \[Rho]g} /. ff[#] & /@ Range[0, 1, 1/10],
    
13. TableHeadings -> {T, \[Rho]l, \[Rho]g}]
    

复制代码

zsmiup

TBE 您好：

运行您的code 得到了 和您的结果相似的解（有可以忽略的tiny difference)， 但是当我给 这个方程组，加上系数，（1/8 和 8/3) 的时候
mathematica 说，100 次得iterations ，也找不到 一个比较好的解，这是说， mathematica不适合解这个方程，还是说这个方程组，本身有点问题。
谢谢
FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>

加上系数的代码

1. Clear["Globe'*"]
   
2. ff[T_] :=
   
3. Module[{sol},
   
4.   sol = FindRoot[{1/
   
5.        8 (3 - \[Rho]g) (3 - \[Rho]l) (\[Rho]g + \[Rho]l) == T,
   
6.      (\[Rho]l - \[Rho]g) (6 - \[Rho]g - \[Rho]l) ==
   
7.       8/3 T*Log[(\[Rho]l (3 - \[Rho]g))/(\[Rho]g (3 - \[Rho]l))]}, {{\
   
8. \[Rho]l, Random[]}, {\[Rho]g, Random[]}}]]
   
9. Table[ff[x], {x, 0.01, 1.1, 0.1}]
   

复制代码

zsmiup

谢谢 chungyuandye 的解答

不知道可不可帮我解释下 & 在这儿 是什么意思？ and吗 ？
ff[#] & /@ Range[0, 1, 1/10]

我还问您下 有没有 mathmatica符号的小手册
比如 /. ; := ; /@ ,这样的mathematica火星文 注解

myleader

&是引用的意思，代表第几个变量

入门的话看一下《introduction to programming in mathematica》