Nsolve rantz无法策划

Question

我有一个问题是在Mathematica中绘制一个equotations系统的解决方案。我的equotations系统有两个变量（s12和t）。明确地解决它是不可能的（s12：= f（t）），但我能够得到每个正t的解决方案。但我想要的是在x-achses上的t和在y-achses上的s12（t）的情节。

我最好的理由是，因为我总是通过评论获得单一解决方案 “*Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result*”这对mathematica的无限解决方案无效。

我可能不得不压制这个警告，还是有人有另一个想法？我只需要粗略的情节。

问题如下：

ClearAll["Global`*"];
cinv1 = 40;
cinv2 = 4;
cinv3 = 3;
h2 = 1.4;
h3 = 1.2;
alpha = 0.04;
z = 20;
p = 0.06;
cop1 = 0;
cop2 = 1;
cop3 = 1.5;
l2 = 0.1;
l3 = 0.17;
teta2 = 0.19;
teta3 = 0.1;
co2 = -0.1;

smax = 40;
c = 1;

Plot[Solve[{s12 == ((cinv1 - 
         cinv2) + ((cinv2 - cinv3)*((s12 teta2)/(
          Sqrt[ (teta2 - teta3)] Sqrt[
           c s12^2 teta2 - (2 alpha z)/c]))))/((1/(teta2 - 
           teta3))*((teta2*cop3 - teta3*cop2) + (teta2*h3*l3*E^(p*t) -
            teta3*h2*l2*E^(p*t)))), s12 > 0}, s12, Reals], {t, 0, 10}]

如上所述，当我使用特定的t时，我得到一个解决方案，否则我会收到如下消息：

"*Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result*"
"*Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result*"
"*Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result*"
*"General::stop: "Further output of \!\(\*
StyleBox[
RowBox[{\"Solve\", \"::\", \"ratnz\"}], \"MessageName\"]\) will be suppressed during this calculation""*

非常感谢你的帮助， 安德烈亚斯

Answer 1

该系统有4个解决方案，其中3个在感兴趣的范围内为正：

s2 = Solve[{s12 - ((cinv1 - cinv2) + ((cinv2 - cinv3) ((s12 teta2)/
          (Sqrt[(teta2 - teta3)] Sqrt[c s12^2 teta2 - (2 alpha z)/c]))))/
           ((1/(teta2 - teta3))*((teta2*cop3 - teta3*cop2) + 
           (teta2*h3*l3*E^(p*t) - teta3*h2*l2*E^(p*t))))} == 0, s12];
Plot[s12 /. s2 , {t, 0, 59}]

Answer 2

要添加的重要事实：

上面提出的解决方案是正确的，但它使用复数来解决。上述解决方案中的图表仅显示复数的实部。这可能会对我造成一些混乱。

尽管如此，有一个只有实数的解决方案。由于Mathematica不能用实数来“连续”地解决这种情况，我最终采取了三步法：

1. 我在离散时间点解决了分配问题
2. 我使用ListLinePlot绘制了解决方案。

3. 我使用Interpolation []允许粗略检测其他曲线的插入

   a = Table[NSolve[{s12 - ((cinv1 - cinv2) + ((cinv2 - cinv3)*((s12 teta2)/(\[Sqrt] (teta2 - teta3) \[Sqrt](c s12^2 teta2 - (2 alpha z)/c)))))/ ((1/(teta2 - teta3))*((teta2*cop3 -teta3*cop2) + (teta2*h3*l3*E^(p*t) - teta3*h2*l2*E^(p*t)))) == 0}, s12][[1]], {t, 0, 100}];

   b = Table[t, {t, 0, 100}];

   f1a = s12 /. a;
   
   f1 = Transpose[{b, f1a}];
   
   ceiling1 = ListLinePlot[{f1},
   PlotRange -> {{0, 20}, {0, 40}},PlotStyle -> {Black, Dotted, Thickness[0.003]}];
   

在下一步中，我还需要找到以这种方式创建的多条曲线的交集。要获得粗略估算，我执行了以下操作：

curve1 = Interpolation[f1];
intersec2a = FindRoot[curve1[x2] - t12[x2, l2], {x2, 0}];
intersec2 = x2 /. intersec2a;

希望这有帮助