刚刚接触mathematica，论文里一处绘图想用mathematica制作，但是不知道用什么语句，求助大家！
以下是该问题详情：
\[Alpha] = 1(*参数\[Alpha]变化范围从0.1到1，其他参数不变*)
\[Beta] = 0.8
c = 0.2
\[Phi] = 0.8
h = 1
Subscript[c, h] = 0.1
\[Rho] = 0.1
Subscript[c, E] = 0.2
e = 0.3
\[Gamma] = 1.5
Subscript[\[Epsilon], D] = -0.1
Subscript[\[Epsilon], U] = 0.1
(*导入到黎卡提方程组（eq1，eq2,eq3,eq4,eq5,eq6）Subscript[中得到M, 1],Subscript[M, \
2],Subscript[M, 3],Subscript[R, 1],Subscript[R, 2],Subscript[R, \
3]（{Subscript[M, 1],Subscript[M, 2],Subscript[M, 3],Subscript[R, \
1],Subscript[R, 2],Subscript[R, 3]}会随着\[Alpha]值的变化而变化）*)
eq1 = 1/4 ((\[Alpha]^2 \[Phi])/\[Beta] - 4 \[Phi] Subscript[c, h] - (
2 \[Beta] Subscript[M, 1] Subscript[R,
1] (1 + Subscript[\[Epsilon],
D])^2)/(-1 + \[Phi]) - (\[Beta] \[Phi]
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\) (1 +
Subscript[\[Epsilon], D])^2)/(-1 + \[Phi])^2 -
2 Subscript[M,
1] (\[Alpha] + \[Rho] + \[Alpha] Subscript[\[Epsilon], D]) + (2
\!\(\*SubsuperscriptBox[\(M\), \(1\), \(2\)]\) (1 +
Subscript[\[Epsilon], U])^2)/h);
eq2 = 1/2 (-c \[Alpha] - (\[Beta] (1 + Subscript[\[Epsilon],
D]) ((-1 + \[Phi]) Subscript[M, 1] Subscript[R,
2] (1 + Subscript[\[Epsilon], D]) +
Subscript[R,
1] (c (-1 + \[Phi]) + \[Phi] Subscript[R,
2] (1 + Subscript[\[Epsilon], D]))))/(-1 + \[Phi])^2 - (
2 \[Gamma] Subscript[c, E] Subscript[M,
1] (1 + Subscript[\[Epsilon], U]))/h +
Subscript[M,
2] (-\[Alpha] -
2 \[Rho] - \[Alpha] Subscript[\[Epsilon],
D] - (\[Beta] Subscript[R,
1] (1 + Subscript[\[Epsilon], D])^2)/(-1 + \[Phi]) + (
2 Subscript[M, 1] (1 + Subscript[\[Epsilon], U])^2)/h));
eq3 = 1/4 ((2 \[Gamma]^2
\!\(\*SubsuperscriptBox[\(c\), \(\[ExponentialE]\), \(2\)]\))/h -
4 \[Rho] Subscript[M, 3] - (
2 \[Beta] Subscript[M, 2] Subscript[R,
2] (1 + Subscript[\[Epsilon],
D])^2)/(-1 + \[Phi]) - (\[Beta] Subscript[R,
2] (1 + Subscript[\[Epsilon],
D]) (2 c (-1 + \[Phi]) + \[Phi] Subscript[R,
2] (1 + Subscript[\[Epsilon], D])))/(-1 + \[Phi])^2 + (2
\!\(\*SubsuperscriptBox[\(M\), \(2\), \(2\)]\) (1 +
Subscript[\[Epsilon], U])^2)/h + (
4 Subscript[c,
E] (e h - \[Gamma] Subscript[M,
2] (1 + Subscript[\[Epsilon], U])))/h);
eq4 = 1/4 (-((\[Alpha]^2 (-1 + \[Phi]))/\[Beta]) +
4 (-1 + \[Phi]) Subscript[c, h] +
Subscript[R,
1] (-((\[Beta] Subscript[R,
1] (1 + Subscript[\[Epsilon], D])^2)/(-1 + \[Phi])) -
2 (\[Alpha] + \[Rho] + \[Alpha] Subscript[\[Epsilon], D]) + (
4 Subscript[M, 1] (1 + Subscript[\[Epsilon], U])^2)/h));
eq5 = -((Subscript[R,
2] (\[Beta] Subscript[R,
1] (1 + Subscript[\[Epsilon],
D])^2 + (-1 + \[Phi]) (\[Alpha] +
2 \[Rho] + \[Alpha] Subscript[\[Epsilon], D])))/(
2 (-1 + \[Phi]))) - (\[Gamma] Subscript[c, E] Subscript[R,
1] (1 + Subscript[\[Epsilon], U]))/h + (
Subscript[M, 2] Subscript[R, 1] (1 + Subscript[\[Epsilon], U])^2)/
h + (Subscript[M, 1] Subscript[R,
2] (1 + Subscript[\[Epsilon], U])^2)/h;
eq6 = -\[Rho] Subscript[R, 3] - (\[Beta]
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\) (1 +
Subscript[\[Epsilon], D])^2)/(4 (-1 + \[Phi])) + (
Subscript[R,
2] (1 + Subscript[\[Epsilon], U]) (-\[Gamma] Subscript[c, E] +
Subscript[M, 2] (1 + Subscript[\[Epsilon], U])))/h;
Y = (h \[Beta] Subscript[R, 2] (1 + Subscript[\[Epsilon], D])^2 -
2 \[Gamma] (1 - \[Phi]) Subscript[c,
E] (1 + Subscript[\[Epsilon], U]) +
2 (1 - \[Phi]) Subscript[M, 2] (1 + Subscript[\[Epsilon], U])^2)/(
h (1 + Subscript[\[Epsilon],
D]) (\[Alpha] (1 - \[Phi]) - \[Beta] Subscript[R,
1] (1 + Subscript[\[Epsilon], D])) -
2 (1 - \[Phi]) Subscript[M, 1] (1 + Subscript[\[Epsilon], U])^2);
W = NSolve[{eq1 == 0, eq2 == 0, eq3 == 0, eq4 == 0, eq5 == 0,
eq6 == 0, Y > 0}, {Subscript[M, 1], Subscript[M, 2], Subscript[M,
3], Subscript[R, 1], Subscript[R, 2], Subscript[R, 3]},
Reals];(*只取能令Y>0 （同时满足黎卡提方程组）的实数解*)
Y = (h \[Beta] Subscript[R, 2] (1 + Subscript[\[Epsilon], D])^2 -
2 \[Gamma] (1 - \[Phi]) Subscript[c,
E] (1 + Subscript[\[Epsilon], U]) +
2 (1 - \[Phi]) Subscript[M, 2] (1 + Subscript[\[Epsilon], U])^2)/(
h (1 + Subscript[\[Epsilon],
D]) (\[Alpha] (1 - \[Phi]) - \[Beta] Subscript[R,
1] (1 + Subscript[\[Epsilon], D])) -
2 (1 - \[Phi]) Subscript[M,
1] (1 + Subscript[\[Epsilon], U])^2) //.
W;(*（结果输出每个{\[OpenCurlyDoubleQuote]\[Alpha]=\
\[CloseCurlyDoubleQuote]，\[OpenCurlyDoubleQuote]Y=\
\[CloseCurlyDoubleQuote]}）并且绘图（\[Alpha]横坐标，Y纵坐标）*)

代码如下：
\[Alpha]=1(*参数\[Alpha]变化范围从0.1到1，其他参数不变*)
\[Beta]=0.8
c=0.2
\[Phi]=0.8
h=1
Subscript[c, h]=0.1
\[Rho]=0.1
Subscript[c, E]=0.2
e=0.3
\[Gamma]=1.5
Subscript[\[Epsilon], D]=-0.1
Subscript[\[Epsilon], U]=0.1
(*导入到黎卡提方程组（eq1，eq2,eq3,eq4,eq5,eq6）Subscript[中得到M, 1],Subscript[M, 2],Subscript[M, 3],Subscript[R, 1],Subscript[R, 2],Subscript[R, 3]（{Subscript[M, 1],Subscript[M, 2],Subscript[M, 3],Subscript[R, 1],Subscript[R, 2],Subscript[R, 3]}会随着\[Alpha]值的变化而变化）*)
eq1=1/4 ((\[Alpha]^2 \[Phi])/\[Beta]-4 \[Phi] Subscript[c, h]-(2 \[Beta] Subscript[M, 1] Subscript[R, 1] (1+Subscript[\[Epsilon], D])^2)/(-1+\[Phi])-(\[Beta] \[Phi] Subsuperscript[R, 1, 2] (1+Subscript[\[Epsilon], D])^2)/(-1+\[Phi])^2-2 Subscript[M, 1] (\[Alpha]+\[Rho]+\[Alpha] Subscript[\[Epsilon], D])+(2 Subsuperscript[M, 1, 2] (1+Subscript[\[Epsilon], U])^2)/h);
eq2=1/2 (-c \[Alpha]-(\[Beta] (1+Subscript[\[Epsilon], D]) ((-1+\[Phi]) Subscript[M, 1] Subscript[R, 2] (1+Subscript[\[Epsilon], D])+Subscript[R, 1] (c (-1+\[Phi])+\[Phi] Subscript[R, 2] (1+Subscript[\[Epsilon], D]))))/(-1+\[Phi])^2-(2 \[Gamma] Subscript[c, E] Subscript[M, 1] (1+Subscript[\[Epsilon], U]))/h+Subscript[M, 2] (-\[Alpha]-2 \[Rho]-\[Alpha] Subscript[\[Epsilon], D]-(\[Beta] Subscript[R, 1] (1+Subscript[\[Epsilon], D])^2)/(-1+\[Phi])+(2 Subscript[M, 1] (1+Subscript[\[Epsilon], U])^2)/h));
eq3=1/4 ((2 \[Gamma]^2 Subsuperscript[c, E, 2])/h-4 \[Rho] Subscript[M, 3]-(2 \[Beta] Subscript[M, 2] Subscript[R, 2] (1+Subscript[\[Epsilon], D])^2)/(-1+\[Phi])-(\[Beta] Subscript[R, 2] (1+Subscript[\[Epsilon], D]) (2 c (-1+\[Phi])+\[Phi] Subscript[R, 2] (1+Subscript[\[Epsilon], D])))/(-1+\[Phi])^2+(2 Subsuperscript[M, 2, 2] (1+Subscript[\[Epsilon], U])^2)/h+(4 Subscript[c, E] (e h-\[Gamma] Subscript[M, 2] (1+Subscript[\[Epsilon], U])))/h);
eq4=1/4 (-((\[Alpha]^2 (-1+\[Phi]))/\[Beta])+4 (-1+\[Phi]) Subscript[c, h]+Subscript[R, 1] (-((\[Beta] Subscript[R, 1] (1+Subscript[\[Epsilon], D])^2)/(-1+\[Phi]))-2 (\[Alpha]+\[Rho]+\[Alpha] Subscript[\[Epsilon], D])+(4 Subscript[M, 1] (1+Subscript[\[Epsilon], U])^2)/h));
eq5=-((Subscript[R, 2] (\[Beta] Subscript[R, 1] (1+Subscript[\[Epsilon], D])^2+(-1+\[Phi]) (\[Alpha]+2 \[Rho]+\[Alpha] Subscript[\[Epsilon], D])))/(2 (-1+\[Phi])))-(\[Gamma] Subscript[c, E] Subscript[R, 1] (1+Subscript[\[Epsilon], U]))/h+(Subscript[M, 2] Subscript[R, 1] (1+Subscript[\[Epsilon], U])^2)/h+(Subscript[M, 1] Subscript[R, 2] (1+Subscript[\[Epsilon], U])^2)/h;
eq6=-\[Rho] Subscript[R, 3]-(\[Beta] Subsuperscript[R, 2, 2] (1+Subscript[\[Epsilon], D])^2)/(4 (-1+\[Phi]))+(Subscript[R, 2] (1+Subscript[\[Epsilon], U]) (-\[Gamma] Subscript[c, E]+Subscript[M, 2] (1+Subscript[\[Epsilon], U])))/h;
Y=(h \[Beta] Subscript[R, 2] (1+Subscript[\[Epsilon], D])^2-2 \[Gamma] (1-\[Phi]) Subscript[c, E] (1+Subscript[\[Epsilon], U])+2 (1-\[Phi]) Subscript[M, 2] (1+Subscript[\[Epsilon], U])^2)/(h (1+Subscript[\[Epsilon], D]) (\[Alpha] (1-\[Phi])-\[Beta] Subscript[R, 1] (1+Subscript[\[Epsilon], D]))-2 (1-\[Phi]) Subscript[M, 1] (1+Subscript[\[Epsilon], U])^2);
W=NSolve[{eq1==0,eq2==0,eq3==0,eq4==0,eq5==0,eq6==0,Y>0},{Subscript[M, 1],Subscript[M, 2],Subscript[M, 3],Subscript[R, 1],Subscript[R, 2],Subscript[R, 3]},Reals];(*只取能令Y>0 （同时满足黎卡提方程组）的实数解*)
Y=(h \[Beta] Subscript[R, 2] (1+Subscript[\[Epsilon], D])^2-2 \[Gamma] (1-\[Phi]) Subscript[c, E] (1+Subscript[\[Epsilon], U])+2 (1-\[Phi]) Subscript[M, 2] (1+Subscript[\[Epsilon], U])^2)/(h (1+Subscript[\[Epsilon], D]) (\[Alpha] (1-\[Phi])-\[Beta] Subscript[R, 1] (1+Subscript[\[Epsilon], D]))-2 (1-\[Phi]) Subscript[M, 1] (1+Subscript[\[Epsilon], U])^2)//.W;(*（结果输出每个{\[OpenCurlyDoubleQuote]\[Alpha]=\[CloseCurlyDoubleQuote]，\[OpenCurlyDoubleQuote]Y=\[CloseCurlyDoubleQuote]}）并且绘图（\[Alpha]横坐标，Y纵坐标）*)

此问题已在别处被解决