题目

求大神帮忙画mathematica的图.我找不出来哪儿错了.C和c不一样哦~
B3 = 1/8 E^(-2 r) (-2 + (-1 + E^(4 r))/(I - 2 l)^2 - (
4 E^(4 r))/(-I + l)^2 - (
4 E^(4 r))/(I + l)^2 + (-1 + E^(4 r))/(I + 2 l)^2 + (
2 HurwitzZeta[2,1/Tc])/Tc^2 + (6 E^(4 r) HurwitzZeta[2,1/Tc])/
Tc^2 + 6 E^(4 r) (-1 + HurwitzZeta[2,1/Tc]/Tc^2) + (
2 PolyGamma[1,1/Tc])/Tc^2 - (
8 E^(4 r) PolyGamma[1,(1 - I l)/Tc])/Tc^2 - (
8 E^(4 r) PolyGamma[1,(1 + I l)/Tc])/Tc^2 + (
2 (-1 + E^(4 r)) PolyGamma[1,(1 - 2 I l)/Tc])/Tc^2 + (
2 (-1 + E^(4 r)) PolyGamma[1,(1 + 2 I l)/Tc])/Tc^2)
Tc = 0.8
ct = l
a = 0.01
r = 0.31
Plot3D[Max[0,
2 E^(-4 a*B3)*abs[[Alpha]*Sqrt[1 - [Alpha]^2]]],{[Alpha],0,
1},{l,0,20},
AxesLabel -> {[Alpha]^2,
"!(*SubscriptBox["[Omega]","c"])t",
"!(*SubscriptBox["C","[CapitalPsi]"])(t)"},
LabelStyle -> Directive[Bold],AxesStyle -> Directive[14]]

提问时间：2021-05-03

答案

Clear["`*"]Tc = 0.8;a = 0.01;r = 0.31;B3[l_] := 1/8 E^(-2 r) (-2 + (-1 + E^(4 r))/(I - 2 l)^2 - (4 E^(4 r))/(-I + l)^2 - (4 E^(4 r))/(I + l)^2 + (-1 + E^(4 r))/(I + 2 l)^2 + (2 HurwitzZeta[2,1/Tc])/Tc...