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since NMinimize is not working I've tried Plot3d and I've got an empty plot

A normal plot of the function seems to be working fine. I'm also pretty certain that there is no infinity or so. 
  (* ::Package:: *)
 smoothMax[tx_, tm_, ts_]:=tm + Log[1 + Exp[(tx - tm)*ts]]/ts
 smoothMin[tx_, tm_, ts_]:=tm - Log[1 + Exp[(tm - tx)*ts]]/ts
 glue[tx_, ts_]:=((Exp[tx*ts] - Exp[-tx*ts])/(Exp[tx*ts] + Exp[-tx*ts] + 1)+1)/2
 current[VD_]:=(csat*(Exp[VD/vte] - (1)) + cktCKTgmin*VD)
 charge1[VD1_]:=(czero*hereDIOtF1 + czero/modelDIOf2*(modelDIOf3*(VD1 - hereDIOtDepCap)+ (modelDIOgradingCoeff/(    modelDIOjunctionPot+ modelDIOjunctionPot))*(VD1*VD1 - hereDIOtDepCap*hereDIOtDepCap)))
 charge0[VD0_]:=(modelDIOjunctionPot*czero*(1- Exp[(1 - modelDIOgradingCoeff)*Log[1 - VD0/modelDIOjunctionPot]]) /(1 - modelDIOgradingCoeff))
 charge2[x_]:=glue[-(x - hereDIOtDepCap), gs]*charge0[smoothMin[x, hereDIOtDepCap, gs]]+ glue[x - hereDIOtDepCap, gs]*charge1[smoothMax[x, hereDIOtDepCap, gs]]
 charge[VD_]:=charge3[current[VD]] + charge2[VD]
 chargecurrent[v2_, v2t_]:=(D[charge[v22], v22]/.v22->v2)*v2t
 v[t_, omega_]:=Sin[omega*t]/2
 rest11[t_, i1_, v2_, v2t_, omega_, current1_, chargecurrent1_, v_]:=(i1 - (v - v2)/rs)^2+((v - v2)/rs - current1 - chargecurrent1)^2
 i1[t_, t0_, a1_, b1_, c1_, d1_]:=a1+b1*(t-t0) + c1*(t - t0)^2 + d1*(t - t0)^3
 v2[t_, t0_, a3_, b3_, c3_, d3_]:=a3+b3*(t-t0) + c3*(t - t0)^2 + d3*(t - t0)^3
 v2t[t_, t0_, a3_, b3_, c3_, d3_]:=D[v2[tt, t0, a3, b3, c3, d3], tt]/.tt->t
 rest22[t_, t0_, a1_, b1_, c1_, d1_, a3_, b3_, c3_, d3_, omega_]:=rest11[
     i1[t, t0, a1, b1, c1, d1],
     v2[t, t0, a3, b3, c3, d3],
     v2t[t, t0, a3, b3, c3, d3],
         v2[t, t0, a3, b3, c3, d3]
     chargecurrent[v2[t, t0, a3, b3, c3, d3], v2t[t, t0, a3, b3, c3, d3]],
     v[t, omega]
 SERIES1[a1_, b1_, c1_, d1_, a3_, b3_, c3_, d3_, t1_, h_]:=Integrate[
         rest22[t, t1, a1, b1, c1, d1, a3, b3, c3, d3, omega],
         {t, t1, 8}],
     {t, t1, t1 + h}];
 For[t1=0.0, t1 < 10^-5, t1 = t1 + h,
         SERIES1[a1, b11, c1, d1, a3, b31, c3, d3, t1, h/100]/h*100,
         {b11, b31}
         SERIES1[a1, b1, c11, d1, a3, b3, c31, d3, t1, h/10]/h*10,
         {c11, c31}
         SERIES1[a1, b1, c1, d11, a3, b3, c3, d31, t1, h]/h,
         {d11, d31}
    Print["t1=", t1];
    Print[",a1=", a1, ",b1=", b1, ",c1=", c1, ",d1=", d1];
    Print[",a2=", a2, ",b2=", b2, ",c2=", c2, ",d2=", d2];
    Print[",a3=", a3, ",b3=", b3, ",c3=", c3, ",d3=", d3];
    Plot[i1[t, t1, a1, b1, c1, d1], {t, t1, t1+h}];
    p = Append[p, {a1+b1*(t-t1)+c1*(t-t1)^2+d1*(t-t1)^3, t1<t<t1+h}],
    p = {{a1+b1*(t-t1)+c1*(t-t1)^2+d1*(t-t1)^3, t1<t<t1+h}}];
    (*Print[Plot[Piecewise[p], {t, 0, t1+h}]];*)
    a1=a1 + b1*h + c1*h^2 + d1*h^3;
    a3=a3 + b3*h + c3*h^2 + d3*h^3;
    Plot[Piecewise[p], {t, 0, 10^-5}]

SERIES2[b11_, b31_]:=SERIES1[a1, b11, c1, d1, a3, b31, c3, d3, 0, h]
Plot3D[SERIES2[b11, b31]*10^11, {b11, -0.0001, 0.0001}, {b31, -0.0001, 0.0001}]
Plot[{SERIES2[b11, -0.001], SERIES2[b11, 0], SERIES2[b11, 0.001]}, {b11, -0.001, 0.001}]

SERIES2[0, 0]

uncommenting the Plot command gives a useful 2d-plot:

and I think I found a workaround:

Plot3D[SERIES2[b11, b31]*10^11-6.07933, {b11, -0.0001, 0.0001}, {b31, -0.0001, 0.0001}]

gives a non empty plot. Potentially 6.07933 is interpreted as a magic number?!
POSTED BY: Peter Foelsche
3 years ago
There's nothing very magical about the number 6.07933. This looks like the problem caused by floating point precision. What subtracting 6.07933 does is it brings your results very close to zero. Floating point numbers are not evenly spaced and are basically more dense near zero than farther away from it  (except for at underflow). Translating numerical computations closer to 0 can sometimes help them by providing (I'm not sure this is reall the right word but whatever, I'll say it) precision.
This is probably why Plot3D is able to handle the workaround but not the original input.

A more general solution would be to use abitrary precision arithmetic using WorkingPrecision in Plot3D or to rewrite the function being plotted so that it varies enough to not require very high precision to notice the differences. 
POSTED BY: Sean Clarke
3 years ago

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