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I have the following equation which I want to solve:

$$ I_D = [Li_2(-e^{V_D-I_D})-Li_{2}(e^{I_D})] $$

Here $Li_2(x)$ is the PolyLog function of order $2$. Is there a way to solve this equation iteratively in Mathematica to get $I_D$ as a function of $V_D$.

Edit: I want to solve this equation numerically for real values of $I_D$ and $V_D$.

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2 Answers 2

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A typical solution of the equation

id - PolyLog[2, -Exp[vd - id]] - PolyLog[2, Exp[id]] == 0

can be obtained by plotting this expression.

ReImPlot[(id - PolyLog[2, -Exp[vd - id]] - PolyLog[2, Exp[id]]) /. vd -> .5, id, -1, 1}, 
    ImageSize -> Large, AxesLabel -> {id, None}, LabelStyle -> {15, Bold, Black}]

enter image description here

Visibly, there is a branch point at id = 0, consistent with the documentation of PolyLog. A small amount of experimentation shows that the zero of the curve shown moves toward the branch point as vd increases. Consequently, there is no solution for vd greater than

FindRoot[(id - PolyLog[2, -Exp[vd - id]] - PolyLog[2, Exp[id]]) /. id -> 0, {vd, -.87}]
(* {vd -> 0.872676} *)

at least for the principal value of PolyLog. With this information, a plot of id as a function of vd is obtained by

Plot[id /. FindRoot[(id - PolyLog[2, -Exp[vd - id]] - PolyLog[2, Exp[id]]), {id, 01}], 
    {vd, -1, .872}, ImageSize -> Large, AxesLabel -> {vd, id}, LabelStyle -> {15, Bold, Black}]

enter image description here

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  • $\begingroup$ Thank you, this is wonderful. Can you check out my recent question: mathematica.stackexchange.com/questions/240828/… and give suggestion as to how to solve this? $\endgroup$ Mar 1, 2021 at 0:50
  • $\begingroup$ @Indeterminate I see that your new question received an answer a few minutes ago. Does it meet your needs? If so, it is good practice to upvote and accept it. $\endgroup$
    – bbgodfrey
    Mar 1, 2021 at 1:46
  • $\begingroup$ thanks I have commented on that answer, it actually doesn't solve my problem completely. And thanks! I have also upvoted your answer and accepted it. $\endgroup$ Mar 1, 2021 at 3:22
  • $\begingroup$ My bad, I am still new, so I assumed upvoting it means accepting it. I have accepted it. Thanks for bringing it to my notice. $\endgroup$ Mar 1, 2021 at 4:44
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You can investigate what your function does when you iterate it by plotting how it updates points in the complex $I_D$ plane, and I find that using VectorPlot to plot a vector field is a useful way of visualising this.

Define your function $f(I_D,V_D)$ - i.e. the righthand side of your equation.

f[id_, vd_] := PolyLog[2, -Exp[vd - id]] - PolyLog[2, Exp[id]];

Create an animation to explore what happens as you vary $V_D$.

With[{f0 = 5, v0 = 3, dv = 0.5, p = 25}, 
  Animate[
    VectorPlot[
      ReIm[f[idR + I idI, vd] - (idR + I idI)],
      {idR, -f0, f0}, {idI, -f0, f0},
      VectorPoints -> p
    ],
    {vd, -v0, v0, dv}
  ]
]

This plots a vector field of the update $f(I_D,V_D)-I_D$ in the complex $I_D$ plane, using arrows to show update direction, and arrow colour to show update magnitude. For $V_D=0$ this looks like the plot below.

enter image description here

Points in the complex $I_D$ plane that map to themselves - i.e. are solutions of your equation - show up prominently in this sort of plot. For instance, in the plot above you can see a couple of "vortices", where the flow circulates around points that map to themselves.

This and other clues can give you a feel for how your function moves points around in the complex plane, and in particular the location of points that map to themselves - i.e. the various solution-branches that you seek.

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  • $\begingroup$ Thanks, that is a great visualization. Is there a numerical way to solve this? (Only real values of $I_D$ and $V_D$) $\endgroup$ Feb 26, 2021 at 19:11
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    $\begingroup$ Because there is more than one solution-branch this is a non-trivial problem. For instance, for $V_D=0$ FindRoot[id == f[id, 0], {id, 0}] gives {id -> -1.00844 - 1.47085 I}, which is the lower of the pair of "vortices" in the above visualisation. If you then plot how this solution depends on $V_D$, the plot will switch (mid-plot) between different solution-branches. Play around with the above visualisation to get a good feel for where the solutions are, then you will be able to give a root-finding approach a guiding "steer" towards your preferred solution-branch. $\endgroup$ Feb 26, 2021 at 19:34

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