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I am fairly new to mathematica so I need a little bit of help. I need to plot the zeros of an equation containing confluent hypergeometric functions. The equation i need to solve is given by the following equation:

eq[j_, δ_, ϵ_, α_, Z_] = 
Exp[I*Sqrt[ϵ^2 - δ^2]]*((j - \
(Z*α*δ)/(I*Sqrt[ϵ^2 - δ^2]))*
 Hypergeometric1F1[
  Sqrt[j^2 - (Z*α)^2] + (Z*α*ϵ)/(I*
      Sqrt[ϵ^2 - δ^2]), 
  1 + 2*Sqrt[j^2 - (Z*α)^2], -2*I*
   Sqrt[ϵ^2 - δ^2]] + (Sqrt[
    j^2 - Z^2 α^2] + (Z*α*ϵ)/(I*
      Sqrt[ϵ^2 - δ^2]))*
 Hypergeometric1F1[
  1 + Sqrt[
    j^2 - (Z*α)^2] + (Z*α*ϵ)/(I*
      Sqrt[ϵ^2 - δ^2]), 
  1 + 2*Sqrt[j^2 - (Z*α)^2], -2*I*
   Sqrt[ϵ^2 - δ^2]]) 

It's a rather complex function. The key is that i need to plot the zeros of this function (zeros in epsilon domain) and that for different values of delta.

So I tried this code:

Plot[(ϵ /. 
Part[Solve[
  eq[1/2, δ, ϵ, 0.25, 1] == 0 && 
   0 < Abs@ϵ < 10, ϵ, Complexes], 
 3])/δ, {δ, 0, 10}]

I have to take the third piece of the list because I only need positive values op epsilon. The problem with this is that it is taking forever to run. I first tried Findroot but the fact that it needs a start value is a huge problem.

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1 Answer 1

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If I understand you correctly. You are looking for a function epsilon[delta] where epsilon are the roots of the equation eq.

eq[j_, δ_, ϵ_, α_, Z_] = 
  Exp[I*Sqrt[ϵ^2 - δ^2]]*((j -  \
(Z*α*δ)/(I*Sqrt[ϵ^2 - δ^2]))*
      Hypergeometric1F1[
       Sqrt[j^2 - (Z*α)^2] + (Z*α*ϵ)/(I*
           Sqrt[ϵ^2 - δ^2]), 
       1 + 2*Sqrt[j^2 - (Z*α)^2], -2*I*
        Sqrt[ϵ^2 - δ^2]] + (Sqrt[
         j^2 - Z^2 α^2] + (Z*α*ϵ)/(I*
           Sqrt[ϵ^2 - δ^2]))*
      Hypergeometric1F1[
       1 + Sqrt[
         j^2 - (Z*α)^2] + (Z*α*ϵ)/(I*
           Sqrt[ϵ^2 - δ^2]), 
       1 + 2*Sqrt[j^2 - (Z*α)^2], -2*I*
        Sqrt[ϵ^2 - δ^2]]);

x1 = Range[0, 0.9, 0.1];
sol1 = ϵ /. 
     FindRoot[eq[1/2, #, ϵ, 0.25, 1] == 0, {ϵ, 1}] & /@ x1//Chop

{1.72952, 1.71889, 1.71227, 1.7097, 1.71117, 1.71665, 1.72612, 1.7395, 1.75672, 1.77768}

x2 = Range[1, 10, 0.1];
sol2 = ϵ /.FindRoot[eq[1/2, #, ϵ, 0.25, 1] == 0, {ϵ, 0}] & /@ x2 //Chop

{1.80227, 1.83038, 1.86186, 1.89658, 1.93441, 1.97518, 2.01877, \ 2.06501, 2.11376, 2.16489, 2.21826, 2.27373, 2.33118, 2.39048, \ 2.45153, 2.5142, 2.57841, 2.64404, 2.71101, 2.77924, 2.84864, \ 2.91913, 2.99065, 3.06313, 3.13651, 3.21073, 3.28573, 3.36147, \ 3.4379, 3.51497, 3.59265, 3.67089, 3.74966, 3.82892, 3.90866, \ 3.98882, 4.0694, 4.15035, 4.23167, 4.31332, 4.39529, 4.47756, 4.5601, \ 4.6429, 4.72595, 4.80922, 4.89271, 4.9764, 5.06028, 5.14434, 5.22856, \ 5.31294, 5.39746, 5.48212, 5.56691, 5.65182, 5.73684, 5.82197, \ 5.9072, 5.99252, 6.07792, 6.16341, 6.24897, 6.3346, 6.4203, 6.50606, \ 6.59188, 6.67775, 6.76368, 6.84965, 6.93566, 7.02172, 7.10782, \ 7.19395, 7.28011, 7.36631, 7.45253, 7.53878, 7.62506, 7.71136, \ 7.79768, 7.88402, 7.97039, 8.05676, 8.14316, 8.22957, 8.316, 8.40243, \ 8.48888, 8.57535, 8.66182}

data = Thread@{x1, sol1}~Union~Thread@{x2, sol2};
ListLinePlot[data, AxesLabel -> {delta, epsilon}]

enter image description here

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  • $\begingroup$ @Robbe I have re-edit my answer. I hope it is clear. $\endgroup$
    – user31001
    Commented Sep 14, 2015 at 12:24

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