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I have an equation and I have tried solving it to find values of $b$ but it doesn't give me the same figure as in the book Foundations for guided-wave optics by Chin-Lin Chen.

The equation that I have tried to solve is Equation 3.14 in the book: $\newcommand{\Ai}{\operatorname{Ai}}$ $\newcommand{\Bi}{\operatorname{Bi}}$ $$ \frac{V^{1/3} \sqrt{b} \Ai(V^{2/3}b)+\Ai'(V^{2/3}b)}{V^{1/3} \sqrt{b} \Bi(V^{2/3}b)+\Bi'(V^{2/3}b)}= \frac{V^{1/3} \sqrt{a+b} \Ai[-V^{2/3}(1-b)]-\Ai'[-V^{2/3}(1-b)]}{V^{1/3} \sqrt{a+b} \Bi[-V^{2/3}(1-b)]-\Bi'[-V^{2/3}(1-b)]} $$

I have tried on it as follows:

a = 10; m = 1;
bValue = Table[
   FindRoot[((
      Subscript[V, i]^(1/3)*Sqrt[b]*AiryAi[Subscript[V, i]^(2/3)*b] + 
       AiryAiPrime[Subscript[V, i]^(2/3)*b])/(
      Subscript[V, i]^(1/3)*Sqrt[b]*AiryBi[Subscript[V, i]^(2/3)*b] + 
       AiryBiPrime[Subscript[V, i]^(2/3)*b])) - ((
      Subscript[V, i]^(1/3)*Sqrt[a + b]*
        AiryAi[-Subscript[V, i]^(2/3)*(1 - b)] - 
       AiryAiPrime[-Subscript[V, i]^(2/3)*(1 - b)])/(
      Subscript[V, i]^(1/3)*Sqrt[a + b]*
        AiryBi[-Subscript[V, i]^(2/3)*(1 - b)] - 
       AiryBiPrime[-Subscript[V, i]^(2/3)*(1 - b)])) - m Pi, {b, 
     0.037309}], {Subscript[V, i], 7.7, 10.2, 0.1}];
VV = Table[i, {i, 7.7, 10.2, 0.1}];
bValues = Re[b /. bValue]
bValuesList = Transpose[{Join[{b}, bValues]}]
bGrid = Grid[bValuesList, Frame -> All] 
bAgainstVdata = Partition[Riffle[VV, bValues], 2]

I have tried once add +m Pi and another trying add -m Pi where the m is the mode order and both of them did not give me the figure 3.4 in the book:

enter image description here

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  • $\begingroup$ I tried to fix your post. You have lots of $$ around code, which I have no idea why. Also latex is not code, so it should start from left edge of screen. If you move it to the right, it becomes code. You still have Latex code inside Mathematica commands. For example, you have FindRoot[LATEX CODE] which I can't fix this. I think you should post your code again using InputForm $\endgroup$
    – Nasser
    Commented Aug 9, 2022 at 9:19
  • $\begingroup$ Please notice that Mathematica syntax is very different from LaTeX syntax. For instance in LaTeX you write \frac23 and in Mathematica you write 2/3. Please edit your post accordingly. $\endgroup$
    – yarchik
    Commented Aug 9, 2022 at 10:07

1 Answer 1

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You've misunderstood the meaning of $m$ in the book. The following gives the desired plot:

ContourPlot[
 Table[(V^(1/3) Sqrt[b] AiryAi[V^(2/3) b] + AiryAiPrime[V^(2/3) b])/(
    V^(1/3) Sqrt[b] AiryBi[V^(2/3) b] + AiryBiPrime[V^(2/3) b]) == (
    V^(1/3) Sqrt[a + b] AiryAi[-V^(2/3) (1 - b)] - AiryAiPrime[-V^(2/3) (1 - b)])/(
    V^(1/3) Sqrt[a + b] AiryBi[-V^(2/3) (1 - b)] - 
     AiryBiPrime[-V^(2/3) (1 - b)]), {a, {10, 100}}] // Evaluate, {V, 0, 20}, {b, 0, 
  1}]

enter image description here

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  • $\begingroup$ can you explain to me about] of m ?!! How you got for m=0,1,2,3?! $\endgroup$
    – A.J.H
    Commented Aug 9, 2022 at 13:00
  • $\begingroup$ You can (very roughly) understand m as the index of the solution of the given equation. For more info please check the document of ContourPlot, once you understand the meaning of my code, you should get a general idea about what the m is. Also, though the discussion for meaning of m and bV diagram is brief in Chapter 3, it's very detailed in Chapter 2 (for the analysis of a similar case), if you have difficulty in understanding them, please consider re-reading Chapter 2 carefully. @A.J.H $\endgroup$
    – xzczd
    Commented Aug 9, 2022 at 13:30
  • $\begingroup$ In fact i am mathematician researcher not physics researcher , but I'm trying to practice equations so that I can be more good on mathematica .... many thanks to you $\endgroup$
    – A.J.H
    Commented Aug 9, 2022 at 14:02

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