I have to solve a transcendental equation for a parameter, say $\beta$. Now, the $\beta$ has a range from $ik$ to $k$ where $i$ is the usual imaginary root $\sqrt{-1}$ and $k$ is a real number. Problem is, the transcendental equation has multiple solutions, and so I cannot guess what will be the proper choice for the initial value of $\beta$ in FindRoot
. I could do it if $\beta$ ranges from $p$ to $q$ where $p$ and $q$ are reals by plotting the transcendental equation's lhs and rhs. However, I don't know how to plot complex ranges.
Is there any way to guess the initial values?
Code I used:
e1 = 1; e2 = -1; e0 = 8.854*^-12; mu0 = 1.257*^-6; c = 3.0*^8; w = 2*Pi*2*^14;
k = w/c; a = 600*^-9; k1 = w*Sqrt[e1]/c; k2 = w*Sqrt[e2]/c; b = p + q I;
u = Sqrt[k1^2 - b^2]; ww = Sqrt[b^2 - k2^2]; v = 1;
t1 = (BesselJ[v - 1, u a] - BesselJ[v + 1, u a])/(2*u*BesselJ[v, u a]);
t2 = -(BesselK[v - 1, ww a] + BesselK[v + 1, ww a])/(2*ww* BesselK[v, ww a]);
x = (e1 t1 + e2 t2) (t1 + t2); y = ((b*v)/(k*a))*(1/u^2 + 1/ww^2);
ClickPane[ ContourPlot[{Re[x - y^2], Im[x - y^2]}, {p, 0, k}, {q, 0, k}],
(xycord = #) & ] Dynamic[xycord]
Plot[{x,y},{beta,p,q}]
Then from the plot i get a feel for good values. But i*k to k (where i is the usual imaginary root sqrt(-1) and k is a real number) is the allowed range for beta and i cannot writePlot[{x,y},{beta,i*k,k}]
it surely gives me error! By usingFindRoot[x==y,{beta,_initial value_}]
I get solution though but changing initial value just a little beta changes completely. $\endgroup$