Why does disappear my plot when i zoom in in the interest zone?

Question

Pablo Faus Llopis 2022-12-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1870762-why-does-disappear-my-plot-when-i-zoom-in-in-the-interest-zone

评论： Pablo Faus Llopis 2022-12-7

This is my code

% CONSTANTS
mu=4*pi*10^-7;
c=3E8;
d=6E-6;
n1=1.45125;
n0=1.44725;
n12=n1^2;
n02=n0^2;
n=(n0/n1)^2;
p=2*pi/c;
NA=sqrt((n1^2)-(n0^2));
% DEFINITION OF MY FUNCTION
beta1= @(k,b) (1/c)*((b)+((k.*n02*(n12-b.^2)*sec(k.*sqrt(n12-b.^2))^2)/(((b.*tan(k.*sqrt(n12-b.^2)))/(sqrt(n12-b.^2)))-(k.*b.*sec(k.*sqrt(n12-b.^2))^2)-((b.*n12)/(sqrt(b.^2-n02))))));
% REPRESENTATION 
fimplicit(beta1, [0 20*pi 0 500])

when i zoom in the interest zone (x (0,10),y(0,10), the plot disappears, why does this happen?

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Answer 1

Walter Roberson 2022-12-4

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https://ww2.mathworks.cn/matlabcentral/answers/1870762-why-does-disappear-my-plot-when-i-zoom-in-in-the-interest-zone#answer_1119862

编辑：Walter Roberson 2022-12-4

在 MATLAB Online 中打开

Your code assumes that there are continuous solutions for at least a subrange of [0 20*pi]

% CONSTANTS

mu=4*pi*10^-7;

c=3E8;

d=6E-6;

n1=1.45125;

n0=1.44725;

n12=n1^2;

n02=n0^2;

n=(n0/n1)^2;

p=2*pi/c;

NA=sqrt((n1^2)-(n0^2));

% DEFINITION OF MY FUNCTION

syms b k

beta1 = matlabFunction( (1/c)*((b)+((k.*n02*(n12-b.^2)*sec(k.*sqrt(n12-b.^2))^2)/(((b.*tan(k.*sqrt(n12-b.^2)))/(sqrt(n12-b.^2)))-(k.*b.*sec(k.*sqrt(n12-b.^2))^2)-((b.*n12)/(sqrt(b.^2-n02)))))), 'vars', [k,b])

beta1 = function_handle with value:

@(k,b)b.*3.333333333333333e-9+(k.*1.0./cos(k.*sqrt(-b.^2+2.1061265625)).^2.*(b.^2-2.1061265625).*6.981775208333333e-9)./(b.*1.0./sqrt(b.^2-2.0945325625).*2.1061265625+b.*k.*1.0./cos(k.*sqrt(-b.^2+2.1061265625)).^2-b.*tan(k.*sqrt(-b.^2+2.1061265625)).*1.0./sqrt(-b.^2+2.1061265625))

Kvals = linspace(0, 6, 500).';

Bs = arrayfun(@(k) vpasolve(beta1(k,b),b), Kvals);

plot(Kvals, [real(Bs), imag(Bs)]); yline(0);

legend({'real', 'imaginary'})

For the range up to 6, the only solutions are roughly 10 individual points where the imaginary part of the vpasolve() are 0. Not a continuous line, just individual points

Kvals2 = linspace(6, 50*pi, 250).';

Bs2 = arrayfun(@(k) vpasolve(beta1(k,b),b), Kvals2);

plot(Kvals2, [real(Bs2), imag(Bs2)]); yline(0);

legend({'real', 'imaginary'})

Above 6 there appear to be a number of individual-point solutions.

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Walter Roberson 2022-12-4

在 MATLAB Online 中打开

One can make the hypothesis that each solution should be "close to" the previous one.

% CONSTANTS

mu=4*pi*10^-7;

c=3E8;

d=6E-6;

n1=1.45125;

n0=1.44725;

n12=n1^2;

n02=n0^2;

n=(n0/n1)^2;

p=2*pi/c;

NA=sqrt((n1^2)-(n0^2));

% DEFINITION OF MY FUNCTION

syms b k

beta1 = matlabFunction( (1/c)*((b)+((k.*n02*(n12-b.^2)*sec(k.*sqrt(n12-b.^2))^2)/(((b.*tan(k.*sqrt(n12-b.^2)))/(sqrt(n12-b.^2)))-(k.*b.*sec(k.*sqrt(n12-b.^2))^2)-((b.*n12)/(sqrt(b.^2-n02)))))), 'vars', [k,b])

beta1 = function_handle with value:

@(k,b)b.*3.333333333333333e-9+(k.*1.0./cos(k.*sqrt(-b.^2+2.1061265625)).^2.*(b.^2-2.1061265625).*6.981775208333333e-9)./(b.*1.0./sqrt(b.^2-2.0945325625).*2.1061265625+b.*k.*1.0./cos(k.*sqrt(-b.^2+2.1061265625)).^2-b.*tan(k.*sqrt(-b.^2+2.1061265625)).*1.0./sqrt(-b.^2+2.1061265625))

Kvals = linspace(0, 6, 500).';

Kvals(1) = []; %0 is a special case

N = length(Kvals);

Bs = zeros(N,1);

guess = 1+1i;

for idx = 1 : N

sol = vpasolve(beta1(Kvals(idx), b), guess);

if isempty(sol)

Bs(idx) = nan;

else

Bs(idx) = sol;

guess = sol;

end

plot(Kvals, [real(Bs), imag(Bs)]); yline(0);

legend({'real', 'imaginary'})

The solutions for the fimplicit are the locations where the orange-ish line crosses 0 -- only two points in this range.

Why is the previous plot so messy then? Well that means there are multiple point-wise solutions, and that should perhaps be investigated more -- but even so, the solutions are still point-wise, not continuous curves.

Pablo Faus Llopis 2022-12-7