Find all possible roots of transcendental function

Question

University 2023-9-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2018751-find-all-possible-roots-of-transcendental-function

评论： Sam Chak 2023-9-9

Please can help with how find all the possible roots of q for xi and u and use the results to compute tau?

clear 
clc
close all
%% Model parameters
alpha3 = -0.001104; eta1 = 0.0240; K1 = 6*10^(-12);              
gamma1 = 0.1093; chi_a = 1.219*10^(-6); d = 0.0002;
H_c  =   pi*sqrt(K1/chi_a)/d;                    
alpha = 1-((alpha3).^2/(gamma1*eta1));
xi =linspace(-0.3, 0.3, 81);              
u = linspace(0, 3, 81);                   
for iu=1:length(u)
    for  ixi=1:length(xi)
        q(iu, ixi) - (1-alpha)*tan(q) + (1/gamma1)*((alpha3*ixi/eta1)*tan(q) + chi_a*(iu*H_c)^2*q)*(alpha*gamma1*  (4*K1*q(iu, ixi).^2/d.^2  - (alpha3*ixi)/eta1  - chi_a*(iu*H_c).^2)^(-1));
        tau(iu, ixi) = (alpha*gamma1*(4*K1*q(iu, ixi).^2/d.^2  - (alpha3*ixi)/eta1  - chi_a*(iu*H_c).^2)^(-1)
    end 
end

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James Tursa 2023-9-8

Please post an image of the function you are trying to find the roots of, rather than have us trying to decipher your code.

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Answer 1

Torsten 2023-9-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2018751-find-all-possible-roots-of-transcendental-function#answer_1305901

编辑：Torsten 2023-9-8

在 MATLAB Online 中打开

Really all zeros ? Then you cannot save them in any array since for each u and xi, you get an infinite number of them. Is the function correct ? There were some strange settings therein: the "q(iu, ixi) -" at the beginning and the use of ixi and iu instead of xi(ixi) and u(iu) in the function definition.

%% Model parameters

alpha3 = -0.001104; eta1 = 0.0240; K1 = 6*10^(-12);

gamma1 = 0.1093; chi_a = 1.219*10^(-6); d = 0.0002;

H_c = pi*sqrt(K1/chi_a)/d;

alpha = 1-((alpha3).^2/(gamma1*eta1));

fun = @(q,xi,u)(1-alpha).*tan(q) + (1./gamma1).*((alpha3.*xi./eta1).*tan(q) + chi_a.*(u*H_c).^2.*q).*(alpha.*gamma1.*(4.*K1.*q.^2/d.^2-(alpha3.*xi)./eta1-chi_a.*(u.*H_c).^2).^(-1));

xi = 0.2;

u = 1.5;

hold on

q = 0:0.01:pi/2-0.1;

plot(q,fun(q,xi,u),'b')

q = pi/2+0.1:0.01:3*pi/2-0.1;

plot(q,fun(q,xi,u),'b')

q = 3*pi/2+0.1:0.01:5*pi/2-0.1;

plot(q,fun(q,xi,u),'b')

q = 5*pi/2+0.1:0.01:7*pi/2-0.1;

plot(q,fun(q,xi,u),'b')

hold off

grid on

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University 2023-9-9

Maple1.png

Thank you Sam. The maple code is a bit long. Unless, you want the complete code. I have attached the contour plot of tau with respect to xi and u

Sam Chak 2023-9-9

@University, Based on the contour plot, it is most likely that the circled region represents the zero-crossing region where the roots are located. Therefore, the solution should appear as a curved line. However, if we examine the presumed implicit function

, which has yet to be clarified, it is possible that this region corresponds to where discontinuities occur due to the tangent function in

.