how to solve transcedental equation in matlab

Question

alburary daniel 2018-6-13

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https://ww2.mathworks.cn/matlabcentral/answers/405381-how-to-solve-transcedental-equation-in-matlab

编辑： Walter Roberson 2018-7-27

I am practicing solving the next transcendental equation in matlab

(a/c)*sqrt((b*m1)^2-(p*c)^2)-atan( sqrt(( (p*c)^2-(b*m2)^2  ) /( (b*m1)^2-(p*c)^2  ) ) )-atan( sqrt(( (p*c)^2-(b*m3)^2  ) /( (b*m1)^2-(p*c)^2  ) ) ) == r*pi

here

a=1x10^-6;
c= 3x10^8;
m1=2.2;
m2=1.5;
m3=1;

I was trying to plot "p" vs "b" where "b" runs from 0 to 3x10^15 and r is a parameter that takes values of 0, 1 and 2. I already tried all day but I cannot find solution, I tried with fzero(fun,xo) without success, can you give any suggestion?

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alburary daniel 2018-6-15

without success today, Can you help me?

Walter Roberson 2018-7-27

编辑：Walter Roberson 2018-7-27

Please do not close questions that have an answer. You were informed about that before https://www.mathworks.com/matlabcentral/answers/405170-how-to-solve-transcendental-equations-in-matlab#comment_578186

I spent several hours working on your question, and it is very frustrating that my answer just disappeared.

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Answer 1

Walter Roberson 2018-6-15

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/405381-how-to-solve-transcedental-equation-in-matlab#answer_324774

在 MATLAB Online 中打开

I got code to work... but the trend is exactly opposite of what you are looking for.

Q = @(v) sym(v, 'r');
arctan = @atan;
a = Q(1*10^-6);
c = Q(3*10^8);
m1 = Q(2.2);
m2 = Q(1.5);
m3 = Q(1);
syms b p r
eqn = (a/c)*sqrt((b*m1)^2-(p*c)^2)-atan( sqrt(( (p*c)^2-(b*m2)^2  ) /( (b*m1)^2-(p*c)^2  ) ))-atan( sqrt(( (p*c)^2-(b*m3)^2  ) /( (b*m1)^2-(p*c)^2  ) ) ) - r*pi;
B0min = Q(3000000000000000)*sqrt(Q(259))*arctan(5*sqrt(Q(259))*sqrt(Q(5))*(1/Q(259)))*(1/Q(259));
B0max = Q(3*10^15);
B0 = linspace(B0min, B0max, 100);
nB0 = length(B0);
R = 0 : 2;
nR = length(R);
ps = zeros(nB0, nR, 'sym');
for ridx = 1 : nR
    this_r = R(ridx);
  eqnr = subs(eqn, r, this_r);
    for K = 1 : nB0
        this_b = B0(K);
        this_eqn = subs(eqnr, b, this_b);
          sols = vpasolve(this_eqn, p, [this_b/200000000, 11*this_b/1500000000]);
          if isempty(sols)
              fprintf('No symbolic solution for eqn, r = %d, b = %g\n', this_r, this_b);
              sols = nan;
          else
              % fprintf('symbolic okay for eqn, r = %d, b = %g\n', this_r, this_b);
          end
          ps(K, ridx) = sols;
      end
  end
plot(ps, B0)
xlabel('p')
ylabel('b')
legend( sprintfc('r = %d', R) )

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Walter Roberson 2018-6-22

在 MATLAB Online 中打开

Here is an updated version of the code:

Q = @(v) sym(v, 'r');
arctan = @atan;
a = Q(1*10^-6);
c = Q(3*10^8);
m1 = Q(2.2);
m2 = Q(1.5);
m3 = Q(1);
syms b p r
eqn = (a/c)*sqrt((b*m1)^2-(p*c)^2)-atan( sqrt(( (p*c)^2-(b*m2)^2  ) /( (b*m1)^2-(p*c)^2  ) ))-atan( sqrt(( (p*c)^2-(b*m3)^2  ) /( (b*m1)^2-(p*c)^2  ) ) ) - r*pi;
%the following B0min is for 2.2, 1.5, 1
%B0min = Q(3000000000000000)*sqrt(Q(259))*arctan(5*sqrt(Q(259))*sqrt(Q(5))*(1/Q(259)))*(1/Q(259));
B0min = arctan(sqrt((m2-m3)*(m2+m3)/((m1-m2)*(m1+m2))))*c/(a*sqrt((m1-m2)*(m1+m2)));
B0max = Q(3*10^15);
B0 = linspace(B0min, B0max, 100);
nB0 = length(B0);
R = 0 : 2;
nR = length(R);
ps = zeros(nB0, nR, 'sym');
for ridx = 1 : nR
    this_r = R(ridx);
    eqnr = subs(eqn, r, this_r);
    badrun = false;
    for K = 1 : nB0
        this_b = B0(K);
        this_eqn = subs(eqnr, b, this_b);
        sols = vpasolve(this_eqn, p, [this_b*m2/c, this_b*m1/c]);
        % "the necesary condition to find the roots is that m2*b<p<m1*b for any value of b"
        %sols = vpasolve(this_eqn, p, [this_b*m2, this_b*m1]);
          if isempty(sols)
              if ~badrun
                  fprintf('No symbolic solution for eqn starting from r = %d, b = %g\n', this_r, this_b);
                  badrun = true;
              end
              sols = nan;
          else
              if badrun
                  fprintf('symbolic okay again starting from r = %d, b = %g\n', this_r, this_b);
              elseif K == 1
                  fprintf('symbolic okay starting from r = %d, b = %g\n', this_r, this_b);
              end
              badrun = false;
          end
          ps(K, ridx) = sols;
      end
      if badrun
          fprintf('No symbolic solution for eqn through to r = %d, b = %g\n', this_r, this_b);
          badrun = false;
      else
          fprintf('symbolic okay through to r = %d, b = %g\n', this_r, this_b);
      end
  end
plot(ps, B0)
xlabel('p')
ylabel('b')
legend( sprintfc('r = %d', R) )

alburary daniel 2018-6-23

well, here n1=m1 and n2=m2=m3

Walter Roberson 2018-6-24

I do not know. I notice that they are using different c values in the diagram, but you set up your equation with only one c value.

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how to solve transcedental equation in matlab

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