Asked by alburary daniel
on 13 Jun 2018 at 1:53

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?

Answer by Walter Roberson
on 15 Jun 2018 at 3:19

Accepted Answer

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) )

Walter Roberson
on 22 Jun 2018 at 17:17

It is not clear how the variable names there relate to your m1, m2, m3 ?

alburary daniel
on 23 Jun 2018 at 14:50

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

Walter Roberson
about 23 hours ago

Sign in to comment.

Opportunities for recent engineering grads.

Apply Today
## 4 Comments

## Walter Roberson (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/405381-how-to-solve-transcedental-equation-in-matlab#comment_578095

My tests suggest that for r = 0, that the smallest roots are at b = +/- 3000000000000000*sqrt(259)*arctan(5*sqrt(259)*sqrt(5)*(1/259))*(1/259) and p = +/- (b/200000000), for a total of two roots for that b.

It appears that there might perhaps be one primary root for each b larger than that; I have not chased to see if the negative is also a root. I do not have a characterization of the roots for larger b. At most p about 2.182E7 for b = 3E15. Possibly

approximatelyp = sqrt(b/(2*pi)) being slightly too high.## alburary daniel (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/405381-how-to-solve-transcedental-equation-in-matlab#comment_578197

In fact, I looking to reproduce a similar result, maybe could be of help. The solution are the blue lines. Red lines and black lines are other equations

## Walter Roberson (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/405381-how-to-solve-transcedental-equation-in-matlab#comment_578418

You ask about plotting p vs b where b runs from 0 to 3E15, but the axis that runs to about 3E15 on the plot is the left hand axis. That suggest that the plot is b vs p instead of p vs p.

It is possible to demonstrate that if a root exists at all, that it is in the range p = b/200000000 to 11*b/1500000000 so you could use those as the bounds on your fzero.

## alburary daniel (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/405381-how-to-solve-transcedental-equation-in-matlab#comment_578922

without success today, Can you help me?

Sign in to comment.