solving transcendental equation numerically

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vijay
vijay on 16 Jan 2014
Commented: Azzi Abdelmalek on 16 Jan 2014
I am trying to solve the 2 transcendental equations for 2 variables A,M for the given L
PBAR = 0;
L = [0.1,0.5,1.0,1.5,2.0];
equation1 = A^3 - L*A^2.*(sqrt(M.^2-1) + M.^2.*acos(1./M)) - PBAR;
equation2 = L*A^2/2*(sqrt(M^2-1) + (M^2-2)*acos(1/M)) + 4*L^2*A/3*(sqrt(M^2-1)*acos(1/M)-M+1)-1;
can any one help me how to solve it numerically

Answers (2)

Mischa Kim
Mischa Kim on 16 Jan 2014
Edited: Mischa Kim on 16 Jan 2014
Hello vijay, what are the equations equal to? Zero? In other words,
0 = A^3 - L*A^2.*(sqrt(M.^2 - 1) + M.^2.*acos(1./M)) - PBAR;
0 = L*A^2/2*(sqrt(M^2 - 1) + (M^2 - 2)*acos(1/M)) + 4*L^2*A/3*(sqrt(M^2 - 1)*acos(1/M) - M+1)-1;
If so, this is a root-finding problem: find A and M such that the two equations are satisfied. There is plenty of literature on solving systems of non-linear equations.
Try Newton-Raphson. The challenge you might run into is to find good starting values for the search, such that the algorithm coverges properly. Also be aware that there could be multiple soulutions to your problem.

Azzi Abdelmalek
Azzi Abdelmalek on 16 Jan 2014
M=sym('M',[1,5])
A=sym('A',[1 5])
PBAR = 0;
L = [0.1,0.5,1.0,1.5,2.0];
equation1 = A.^3 - L.*A^.2.*(sqrt(M.^2-1) + M.^2.*acos(1./M)) - PBAR;
equation2 = L.*A.^2/2.*(sqrt(M.^2-1) + (M^.2-2).*acos(1./M)) + 4*L.^2.*A/3.*(sqrt(M.^2-1).*acos(1./M)-M+1)-1;
solve([equation1;equation2])
  4 Comments
vijay
vijay on 16 Jan 2014
there are solutions for different values of L
for eg:
for L = 5.0
the values for A = 1.800; M = 1.01574;
but i am not able to solve the equations
Azzi Abdelmalek
Azzi Abdelmalek on 16 Jan 2014
syms A M
PBAR = 0;
L = [0.1,0.5,1.0,1.5,2.0];
for k=1:numel(L)
equation1 = A.^3 - L(k).*A^.2.*(sqrt(M.^2-1) + M.^2.*acos(1./M)) - PBAR;
equation2 = L(k).*A.^2/2.*(sqrt(M.^2-1) + (M^.2-2).*acos(1./M)) + 4*L(k).^2.*A/3.*(sqrt(M.^2-1).*acos(1./M)-M+1)-1;
sol=solve([equation1;equation2]);
M1(k,1)=sol.M
A1(k,1)=sol.A
end

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