solving transcendental equation numerically

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vijay
vijay 2014-1-16
评论： Azzi Abdelmalek 2014-1-16
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

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回答（2 个）

Mischa Kim
Mischa Kim 2014-1-16
编辑：Mischa Kim 2014-1-16
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 2014-1-16
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])
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vijay
vijay 2014-1-16
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 2014-1-16
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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