How come an explicit solution cannot be found using solve?

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Veronica
Veronica 2013-2-23
The error is:
Warning: 362 equations in 1 variables.
> In /Applications/MATLAB_R2012a.app/toolbox/symbolic/symbolic/symengine.p>symengine at 54
In mupadengine.mupadengine>mupadengine.evalin at 97
In mupadengine.mupadengine>mupadengine.feval at 150
In solve at 160
In Hyper_HW2_P3 at 39
Warning: Explicit solution could not be
found.
> In solve at 169
In Hyper_HW2_P3 at 39
The code is:
% Given parameters of geoid for Earth
U = 6.263685953707e7; % potential [m^2/s^2]
GM = 3.986005e14; % gravitational constant [m^3/s^2]
J_2 = 1.08263e-3; % Jeffrey constants
J_3 = 2.532153e-7;
J_4 = 1.6109876e-7;
w = 7.292115147e-5; % angular acceleration [rad/s]
R_e = 6.378135e6; % equatorial radius [m]
R_p = 6.3567506e6; % polar radius [m]
phi = [0:180]; % angles of colatitude [deg]
% Relations for Legendre Polynomials for scalar potential equation
P_2 = (1/2).*(3.*cosd(phi).^2-1);
P_3 = (1/2).*(5.*cosd(phi).^3 - 3.*cosd(phi));
P_4 = (1/8).*(35.*cosd(phi).^3 - 30.*cosd(phi).^2 + 3);
% Determine R_RE from ellipse equation
R_RE = (R_e*R_p)./(sqrt(R_p^2.*sind(phi).^2 + R_e^2.*cosd(phi).^2));
% Determine potential from the centrifugal force to include the effects of
% rotation for the geoid
syms R_GEOID
U_c = -(1/2).*w^2.*R_GEOID^2.*sind(phi).^2;
% Analytically solve for R_GEOID from potential function
P2 = (R_e/R_GEOID)^2*J_2*P_2;
P3 = (R_e/R_GEOID)^3*J_3*P_3;
P4 = (R_e/R_GEOID)^4*J_4*P_4;
S = solve(-(1/2).*w^2.*R_GEOID.^2.*sind(phi).^2 == U_c,...
(GM./R_GEOID).*(1 - (P2 + P3 + P4)) == U )
Thanks!

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采纳的回答

Walter Roberson
Walter Roberson 2013-2-23
When you solve() a matrix, it tries to solve all of the entries simultaneously. It does not consider the elements to be a set of related problems that are to be solved for individually with an array of different results.
What you should do is leave phi as a symbolic parameter instead of making it numeric, and solve() for that, giving you a formula (or set of formulae) that is parameterized on phi. You can then subs() the actual numeric phi values in for the list of solutions.
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Walter Roberson
Walter Roberson 2013-2-24
Hmmmm -- and if you do that assignment like that, does
double(T)
work?
Have a look at a few of the elements of T, and double() them individually to see if double() is trying to return multiple values. If so then
for K = 1 : length(T)
RG{K} = double(T(K));
end
If you want to select down to real roots you can then
rRG = cellfun(@(C) C(imag(C)==0), 'Uniform', 0);
And if you are lucky exactly one element per index will remain and then you would
rRGv = cell2mat(rRG);
plot(phis, rRGv)
My analysis so far suggests that 1 of the roots is always real (for real values of b), and that the other 4 roots have non-zero imaginary parts that approach 0 in the limiting case as b approaches infinity (i.e., only one real root until b = infinity)

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