Solving a system of equations with dependent variables symbolically

Warning: Solutions are parameterized by the symbols: u. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

sol = struct with fields:

a: -(s^7 + 3*s^6*u*x - 4*s^6*u*z + 4*s^6*y*z + s^5*u^4 + s^5*u^3*y + 3*s^5*u^2*x^2 - 5*s^5*u^2*x*z + 2*s^5*u^2*z^2 + s^5*u^2 + 11*s^5*u*x*y*z - 15*s^5*u*y*z^2 + s^5*u*y - 4*s^5*x^2*z^2 + 3*s^5*x^2 + 7*s^5*x*z^3 - 5*s^5*x*z + 6*s^5*y^2*z^2 - 3*s^5*… b: (- s^5*u^3 - s^5*u^2*y + 3*s^5*u*x^2 - 11*s^5*u*x*z + 9*s^5*u*z^2 - s^5*u - s^5*y + 2*s^4*u^4*x - 3*s^4*u^4*z + s^4*u^3*x*y - 6*s^4*u^3*y*z + 5*s^4*u^2*x^3 - 19*s^4*u^2*x^2*z + 22*s^4*u^2*x*z^2 + 3*s^4*u^2*x - 4*s^4*u^2*y^2*z - 8*s^4*u^2*z^3 - … c: (s^6*u - s^5*u^2*x + s^5*u^2*z + 4*s^5*u*y*z - s^5*x + s^5*z + s^4*u^3*x^2 - 3*s^4*u^3*x*z + 2*s^4*u^3*z^2 - 4*s^4*u^2*x*y*z + 4*s^4*u^2*y*z^2 - s^4*u*x^4 + 10*s^4*u*x^3*z - 34*s^4*u*x^2*z^2 + 4*s^4*u*x^2 + 42*s^4*u*x*z^3 - 11*s^4*u*x*z + 6*s^4… d: u

sol.a, sol.b,sol.c, sol.d

ans =

u

11 个评论
显示 9更早的评论隐藏 9更早的评论

Walter Roberson 2023-2-26

在 MATLAB Online 中打开

syms a b c d e f g

syms x y z w u v s

syms cte1 cte2

eqns = [x == a *cte1 - e * c, y == - (b * e) / cte2, z == (a * g) / cte2, w == d * g - b * f, u == (a * e) / cte2, v == cte1 * b + e * d, s == a * f + c * g];

a_partial = solve(eqns(1), a)

a_partial =

eqns2 = subs(eqns(2:end), a, a_partial);

b_partial = solve(eqns2(1), b)

b_partial =

eqns3 = subs(eqns2(2:end), b, b_partial);

c_partial = solve(eqns3(end), c)

c_partial =

eqns4 = subs(eqns3(1:end-1), c, c_partial);

d_partial = solve(eqns4(end), d)

d_partial =

eqns5 = subs(eqns4(1:end-1), d, d_partial);

e_partial = solve(eqns5(end), e)

e_partial =

eqns6 = subs(eqns5(1:end-1), e, e_partial); %now has two rows because two e solutions

f_partial_a_sol = solve(eqns6(1,1), f, 'returnconditions', true)

f_partial_a_sol = struct with fields:

f: [2×1 sym] parameters: [1×0 sym] conditions: [2×1 sym]

f_partial_a = f_partial_a_sol.f

f_partial_a =

f_partial_a_sol.conditions

ans =

f_partial_b_sol = solve(eqns6(2,1), f, 'returnconditions', true)

f_partial_b_sol = struct with fields:

f: [2×1 sym] parameters: [1×0 sym] conditions: [2×1 sym]

f_partial_b = f_partial_b_sol.f

f_partial_b =

f_partial_b_sol.conditions

ans =

eqns7_a = subs(eqns6(1,2:end), f, f_partial_a)

eqns7_a =

eqns7_b = subs(eqns6(2,2:end), f, f_partial_b)

eqns7_b =

g_partial_aa = solve(eqns7_a(1,1), g)

g_partial_aa =

g_partial_ab = solve(eqns7_a(2,1), g)

g_partial_ab =

g_partial_ba = solve(eqns7_b(1,1), g)

g_partial_ba =

g_partial_bb = solve(eqns7_b(2,1), g)

g_partial_bb =

You can now back-substitute the four different branches -- each f has two branches and each of those leads to two different g.

After you get all the way back to a, b, c, d, e, f, g coefficients, you might want to try to prove that all of the outputs are real-valued. That might be a bit tricky, especially without knowing the signs of x y z w u v s cte1 cet2 .

You could possibly cut several steps off of the process by solving eqns([1 2 end-2:end]) for [a b c d e] in one step and then go after f and g.

xav 2023-2-27