solving a system of nonlinear equations in terms of parameter (essentially rearrangement)

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Seyoung Won 2019-8-26

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https://ww2.mathworks.cn/matlabcentral/answers/477599-solving-a-system-of-nonlinear-equations-in-terms-of-parameter-essentially-rearrangement

编辑： Torsten 2019-8-26

I'm having a trouble solving a system of equations with multiple parameters and unknowns

syms a b c k t p1 p2 v1 v2 x
eqn_x = x == a + ((p2-v2)-(p1-v1))/(2*t*(1-a-b)) + (1-a-b)/2 ;
eqn_p1 = p1 == c - k + (1/3)*(v1-v2) + t*(1-a-b)*(1+(a-b-4*v1-2*v2)/3) ;
eqn_p2 = p2 == c - k + (1/3)*(v2-v1) + t*(1-a-b)*(1+(a-b-4*v2-2*v1)/3) ;
eqn_v1 = v1 == (c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)) ;
eqn_v2 = v2 == (c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)) ;
eqns = [eqn_x, eqn_p1, eqn_p2, eqn_v1, eqn_v2] ;
S = solve(eqns, [p1 p2 v1 v2 x]) ;

This is my code, the goal is to get v1 and v2 term rearranged in terms of only a, b, c, k, & t terms.

there are 5 functions and 5 unknowns. though it is not a system of linear equation because v1 and v2 term depends on quadratic function of x, I guess there is a possible rearrangement.

But what I get is p1, p2, v1, v2, x being [0x1 sym]. How do I get those rearrangement?

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Answer 1

Torsten 2019-8-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/477599-solving-a-system-of-nonlinear-equations-in-terms-of-parameter-essentially-rearrangement#answer_389067

Does this work ?

syms a b c k t x
eqn = x == a + (((c - k + (1/3)*(((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)))-((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))) + t*(1-a-b)*(1+(a-b-4*((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)))-2*((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b))))/3))-((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b))))-((c - k + (1/3)*(((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))-((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)))) + t*(1-a-b)*(1+(a-b-4*((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))-2*((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b))))/3)) -((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))))/(2*t*(1-a-b)) + (1-a-b)/2 ;
S = solve(eqn,x)

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Seyoung Won 2019-8-26

no it doesn't.... and I think mere substitution wouldn't work because even if one put x terms into v_1 and v_2, at the end of substitution v_1 and v_2 are simultaneously determined so I think it's supposed to be solving a system of 5 equations.

But I have no idea where my code went wrong, a b c k t are just parameter values and I have 5 equations for 5 unknowns, x p1 p2 v1 v2