Slove function return empty solutions

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Roy 2023-3-16

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https://ww2.mathworks.cn/matlabcentral/answers/1929805-slove-function-return-empty-solutions

评论： Walter Roberson 2023-3-24

Hello, I'm trying to solve the attached syntax, but the aolve function return empty solutions. Please help.

syms V_1 V_2 x_1 x_2 r
pi1 = (V_1) * (x_1^r/(x_1^r+x_2^r)) - x_1 
pi2 = (V_2) * (x_2^r/(x_1^r+x_2^r)) - x_2
dpi1dx = diff(pi1, x_1) 
dpi2dx = diff(pi2, x_2)
s = solve(dpi1dx==0, dpi2dx==0, x_1, x_2)

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Roy 2023-3-24

Please help

Walter Roberson 2023-3-24

I am very busy these days.

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回答（2 个）

Walter Roberson 2023-3-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1929805-slove-function-return-empty-solutions#answer_1194625

Use dsolve for differential equations

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Walter Roberson 2023-3-21

在 MATLAB Online 中打开

If you set x_2 to be a constant multiple, c, of x_1, then

syms x_1 x_2

syms c r V_1 V_2 positive

pi1 = (V_1) * (x_1^r/(x_1^r+x_2^r)) - x_1

pi1 =

pi2 = (V_2) * (x_2^r/(x_1^r+x_2^r)) - x_2

pi2 =

dpi1dx = diff(pi1, x_1)

dpi1dx =

dpi2dx = diff(pi2, x_2)

dpi2dx =

eqns = subs([dpi1dx==0, dpi2dx==0], [x_2], [c*x_1])

eqns =

seqns = simplify(eqns)

seqns =

sol = solve(seqns(2), x_1, 'returnconditions', true)

sol = struct with fields:

x_1: (V_2*c^r*r)/(c + c*c^(2*r) + 2*c*c^r) parameters: [1×0 sym] conditions: r < 1/2 | 1/2 <= r

simplify(sol.conditions)

ans =

symtrue

simplify(sol.x_1, 'steps', 20)

ans =

seqns2 = simplify(subs(seqns(1), x_1, sol.x_1), 'steps', 20)

seqns2 =

which is to say that if x_1 and x_2 have a particular ratio, then V_1 and V_2 must have the same ratio.

Or you could read this as saying that if you know the ratio of V_1 and V_2 then x_1 and x_2 must have the same ratio, and x_1 is as given in ans above.

Notice that with V_2 == V_1 * c then V_2 / c is V_1 so ans could be rewritten as V_1 * c^r * r / (c^r+1)^2

Walter Roberson 2023-3-21

The problem is not solveable for most r .

For example for r = 3/2 then the solutions are

RootOf(4*Z^3*x_2^(3/2) + 2*Z^6 - 3*Z*x_2^(3/2)*V_1 + 2*x_2^3,Z)^2

which is the set of Z such that the expression 4*etc becomes 0. But notice the Z^6 part -- so you would need the closed-form solution for a degree 6 polynomial, and such solutions only exist if the expression can be factored into polynomials of degree 4 or lower.

If r = N/4 for odd integer N, then you need to solve something of degree either 2*N+4 (for small N) or degree 2*N (starting at N = 5). r = 1/5 and r = 3/5 are tractable (but long!!), the other N/5 are not tractable.

Roy 2023-3-21