SOLVE-ONLY one solution

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SYML2nd 2023-2-17

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https://ww2.mathworks.cn/matlabcentral/answers/1914545-solve-only-one-solution

评论： Walter Roberson 2023-2-19

I am trying to solve a system of two equations which I know they intersect in three points. I am using this code (which I simplified)

syms x y
[x_sol,y_sol] = solve(Eqn1,Eqn);

Why it displays only the first solution and not the three solutions?

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Torsten 2023-2-17

Maybe if we knew the equations and the code you use, we could give an answer.

Torsten 2023-2-19

在 MATLAB Online 中打开

Use

[x_sol, y_sol] = vpasolve([a*exp(-b*x) == y, c*x^3 + d*x^2 + e*x + f == y], [x, y]);

instead of

[x_sol, y_sol] = solve([a*exp(-b*x) == y, c*x^3 + d*x^2 + e*x + f == y], [x, y]);

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Answer 1

Walter Roberson 2023-2-17

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https://ww2.mathworks.cn/matlabcentral/answers/1914545-solve-only-one-solution#answer_1174185

solve() will return multiple solutions only if the equations are simultaneous multinomials -- though in some cases it is able to convert trig equations into internal polynomials.

It does not even return multiple solutions to equations involving exp() in terms of the LambertW function when those are available.

In all other cases, it will return a single closed-form solution if it can find one, and otherwise a single numeric solution.

In some cases, especially ones involving trig functions, it is able to compute parameterized additional solutions, if you request 'returnconditions', true

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SYML2nd 2023-2-19

编辑：SYML2nd 2023-2-19

I tried but it does not work. Does it exists an other method (not using 'solve' maybe) that does not have this problem?

Thanks

Walter Roberson 2023-2-19

You could try solving iteratively, solve one equation for one variable, substitute the result into the remaining equations, and so on. Depending on what the equations are like, you might find a step for which there are multiple solutions, in which case you can iterate over all of the possibilities, hoping that they will lead to valid solutions.

Unfortunately it does not take all that much before formulas are too complicated to make progress on.

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