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You have a complicated non-linear system

Kitonum — Tue, 10 Apr 2012 19:20:09 Z

You have a complicated non-linear system of 4 equations with 4 unknowns and three parameters. Maple is usually not solve these systems symbolically. This system can be solved numerically, but it must be given the values of all parameters, such as:

restart;

eq1:=A2*t2*exp(zt*t2*om)*sin(t2*wd)+A3*t3*sin(t3*wd)*exp(zt*t3*om) = 0;

eq2:=A2*t2*exp(zt*t2*om)*cos(t2*wd)+A3*t3*cos(t3*wd)*exp(zt*t3*om) = 0;

eq3:= A2*exp(zt*t2*om)*sin(t2*wd)+A3*sin(t3*wd)*exp(zt*t3*om) = 0;

eq4:=1-A2-A3+A2*exp(zt*t2*om)*cos(t2*wd)+A3*cos(t3*wd)*exp(zt*t3*om) = 0;

zt:=1/2: wd:=2: om:=3:

s1:=fsolve({eq1,eq2,eq3,eq4},{t2=0..infinity, t3=0..infinity, A2=0..infinity,A3=0..infinity});

Symbolic solution

Markiyan Hirnyk — Tue, 10 Apr 2012 21:42:06 Z

This system of transcendental equations can be reduced to a polynomial system as follows.

restart;
eq1:=A2*t2*exp(zt*t2*om)*sin(t2*wd)+A3*t3*sin(t3*wd)*exp(zt*t3*om) = 0:
eq2:=A2*t2*exp(zt*t2*om)*cos(t2*wd)+A3*t3*cos(t3*wd)*exp(zt*t3*om) = 0:
eq3:= A2*exp(zt*t2*om)*sin(t2*wd)+A3*sin(t3*wd)*exp(zt*t3*om) = 0:
eq4:=1-A2-A3+A2*exp(zt*t2*om)*cos(t2*wd)+A3*cos(t3*wd)*exp(zt*t3*om) = 0:

> Eq1 := eval(eq1, [exp(zt*t2*om) = a/A2, exp(zt*t3*om) = b/A3, sin(t2*wd) = c, cos(t2*wd) = d, sin(t3*wd) = e, cos(t3*wd) = f]);

t2 a c + t3 e b = 0
> Eq2 := eval(eq2, [exp(zt*t2*om) = a/A2, exp(zt*t3*om) = b/A3, sin(t2*wd) = c, cos(t2*wd) = d, sin(t3*wd) = e, cos(t3*wd) = f]);

t2 a d + t3 f b = 0

> Eq3 := a^2*t2^2 = b^2*t3^2;


                               a ^2* t2 ^2 = b^2* t3 ^2
> Eq4 := c^2+d^2 = 1;

                                                                 c^2 + d^2 = 1
> Eq5 := e^2+f^2 = 1;

                                                                  e^2 + f^2 = 1
> Eq6 := a*c+b*e = 0;

                                a c + b e = 0

> Eq7 := 1-A2-A3+a*d+b*f = 0;

                         1 - A2 - A3 + a d + b f = 0
>Eq8 := c/d = e/f;

c/d = e/f

Then we solve the polynomial system of 8 equtions in 10 unknowns:

> SolveTools:-PolynomialSystem({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, Eq7, Eq8}, {A2, A3, a, b, c, d, e, f, t2, t3});

{ A2 = 1 - A3 + a + b, c = 0, d = 1, e = 0, f = 1, t2 = - t3*b/a}

This implies

> A := solve(sin(t3*wd) = 0, t3, allsolutions);#c=0

                                   Pi _Z1
                                   ------
                                     wd
> B := solve(sin(t2*wd) = 0, t2, allsolutions);# e=0

                                   Pi _Z2
                                   ------
                                     wd
> C := solve(eval(t2 = -t3*b/a, [t3 = A, t2 = B]), a);# a is eliminated



The solution depends on _Z1, _Z2 , which must be posint, and the parameter b. At last,

> solve({A3 >= 0, eval(1-A3+a+b, a = C) >= 0}, b);# A3 >= 0 and A2 >= 0

Explanation

Markiyan Hirnyk — Wed, 11 Apr 2012 20:00:24 Z

In answer to

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Answer: Symbolic solution
Answer Url: http://www.mapleprimes.com/questions/132830-Solve-Trigonometric-Equations#comment132863
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Thank you for your comments.
I am looking for the exact solution.
So, I have questions about your formulation.

I think that you used the trigonometric identity for Eq4 and Eq5.
But, I do not understand about Eq3 and Eq8.
Please explain two formulas in a little more detail.

Specillay, I am not an expert on maple.
Do you have any other way to be suggested?

Thanks

The equation eq1 is equivalent to t2*A2*exp(zt*t2*om)*sin(t2*wd) = -t3*A3*exp(zt*t3*om)*sin(t3*wd), which implies the squared equation t2^2*A2^2*exp(2*zt*t2*om)*sin(t2*wd)^2=t3^2*A3^2*exp(2*zt*t3*om)*sin(t3*wd)^2 . (1)

Similarly eq2 implies

t2^2*A2^2*exp(2*zt*t2*om)*cos(t2*wd)^2=t3^2*A3^2*exp(2*zt*t3*om)*cos(t3*wd)^2. (2)

Adding the LHS and RHS of (1) and (2) and taking into account sin(x)^2+cos(x)^2=1, we obtain Eq3. The Eq8 is obtained through dividing (2) by (1). See ?SolveTools[PolynomialSystem] for more info concerning the solution.

I don't see another way to solve the system under consideration symbolically.

Adjustment

Markiyan Hirnyk — Sat, 14 Apr 2012 09:30:25 Z

The end of my solution should be corrected as the variable b is not a parameter, b = A3*exp(zt*Pi*t3*om/wd) =
A3*exp(zt*Pi*_Z1*om/wd). In view of it
> C := eval(solve(eval(t2 = -t3*b/a, [t3 = A, t2 = B]), a), b = A3*exp(zt*Pi*_Z1*om/wd));


                           _Z1*A3*exp(zt*Pi*_Z1*om/wd)/_Z2
                         -
> solve(eval(A2 = 1-A3+a+b, [a = C, b = A3*exp(zt*Pi*_Z1*om/wd)]), A3);



If 0 < A2, A2 < 1, then the condition _Z1 >= _Z2 is sufficient for A3 > 0.