How to solve 4 transcendental equations of 4 unknown values?

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LAKKIMSETTI SUNANDA 2021-2-2

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https://ww2.mathworks.cn/matlabcentral/answers/734140-how-to-solve-4-transcendental-equations-of-4-unknown-values

评论： Walter Roberson 2021-3-7

Hi. I have a problem in solving the equations. I have four equations and four unknowns and I have to find those 4 unknown variables. The equations are as follows:

syms Y2 Y3 l2 l3;
2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1)==0;
2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3)==0;
b1== (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));
b3== (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));

Here Y1, b1, b3, l1, lambda are known values and the unknown variables are Y2, Y3, l2, l3. The above t1 ,t2 ,t3 are in terms of l2 and l3.

Y1=0.02;
l1 = 0.0105;
lambda =[0.0426  0.0400  0.0405  0.0420];
b1=0.1079;
b3=0.03189;
t2_1=(2*pi/lambda(1))*l2; 
t3_1=(2*pi/lambda(1))*l3; 
t2_3=(2*pi/lambda(3))*l2; 
t3_3=(2*pi/lambda(3))*l3; 
t1_1=(2*pi/lambda(1))*l1;
t1_3=(2*pi/lambda(3))*l1;

I tried using vpasolve and also defining them as functions for plotting contour graphs. But I'm not able to find a solution of all positive values. Please help me with this. Thank you.

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Answer 1

darova 2021-2-2

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https://ww2.mathworks.cn/matlabcentral/answers/734140-how-to-solve-4-transcendental-equations-of-4-unknown-values#answer_613195

在 MATLAB Online 中打开

Use syms and solve

syms Y2 Y3 l2 l3
eq1 = 2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1);
eq2 = 2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3);
eq3 = b1 - (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));
eq4 = b3 - (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));
solve([eq1 eq2 eq3 eq4])

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Walter Roberson 2021-2-3

编辑：Walter Roberson 2021-2-3

在 MATLAB Online 中打开

syms Y1 l1 b1 b3

lambda = sym('lambda', [1 3])

lambda =

syms Y2 Y3 l2 l3

t2_1=(2*pi/lambda(1))*l2;

t3_1=(2*pi/lambda(1))*l3;

t2_3=(2*pi/lambda(3))*l2;

t3_3=(2*pi/lambda(3))*l3;

t1_1=(2*pi/lambda(1))*l1;

t1_3=(2*pi/lambda(3))*l1;

eq1 = 2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1);

eq2 = 2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3);

eq3 = b1 - (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));

eq4 = b3 - (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));

eqn = [eq1; eq2; eq3; eq4];

char(eqn)

ans = '[2*Y1*tan((2*l1*pi)/lambda1) + Y2*tan((2*l2*pi)/lambda1) + Y3*tan((2*l3*pi)/lambda1); 2*Y1*tan((2*l1*pi)/lambda3) + Y2*tan((2*l2*pi)/lambda3) + Y3*tan((2*l3*pi)/lambda3); b1 - (2*Y1*l1*pi)/lambda1 + (Y2*l2*pi*cos((2*l1*pi)/lambda1)^2)/(lambda1*cos((2*l2*pi)/lambda1)^2) + (Y3*l3*pi*cos((2*l1*pi)/lambda1)^2)/(lambda1*cos((2*l3*pi)/lambda1)^2); b3 - (2*Y1*l1*pi)/lambda3 + (Y2*l2*pi*cos((2*l1*pi)/lambda3)^2)/(lambda3*cos((2*l2*pi)/lambda3)^2) + (Y3*l3*pi*cos((2*l1*pi)/lambda3)^2)/(lambda3*cos((2*l3*pi)/lambda3)^2)]'

sol1 = solve(eqn(1:2), [Y2,Y3]);

sol1.Y2

ans =

sol1.Y3

ans =

eqn2 = subs(eqn(3:4), {Y2,Y3}, {sol1.Y2, sol1.Y3})

eqn2 =

symvar(eqn2)

ans =

We have reduced the system to two equations in l2 and l3, given Y1, b1, b3, l1, lambda1, lambda3 . With numeric values of those, there would be the potential for vpasolve() to find a solution. However, you will not be able to find a symbolic solution.

darova 2021-2-3

@Walter Roberson he has 4 uknowns

@LAKKIMSETTI SUNANDA I suggest you to express Y1 and use isosurface

Walter Roberson 2021-2-3

I showed how to reduce the four equations into two equations. Once you have solutions to the two equations you can back-substitute to find the other two values.

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Answer 2

Walter Roberson 2021-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/734140-how-to-solve-4-transcendental-equations-of-4-unknown-values#answer_614067

在 MATLAB Online 中打开

format long g

Q = @(v) sym(v);

syms Y2 Y3 l2 l3

Y1 = Q(0.02);

l1 = Q(0.0250);

lambda =[Q(0.1035), Q(0.0999), Q(0.0966)];

b1 = Q('11698461397958468980052614032629')/Q(10)^32;

b3 = Q('12515236662420125800636060478832')/Q(10)^32;

t2_1=(2*pi/lambda(1))*l2;

t3_1=(2*pi/lambda(1))*l3;

t2_3=(2*pi/lambda(3))*l2;

t3_3=(2*pi/lambda(3))*l3;

t1_1=(2*pi/lambda(1))*l1;

t1_3=(2*pi/lambda(3))*l1;

eq1 = 2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1);

eq2 = 2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3);

eq3 = b1 - (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));

eq4 = b3 - (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));

eqn = [eq1; eq2; eq3; eq4];

char(eqn)

ans = '[tan((100*pi)/207)/25 + Y2*tan((4000*l2*pi)/207) + Y3*tan((4000*l3*pi)/207); Y2*tan((10000*l2*pi)/483) - tan((233*pi)/483)/25 + Y3*tan((10000*l3*pi)/483); (2000*Y2*l2*pi*cos((100*pi)/207)^2)/(207*cos((4000*l2*pi)/207)^2) - (2*pi)/207 + (2000*Y3*l3*pi*cos((100*pi)/207)^2)/(207*cos((4000*l3*pi)/207)^2) + 11698461397958468980052614032629/100000000000000000000000000000000; (5000*Y2*l2*pi*cos((233*pi)/483)^2)/(483*cos((10000*l2*pi)/483)^2) - (5*pi)/483 + (5000*Y3*l3*pi*cos((233*pi)/483)^2)/(483*cos((10000*l3*pi)/483)^2) + 782202291401257862539753779927/6250000000000000000000000000000]'

sol1 = solve(eqn(1:2), [Y2,Y3]);

sol1.Y2

ans =

sol1.Y3

ans =

eqn2 = subs(eqn(3:4), {Y2,Y3}, {sol1.Y2, sol1.Y3})

eqn2 =

N = 10;

sols = zeros(N,4);

for K = 1 : N

initial = randn(2,1); %COLUMN

sol2 = vpasolve(eqn2, [l2, l3], initial);

backY2 = subs(sol1.Y2, sol2);

backY3 = subs(sol1.Y3, sol2);

sols(K,:) = vpa([backY2, backY3, sol2.l2, sol2.l3]);

end

sols

sols = 10×4

0.126485933403849 -0.167119007948023 -0.220120421257385 0.383807863154484 0.432341841059124 -0.233227313778012 1.13396636173026 0.796286515669898 -2.21198116937819 2.13838491109613 1.69346671322037 1.0702137362349 0.626751243150408 0.493888638447776 -1.66870698078474 1.9101114699078 -0.676916970037412 -0.429495987828577 0.773842328718426 0.22518335023655 0.17078702060627 -0.898235237721649 0.499807867212554 1.24946279610679 -0.872625762873269 -0.405122824760962 0.37514697209635 0.823405507279456 -0.944817212856229 -0.143244176682331 0.474651563810915 1.05004207412011 0.77749208639984 1.12244937333127 1.64897703075677 -2.02387843393713 0.43109071095526 -0.441555710731303 2.07339632941486 0.949422892891187

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Walter Roberson 2021-3-5

在 MATLAB Online 中打开

syms Y2 Y3 l2 l3;
Y1=0.02;
l1 = 0.0105;
lambda =[0.0426  0.0400  0.0405  0.0420];
b1=0.1079;
b3=0.03189;
t2_1=(2*pi/lambda(1))*l2; 
t3_1=(2*pi/lambda(1))*l3; 
t2_3=(2*pi/lambda(3))*l2; 
t3_3=(2*pi/lambda(3))*l3; 
t1_1=(2*pi/lambda(1))*l1;
t1_3=(2*pi/lambda(3))*l1;
eqn = [
2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1)==0;
2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3)==0;
b1== (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));
b3== (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));
];
sol1 = solve(eqn(1:2), [Y2,Y3]);
eqn2 = subs(eqn(3:4), {Y2,Y3}, {sol1.Y2, sol1.Y3});
N = 100;
sols = zeros(N,4);
solcount = 0;
for K = 1 : N
   initial = randn(2,1);  %COLUMN
    sol2 = vpasolve(eqn2, [l2, l3], initial);
    if isempty(sol2) || isempty(sol2.l2); continue; end
    backY2 = subs(sol1.Y2, sol2);
    backY3 = subs(sol1.Y3, sol2);
    thissol = vpa([backY2, backY3, sol2.l2, sol2.l3]);
    if ~all(thissol >= 0); continue; end
    solcount = solcount + 1;    
    sols(solcount,:) = thissol;
    break
end
sols = sols(1:solcount,:);
fprintf('Found %d solutions in %d tries\n', solcount, K);
disp(sols)

All of the candidate solutions I found had at least one negative value; I did not find any solutions that were all positive.

LAKKIMSETTI SUNANDA 2021-3-7

@Alex Sha Now this is giving us the positive solutions. So I don't think the equations are wrong as I checked those and derived them by hand.

@Walter Roberson We are now getting solutions for a different input. So it should definitely have positive solutions for the above given input too. Those roots will be very close to 0. So if we are able to consider a range of 0 to 0.2, I think we can find them. I'm still working on that and your help will be highly appreciated.

Walter Roberson 2021-3-7

在 MATLAB Online 中打开

For the first set of inputs, look at pretty(sol1.Y2)

             / 233 pi \    / 4000 pi l3 \      / 100 pi \    / 10000 pi l3 \
          tan| ------ | tan| ---------- | + tan| ------ | tan| ----------- |
             \   483  /    \     207    /      \   207  /    \     483     /
- ----------------------------------------------------------------------------------
  /    / 4000 pi l2 \    / 10000 pi l3 \      / 4000 pi l3 \    / 10000 pi l2 \ \
  | tan| ---------- | tan| ----------- | - tan| ---------- | tan| ----------- | | 25
  \    \     207    /    \     483     /      \     207    /    \     483     / /

and compare to pretty(sol1.Y3)

           / 233 pi \    / 4000 pi l2 \      / 100 pi \    / 10000 pi l2 \
        tan| ------ | tan| ---------- | + tan| ------ | tan| ----------- |
           \   483  /    \     207    /      \   207  /    \     483     /
----------------------------------------------------------------------------------
/    / 4000 pi l2 \    / 10000 pi l3 \      / 4000 pi l3 \    / 10000 pi l2 \ \
| tan| ---------- | tan| ----------- | - tan| ---------- | tan| ----------- | | 25
\    \     207    /    \     483     /      \     207    /    \     483     / /

The denominators are the same and the numerators have the same form except that one involves l2 and the other involves l3, and there is a negative sign involved.

In order for both of them to be positive, both numerators must have the same sign, and that sign has to be the same as the denominator. Which implies that the sign of the terms with l3 must be negative of the sign of the terms with l2 for the same structure.

This is not impossible, but it is going to take some work to find.

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How to solve 4 transcendental equations of 4 unknown values?

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How to solve 4 transcendental equations of 4 unknown values?

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