Not able to get real solution through syst solve

Question

Sabrina Garland 2023-8-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2004777-not-able-to-get-real-solution-through-syst-solve

编辑： Torsten 2023-8-7

I have been trying to solve the following equation and following is my code -

syms y t

f1 = -t^3/(3*(1 - y)^(2/3)) + (3 - t^2)^2/(6*(2 - t)*y^(2/3)) + (2*t)/(3*y^(2/3)) - (3 - 2*t)^2/(6*(2 - t)*(1 - y)^(2/3));

f2 = t - ((nthroot(y, 3)*(2/3)- (nthroot(1-y, 3)*((3 - 2*t)/(3*(2-t)))))/(nthroot(y, 3)*(((3-t^2))/(3*(2-t))) - nthroot(1-y, 3)*(t/3)));

[soly, solt] = solve([f1==0,f2==0],[y,t]);

Warning: Unable to solve symbolically. Returning a numeric solution using vpasolve.

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2];

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

solyt = unique(sol.','rows','stable').'

solyt =

but I am only able to get complex solution. When I solved it through wolfram alpha I got real solution of x= 1 ^ y= 0.8132. What is wrong with my code?

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Star Strider 2023-8-4

What function did you give to Wolfram Alpha?

Please post the Wolfram Alpha URL that contains the expression you gave it to solve. I expect that there are differrences between it and the posted MATLAB code.

Sabrina Garland 2023-8-4

Screenshot_20230805-044806.jpg

Please see the attached image. I solved through system of equation. Isn't this what my Matlab code is doing as well?

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Answer 1

Torsten 2023-8-4

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2004777-not-able-to-get-real-solution-through-syst-solve#answer_1283507

编辑：Torsten 2023-8-4

在 MATLAB Online 中打开

Always plot the functions before solving for common points and set initial values for the variables according to what you see in the plot.

syms y t

f1 = -t.^3./(3*(1 - y).^(2/3)) + (3 - t.^2)^2./(6*(2 - t).*y.^(2/3)) + (2*t)./(3*y.^(2/3)) - (3 - 2*t).^2./(6*(2 - t).*(1 - y).^(2/3));

f2 = t - ((nthroot(y, 3)*(2/3)- (nthroot(1-y, 3).*((3 - 2*t)./(3*(2-t)))))./(nthroot(y, 3).*(((3-t.^2))./(3*(2-t))) - nthroot(1-y, 3).*(t/3)));

fimplicit(matlabFunction(f1))

hold on

fimplicit(matlabFunction(f2))

[soly, solt] = vpasolve([f1==0,f2==0],[y,t],[0.5 1]);

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2];

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

solyt = unique(sol.','rows','stable').'

solyt =

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Torsten 2023-8-5

编辑：Torsten 2023-8-5

在 MATLAB Online 中打开

Again: Plot the functions and choose the initial guess according to where the plots intersect. There is no other way in MATLAB, and I already gave you the code to use in my above answer.

syms y t

f1 = -0.3*t.^3/(3*(1 - y).^(0.7)) + 0.15*(3 - t.^2)^2./((2 - t).*y.^(0.7)) + (0.6*t)./(y.^(0.7)) - 0.15*(3 - 2*t).^2./((2 - t).*(1 - y).^(0.7));

f2 = t - (((y.^(0.7))*(2/3)- (((1-y).^(0.7)).*((3 - 2*t)./(3*(2-t)))))./((y.^(0.7)).*(((3-t.^2))./(3*(2-t))) - ((1-y).^(0.7)).*(t/3)));

fimplicit(matlabFunction(f1))

hold on

fimplicit(matlabFunction(f2))

[soly, solt] = vpasolve([f1==0,f2==0],[y,t],[0.3 -0.5]);

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2];

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

%Columns of solyt are solutions for y and t. Thus there are 9 (y,t) pairs %found that satisfy f1==0 and f2==0.

solyt = unique(sol.','rows','stable').'

solyt =

Torsten 2023-8-5

编辑：Torsten 2023-8-5

在 MATLAB Online 中打开

Because you modified your functions within this post, I don't know if I'm still up-to-date.

syms y t

f1 = -t.^3./(4*(1 - y).^(3/4)) + (3 - t.^2)^2./(8*(2 - t).*y.^(3/4)) + (t)./(2*y.^(3/4)) - (3 - 2*t).^2./(8*(2 - t).*(1 - y).^(3/4));

f2 = t - (((y.^(1/4))*(2/3)- (((1-y).^(1/4)).*((3 - 2*t)./(3*(2-t)))))./((y.^(1/4)).*(((3-t.^2))./(3*(2-t))) - ((1-y).^(1/4)).*(t/3)));

fimplicit(matlabFunction(f1))

hold on

fimplicit(matlabFunction(f2))

[soly, solt] = vpasolve([f1==0,f2==0],[y,t],[0.81 1]);

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2];

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

solyt = unique(sol.','rows','stable').'

solyt =

[soly, solt] = vpasolve([f1==0,f2==0],[y,t],[0.9 0.8]);

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2];

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

solyt = unique(sol.','rows','stable').'

solyt =

[soly, solt] = vpasolve([f1==0,f2==0],[y,t],[0.05 -1]);

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2];

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

solyt = unique(sol.','rows','stable').'

solyt =

Sabrina Garland 2023-8-7

编辑：Walter Roberson 2023-8-7

在 MATLAB Online 中打开

@Torsten Please help.

When I changed the power in f2 function from 0.7 to 0.3, I am still getting same answer of t=1 and y=0.9038.

syms y t

f1 = -0.3*t.^3/(3*(1 - y).^(0.7)) + 0.15*(3 - t.^2)^2./((2 - t).*y.^(0.7)) + (0.6*t)./(y.^(0.7)) - 0.15*(3 - 2*t).^2./((2 - t).*(1 - y).^(0.7));

f2 = t - (((y.^(0.3))*(2/3)- (((1-y).^(0.3)).*((3 - 2*t)./(3*(2-t)))))./((y.^(0.3)).*(((3-t.^2))./(3*(2-t))) - ((1-y).^(0.3)).*(t/3)));

fimplicit(matlabFunction(f1))

hold on

fimplicit(matlabFunction(f2))

[soly, solt] = vpasolve([f1==0,f2==0],[y,t],[0.7 0.9])

soly =

0.9038609440509542648033735347765

solt =

1.0

ynum = vpa(soly);

tnum = vpa(solt);

res1 = arrayfun(@(i)vpa(subs(f1,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res2 = arrayfun(@(i)vpa(subs(f2,[y t],[ynum(i) tnum(i)])),1:size(ynum,1));

res = [res1;res2]

res =

sol = [];

for i = 1:size(res,2)

if abs(res(:,i)) < 1e-10

sol = [sol,[ynum(i);tnum(i)]];

end

%Columns of solyt are solutions for y and t. Thus there are 9 (y,t) pairs

%found that satisfy f1==0 and f2==0.

solyt = unique(sol.','rows','stable').'

solyt =

Why is Matlab mobile still giving me figures with multiple colors and now producing same answer when powers are changed. I am stuck. Please help

Walter Roberson 2023-8-7

vpasolve() only returns more than one solution under the conditions that:

the number of equations is the same as the number of free variables; and
the inputs are all polynomials in the free variables

If you have non-linear equations that are not polynomials, then vpasolve() will only ever return one solution... so much of your code is not doing anything useful.

Torsten 2023-8-7

编辑：Torsten 2023-8-7

@Sabrina Garland

During the discussion, you used 3 different exponents (0.7,2/3 and 0.75) so that I don't know any more which results you address when you ask a question.

The display of the implicit graphs in different colors - I cannot explain it if you really get it with the code above.

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