Plotting two non linear equations in same graph

I have following two equations

y - ((t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 146*t^4 - 88*t^3 - 116*t^2 + 144*t + 81)/(t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 162*t^4 - 184*t^3 + 100*t^2 - 72*t + 162))==0,

t - ((sqrt(y)*(2/3)- (sqrt(1 - y)*((3 - 2*t)/(3(2-t)))))/(sqrt(y)*(((3-t^2))/(3*(2-t))) - sqrt(1-y)*(t/3))) ==0

but I can't seem to plot it. Here's what I did -

t= linspace(0,1);
y= linspace(0.5,1);
[X, Y] = meshgrid(t, y);
f1= @(t,y)(y - ((t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 146*t^4 - 88*t^3 - 116*t^2 + 144*t + 81)/(t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 162*t^4 - 184*t^3 + 100*t^2 - 72*t + 162)));
f2 = @(t,y)(t - ((sqrt(y)*(2/3)- (sqrt(1 - y)*((3 - 2*t)/(3*(2-t)))))/(sqrt(y)*(((3-t^2))/(3*(2-t))) - sqrt(1-y)*(t/3))));
surf(X, Y, f1(X, Y)) ;

Answer 1

[Torsten]

t= linspace(0,1);
y= linspace(0.5,1);
[T, Y] = meshgrid(t, y);
f1 = @(t,y)(y - ((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162)));
f2 = @(t,y)(t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3))));

surf(T,Y,f1(T,Y))
hold on
surf(T,Y,f2(T,Y))

[Sabrina Garland]

Thank you @Torsten. If have to make legend of this graph ie want 2d version and not 3d. What should I do?

[Torsten]

You mean plot y over t as

t = linspace(0,1);
f1 = @(t)((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162));
figure(1)
plot(t,f1(t))
xlabel('t')
ylabel('y')

f2 = @(t,y)t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3)));
figure(2)
fimplicit(f2)
xlabel('t')
ylabel('y')

[Sabrina Garland]

Can't I have both on same graph?

[Torsten]

t = linspace(-3.5,3);
f1 = @(t)((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162));
f2 = @(t,y)t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3)));
plot(t,f1(t))
hold on
fimplicit(f2,[-3.5 3 0 1])
hold off
xlabel('t')
ylabel('y')

[Sabrina Garland]

Thank you!

[Sabrina Garland]

I solved the system the following way

syms t y [soly, solt] = solve(y - ((t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 146*t^4 - 88*t^3 - 116*t^2 + 144*t + 81)/(t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 162*t^4 - 184*t^3 + 100*t^2 - 72*t + 162))==0, t - ((sqrt(y)*(2/3)- (sqrt(1 - y)*((3 - 2*t)/(3*(2-t)))))/(sqrt(y)*(((3-t^2))/(3*(2-t))) - sqrt(1-y)*(t/3))) ==0)

Is it correct?

[Torsten]

syms y t
f1 = y - ((t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 146*t^4 - 88*t^3 - 116*t^2 + 144*t + 81)/(t^8 - 8*t^7 + 36*t^6 - 96*t^5 + 162*t^4 - 184*t^3 + 100*t^2 - 72*t + 162));
f2 = t - ((sqrt(y)*(2/3)- (sqrt(1 - y)*((3 - 2*t)/(3*(2-t)))))/(sqrt(y)*(((3-t^2))/(3*(2-t))) - sqrt(1-y)*(t/3)));
[soly, solt] = solve([f1==0,f2==0],[y,t]);

Warning: Possibly spurious solutions.

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
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 = 

size(solyt)

ans = 1×2

2 9

ysol = solyt(1,:);
ysol = ysol(:)

ysol = 

tsol = solyt(2,:);
tsol = tsol(:)

tsol = 

[Sabrina Garland]

y can only have range of 0.5 to 1. How is it giving ans of 2 and 9. And when I solved through wolfram alpha I got t=1 and y= 0.990099, why Matlab is giving the opposite result. Even through substitution manually that make sense

[Torsten]

The first row of "solyt" are the y-values, the second row of "solyt" are the corresponding t-values.

Should be obvious from the command

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

And only from the aspect of real solutions, y is restricted to 0 <= y <= 1, not 0.5 <= y <= 1. Maybe you are talking about physical restrictions that I don't know of.

[Sabrina Garland]

Okay got it! But why the plot shows y to be 1 and t to be 0.990099. And when I plot the graph I am getting 3d figure (with same command as yours)

[Torsten]

We are talking about three different things in this discussion:

First thing is to plot

z = f1(t,y) = y - ((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162))
z = f2(t,y) = t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3)))

This gives two surface plots of z over t and y (plot in a 3d-frame)

Second thing is to solve

y - ((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162)) == 0
t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3))) == 0

for y. This will give two line plots of y over t (plot in a 2d-frame).

Third thing is to explicitly solve

y - ((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162)) == 0
t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3))) == 0

for t and y. This will give a handful of coordinate pairs (t,y) that satisfy both equations.

These three different interpretations are done in my three responses. I'm not certain what you are aiming at.

Points where the blue and the red curves in the line plots from the second approach meet should occur as coordinate pairs in the third part. I'm not sure which plot you mean when you say:

But why the plot shows y to be 1 and t to be 0.990099.

If I can trust my old eyes, it shows exactly the opposite.

And when I plot the graph I am getting 3d figure (with same command as yours)

Which graph do you mean ?

[Sabrina Garland]

Second thing...

[Sabrina Garland]

In the Attached graph, intersection happens at t=0.990099 and y= 1

[Torsten]

No. It happens at t = 1 and y = something below 1.

Or look again at the graphs here. t is abscissa and y is ordinate !

t = linspace(-3.5,3);
f1 = @(t)((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162));
f2 = @(t,y)t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3)));
plot(t,f1(t))
hold on
fimplicit(f2,[-3.5 3 0 1])
hold off
xlabel('t')
ylabel('y')

Answer 2

[Voss]

Use element-wise operations (.^, ./, .*)

t= linspace(0,1);
y= linspace(0.5,1);
[X, Y] = meshgrid(t, y);
f1= @(t,y)(y - ((t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 146*t.^4 - 88*t.^3 - 116*t.^2 + 144*t + 81)./(t.^8 - 8*t.^7 + 36*t.^6 - 96*t.^5 + 162*t.^4 - 184*t.^3 + 100*t.^2 - 72*t + 162)));
f2 = @(t,y)(t - ((sqrt(y)*(2/3)- (sqrt(1 - y).*((3 - 2*t)./(3*(2-t)))))./(sqrt(y).*(((3-t.^2))./(3*(2-t))) - sqrt(1-y).*(t/3))));
surf(X, Y, f1(X, Y)) ;

https://www.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html