Plotting two non linear equations in same graph

Question

Sabrina Garland 2023-7-28

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2001987-plotting-two-non-linear-equations-in-same-graph

编辑： Torsten 2023-7-30

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)) ;

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Answer 1

Torsten 2023-7-28

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2001987-plotting-two-non-linear-equations-in-same-graph#answer_1280307

在 MATLAB Online 中打开

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))

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Torsten 2023-7-29

编辑：Torsten 2023-7-29

在 MATLAB Online 中打开

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

%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 =

Torsten 2023-7-29

编辑：Torsten 2023-7-30

在 MATLAB Online 中打开

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 2023-7-30

image.png

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

Torsten 2023-7-30

编辑：Torsten 2023-7-30

在 MATLAB Online 中打开

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')

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Answer 2

Voss 2023-7-28

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2001987-plotting-two-non-linear-equations-in-same-graph#answer_1280302

在 MATLAB Online 中打开

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