solving equation in matlab

I want to solve this equation in matlab to get the values of y knowing that

 x =[0:0.1:7];  
 eq1 = ((x/7)^2*sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)^2*sqrt(abs(y+3/7*sqrt(33))/(y+3/7*sqrt(33)))-1);

then draw x and y together.

[Walter Roberson]

is eq1 a constant? Is that constant 0 ? That is, is eq1 to be solved for the y that make eq1 0 when used with those x ?

[Esraa El-Basha]

Yes, eq1 = 0 and i need to get all the values of y substituting by this range of x. and then draw x and y together.

Answer 1

[Malhar Ashtaputre]

Hello Esraa El-Basha,

Try to make the equation in following form.

y = function(eq1,x)

Then define x and eq1=0, then try running.

Answer 2

[Walter Roberson]

syms x y
eq1 = ((x/7)^2*sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)^2*sqrt(abs(y+3/7*sqrt(33))/(y+3/7*sqrt(33)))-1);
fimplicit(eq1, [-8 8 -3 3])

[John BG]

fimplicit returns a result that is all zeros except a single NaN

z=fimplicit(eq1, [-8 8 -3 3])
z.ZData

Then one realizes that it's a complex function, so solving complex functions requires to check both real and imaginary parts, or to work with abs.

Also, one wonders if the graph that fimplicit returns is of any use:

Answer 3

[John BG]

Hi Esraa

this is John BG <mailto:[e-mail] [e-mail]>

1. Check Real and Imaginary

Since the equation is complex, it's useful to check both real and imaginary parts of the equation:

clear all;clc;close all
 x =[0:0.1:7];
 y =[0:0.1:7];
[X Y]=meshgrid(x,y);
Z = ((X/7).^2*sqrt(abs(abs(X)-3)./(abs(X)-3))+(Y/3).^2*sqrt(abs(Y+3/7*sqrt(33))./(Y+3/7*sqrt(33)))-1);
figure(1);surf(X,Y,abs(Z));xlabel('X');ylabel('Y');

.

.

figure(2);surf(X,Y,imag(Z));xlabel('X');ylabel('Y');

.

.

the apparent minimum value is

min(min(Z)) 
ans =
  22.835714285714282

for

   x =[0:0.001:7];
   y =[0:0.001:7];
  ..
  min(min(Z))
   =
       2.332833357142857e+03

For abs(x)<3 the solution is complex

When x=>3 then the solution doesn't get any smaller than min(min(Z)) that seems to be 23.32

2. Asymptotic behavior apparently changes minimum

When changing x y ranges, such minimum value changes too

x =[-7:0.1:7];
 y =[-7:0.1:7];
[X Y]=meshgrid(x,y);
Z = ((X/7).^2*sqrt(abs(abs(X)-3)./(abs(X)-3))+(Y/3).^2*sqrt(abs(Y+3/7*sqrt(33))./(Y+3/7*sqrt(33)))-1);
figure(1);surf(X,Y,abs(Z));xlabel('X');ylabel('Y');
figure(2);surf(X,Y,imag(Z));xlabel('X');ylabel('Y');

.

now

min(min(Z)) 
=
  46.671428571428571

I tried smaller steps than 0.001, for the given range, which slowed down things.

Then when narrowing the range

x =[0:0.0001:.5];
 y =[0:0.0001:.5];
..
min(min(Z))
 =
 -0.001044693333332 + 8.505952551020407i

3. Simplify

Stepping back on realizes that your the equation is the same, regarding zeros search, as

X^2*1./sqrt(abs(abs(X)-3))+

Y^2*1./(Y+3/7*sqrt(33)))-1)

these 2 terms will not zero for any both large values of X Y.

When approaching x=3 or y=-3/7*sqrt(33) the result rockets.

Now try this

 x =[-5:0.01:5];
 y =[-5:0.01:5];
[X Y]=meshgrid(x,y);
Z=X.^2*1./sqrt(abs(abs(X)-3))-Y.^2*1./(Y+3/7*sqrt(33))+1;
figure(1);surf(X,Y,10*log10(abs(Z)));xlabel('X');ylabel('Y');

.

that resembles the 'parenthesis like' curves obtained with fimplicit,

however, again increasing the range one realizes that the solutions are not a simple circle or ellipse but these two conic-like notch curves, plus the nearly flat notch running along X axis :

.

.

Therefore the solution to your equation are the zeros of

Z=X.^2*1./sqrt(abs(abs(X)-3))-Y.^2*1./(Y+3/7*sqrt(33))+1

Esraa

if you find this answer useful would you please be so kind to consider marking this answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>