With k（0<k<0.5） and u（-1<u<1) as independent variables and x and y as dependent variables, how do you graph x and y, respectively, for a system of implicit function equations?

0<k<0.5

-1<u<1

(1-k).*(x+1)./((x+1).^2+y.^2).^1.5+k.*(x+1)./((x+1).^2+(y-u).^2).^1.5-(x-1)./((x-1).^2+y.^2).^1.5==0

(1-k).*(y)./((x+1).^2+y.^2).^1.5-k.*(y-u)./((x+1).^2+(y-u).^2).^1.5-(y)./((x-1).^2+y.^2).^1.5==0

Answer 1

[John D'Errico]

You have 2 parameters, k and u. If you were to fix those parameters, then you could solve for the solutions. There would be zero, one, or more such solutions, since this is a nonlinear system of equations, in two unknowns. For some sets of the parameters, the solutions may become complex, since you have sqrts in there. And there appear to be some singularities too.

But you want to vary k and u now, and see how the solutions themselves vary, essentially as a function of k and u.

Worse, if you leave u and k as symbolic unknowns, then it is very likely there is no analytical solution for x and y, as a function of u and k. I can try and see what solve says.

 syms k u x y real
 solve((1-k).*(x+1)./((x+1).^2+y.^2).^1.5+k.*(x+1)./((x+1).^2+(y-u).^2).^1.5-(x-1)./((x-1).^2+y.^2).^1.5==0,...
   (1-k).*(y)./((x+1).^2+y.^2).^1.5-k.*(y-u)./((x+1).^2+(y-u).^2).^1.5-(y)./((x-1).^2+y.^2).^1.5==0, ...
   [x,y],returnconditions = true)

(It fails if I try to do so using the Answers MATLAB version, but it is time limited.) Unfortunately, when I try that on my own machine, even without limiting the time it fails to return a solution. That means solve cannot find a solution. No surprise at all.

Can you do anything? Does the problem even have a solution? I'll pick two values of k and u.

E1 = @(k,u,x,y) (1-k).*(x+1)./((x+1).^2+y.^2).^1.5+k.*(x+1)./((x+1).^2+(y-u).^2).^1.5-(x-1)./((x-1).^2+y.^2).^1.5;
E2 = @(k,u,x,y) (1-k).*(y)./((x+1).^2+y.^2).^1.5-k.*(y-u)./((x+1).^2+(y-u).^2).^1.5-(y)./((x-1).^2+y.^2).^1.5;
fimplicit(@(x,y) E1(0.25,0.5,x,y),'r-')
hold on
fimplicit(@(x,y) E2(0.25,0.5,x,y),'b-')
grid on
hold off

And that does not look at all promising. There are at least 6 solutions, thus the points where the red and blue curves cross. And if I cary k and u, as I said, the number of solutions will surely vary.

fimplicit(@(x,y) E1(0.1,-0.3,x,y),'r-')
hold on
fimplicit(@(x,y) E2(0.1,-0.3,x,y),'b-')
grid on
hold off

And this time, it looks like 4 solutions, though possibly other stuff may happen outside the region fimplicit chose to plot.

Not all problems have a simple solution. Sorry.