Find critical points of a function with two variables

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Ali Mortazavi
Ali Mortazavi 2017-7-31
回答： Walter Roberson 2017-7-31
I'm trying to find critical points to this function but it is so long that when i try to run this:
syms x y z
eqns = (120./((3*x+12).^2+(3*y+8).^2+20))-(240./((3*x-9).^2+(3*y+1).^2+10))-(360./((3*x+12).^2+(3*y-1).^2+13))+(480./((3*x-9).^2+(3*y-8).^2+17));
dx = diff(eqns,x)== 0 %=(240(18x-54))/((3x-9)^2+(3y+1)^2+10)^2-(480(18x 54))/((3x-9)^2+(3y-8)^2+17)^2+(360(18x+72))/((3x+12)^2+(3y-1)^2+13)^2-(120(18x+72))/((3x+12)^2+(3y+8)^2+20)^2
dy = diff(eqns,y)== 0% =(240(18y+6))/((3x-9)^2+(3y+1)^2+10)^2+(360(18y-6))/((3x+12)^2+(3y-1)^2+13)^2-(480(18y-48))/((3x-9)^2+(3y-8)^2+17)^2-(120(18y+48))/((3x+12)^2+(3y+8)^2+20)^2
[xcr,ycr]=solve(dx,dy); [xcr,ycr]
I get a spinning blue ball that goes forever
the domain is -10<x<10 -10<y<10
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回答（1 个）

Walter Roberson
Walter Roberson 2017-7-31
The key is two-fold:
syms x y real
eqns = (120./((3*x+12).^2+(3*y+8).^2+20))-(240./((3*x-9).^2+(3*y+1).^2+10))-(360./((3*x+12).^2+(3*y-1).^2+13))+(480./((3*x-9).^2+(3*y-8).^2+17));
dx = diff(eqns,x);
dy = diff(eqns,y);
sol = vpasolve([dx,dy])
This is fairly fast, and gives sol.x and sol.y with 6 solutions each. You can then
subs(dx,{x,y},{sol.x,sol.y})
subs(dy,{x,y},{sol.x,sol.y})
and see that the derivatives are very close to 0. The exception is at
x = -4.4546625928986563919737615802286
y = -9.5289449194003238929219876399778
which is a point at which the derivatives are steep enough to give numeric problems. This also happens to be the location that my other tests had been finding. It appears that point is a saddle point.
FF = subs(eqns,sol);
fsurf(eqns,[-10 10 -10 10]);
hold on
scatter3(sol.x,sol.y,FF,30,'r+')
You can see two local minima, two local maxima, and 2 saddle points. (I would not be surprised if there were more saddle points that this process does not locate.)

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