How to rewrite non linear function?

Hello!

I have a problem! This is my first question here, so sorry if it's already answered :)

I have a non linear function

f=@(x,y)[x^3+10*x-y-5; x+y^3-10*y+1]

and I need it use in fsolve, but it won't work if the function isn't writen as

f=[x(1)^3+10*x(1)-x(2)-5; x(1)+x(2)^3-10*x(2)+1]

Can it be somehow rewriten it to the correct form? Is there any way? I can't just rewrite the function because then the rest of the code won't work and subs won't work with x(1) and x(2).

Thank you!

Answer 1

[Walter Roberson]

initial_conditions = randn(1,2)

initial_conditions = 1×2

-0.1679 1.7828

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f=@(x,y)[x^3+10*x-y-5; x+y^3-10*y+1]

f = function_handle with value:
    @(x,y)[x^3+10*x-y-5;x+y^3-10*y+1]

objective = @(XY) f(XY(1), XY(2))

objective = function_handle with value:
    @(XY)f(XY(1),XY(2))

fsolve(objective, initial_conditions)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

ans = 1×2

0.5024 0.1506

and subs won't work with x(1) and x(2)

That implies you have access to the Symbolic Toolbox. If so then

syms x y
objective = matlabFunction(f(x,y), 'vars', {[x, y]})

objective = function_handle with value:
    @(in1)[in1(:,1).*1.0e+1-in1(:,2)+in1(:,1).^3-5.0;in1(:,1)-in1(:,2).*1.0e+1+in1(:,2).^3+1.0]

and that will produce an anonymous function that expects vector input.

... Or you could

syms x y
sol = solve( f(x,y) );
mask = isAlways(imag(sol.x) == 0) & isAlways(imag(sol.y) == 0);
xsol = sol.x(mask)

xsol = 

−root(σ1,z,1)3+10 root(σ1,z,1)−1−root(σ1,z,4)3+10 root(σ1,z,4)−1−root(σ1,z,7)3+10 root(σ1,z,7)−1where  σ1=z9−30 z7+3 z6+300 z5−60 z4−987 z3+300 z2−129 z+16

ysol = sol.y(mask)

ysol = 

root(σ1,z,1)root(σ1,z,4)root(σ1,z,7)where  σ1=z9−30 z7+3 z6+300 z5−60 z4−987 z3+300 z2−129 z+16

double([xsol,ysol])

ans = 3×2

0.5024 0.1506 0.7627 3.0702 0.1775 -3.2196

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This shows you that there are three real solutions. Which one you will get with fsolve() will depend upon your starting conditions.