How to solve a system of two non-linear equations symbolically

I have the following equations:

where is a constant. I want to find both and as a function of A. I tried solving it symbolically, which does give a result, but I do not know how to interpret or simplify it.

  syms L0
  syms L1
  syms A
  a=2/3;
  [L0 L1]= solve(L0 -1 +a*L1^(a-1) -a*A*L0^(a-1)==0, L1 -1 -a*L1^(a-1) +a*A*L0^(a-1)==0)
 

Plotting the Solution numerically worked with the following code, and led to the result I expected, but do you have an idea how I could get a symbolical solution (or approximation if necessary), for the variables and as a function of A? I'd need it to do further calculations (by hand).

Thanks a lot, your help is appreciated!

Answer 1

[Walter Roberson]

 sol = solve(L0 -1 +a*L1^(a-1) -a*A*L0^(a-1)==0, L1 -1 -a*L1^(a-1) +a*A*L0^(a-1)==0,'returnconditions',true);
 

sol =

struct with fields:

L0: [28×1 sym]

L1: [28×1 sym]

parameters: [1×1 sym]

conditions: [28×1 sym]

>> sol.conditions

ans =

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A == 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

A ~= 0

So what you have is two sets of solutions. The first 13 solutions are relevant when A == 0. The first is 0, and the next 12 solutions are the cube of the 12 roots of the polynomial z^12 - 5*z^9 + 9*z^6 - 7*z^3 + 46/27 . The 15 solutions after that are the cubes of the 15 roots of the polynomial z^15 - 5*z^12 - 2*A*z^11 + 9*z^9 + 8*A*z^8 + (4*A^2*z^7)/3 - 7*z^6 - 10*A*z^5 - 4*A^2*z^4 - (z^3*(8*A^3 - 46))/27 + 4*A*z^2 + (8*A^2*z)/3 + (16*A^3)/27 .

If your is intended to indicate the open set, then A cannot be 0, and you can

 syms L0 L1 A
 assume(A>0 & A<1)
 sol = solve(L0 -1 +a*L1^(a-1) -a*A*L0^(a-1)==0, L1 -1 -a*L1^(a-1) +a*A*L0^(a-1)==0)

to get just the 15 solutions

You can convert that polynomial into a roots() call:

 roots([1, 0, 0, -5, -2*A, 0, 9, 8*A, (4*A^2)/3, -7, + -10*A, -4*A^2, 46/27 - (8*A^3)/27, 4*A, (8*A^2)/3, (16*A^3)/27])

It appears to me that the 3rd and 6th and 15th root are real-valued. I am not sure at the moment if there are any particular real points for specific A values; I am seeing artifacts due to the way that roots are numbered.

[Franziska]

thanks for the help!

With the roots() call I get the same 15 solutions... How did that help?

I guess there simply doesn't exist a solution that is 'easy' enough to keep working with it 'by hand'/pen and paper calculations? Do you have any idea on some helpful approximation method that might help to find a symbolical solution that I can work with?

[Walter Roberson]

Most people give out on pen and paper calculations beyond exact solutions for quadratics, as the exact solutions for cubics are a bit involved, and the exact solutions for quartics are rather messy.

Are you asking for an approximation as a quadratic?

[Walter Roberson]

"With the roots() call I get the same 15 solutions... How did that help? "

With the roots call you get 15 numeric solutions.

>> A=1/3;roots([1, 0, 0, -5, -2*A, 0, 9, 8*A, (4*A^2)/3, -7, + -10*A, -4*A^2, 46/27 - (8*A^3)/27, 4*A, (8*A^2)/3, (16*A^3)/27])

ans =

            1.3086285136187 +                     0i
           1.13534173484129 +     0.166598557544465i
           1.13534173484129 -     0.166598557544465i
           0.86739394608367 +                     0i
         -0.638326898150363 +      1.12185507902337i
         -0.638326898150363 -      1.12185507902337i
         -0.345631679731567 +      1.05445786581201i
         -0.345631679731567 -      1.05445786581201i
         -0.646415333735405 +     0.843949170383418i
         -0.646415333735405 -     0.843949170383418i
         -0.233517483108016 +     0.717352144678334i
         -0.233517483108016 -     0.717352144678334i
         -0.408011319404886 +                     0i
         -0.155455910264676 +    0.0557721334189132i
         -0.155455910264676 -    0.0557721334189132i

This does not require the symbolic toolbox.

There are no closed form solutions for degree 15 polynomials.

Perhaps you would be willing to live with the approximation that

         L0_3 = 0.4292*A.^3 - 0.8521*A.^2 + 0.02415*A + 0.002639  %values < 0 decreasing
         L0_6 = 0.002639*A.^3 - 0.2629*A.^2 + 0.7519*A + 0.4282   %values > 0.4
         L0_15 = -0.00775*A.^3 + 0.06886*A.^2 + 0.1602*A +  2.18  %values > 2
         

In each case, the corresponding L1 value is 2 minus the L0 value. If you were expecting both L0 and L1 to be positive, then the 6th root is the only solution.

[Franziska]

Adding assumptions actually helped, but I still struggle interpreting. The Assumptions that must be true for the variables are:

I solved the equation for using the assumptions and got a unique solution:

a=2/3;
syms A
syms L0
syms L1

assume(A>0 & 0<L0 & L0<2)
Sol= solve(L0 - 1 +a*(2-L0)^(a-1) -a*A*L0^(a-1)==0, L0, 'ReturnConditions', true)
Sol.L0
Sol.conditions
Sol.parameters

This gave me the solution for .

I do not understand where this comes from and how to interpret it. (I am new to matlab)

Anyway thank you so much for your help!

Answer 2

[darova]

Use matlabFunction() to make function handle from symboliv expression

L0 = matlabFunction(L0);
L1 = matlabFunction(L1);
A = 0.5:0.01:2;
plot(A,L0(A),A,L1(A))