![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225315/image.png)
How to solve a system of two non-linear equations symbolically
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I have the following equations:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225158/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225159/image.png)
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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225160/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225161/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225162/image.png)
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).
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225161/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225162/image.png)
Thanks a lot, your help is appreciated!
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采纳的回答
Walter Roberson
2019-6-18
编辑:Walter Roberson
2019-6-20
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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/225315/image.png)
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.
4 个评论
Walter Roberson
2019-6-20
"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.
更多回答(1 个)
darova
2019-6-18
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))
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