How to Solve These Equations In MATLAB

[EngWiPy]

Hi,

After some programming in MATLAB I arrived to the following equations:

 mu1 + 3*mu2 - 1/(L1 - U1)
       2*mu1 - 1/(L2 - U2)
              -1/(L1 - U1)
       2*mu2 - 1/(L2 - U2)
             U1 + 2*U2 - 2
           3*U1 - 2*L2 + 5

which all equal to zero. How can I solve these 6 equations with 6 unknowns in MATLAB?

Thanks in advance

 

[jhae2.718]

If the equations are defined using symbolic variables, you should be able to just use solve().

 

[EngWiPy]

If the equations are defined using symbolic variables, you should be able to just use solve().

Can you please tell me the syntax? It is not working with me.

Thanks

 

[jhae2.718]

Here's the MathWorks' documentation: http://www.mathworks.com/help/toolbox/symbolic/solve.html"

solve(eq1, eq2, ..., eqn, var1, var2, ..., varn)

Basically, solve() takes either functions defined in strings (e.g. 'x^2 + y^2') or functions/anonymous functions, and solves for the sym variables (FYI you can check variable type using class() ). You can optionally specify which variable to solve for.

So, IIRC, for a simple system of quadratic equations, you could do:

syms x;
f = x.^2 + 2*x + 1;
g = -x.^2 - 2*x - 1; 
roots = solve(f, g);

Those examples should work, but my brain fails me after 11 PM, so my apologies if they don't...(unfortunately, I don’t have MATLAB to check right now. C'mon, guys: release r2011a!)

 

[SteamKing]

S David: In looking at the eqns. in the OP, you have -1/(L1 - U1) = 0.
This implies (L1 - U1) = -1/0. I think there is some inconsistency in the set
of those six equations.

 

[EngWiPy]

S David: In looking at the eqns. in the OP, you have -1/(L1 - U1) = 0.
This implies (L1 - U1) = -1/0. I think there is some inconsistency in the set
of those six equations.

You are absolutely right. This means either L1-U1=infinity, or -1=0, where both are impossible because in my problem, both L1 and U1 are dimensions of a rectangle in a polyhedron. I think I have to re-check my analysis.

jhae2.718: when I tried to solve the equations you gave, MATLAB gives me

syms x y;
f = x.^2 + 2*y + 1;
g = -x.^2 - 2*y - 1; 
roots = solve(f, g) 

roots = 

    x: [2x1 sym]
    y: [2x1 sym]

assuming that you have two variables not one, because you have two equations. right? What is this output and why?

Thanks

 

[jhae2.718]

Yes, that's because of multiple variables. If you type roots.x into the Command Window, it will display x, and likewise for roots.y.

You could also do:

[x y] = solve(f, g);

which would give you x and y as separate variables.

 

[EngWiPy]

Yes, that's because of multiple variables. If you type roots.x into the Command Window, it will display x, and likewise for roots.y.

You could also do:

syms x y;
f = x.^2 + 2*y + 1;
g = -x.^2 - 2*y - 1; 
[x y] = solve(f, g);

which would give you x and y as separate variables.

Now I got:

 syms x y;
f = x.^2 + 2*y + 1;
g = -x.^2 - 2*y - 1; 
[x y] = solve(f, g)
 
x =
 
  (- 2*z - 1)^(1/2)
 -(- 2*z - 1)^(1/2)
 
 
y =
 
 z
 z

what is z here?

 

[jhae2.718]

I created a pretty bad example last night, since g = -f.

What is happening is that MATLAB is assuming that f : x,y \mapsto z and g : x,y \mapsto z in an xyz coordinate system.

The system x+2y=0, 3x+y=1 might be a better example.

I believe that in solve() you can set syms equal to zero as in solve(f=0, g=0).

 

[EngWiPy]

I created a pretty bad example last night, since g = -f.

What is happening is that MATLAB is assuming that f : x,y \mapsto z and g : x,y \mapsto z in an xyz coordinate system.

when I solve my problem, I get the same variable z in the solution. What does this mean exactly?

 

[jhae2.718]

What was the system you were working with?

solve() should work if a set of solutions to the system exists. The example I posted used essentially the same equation twice, so there was an infinitude of solutions.

Try this, and you should see a working example (finally, and sorry!):

syms x y;
f = 3*x - 2*y -1;
g = x + 2*y;
[x y] = solve(f, g);

 

[EngWiPy]

What was the system you were working with?

solve() should work if a set of solutions to the system exists. The example I posted used essentially the same equation twice, so there was an infinitude of solutions.

Try this, and you should see a working example (finally, and sorry!):

syms x y;
f = 3*x - 2*y -1;
g = x + 2*y;
[x y] = solve(f, g);

After minor modification, I got the following equations ans solution:

 U1 - L1 - (L1 - U1)*(mu1 + 3*mu2)
           U2 - L2 - 2*mu1*(L2 - U2)
           U1 - L1 - 6*mu3*(L1 - U1)
 U2 - L2 - (L2 - U2)*(2*mu2 + 4*mu3)
                       U1 + 2*U2 - 2
                     3*U1 - 2*L2 + 5
                   - 6*L1 - 4*L2 - 6
 

roots = 

     L1: [3x1 sym]
     L2: [3x1 sym]
     U1: [3x1 sym]
     U2: [3x1 sym]
    mu1: [3x1 sym]
    mu2: [3x1 sym]
    mu3: [3x1 sym]

where if you type roots.L1 for example it will give you a solution of z variable.

 

[jhae2.718]

I believe z is any arbitrary number; that is, there exists multiple solutions to the equation. You can try different vales for z using the subs() command; you'll probably have to determine the solution based on the problem you are solving.

 

[EngWiPy]

I believe z is any arbitrary number; that is, there exists multiple solutions to the equation. You can try different vales for z using the subs() command; you'll probably have to determine the solution based on the problem you are solving.

But I have 7 equations and 7 variables. There must be a distinct solution. Right?

 

[jhae2.718]

Are there any constraints on the variables (e.g., positive and real)? If so you could declare those assumptions using the real and positive keywords for http://www.mathworks.com/help/toolbox/symbolic/syms.html".

As far as seven equations and seven variables leading to a unique solution, I don't think that's necessarily true. A mathematician could probably answer that more definitively, though.

I have found using MATLAB to evaluate symbolic expressions to be rather frustrating at times.