CODEWITHSUNDEEP
matlabnumerical-methods
maxematician
maxematician

Reputation: 199

How can I get MATLAB to numerically solve a particularly nasty system of equations?

I have four parameters, t1,t2, theta1, and theta2. I want to find a solution (there can potentially be infinitely many) to the following system of equations:

t1*cos(theta1) + t2*cos(theta2) + 2*t1*t2*sin(theta1 + theta2) = t1*sin(theta1) + t2*sin(theta2) + 2*t1*t2*cos(theta1 + theta2) + 1

t1*cos(theta1) + t2*cos(theta2) + t1*sin(theta1) + t2*sin(theta2) = 2*t1*t2*sin(theta1 + theta2) + 2*t1*t2*cos(theta1 + theta2) + 1

2*t1^2*t2^2 + sin(theta1)*t1*t2^2 + sin(theta1 + theta2)*t1*t2 + 1 = sin(theta2)*t1^2*t2 + t1^2 + sin(theta1)*t1 + t2^2

Since there are more parameters than equations, we have to impose further restrictions to identify a specific solution to this system. Right now I don't really care which solution is picked, as long as it isn't trivial (like all the variables equaling zero).

My current method is to set the third equation to 0.5, solving for t1 and t2 in terms of theta1 and theta2, and substituting these expressions into the first two equations to solve for theta1 and theta2. However, MATLAB is trying to do this symbolically, which is extremely time-consuming. Is there a way I can get some approximate solutions to this system? I can't exactly plot the surfaces represented by these equations and look at their intersection, because each side of the equations involves more than two parameters, which means it can't be visualized in three dimensions.

Upvotes: 2

Views: 393

Answers (1)

rlbond
rlbond

Reputation: 67759

You can use fsolve to do this.

First, you need to create anonymous functions for the residuals of the three equations. I'm going to create a vector x where x = [t1; t2; theta1; theta2]. That makes the residuals:

r1 = @(x) x(1)*cos(x(3)) + x(2)*cos(x(4)) + 2*x(1)*x(2)*sin(x(3) + x(4)) - (x(1)*sin(x(3)) + x(2)*sin(x(4)) + 2*x(1)*x(2)*cos(x(3) + x(4)) + 1);

r2 = @(x) x(1)*cos(x(3)) + x(2)*cos(x(4)) + x(1)*sin(x(3)) + x(2)*sin(x(4)) - (2*x(1)*x(2)*sin(x(3) + x(4)) + 2*x(1)*x(2)*cos(x(3) + x(4)) + 1);

r3 = @(x) 2*x(1)^2*x(2)^2 + sin(x(3))*x(1)*x(2)^2 + sin(x(3) + x(4))*x(1)*x(2) + 1 - (sin(x(4))*x(1)^2*x(2) + x(1)^2 + sin(x(3))*x(1) + x(2)^2);

Then, we need to make a single function that is a vector of those three residuals, since fsolve tries to get this vector to be zero:

r = @(x) [r1(x); r2(x); r3(x)]

Now, we call fsolve. I picked an arbitrary starting point:

>> [x, fval, exitflag] = fsolve(r, [0.5,0.5,pi/4,pi/4])
Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt algorithm instead. 
> In fsolve at 287 

Equation solved.

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

<stopping criteria details>


x =

    0.9654    0.5182    0.7363    0.7344


fval =

   1.0e-10 *

   -0.0090
    0.2743
   -0.0181


exitflag =

     1

You can ignore the warning. x is the four values you were looking for. fval is the value of the residuals. exitFlag == 1 means we found a root.

Upvotes: 2

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