Solving System of Equations with Symbolic Toolbox

I'm trying to solve a series of equations using the symbolic math toolbox, and I'm not sure why two methods aren't equivalent. Basically, I'm using the solution from the 1st equation in the 2nd, the solutions for the 1st and 2nd in the 3rd, and so forth. When I write it out by hand, it looks something like:

Fist Step:

Solution(1) = solve(Equation(1),X(1));

Second Step:

X(1)=Cut and paste of Solution(1);
Solution(2)=solve(Equation(2),X(2));

Third Step:

X(2)=Cut and paste of Solution(2);
X(1)=Cut and paste of Solution(1);
Solution(3)=solve(Equation(3),X(3));

Fourth Step:

X(3)=Cut and paste of Solution(3);
X(2)=Cut and paste of Solution(2);
X(1)=Cut and paste of solution(1);
Solution(4)=solve(Equation(4),X(4));

When I do this, the updated values of X(3) or X(2) get carried down into X(2) and X(1). I would like to continue this out for much longer than would be feasible to cut and paste data, so I tried putting it into a loop like this:

for n=1:4
    for m=(n-1):(-1):1
        X(m)=Solution(m);
    end
    Solution(n)=solve(Equation(n),X(n));
end

It seems like the inner loop should be updating the X values in decending order as was done with the cutting and pasting of the solutions, but that is not happening. Is there a reason why using X(m)=Solution(m) is not equivalent to X(m)=Cut and paste of Solution(m). I'm new to using the symbolic toolbox, so I'm not all the familiar with it.

Thanks,

Jon

Answer 1

[Torsten]

Say you want to solve

eqn1 = expr1 == 0;
...
eqnn = exprn == 0;

Then according to what you are trying to do you can proceed as follows:

sol1  = solve(expr1 == 0,x1);
expr2 = subs(expr2,x1,sol1);
sol2  = solve(expr2 == 0,x2);
expr3 = subs(expr3,[x1,x2],[sol1,sol2])
sol3  = solve(expr3 == 0,x3)
...

Is it this what you were asking for ?

[Jon LeBlanc]

Thanks! I just had to tweak it slightly, since subbing multiple variables at a time still left some instances of the substituted variable. I ended up using:

sol1  = solve(expr1 == 0,x1);
expr2 = subs(expr2,x1,sol1);
sol2  = solve(expr2 == 0,x2);
expr3 = subs(expr3,x1,sol1);
expr3 = subs(expr3,x2,sol2);
sol3  = solve(expr3 == 0,x3);

Thanks again for your help!

Answer 2

[Walter Roberson]

E = Equation;
partial.(X(1)) = solve(E(1), X(1));
for n = 2 : min(length(Equation), length(X));
    E = subs(E(2:end), partial);
    sol = solve(E(1), X(n));
    if isempty(sol); break; end
    partial.(X(n)) = sol;
end

Afterwards you are left with a struct of partial solutions, each one depending on fewer and fewer variables. Afterwards, you need to back-substitute.

The above code is not robust for the possibility that you might get multiple solutions for a variable. In such a case, you need to iterate over each of the values, creating a "branch" that assumes each of the values in turn -- and the branches might lead to other branches, not necessarily the same number of solutions in each branch.