Peculiar Result with Solve (Symbolic Math Toolbox) with Three Equations and Three Unknowns. Can Anyone Explain?

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Paul 2021-1-9

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评论： David Goodmanson 2021-1-11

Suppose I have three equations and three unknowns as so:

>> syms X1 X2 X3 real
>> e1 = X1/(X1 + X2 + X3) == 0.1;
>> e2 = X2/(X1 + X2 + X3) == 0.4;
>> e3 = X3/(X1 + X2 + X3) == 0.5;

Now solve the equations and note the condition on the parameter of the solution x ~= 0.

>> sol=solve([e1,e2,e3],[X1,X2,X3],'ReturnConditions',true);
>> [sol.X1 sol.X2 sol.X3]
 
ans =
 
[ x/5, (4*x)/5, x]
 
>> sol.conditions
 
ans =
 
in(x, 'real') & x ~= 0

This solution makes perfect sense. Now don't specify ReturnConditions

>> sol=solve([e1,e2,e3],[X1,X2,X3],'ReturnConditions',false);
>> [sol.X1 sol.X2 sol.X3]
 
ans =
 
[ 1/5, 4/5, 1]

sol returns one possible solution. So far so good. But now change the RHS of the equations:

>> e1 = X1/(X1 + X2 + X3) == 0;
>> e2 = X2/(X1 + X2 + X3) == 0;
>> e3 = X3/(X1 + X2 + X3) == 1;
>> sol=solve([e1,e2,e3],[X1,X2,X3],'ReturnConditions',true);
>> [sol.X1 sol.X2 sol.X3]
 
ans =
 
[ 0, 0, x]
 
>> sol.conditions
 
ans =
 
in(x, 'real')

The solution would make sense, except that the condtions on the parameter do not include x ~= 0. But that's an important condition. Because if x = 0, then sol.X1 = sol.X2 = sol.X3 = 0, which doesn't make sense as an allowable solution. Now don't speciify ReturnConditions

>> sol=solve([e1,e2,e3],[X1,X2,X3],'ReturnConditions',false);
>> [sol.X1 sol.X2 sol.X3]
 
ans =
 
[ 0, 0, 0]

Whoa. That result can't be correct, can it?

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Answer 1

David Goodmanson 2021-1-9

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编辑：David Goodmanson 2021-1-9

Hi Paul,

certainly in the first case,

[eps/5, 4eps/5,eps]

is a solution for any nonzero eps, and if you let eps--> 0 and invoke continuity, then [0 0 0] can be seen as a solution. Similarly, you can take a look at

[eps/5, 4eps/5,eps] /2eps =? [.1 .4 .5]

which is of the form 0/0 as eps ---> 0. Then L'Hopital's rule says that the equality is correct.

In the second case, consider solutions to

X1/(X1 + X2 + X3) = a
X2/(X1 + X2 + X3) = b
X3/(X1 + X2 + X3) = c

where a,b,c are assumed all nonnegative, a+b+c = 1 (required) and c~=0.

Then for X3~= 0 the solution is

[X1 X2 X3] = [a/c b/c 1] X3

because in that case the equations become

(X3/X3) (a/c)/(a/c+b/c+1) = a 
(X3/X3) (b/c)/(a/c+b/c+1) = b 
(X3/X3) 1/(a/c+b/c+1) = c 

Now for X3 --> 0 (and any c~=0) the continuity / L'Hopital argument can be used on (X3/X3). In particular this covers the case [a b c] = [0 0 1], which produces [X1 X2 X3] = [0 0 0]. It's also good for the first case [a b c] = [.1 .4 .5] and every other case with c~=0..

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Paul 2021-1-9

编辑：Paul 2021-1-9

Hi David,

Thanks for the response. However, I'm still not fully understanding the results from solve for the second case.

To be clear, I don't have an objection to x = 0 for the first case, because the parameterized solution to the first case explicitly excluded x = 0 from the parameterized solution as seen in the the returned value of sol.conditions.So in the first case there's no need in my view to consider limiting or continuity arguments, because for every non-zero value of the parameter x the parameterized solution is well defined. With ReturnConditions == false, solve returned a particular solution for x = 1, which is perfectly reasonable.

On the other hand, in the second case the solution does not explicitly exclude x = 0 from the parameterized solution in sol.conditions. Why not? Why does x~= 0 show up as a sol.conditions for the first case, but not the second? I don't see fhe fundamental difference between the two cases. For the second case (based on the first), I expected the parameterized solution to be of the form:

[0,0,x]

with sol.conditions to include the constraint x ~= 0, and the particular soluton to be [0, 0, 1], which actually satisfies the equations.

From a Symbolic Toolbox perspective, do you have any idea why solve treats these two cases differently? After all, if I understand your answer, you're suggesting by a continuity argument that [0 0 0] "can be seen as an answer" for the first case. But in the first case, solve saw fit to explicitly exclude that solution from the parameterized solution space. But, in the second case it did not. Why the difference?

Let's look at a third case, which is the most general:

>> clear all
>> syms X1 X2 X3 real
>> syms a b c real
>> e1 = X1/(X1 + X2 + X3) == a;
>> e2 = X2/(X1 + X2 + X3) == b;
>> e3 = X3/(X1 + X2 + X3) == c;
>> sol=solve([e1,e2,e3],[X1,X2,X3],'ReturnConditions',true);
>> sol.conditions
 
ans =
 
                  in(x, 'real') & a + b == 1 & b ~= 0 & c == 0 & x ~= 0
                  in(x, 'real') & a + c == 1 & b == 0 & c ~= 0 & x ~= 0
                  in(x, 'real') & b + c == 0 & a == 1 & c ~= 0 & x ~= 0
 in(x, 'real') & b + c ~= 0 & b ~= 0 & c ~= 0 & x ~= 0 & a + b + c == 1
                      in(x, 'real') & a == 1 & b == 0 & c == 0 & x ~= 0

Solve returned five parameterized solutions based on five conditions. Note that all five conditions explicitly specify x ~= 0. The particular condtion we are interested for the second case with a = b = 0 and c = 1 is the second condition, for which the parameterized solution is

 
>> [sol.X1(2) sol.X2(2) sol.X3(2)]
 
ans =
 
[ -(x*(c - 1))/c, 0, x]

This solution yields what I expect if c = 1, namely [0 0 x]. [0 0 0] is not an admissilble solutiion.

The first case I posted had a = .1, b = 0.4, c = 0.5, which corresponds to the fourth condition

>> [sol.X1(4) sol.X2(4) sol.X3(4)]
 
ans =
 
[ -(x*(b + c - 1))/c, (b*x)/c, x]

which for the given values of a,b, and c corresponds exactly to the solution I showed for the first case in the original question.

I still don't see how solve can return [0 0 0] as a solution for the second case, in light of the results from the third case and also because the doc page for solve implies the (non-empty) returned solutions explicitly satisfy the equations (no limiting arguments needed), at least how I read it.

Paul 2021-1-11

编辑：Paul 2021-1-11

David,

Thanks for your thoughtful response (as usual). For me, I wannt solve() to return solutions that explicitly satiisfy the equations. And I certainly want consistency.

Regarding your example, the equations can be expressed as; M*X = 0 where

M(a, b, c) =
 
[ a - 1,     a,     a]
[     b, b - 1,     b]
[     c,     c, c - 1]
 

So the only solutions are X=0 and X being in the null space of M should it have dim > 0 (depending on the values of a,b and c. Certainly a = 0, b =0, c=1 results in a null space of dim = 1 (are there any M with a non-emply null space that doesn't have two of the paramters equal zero on the other = 1?). In any case, there are clearly condtions that should result in more than one solution. But

>> clear all
>> syms X1 X2 X3 a b c real
>> e1 = X1 == a*(X1 + X2 + X3);
>> e2 = X2 == b*(X1 + X2 + X3);
>> e3 = X3 == c*(X1 + X2 + X3);
>> sol=solve([e1,e2,e3],[X1,X2,X3],'ReturnConditions',true);
>> [sol.X1 sol.X2 sol.X3]
 
ans =
 
[ 0, 0, 0]
 
>> sol.conditions
 
ans =
 
a + b ~= 1 & a ~= 1 & a + b + c ~= 1
 

solve() returns only the zero solution but then puts conditions on it, even though that zero solution should apply for any values of a,b, and c (shouldn't it?).

David Goodmanson 2021-1-11

Paul.

Any a,b,c such that a+b+c=1 will have a nonempty null space, so both (0,0,1) and your original example (0.1 0.4 0.5) work. (I keep trying to advance the idea that your two examples are more alike than they are different).

In line with this result, note that

det(M(a,b,c)) = a+b+c-1

so again a+b+c=1 --> det(M) = 0 --> nonempty null space.

For M*X=0, all that is required to force X = [0 0 0] to be the only solution is to have a nonzero determinant, a+b+c~=1 and I believe the other two requirements set by the solver, a+b~=1 & a~=1 are not required. It appears that wrangling solve into 100% conistency could be difficult.

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