# Is there a solution for these three differential equations?

## Summary:

I have given three differential equations with three unknowns and I cannot find a numerical solution with the 'fsolve' command in Maple. It seems like there is no solution at all. The question is whether this is the case, and if so, why.
I have three differential equations with three unknowns ##p##, ##q## and ##r##:

$$\displaystyle {\frac {\partial }{\partial p}}\sum _{k=1}^{5}f_{{k}}\ln \left( P \left( X=k \right) \right) =0$$,
$$\displaystyle {\frac {\partial }{\partial q}}\sum _{k=1}^{5}f_{{k}}\ln \left( P \left( X=k \right) \right) =0$$,
$$\displaystyle {\frac {\partial }{\partial r}}\sum _{k=1}^{5}f_{{k}}\ln \left( P \left( X=k \right) \right) =0$$
with
$$\displaystyle P \left( X=k \right) \, = \,{q}^{k-1}r+2\, \left( k-1 \right) {q}^{k-2} \left( 1-p-q-r \right) p+ \left( k-1 \right) {q}^{k-2}{p}^{2}+ \left( k-1 \right) {q}^{k-2} \left( 1-p-q-r \right) r\\ \mbox{}+ \left( k-1 \right) {q}^{k-2}pr+1/2\, \left( k-2 \right) \left( k-1 \right) {q}^{k-3} \left( 1-p-q-r \right) ^{3}+1/2\, \left( k-2 \right) \left( k-1 \right) {q}^{k-3} \left( 1-p-q-r \right) ^{2}p\\ \mbox{}+1/2\, \left( k-2 \right) \left( k-1 \right) {q}^{k-3} \left( 1-p-q-r \right) ^{2}r$$
and
##\displaystyle f_{{1}}\, = \,0##, ##\displaystyle f_{{2}}\, = \,26##, ##\displaystyle f_{{3}}\, = \,111##, ##\displaystyle f_{{4}}\, = \,17## and ##\displaystyle f_{{5}}\, = \,2##.

I can't find a solution for ##p##, ##q## and ##r##. Is there a solution at all?

If I replace ##\displaystyle f_{{1}}\, = \,0##, ##\displaystyle f_{{2}}\, = \,26## with ##\displaystyle f_{{1}}\, = \,1##, ##\displaystyle f_{{2}}\, = \,25## then I get a solution:

##p = 0.08557##, ##q = 0.05161##, ##r = 0.00641##.

Delta2

Delta2
Homework Helper
Gold Member
These are not differential equations because in DEs the unknowns are functions (of time or some other independent variable). Here your unknowns seem to be numbers and the equations are algebraic. We just have to calculate some derivatives to find the final algebraic expression of the algebraic equation. So you should try a command in Maple that numerically solves algebraic equations.

Be that as it may, there are three equations here with three unknowns ##p##, ##q## and ##r##. And of course I first used the 'solve' command to solve algebraically the equations in p, q and r. But that didn't work in Maple. So I tried to find a numeric solution with the command 'fsolve'. But that didn't work either and the question remains: why. Besides, you did not answer the question why I do get a solution when I change ##\displaystyle f_{{1}}\, =\,0## and ##\displaystyle f_{{2}}\, =\,26## to ##\displaystyle f_{{1}}\, =\,1## and ##\displaystyle f_{{2}}\, =\,25##.

Delta2
Homework Helper
Gold Member
Well the full algebraic expression of the equations (after calculating the derivatives) is polynomial , however we seem to have 3 very complex polynomials with variables p,q,r and its no surprise to me that maple cant find a solution with algebraic expression (i mean a solution containing radicals , powers e.t.c). I think to this relates the fact that ##k## , which seems to be the variable controlling the degree of the polynomials, goes high up to 5, maybe if u set ##k## to be maximum 3 you ll get solutions with the solve command.

So numerical solution seems to me is the only working approach. When you change ##f_1## and ##f_2## you get a different system of polynomials for which there might be solutions. I am afraid i cant answer the question as to why when you do the specific change the system has solutions.

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Delta2
Two possible issues: (1) the solution is complex but the program is looking for solutions in the real numbers, (2) the numerical algorithm looking for solutions is starting its search at a particular point and doesn't find distant solutions. Can you control in Maple where the program starts looking for solutions? The one solution you did get does seem to be near zero, which is where I might expect the program to start searching by default.

When I replaced ##\displaystyle f_{{1}}\, = \,0## and ##\displaystyle f_{{2}}\, = \,26## with ##\displaystyle f_{{1}}\, = \,1## and ##\displaystyle f_{{2}}\, = \,25## I got the following parameter estimates ##\hat{p} = 0.08557##, ##\hat{q} = 0.05161## and ##\hat{r} = 0.00641##. Using these estimates as starting values for the estimation based on the original ##f_k## values, Maple has also failed to find a solution.

Office_Shredder
Staff Emeritus