- #1

Ad VanderVen

- 169

- 13

- TL;DR 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##.

$$\displaystyle {\frac {\partial }{\partial p}}\sum _{k=1}^{5}f_{{k}}\ln \left( P \left( X=k \right) \right) =0$$,

$$\displaystyle {\frac {\partial }{\partial q}}\sum _{k=1}^{5}f_{{k}}\ln \left( P \left( X=k \right) \right) =0$$,

$$\displaystyle {\frac {\partial }{\partial r}}\sum _{k=1}^{5}f_{{k}}\ln \left( P \left( X=k \right) \right) =0$$

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##.