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I'm sorry if this is a weird question, but I have a system of differential equations of the form \begin{align} \begin{bmatrix} S_1'=f_1(S_1, S_2, S_3,...S_n) \\ S_2'=f_2(S_1, S_2, S_3,...S_n) \\ S_3'=f_3(S_1, S_2, S_3,...S_n) \\ \vdots \\ S_n'=f_n(S_1, S_2, S_3,...S_n) \end{bmatrix} \end{align}

where $f_i(S_1, S_2, S_3,...S_n)$ is a non-linear function that is a type of logistic model equation. As such it currently stands that $0\leq S_1, S_2, S_3, ..., S_n \leq 1 \forall t>0$.

However I was wondering if I could restrict these differential equations even harder to the point in which all of the values of $S_k$ could sum to $1$, or written in math $$ \sum_{k=1}^n S_k(t) = 1 $$

I looked around, but I couldn't find anything like this. Do any of you have an idea on what I could possibly do to solve this?

Note I only am doing this numerically, an explicit solution to all equations will not be required, so if something is only numerically solvable, that is still good

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  • $\begingroup$ The right-hand sides of the equations are identical, which means that the derivatives of all the $S_i$ are the same, so there are a function $S_0$ and constants $c_1,c_2,\ldots,c_n$ such that $S_{i}=S_0+c_{i}$ for all $i$. Is that really what you mean? $\endgroup$ Commented Feb 9, 2018 at 19:49
  • $\begingroup$ If you did that then one of the systems would be redundant since $S_n' = - \sum_{k\neq 1} S_k'$. $\endgroup$ Commented Feb 9, 2018 at 19:50
  • $\begingroup$ I presume you meant $f_k(....)$ on the $k$th right hand side otherwise @TaneliHuuskonen's comment is more relevant. $\endgroup$ Commented Feb 9, 2018 at 19:51
  • $\begingroup$ This happens if and only one starts from initial conditions such that $$\sum_{k=1}^nS_k(0)=1$$ and if the dynamics is such that, for every $(x_1,\ldots,x_{n-1})$, $$\sum_{k=1}^nf_k(x_1,\ldots,x_{n-1},1-x_1-\cdots-x_{n-1})=0$$ $\endgroup$ Commented Feb 9, 2018 at 20:26

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Note that your constraint defines some codimention 1 surface on which your solutions must lie. In physics we call this a holonomic constraint, i'm sorry, but I am not sure whether this is a right mathematical name. Such a constraint may be used to reduce the dimension of the problem. This is done very easily. You have the constraint $$\sum_{k=1}^{n}S_{k}=1$$ Differentiate the constraint to give $$\sum_{k=1}^{n}S_{k}'(t)=\sum_{k=1}^{n}f_{k}(S_{1}, ..., S_{n})=0$$ This thing in principle can be solved for one of the $S$'s. Say, $S_{n}$, that is $$S_{n}=F(S_{1}, ..., S_{n-1})$$ Thus you have to solve $$S_{1}'=f_{1}(S_{1}, ..., S_{n-1}, F(S_{1}, ..., S_{n-1}))$$ $$....$$ $$S_{n-1}'=f_{n-1}(S_{1}, ..., S_{n-1}, F(S_{1}, ..., S_{n-1}))$$ One equation less. \ P.S. this principles are the hurt of classical mechanics. In classical mechanics, the constraints may be imposed as in your problem, or are due to intrinsic symmetries. If the number of them is $n-1$ for a system with $n$ degrees of freedom, the system is said to be completely integrable, i.e. it can be reduced to a single separable ode, which is always soluble by quadratures. A famous example of scientific importance is a two body problem, the $6$ dimensional configuration space possesses just right number of symmetries to reduce the problem to $1D$ and thus integrate it. Whereas a 3 body problem doesn't....

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