MMP.9:5.2 - Answer about a nonlinear population without a closed mean equation

A finite population has nonnegative member values X_i with X_i_dot=-X_i^2. The available initial information is 0<=X_i(0)<=M and mean m0. The requested output is the mean m(t).

Differentiating the mean gives:

m_dot=-mean(X_i^2)=-m^2-Var(X_i).

Replacing this by m_dot=-m^2 sets the variance contribution to zero. Populations with the same mean can have different variance, so first ask whether a bound already answers the question.

Each member has X_i(t)=X_i(0)/(1+t*X_i(0)) for t>=0. Since X_i(0)<=M, averaging gives the lower bound m0/(1+M*t). The response a/(1+t*a) is concave for nonnegative a and t>=0; the mean of the responses is at most the response of the mean. Thus:

m0/(1+M*t) <= m(t) <= m0/(1+m0*t).

With M=2, m0=1 and t=1, the mean lies between 1/3 and 1/2. A requirement m(1)<=0.55 is established without a variance model or a complete initial distribution.

Change the requirement to m(1)<=0.4. A population whose members all start at 1 has m(1)=1/2. An equally divided population starting at 0 and 2 has m(1)=1/3. Both fit the supplied initial information, so it cannot settle the changed requirement. Information about the initial population or a different acceptable requirement would change the next move. Treating the zero-variance closure as the whole population would conceal this distinction.