MMP.9:5 - Archetypal Grounding

MMP.9:5.1 - Replace many response components by two evolving quantities

Suppose a dimensionless model has an observed x and n hidden response components:

x_dot=-x+sum_i y_i+u(t), y_i_dot=c_i*x-2*y_i,

where the nonnegative c_i sum to one. Initial values and a prescribed input u(t) are supplied. The question asks for x over a finite horizon; each y_i separately is irrelevant to that result.

Retaining x alone leaves the unclosed contribution sum_i y_i. Solve each hidden equation and sum:

sum_i y_i(t)=exp(-2*t)*sum_i y_i(0)+integral_0^t exp(-2*(t-s))*x(s) ds.

All hidden contributions have the same decay kernel. Define v=sum_i y_i. Differentiating the sum gives the two-variable model:

x_dot=-x+v+u(t), v_dot=x-2*v, v(0)=sum_i y_i(0).

The derived equations and initial sum preserve x for every supplied input for which these linear equations have their solution. For n>1 this uses two evolving quantities instead of n+1. The required hidden initial information is their sum. Arbitrarily setting v(0)=0 would already change x_dot(0) when the actual sum is nonzero.

The initial sum requires n terms once. Each subsequent evaluation of the reduced right-hand side uses x, v and u(t); it no longer recomputes n component contributions. This saves repeated arithmetic when those components would otherwise be advanced separately.

Now one component has decay rate 3 instead of 2. Summing produces v_dot=x-2*v-y_1; the old two-variable model has lost a contribution. Retain y_1 separately and update it by y_1_dot=c_1*x-3*y_1, or separate the two decay groups. The change determines which added state is needed. The same grouping can combine any components that share their response kernel; different kernels remain distinct until another justified approximation combines them.

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.

MMP.9:5.3 - Decide whether a fast transient can be omitted

For a dimensionless model with epsilon>0,

x_dot=y, epsilon*y_dot=-y,

the solutions for t>=0 are y(t)=y0*exp(-t/epsilon) and x(t)=x0+epsilon*y0*(1-exp(-t/epsilon)).

Suppose the admitted initial values satisfy abs(y0)<=Y. Replacing the model by constant x0 gives an error at most epsilon*Y for every t>=0. With epsilon=0.01 and Y=1, that is 0.01. It can meet a tolerance 0.02 while failing to establish tolerance 0.001. The coefficient alone is insufficient if the initial class changes: y0=1/epsilon leaves a later change approaching one.

When y0 is known and the use concerns only later times, a different approximation is constant x0+epsilon*y0. Its error is at most epsilon*Y*exp(-t/epsilon). For tolerance 0<eta<epsilon*Y, it meets that tolerance at all times starting from t_start=epsilon*log(epsilon*Y/eta). With eta=0.001 in the preceding case, t_start is about 0.0231. The adjusted initial value has retained the transient’s later effect; it does not reproduce the initial interval.

If y0 is unknown, the adjusted value is also unknown. Its stated range can still yield a useful interval for x. The choice between an early transient calculation, a later approximation and a bound follows the requested result and available initial information.