= Homeostatic temperature model
{title2=$\tau\dot x=d-x-u$}
Here is an explicitly synthetic <mathematical model> of a warming disturbance and its compensatory cooling. It is a teaching analogy for <thermoregulation>, not a measured human heat budget. Write $x=(T-T_r)/\Delta T$, where $T$ is a lumped internal <temperature>, $T_r$ is a fixed reference and $\Delta T>0$ is a chosen <temperature> scale. Let $d$ be an additional heating rate and $u$ an additional cooling rate, both scaled to the same reference rate. All three are <dimensionless variables>, while $\tau>0$ is a <time constant> in the chosen time units. The <ordinary differential equation> is
$$
\tau\dot x=d-x-u,\qquad u=kx,\qquad k\geq0.
$$
The term $-x$ idealizes increased passive <heat transfer> with increased internal <temperature>, around the baseline. The term $-u$ opposes warming through <proportional control> with <feedback gain> $k$. This local signed model does not treat a negative $u$ as negative <sweating>: it would represent decreased cooling relative to baseline. Changes of the environment and multiple body compartments are omitted.
For a constant disturbance $d=D$ and <initial condition> $x(0)=0$, the <exponential function> solution is
$$
x(t)=\frac{D}{1+k}\left(1-e^{-(1+k)t/\tau}\right),\qquad x_*=\frac{D}{1+k},\qquad u_*=\frac{kD}{1+k}.
$$
Substitution checks the <ordinary differential equation> directly. Increasing $k$ reduces the lasting offset and, in this instantaneous model, shortens the response <time constant> to $\tau/(1+k)$. It does not make $x_*$ zero for a nonzero disturbance and finite $k$: a persistent cooling command requires a persistent displacement under <proportional control>. At the <equilibrium point>, the extra heating is balanced by passive loss $x_*$ plus active cooling $u_*$. Zero change of <temperature> means balanced rates, rather than zero rates, consistent with <conservation of energy>.
The first diagram uses $D=1$, $\tau=4$, and a pulse from $t=2$ to $t=10$, comparing $k=0$ with $k=3$. The latter approaches $x_*=1/4$ during the pulse, compared with $1$ without active <feedback control>. After the pulse ends, each displacement decays as $x(10)e^{-(1+k)(t-10)/\tau}$. These are computed curves, not physiological measurements. The $k=0$ curve still has passive loss; it is not a claim that an unregulated body warms without limit.
Real <physiological effectors> take time to respond. To isolate that issue, set $\tau=1$ and replace the instantaneous command by another <ordinary differential equation>:
$$
\dot x=1-x-u,\qquad \dot u=4x-u,\qquad x(0)=u(0)=0.
$$
Here the disturbance remains $1$ from $t=0$ onward and the effector <time constant> is also $1$. The <equilibrium point> is still $(x_*,u_*)=(1/5,4/5)$, but the <characteristic polynomial> is $(\lambda+1)^2+4$. Its <eigenvalues> $-1\pm2i$ give decaying <oscillations> and <asymptotic stability>. The exact response is
$$
x(t)=\frac15+e^{-t}\left(-\frac15\cos(2t)+\frac25\sin(2t)\right),\qquad u(t)=\frac45+e^{-t}\left(-\frac45\cos(2t)-\frac25\sin(2t)\right).
$$
In particular, $\dot x=e^{-t}\cos(2t)$, so the first maximum occurs at $t=\pi/4$, with $x=1/5+(2/5)e^{-\pi/4}\simeq0.3824$. The response exceeds its eventual displacement while remaining above its baseline. The return has the same negative sign as before; the extra state changes the timing. A finite effector <time constant> is distinct from the fixed time lag in the <delayed negative feedback model>.
\Image[/homeostasis-feedback-lag-capacity.png]
{title=Synthetic <homeostasis> responses: offset, lag and limited effector capacity}
{description=Original computed plots from the <homeostatic temperature model>, all in arbitrary units and explicitly labelled synthetic. Upper panel: displacement under the same sustained disturbance. Lower panel: corresponding cooling command. Instantaneous <proportional control>, a finite effector <time constant>, and <effector saturation> give different responses despite opposing the same disturbance. No experimental data or clinical limits are shown.}
Back to article page
