mathematics//dynamical systems//exponential decay

A proportional loss produces an exponential: when the rate of loss is proportional to what remains,

A proportional loss produces an exponential: when the rate of loss is proportional to what remains,

x˙=−λx,λ>0,x(t)=x(0)e−λt.\dot x=-\lambda x,\qquad \lambda>0,\qquad x(t)=x(0)e^{-\lambda t}.x˙=−λx,λ>0,x(t)=x(0)e−λt.

xxx is a level, x˙\dot xx˙ a rate and λ\lambdaλ a parameter with units of inverse time. For x≥0x\ge0x≥0, the more is left, the more is lost per unit time. Over equal intervals the same proportion remains; the same amount is not subtracted.

The general case. Any rate proportional to the quantity itself, x˙=kx\dot x=kxx˙=kx with kkk an arbitrary constant, has the solution

x(t)=Cekt,C=x(0),x(t)=Ce^{kt},\qquad C=x(0),x(t)=Cekt,C=x(0),

because the derivative of CektCe^{kt}Cekt is kkk times itself. CCC is not part of the equation: it is the initial condition, and every CCC is a solution of the same equation. The sign of kkk decides the shape: negative is decay, positive is growth, zero is a constant. The exponential is the function whose rate is proportional to its value, which is exactly why it appears wherever a proportional law does (differential equations).

The time scale. The time constant is τc=1/λ\tau_c=1/\lambdaτc​=1/λ: after τc\tau_cτc​, e−1≈0.368e^{-1}\approx0.368e−1≈0.368 of the initial value remains. The half-life is ln⁡2/λ\ln 2/\lambdaln2/λ. The exponent must be dimensionless.

In a linear tank ah˙=−kha\dot h=-khah˙=−kh the rate is λ=k/a\lambda=k/aλ=k/a: more conductance empties faster, more capacity empties slower. Not every hydraulic discharge is exponential. An outflow proportional to h\sqrt hh​, as in another orifice model, gives a different law. The mechanism decides the shape.

With an input, the deviation from equilibrium decays. If x˙=g0−λx\dot x=g_0-\lambda xx˙=g0​−λx with constant g0g_0g0​,

x∗=g0λ,x(t)=x∗+[x(0)−x∗]e−λt.x^*=\frac{g_0}{\lambda},\qquad x(t)=x^*+[x(0)-x^*]e^{-\lambda t}.x∗=λg0​​,x(t)=x∗+[x(0)−x∗]e−λt.

The level can rise while the deviation shrinks, so a decreasing exponential in a formula does not mean the whole output decreases. With a varying input,

x(t)=x(0)e−λt+∫0te−λ(t−τ)g(τ) dτ.x(t)=x(0)e^{-\lambda t}+\int_0^t e^{-\lambda(t-\tau)}g(\tau)\,d\tau.x(t)=x(0)e−λt+∫0t​e−λ(t−τ)g(τ)dτ.

The kernel says how much of each past contribution remains: this is memory as convolution, see delay and lag.

Decay, self-amplification and saturation. The term +rx+rx+rx self-amplifies if r>0r>0r>0. Adding a quadratic brake,

x˙=rx−γx2,r,γ>0,x≥0,\dot x=rx-\gamma x^2,\qquad r,\gamma>0,\quad x\ge0,x˙=rx−γx2,r,γ>0,x≥0,

the positive equilibrium is x∗=r/γx^*=r/\gammax∗=r/γ and it attracts every x(0)>0x(0)>0x(0)>0. It is not a hard ceiling: from above the level falls toward it, from below it rises. The term −γx2-\gamma x^2−γx2 alone does not create that positive equilibrium: for x(0)>0x(0)>0x(0)>0 it gives x(t)=x(0)/(1+γx(0)t)x(t)=x(0)/(1+\gamma x(0)t)x(t)=x(0)/(1+γx(0)t), a non-exponential decay. "Quadratic" and "saturation" are not synonyms; see feedback loop.