Background: The Integrating Factor University of Massachusetts Amherst
A linear first-order equation y ′ + p y = q y' + p\,y = q y ′ + p y = q μ = e ∫ p \mu = e^{\int p} μ = e ∫ p ( μ y ) ′ (\mu y)' ( μ y ) ′
Every equation solved exactly in From Newton’s Second Law to a Force Balance is linear and first order. This
page derives the one formula that solves all of them, and then shows where each
of those exact solutions comes from.
Let p p p q q q [ 0 , ∞ ) [0, \infty) [ 0 , ∞ ) P ( t ) = ∫ 0 t p ( s ) d s P(t) = \int_0^t p(s)\,\dd s P ( t ) = ∫ 0 t p ( s ) d s
d y d t + p ( t ) y = q ( t ) , y ( 0 ) = y 0 , \frac{\dd y}{\dd t} + p(t)\,y = q(t),
\qquad y(0) = y_0, d t d y + p ( t ) y = q ( t ) , y ( 0 ) = y 0 , has exactly one solution, namely
y ( t ) = e − P ( t ) ( y 0 + ∫ 0 t e P ( s ) q ( s ) d s ) . y(t) = e^{-P(t)}\Bigl(y_0 + \int_0^t e^{P(s)}\,q(s)\,\dd s\Bigr) . y ( t ) = e − P ( t ) ( y 0 + ∫ 0 t e P ( s ) q ( s ) d s ) . Put μ ( t ) = e P ( t ) \mu(t) = e^{P(t)} μ ( t ) = e P ( t ) integrating factor . It is positive and
satisfies μ ′ = p μ \mu' = p\,\mu μ ′ = p μ μ ( 0 ) = 1 \mu(0) = 1 μ ( 0 ) = 1
Every solution has the form (2) If y y y (1)
d d t ( μ y ) = μ y ′ + μ ′ y = μ ( y ′ + p y ) = μ q . \frac{\dd}{\dd t}(\mu y) = \mu y' + \mu' y = \mu\,(y' + p y) = \mu q . d t d ( μ y ) = μ y ′ + μ ′ y = μ ( y ′ + p y ) = μ q . The left side is now an exact derivative. Integrating from 0 to t t t μ ( 0 ) y ( 0 ) = y 0 \mu(0)y(0) = y_0 μ ( 0 ) y ( 0 ) = y 0
μ ( t ) y ( t ) − y 0 = ∫ 0 t μ ( s ) q ( s ) d s , \mu(t)\,y(t) - y_0 = \int_0^t \mu(s)\,q(s)\,\dd s , μ ( t ) y ( t ) − y 0 = ∫ 0 t μ ( s ) q ( s ) d s , and dividing by μ ( t ) > 0 \mu(t) > 0 μ ( t ) > 0 (2)
The formula is a solution. Differentiating (2)
y ′ = − p e − P ( y 0 + ∫ 0 t e P ( s ) q ( s ) d s ) + e − P e P q = − p y + q , y' = -p\,e^{-P}\Bigl(y_0 + \int_0^t e^{P(s)}q(s)\,\dd s\Bigr)
+ e^{-P}e^{P}q = -p\,y + q , y ′ = − p e − P ( y 0 + ∫ 0 t e P ( s ) q ( s ) d s ) + e − P e P q = − p y + q , and at t = 0 t = 0 t = 0 y ( 0 ) = y 0 y(0) = y_0 y ( 0 ) = y 0
Remark 1 (The method in three steps)As a recipe:
Write the equation in the standard form (1) μ = exp ( ∫ p d t ) \mu = \exp\bigl(\int p\,\dd t\bigr) μ = exp ( ∫ p d t )
Multiply through by μ \mu μ ( μ y ) ′ (\mu y)' ( μ y ) ′ μ \mu μ μ ′ = p μ \mu' = p\mu μ ′ = p μ
Integrate both sides and divide by μ \mu μ
The equation of motion ¶ In scaled variables the equation of motion is
ε d v ~ / d t ~ = F ~ ( t ~ ) − v ~ \ve\,\dd\tilde v/\dd\tilde t = \tilde F(\tilde t) - \tilde v ε d v ~ / d t ~ = F ~ ( t ~ ) − v ~ v ~ ( 0 ) = v ~ 0 \tilde v(0) = \tilde v_0 v ~ ( 0 ) = v ~ 0 (1)
d v ~ d t ~ + 1 ε v ~ = 1 ε F ~ ( t ~ ) , p = 1 ε , P ( t ~ ) = t ~ ε , \frac{\dd\tilde v}{\dd\tilde t} + \frac{1}{\ve}\,\tilde v
= \frac{1}{\ve}\,\tilde F(\tilde t),
\qquad p = \frac{1}{\ve}, \quad P(\tilde t) = \frac{\tilde t}{\ve}, d t ~ d v ~ + ε 1 v ~ = ε 1 F ~ ( t ~ ) , p = ε 1 , P ( t ~ ) = ε t ~ , and (2)
v ~ ( t ~ ) = v ~ 0 e − t ~ / ε + 1 ε ∫ 0 t ~ e − ( t ~ − s ) / ε F ~ ( s ) d s . \tilde v(\tilde t) = \tilde v_0\,e^{-\tilde t/\ve}
+ \frac{1}{\ve}\int_0^{\tilde t} e^{-(\tilde t - s)/\ve}\,\tilde F(s)\,\dd s . v ~ ( t ~ ) = v ~ 0 e − t ~ / ε + ε 1 ∫ 0 t ~ e − ( t ~ − s ) / ε F ~ ( s ) d s . The first term is the initial velocity, forgotten on the time ε \ve ε ε − 1 e − ( t ~ − s ) / ε \ve^{-1}e^{-(\tilde t - s)/\ve} ε − 1 e − ( t ~ − s ) / ε 1 − e − t ~ / ε 1 - e^{-\tilde t/\ve} 1 − e − t ~ / ε ε \ve ε (7)
Constant force. The integral is
F ~ ( 1 − e − t ~ / ε ) \tilde F\bigl(1 - e^{-\tilde t/\ve}\bigr) F ~ ( 1 − e − t ~ / ε )
v ~ ( t ~ ) = F ~ + ( v ~ 0 − F ~ ) e − t ~ / ε , \tilde v(\tilde t) = \tilde F + (\tilde v_0 - \tilde F)\,e^{-\tilde t/\ve} , v ~ ( t ~ ) = F ~ + ( v ~ 0 − F ~ ) e − t ~ / ε , which is (6) The Singular Limit . In dimensional variables, with
p = 1 / τ i p = 1/\tau_i p = 1/ τ i q = F / m q = F/m q = F / m (10)
Ramped force. For F ~ ( s ) = 1 + s \tilde F(s) = 1 + s F ~ ( s ) = 1 + s v ~ 0 = 0 \tilde v_0 = 0 v ~ 0 = 0 u = 1 + s u = 1 + s u = 1 + s d g = ε − 1 e − ( t ~ − s ) / ε d s \dd g = \ve^{-1}e^{-(\tilde t - s)/\ve}\,\dd s d g = ε − 1 e − ( t ~ − s ) / ε d s g = e − ( t ~ − s ) / ε g = e^{-(\tilde t - s)/\ve} g = e − ( t ~ − s ) / ε
1 ε ∫ 0 t ~ e − ( t ~ − s ) / ε ( 1 + s ) d s = [ ( 1 + s ) e − ( t ~ − s ) / ε ] 0 t ~ − ∫ 0 t ~ e − ( t ~ − s ) / ε d s = 1 + t ~ − e − t ~ / ε − ε ( 1 − e − t ~ / ε ) . \frac{1}{\ve}\int_0^{\tilde t} e^{-(\tilde t - s)/\ve}(1 + s)\,\dd s
= \Bigl[(1 + s)\,e^{-(\tilde t - s)/\ve}\Bigr]_0^{\tilde t}
- \int_0^{\tilde t} e^{-(\tilde t - s)/\ve}\,\dd s
= 1 + \tilde t - e^{-\tilde t/\ve} - \ve\bigl(1 - e^{-\tilde t/\ve}\bigr) . ε 1 ∫ 0 t ~ e − ( t ~ − s ) / ε ( 1 + s ) d s = [ ( 1 + s ) e − ( t ~ − s ) / ε ] 0 t ~ − ∫ 0 t ~ e − ( t ~ − s ) / ε d s = 1 + t ~ − e − t ~ / ε − ε ( 1 − e − t ~ / ε ) . This is 1 + t ~ − ε + ( ε − 1 ) e − t ~ / ε 1 + \tilde t - \ve + (\ve - 1)e^{-\tilde t/\ve} 1 + t ~ − ε + ( ε − 1 ) e − t ~ / ε (16)
The error of the composite. In the proof of
Theorem 1 The Singular Limit ) the error w = v ~ − v ~ u n i f w = \tilde v - \tilde v_\mathrm{unif} w = v ~ − v ~ unif w ′ + w / ε = − F ~ ′ w' + w/\ve = -\tilde F' w ′ + w / ε = − F ~ ′ w ( 0 ) = 0 w(0) = 0 w ( 0 ) = 0 p = 1 / ε p = 1/\ve p = 1/ ε q = − F ~ ′ q = -\tilde F' q = − F ~ ′ y 0 = 0 y_0 = 0 y 0 = 0 (2)
w ( t ~ ) = − ∫ 0 t ~ e − ( t ~ − s ) / ε F ~ ′ ( s ) d s , w(\tilde t) = -\int_0^{\tilde t} e^{-(\tilde t - s)/\ve}\,\tilde F'(s)\,\dd s , w ( t ~ ) = − ∫ 0 t ~ e − ( t ~ − s ) / ε F ~ ′ ( s ) d s , the formula used there.