Forced ODEs variation of constants and resonance

A homogeneous model describes what a system does without continuing input. Rainfall, emissions or periodic driving supply forcing terms. Variation of constants separates the initial-state response from the accumulated response to forcing. Resonance shows why a seemingly harmless periodic input can produce a qualitatively different solution when natural and forcing frequencies coincide.

1 Variation of constants for a system

For a constant matrix $A$ and continuous forcing $f(t)$, consider $x'=Ax+f(t)$, $x(0)=x_0$. Differentiate $e^{-tA}x(t)$:

$$\frac d{dt}(e^{-tA}x)=e^{-tA}f(t),$$

because $A$ commutes with its exponential. Integration gives

$$x(t)=e^{tA}x_0+\int_0^te^{(t-s)A}f(s)ds.$$

This proves the formula and, by direct differentiation, verifies it. Subtracting two candidate solutions leaves a homogeneous equation with zero initial value, proving uniqueness. A matrix integral means integrating entries or components. No inverse of $A$ is required, so the formula also handles singular matrices and conserved modes.

For constant forcing $b$ and invertible $A$, the equilibrium is $x_*=-A^{-1}b$ and $x(t)=x_*+e^{tA}(x_0-x_*)$. This rewriting is unavailable when $A$ is singular. More importantly, an algebraic equilibrium need not attract trajectories; the behavior is determined by $e^{tA}$. For a real eigenbasis, every direction decays precisely when all eigenvalues are negative. A zero eigenvalue keeps a nondecaying eigenmode, and defective blocks can also produce polynomial factors; signs alone at zero do not classify boundedness.

2 Forced exchange and conserved modes

Take two equal-volume compartments with exchange rate $k>0$, no removal, and constant input $u$ into the first compartment:

$$c_1'=-kc_1+kc_2+u,\qquad c_2'=kc_1-kc_2.$$

Then $s=c_1+c_2$ satisfies $s'=u$, while $d=c_1-c_2$ satisfies $d'=-2kd+u$. Thus

$$s=s_0+ut,\qquad d=\frac{u}{2k}+\left(d_0-\frac{u}{2k}\right)e^{-2kt},\qquad c_1=\frac{s+d}{2},\quad c_2=\frac{s-d}{2}.$$

The concentration difference approaches a constant, but the total keeps growing. There is no equilibrium for $u\ne0$: summing equilibrium equations would demand $0=u$. Claiming every stable-looking exchange system settles to a fixed level would ignore the zero eigenvalue of the conserved total mode.

If both compartments also remove at rate $\mu>0$, the total equation becomes $s'=-\mu s+u$, so the steady total is $u/\mu$. Deriving the aggregate equation before solving the full system is an efficient consistency check.

3 A forced second-order equation

For $y''+ay'+by=g(t)$, write $x=(y,y')^T$. Then

$$x'=\begin{pmatrix}0&1\\-b&-a\end{pmatrix}x+\binom0{g(t)}.$$

The system formula applies and accounts for both initial conditions. For simple forcing, undetermined coefficients gives a shorter calculation, provided overlap with homogeneous solutions is handled correctly.

Example. Solve $y''+2y'+2y=e^{-t}$ with zero position and velocity. Put $y=e^{-t}z$. Differentiation gives $y''+2y'+2y=e^{-t}(z''+z)$, so $z''+z=1$. Initial conditions become $z(0)=z'(0)=0$, hence $z=1-\cos t$ and

$$y=e^{-t}(1-\cos t).$$

A trial $Ce^{-t}$ alone would not satisfy both initial conditions; the homogeneous terms must be retained as needed.

4 Resonance and damping

For the undamped oscillator $y''+\omega_0^2y=F_0\cos\omega t$, with $\omega_0>0$, $\omega\ge0$, and $\omega\ne\omega_0$, a particular solution is

$$y_p=\frac{F_0}{\omega_0^2-\omega^2}\cos\omega t.$$

Substitution proves it. At $\omega=\omega_0$ the denominator vanishes because the proposed forcing shape is itself homogeneous. A new trial gives

$$y_p=\frac{F_0}{2\omega_0}t\sin\omega_0t.$$

Differentiating twice verifies the equation; its envelope grows linearly. This is resonance in the ideal undamped linear model, not a prediction that every real system grows without limit.

With positive damping,

$$y''+2\gamma y'+\omega_0^2y=F_0\cos\omega t,\qquad\gamma>0,$$

use $y_p=A\cos\omega t+B\sin\omega t$. Comparing coefficients gives

$$(\omega_0^2-\omega^2)A+2\gamma\omega B=F_0,\quad (\omega_0^2-\omega^2)B-2\gamma\omega A=0.$$

The denominator is $D=(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2$, giving $A=F_0(\omega_0^2-\omega^2)/D$ and $B=2\gamma\omega F_0/D$. The amplitude is

$$\sqrt{A^2+B^2}=\frac{|F_0|}{\sqrt D}.$$

It stays finite at $\omega=\omega_0$ when damping is positive. For nonzero forcing $F_0\ne0$, the frequency maximizing this response is not generally exactly $\omega_0$: differentiation of $D$ gives a positive-frequency peak at $\sqrt{\omega_0^2-2\gamma^2}$ only when $\gamma<\omega_0/\sqrt2$. Otherwise the largest amplitude over nonnegative frequencies occurs at zero. If $F_0=0$, every forcing frequency has zero response amplitude and the peak-location discussion does not single out a frequency. Natural frequency, damped free frequency and driven response-peak frequency are distinct notions.

5 An integral response formula

For $y''+\omega_0^2y=g(t)$ with $\omega_0>0$, continuous $g$ on an interval containing zero, and zero initial conditions,

$$y(t)=\int_0^t\frac{\sin(\omega_0(t-s))}{\omega_0}g(s)ds.$$

Proof. Differentiate using the endpoint term and differentiation under the integral for continuous $g$. The first derivative is $\int_0^t\cos(\omega_0(t-s))g(s)ds$. Differentiating again gives $g(t)-\omega_0\int_0^t\sin(\omega_0(t-s))g(s)ds=g(t)-\omega_0^2y(t)$. Both initial values vanish. Uniqueness establishes the formula. $\square$

This is a convolution with the impulse-response kernel. It also recovers the resonant solution by inserting a resonant cosine, without guessing a trial form.

6 Graded practice

E1 Foundation. Solve $y'+2y=4e^{-t}$ with $y(0)=1$.

E2 Resonance. Solve $y''+y=\cos t$ with zero initial position and velocity.

E3 Defective forced system. Solve $x_2'=-x_2+1$, $x_1'=-x_1+x_2$ with $x_1(0)=x_2(0)=0$.

E4 Conserved mode. In the forced exchange model, set $k=1$, $u=2$, and initial concentrations zero. Find $c_1,c_2$ and verify their sum.

E5 Damped response. For $\omega_0=3$, $\gamma=1$, $F_0=2$, find the positive response-peak frequency and the amplitude at forcing frequency 3.

7 Full solutions

S1. The integrating factor gives $y=e^{-2t}[1+4(e^t-1)]=4e^{-t}-3e^{-2t}$. Differentiation verifies the ODE and the initial value.

S2. The resonant particular solution is $y=(t/2)\sin t$. Both it and its first derivative vanish at zero, so no extra homogeneous term is needed. Substitution gives $y''+y=\cos t$.

S3. $x_2=1-e^{-t}$. Multiplying the first-coordinate equation by $e^t$ gives $(e^tx_1)'=e^t-1$, so $x_1=1-(1+t)e^{-t}$. The factor $t$ records the coupling through the repeated eigenvalue.

S4. $s=2t$ and $d=1-e^{-2t}$. Thus $c_1=t+(1-e^{-2t})/2$ and $c_2=t-(1-e^{-2t})/2$. Their sum is $2t$ and their difference satisfies $d'=-2d+2$. Although the difference tends to 1, neither concentration tends to a finite steady value.

S5. Since $1<3/\sqrt2$, the peak is at $\sqrt{9-2}=\sqrt7$. At $\omega=3$, $D=4\cdot1\cdot9=36$, giving amplitude $2/6=1/3$. The two frequencies are unequal, illustrating the damping shift.

Summary

Variation of constants separates initial and forced responses without assuming an invertible system matrix. Conserved modes can grow under sustained input even when differences decay. Resonance requires a modified particular solution, while positive damping regularizes the response and separates natural, free-damped and driven-peak frequencies.