Constant-Coefficient Linear SDEs With Forcing

Forcing in a Linear SDE

A deterministic input can change the mean trajectory of a linear stochastic system without changing its additive-noise structure. For constant coefficients, the variation-of-constants formula separates the effects: the initial value is propagated by the homogeneous equation, the forcing contributes an ordinary time integral, and Brownian noise contributes a stochastic convolution. The main calculation here is to evaluate the ordinary integral for several standard forcing functions.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Fix a finite horizon $T>0$. Consider the scalar SDE $dX_t=(aX_t+f(t))\,dt+\sigma\,dB_t$, with $X_0=\xi$, where $a,\sigma\in\mathbb{R}$, $f$ is a deterministic continuous function on $[0,T]$, and $\xi$ is $\mathcal{F}_0$-measurable with $\mathbb{E}[\xi^2]<\infty$. Here $aX_t+f(t)$ is the drift and $\sigma$ is the diffusion coefficient. In particular, the noise is additive: its coefficient does not depend on $X_t$.

Definition: A strong solution is an adapted continuous process $X_t$, defined using the given Brownian motion and initial value, that satisfies the SDE in its integral form. Pathwise uniqueness means that any two such solutions with the same Brownian motion and initial value agree almost surely at every time. These are distinct from weak existence and uniqueness in law, which concern solutions on possibly different probability spaces and equality of their distributions. In this setting, the global Lipschitz existence and uniqueness theorem applies: the drift is Lipschitz in $x$, since $|ax-ay|=|a||x-y|$, and has linear growth on $[0,T]$, since $f$ is bounded there; the diffusion is constant and therefore Lipschitz and of linear growth. Thus a unique strong solution exists and pathwise uniqueness holds.

Variation of Constants

Theorem: Under the stated assumptions, for $0\le t\le T$ the unique solution is $$ X_t=e^{at}\xi+\int_0^t e^{a(t-s)}f(s)\,ds+\sigma\int_0^t e^{a(t-s)}\,dB_s. $$ For each fixed $t$, the Ito integrand $s\mapsto \sigma e^{a(t-s)}$, $0\le s\le t$, is deterministic, hence progressively measurable, and belongs to $\mathcal{L}^2$, since $$ \mathbb{E}\int_0^t \sigma^2e^{2a(t-s)}\,ds =\sigma^2\int_0^t e^{2a(t-s)}\,ds<\infty. $$

Proof. Define $Y_t=e^{-at}X_t$. The Ito product rule with the deterministic, finite-variation factor $e^{-at}$ gives $dY_t=e^{-at}\,dX_t-ae^{-at}X_t\,dt$; the quadratic covariation with this factor is zero. Substituting the SDE cancels the two terms involving $aX_t$, leaving $dY_t=e^{-at}f(t)\,dt+\sigma e^{-at}\,dB_t$. Integrating from $0$ to $t$, and using $Y_0=\xi$, gives $Y_t=\xi+\int_0^t e^{-as}f(s)\,ds+\sigma\int_0^t e^{-as}\,dB_s$. Multiplication by $e^{at}$ gives the displayed formula. Conversely, this calculation can be reversed: define $Y_t$ by the last expression and $X_t=e^{at}Y_t$; the product rule then shows that $X_t$ satisfies the SDE. The deterministic Ito integrand is square-integrable on every finite interval, so the stochastic integral is well-defined. Finally, if $X$ and $\widetilde X$ are two solutions with the same noise and initial value, their difference $D_t=X_t-\widetilde X_t$ satisfies $dD_t=aD_t\,dt$, $D_0=0$. Multiplying by $e^{-at}$ gives $e^{-at}D_t=0$, so they agree for every $t$ almost surely. $\square$

The stochastic convolution is an Ito integral, not an ordinary time integral. For each fixed $t$, it has mean zero: its integrand is in $\mathcal{L}^2$, and the Ito integral is a square-integrable martingale, so its conditional expectation at an earlier time equals its earlier value. In particular, its expectation at time $t$ is zero. Thus if $\mathbb{E}|\xi|<\infty$, the mean is $m(t)=e^{at}\mathbb{E}[\xi]+\int_0^t e^{a(t-s)}f(s)\,ds$. This mean will provide an independent check in the examples.

Explicit Deterministic Forcing Integrals

Write $I_f(t)=\int_0^t e^{a(t-s)}f(s)\,ds$. The following formulas evaluate this part of the solution; they do not change the stochastic convolution.

Proposition: For constant forcing $f(s)=q$, $$ I_f(t)= \begin{cases} q\dfrac{e^{at}-1}{a},&a\ne0,\\[6pt] qt,&a=0. \end{cases} $$ For exponential forcing $f(s)=q e^{\lambda s}$, $$ I_f(t)= \begin{cases} q\dfrac{e^{\lambda t}-e^{at}}{\lambda-a},&\lambda\ne a,\\[6pt] qt e^{at},&\lambda=a. \end{cases} $$ For sinusoidal forcing $f(s)=A\cos(\omega s)+C\sin(\omega s)$, with $\omega>0$, $$ I_f(t)= \frac{A[-a\cos(\omega t)+\omega\sin(\omega t)+a e^{at}] +C[-a\sin(\omega t)-\omega\cos(\omega t)+\omega e^{at}]} {a^2+\omega^2}. $$

Proof. For constant forcing, integrate $q e^{a(t-s)}$ over $s\in[0,t]$; when $a\ne0$ this is $q(e^{at}-1)/a$, and when $a=0$ the integrand is the constant $q$, giving $qt$. For exponential forcing, factor out $q e^{at}$ to obtain $q e^{at}\int_0^t e^{(\lambda-a)s}\,ds$. If $\lambda\ne a$, evaluating the integral gives $q(e^{\lambda t}-e^{at})/(\lambda-a)$; if $\lambda=a$, its integrand is $1$, giving $qt e^{at}$. For the sinusoidal case, direct integration yields $$ \int_0^t e^{a(t-s)}\cos(\omega s)\,ds =\frac{-a\cos(\omega t)+\omega\sin(\omega t)+a e^{at}}{a^2+\omega^2}, $$ $$ \int_0^t e^{a(t-s)}\sin(\omega s)\,ds =\frac{-a\sin(\omega t)-\omega\cos(\omega t)+\omega e^{at}}{a^2+\omega^2}. $$ For example, each identity follows by differentiating its right-hand side with respect to $t$: the derivative is $a$ times that side plus, respectively, $\cos(\omega t)$ or $\sin(\omega t)$, and the side is zero at $t=0$. Those two properties characterize the corresponding convolution, since the difference of two candidates solves $J'=aJ$, $J(0)=0$. Linearity then gives the claimed formula for $A\cos(\omega s)+C\sin(\omega s)$. The denominator is positive because $\omega>0$. $\square$

Worked Forcing Calculations

Worked Example: Constant Forcing

Take $a=-2$, $q=3$, $\sigma=1$, and $\xi=1$. Find the solution at time $t=1$ when $f(s)=3$.

The initial-value term is $e^{-2}\xi=e^{-2}$. Since $a=-2\ne0$, the forcing integral is $$ \int_0^1 e^{-2(1-s)}3\,ds =3\frac{e^{-2}-1}{-2} =\frac{3}{2}(1-e^{-2}). $$ The stochastic convolution is $\int_0^1 e^{-2(1-s)}\,dB_s$, whose integrand is deterministic and square-integrable. Therefore $$ X_1=e^{-2}+\frac{3}{2}(1-e^{-2})+\int_0^1 e^{-2(1-s)}\,dB_s =\frac{3}{2}-\frac{1}{2}e^{-2}+\int_0^1 e^{-2(1-s)}\,dB_s. $$

Check: the stochastic integral has mean zero, so $\mathbb{E}[X_1]=\tfrac32-\tfrac12e^{-2}$. Independently, the mean $m$ solves $m'=-2m+3$, $m(0)=1$. The function $m(t)=\tfrac32-\tfrac12e^{-2t}$ has $m(0)=1$ and $m'(t)=e^{-2t}=-2m(t)+3$, giving the same value at $t=1$.

Worked Example: Resonant and Nonresonant Exponential Forcing

Let $a=-1$, $\xi=0$, and $\sigma=2$. Compare the forcing $f(s)=2e^{-s}$ with $f(s)=3e^{2s}$. The first is resonant because $\lambda=a=-1$; the second is nonresonant because $\lambda=2\ne-1$.

For $f(s)=2e^{-s}$, the resonant formula gives $I_f(t)=2te^{-t}$. Hence $$ X_t=2te^{-t}+2\int_0^t e^{-(t-s)}\,dB_s. $$ At $t=1$, the deterministic contribution is $2e^{-1}$. For $f(s)=3e^{2s}$, the nonresonant formula gives $$ I_f(t)=3\frac{e^{2t}-e^{-t}}{2-(-1)} =e^{2t}-e^{-t}, $$ so the solution is $X_t=e^{2t}-e^{-t}+2\int_0^t e^{-(t-s)}\,dB_s$. At $t=1$, its deterministic contribution is $e^2-e^{-1}$.

Check: in the resonant case the mean $m(t)=2te^{-t}$ satisfies $m(0)=0$ and $m'(t)=2e^{-t}-2te^{-t}=-m(t)+2e^{-t}$. In the nonresonant case $m(t)=e^{2t}-e^{-t}$ satisfies $m(0)=0$ and $m'(t)=2e^{2t}+e^{-t}=-m(t)+3e^{2t}$. Both agree with the mean equation $m'=am+f$, and the stochastic terms have mean zero.

Worked Example: Sinusoidal Forcing

Take $a=-1$, $\omega=2$, $A=3$, $C=2$, $\xi=0$, and $\sigma=1$. Thus $f(s)=3\cos(2s)+2\sin(2s)$. Evaluate the deterministic term and the solution at $t=\pi/4$.

Here $a^2+\omega^2=1+4=5$. The cosine contribution has numerator $-a\cos(2t)+2\sin(2t)+a e^{-t}=\cos(2t)+2\sin(2t)-e^{-t}$. The sine contribution has numerator $-a\sin(2t)-2\cos(2t)+2e^{-t}=\sin(2t)-2\cos(2t)+2e^{-t}$. Multiplying the first by $A=3$, the second by $C=2$, and adding gives $$ I_f(t)=\frac{-\cos(2t)+8\sin(2t)+e^{-t}}{5}. $$ At $t=\pi/4$, $2t=\pi/2$, so $\cos(\pi/2)=0$ and $\sin(\pi/2)=1$. Therefore $I_f(\pi/4)=(8+e^{-\pi/4})/5$, and $$ X_{\pi/4}=\frac{8+e^{-\pi/4}}{5} +\int_0^{\pi/4}e^{-(\pi/4-s)}\,dB_s. $$

Check: differentiating the proposed integral gives $I_f'(t)=(2\sin(2t)+16\cos(2t)-e^{-t})/5$. Adding $I_f(t)$ gives $(10\sin(2t)+15\cos(2t))/5=2\sin(2t)+3\cos(2t)$, which is $-a$ handled through the equation $I_f'=-I_f+f$. Also $I_f(0)=(-1+1)/5=0$. The forcing term therefore satisfies the required initial condition and differential equation.

Common Pitfall and Use of the Formula

Do not divide by a vanishing resonance denominator. A tempting but invalid substitution into the nonresonant exponential formula at $\lambda=a$ gives $q(e^{at}-e^{at})/(a-a)=0/0$, not a forcing contribution of zero. The correction is to return to the integral: when $\lambda=a$, its integrand is $q e^{a(t-s)}e^{as}=q e^{at}$, independent of $s$, so the integral is $qt e^{at}$. This term is generally nonzero for $q\ne0$ and $t>0$.

The forcing integral describes the deterministic response, while the stochastic convolution describes the response to Brownian noise. In particular, one should not replace the Ito integral by an ordinary integral or combine it with the deterministic forcing integral. The explicit separation is useful in applications because it identifies the mean response directly and keeps the random contribution in a standard, square-integrable Ito integral. The formulas also cover edge cases: constant forcing with $a=0$ gives $qt$, and exponential forcing with $\lambda=a$ gives the resonant term.

Check Your Understanding

For each question, keep the deterministic forcing integral separate from the Ito integral.

For $a=0$, $f(s)=q$, and $\xi=x$, write the full solution and identify its deterministic terms.
Evaluate $\int_0^t e^{a(t-s)}q e^{\lambda s}\,ds$ when $\lambda\ne a$, and explain why the formula changes at $\lambda=a$.
For $a=-2$ and $f(s)=\cos s$, use the sinusoidal formula to find the forcing contribution $I_f(t)$.
State why the stochastic convolution in the variation-of-constants formula has mean zero.
For a proposed sinusoidal forcing contribution, which initial value and differential equation can be used to check the calculation?