Variation Of Constants For Scalar Linear SDEs

From a Homogeneous Equation to a Forced Equation

For an ordinary differential equation, an integrating factor removes the term proportional to the unknown function, leaving an integral of the forcing. For a stochastic differential equation, the same strategy uses the solution of the homogeneous equation as a random integrating factor. The product rule, however, has an additional quadratic-covariation term. When the forcing includes additive noise, that term changes the drift integral by $-c_tg_t$.

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 $T$. Assume that $a_t,c_t,f_t,g_t$ are bounded, deterministic, measurable functions on $[0,T]$, and let $X_0$ be a deterministic real number. The same calculation applies under more general adapted-coefficient hypotheses when the displayed integrals are well-defined; the deterministic bounded assumptions give a simple sufficient setting here.

Definition: A solution of the scalar linear SDE on $[0,T]$ is a continuous adapted process satisfying, almost surely for every $t\in[0,T]$, $$ X_t=X_0+\int_0^t(a_sX_s+f_s)\,ds+\int_0^t(c_sX_s+g_s)\,dB_s. $$ The last integral is an Itô integral: its integrand must be progressively measurable and satisfy $\mathbb{E}\int_0^T(c_sX_s+g_s)^2\,ds<\infty$.

For reference, $\mathcal{L}^2$ denotes the progressively measurable processes $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. An Itô integral with integrand in $\mathcal{L}^2$ is a square-integrable martingale: it is integrable, and for $s\le t$, its conditional expectation at time $s$ equals its value at $s$. In this tutorial, deterministic integrals with respect to $ds$ are ordinary Lebesgue integrals.

The Fundamental Solution and Its Reciprocal

Let $\Phi_0=1$ and let $\Phi$ solve the homogeneous equation $d\Phi_t=a_t\Phi_t\,dt+c_t\Phi_t\,dB_t$. Its explicit form is

\Phi_t=\exp\left(\int_0^t\left(a_s-\tfrac12c_s^2\right)ds+\int_0^t c_s\,dB_s\right).

The stochastic integral here has deterministic bounded integrand $c$, so $c\in\mathcal{L}^2$. To check the formula, set $A_t=\int_0^t(a_s-\tfrac12c_s^2)\,ds$ and apply Itô’s formula to $F(t,x)=e^{A_t+x}$ at $x=\int_0^t c_s\,dB_s$. The derivatives are $F_t=(a_t-\tfrac12c_t^2)F$, $F_x=F$, and $F_{xx}=F$. The quadratic-variation correction is $\tfrac12F_{xx}c_t^2dt=\tfrac12F c_t^2dt$, so $d\Phi_t=a_t\Phi_tdt+c_t\Phi_tdB_t$. In particular, $\Phi_t$ is strictly positive.

Write $\Psi_t=\Phi_t^{-1}$. Applying Itô’s formula to $G(t,x)=e^{-A_t-x}$, whose derivatives are $G_t=-(a_t-\tfrac12c_t^2)G$, $G_x=-G$, and $G_{xx}=G$, gives

d\Psi_t=(c_t^2-a_t)\Psi_t\,dt-c_t\Psi_t\,dB_t,\qquad \Psi_0=1.

Boundedness of $a$ and $c$ on the finite interval implies that $\Phi_t$ and $\Psi_t$ have finite moments of every fixed order, uniformly for $t\in[0,T]$. Indeed, their random parts are exponentials of centered Gaussian integrals with variance at most $T\|c\|_\infty^2$, whose exponential moments are finite. Thus $\Psi g$ is progressively measurable and belongs to $\mathcal{L}^2$, so the Itô integral in the solution formula below is well-defined.

Variation of Constants

Theorem: Under the bounded deterministic coefficient assumptions above, the unique solution is $$ X_t=\Phi_t\left[X_0+\int_0^t\Psi_s(f_s-c_sg_s)\,ds+\int_0^t\Psi_sg_s\,dB_s\right]. $$ The term $-c_sg_s$ is the Itô correction produced by the covariation between the diffusion parts of $\Phi$ and the bracketed process.

Proof. Define $Y_t=X_0+\int_0^t\Psi_s(f_s-c_sg_s)\,ds+\int_0^t\Psi_sg_s\,dB_s$, and set $X_t=\Phi_tY_t$. The Itô integrand $\Psi g$ is in $\mathcal{L}^2$, as shown above. The Itô integral is therefore a square-integrable martingale; in particular, its increments have conditional expectation zero given the earlier filtration. Its differential is $dY_t=\Psi_t(f_t-c_tg_t)\,dt+\Psi_tg_t\,dB_t$.

Apply Itô’s formula to the product function $F(u,v)=uv$. Its first derivatives are $F_u=v$, $F_v=u$; its second derivatives are $F_{uu}=F_{vv}=0$, $F_{uv}=1$. Hence the second-order term is the covariation $d[\Phi,Y]_t$. Since the diffusion coefficients of $\Phi$ and $Y$ are $c_t\Phi_t$ and $\Psi_tg_t$, respectively, and $\Phi_t\Psi_t=1$, this covariation is $c_tg_t\,dt$. Consequently $$ dX_t=Y_t(a_t\Phi_tdt+c_t\Phi_tdB_t)+\Phi_t\big(\Psi_t(f_t-c_tg_t)dt+\Psi_tg_tdB_t\big)+c_tg_tdt. $$ Using $\Phi_tY_t=X_t$ and $\Phi_t\Psi_t=1$, the drift is $a_tX_t+(f_t-c_tg_t)+c_tg_t=a_tX_t+f_t$, and the diffusion is $c_tX_t+g_t$. Also $X_0=\Phi_0Y_0=X_0$. Thus the formula satisfies the SDE. The bounded coefficients and the finite-horizon moment bounds for $\Phi,\Psi$ ensure the resulting solution has the required square-integrable diffusion integrand; equivalently, its second-moment bound follows from the standard a priori estimate for globally Lipschitz SDEs.

For uniqueness, let $X$ and $\widetilde X$ be two solutions driven by this same Brownian motion with the same initial value, and put $D_t=X_t-\widetilde X_t$. Then $dD_t=a_tD_tdt+c_tD_tdB_t$ and $D_0=0$. Apply the product rule to $\Psi_tD_t$. Its drift is $(c_t^2-a_t)\Psi_tD_t+a_t\Psi_tD_t-c_t^2\Psi_tD_t=0$, where the last term is the cross-variation of $-c_t\Psi_t\,dB_t$ and $c_tD_t\,dB_t$. Its diffusion is $-c_t\Psi_tD_t+c_t\Psi_tD_t=0$. Therefore $\Psi_tD_t=0$ for all $t$, and positivity of $\Psi_t$ gives $D_t=0$. This proves pathwise uniqueness. $\square$

Pathwise uniqueness means that solutions using the same Brownian motion and initial value agree almost surely. It is distinct from uniqueness in law, which says that all solutions have the same probability distribution. A strong solution is constructed on the given probability space using its Brownian motion; a weak solution allows the probability space and Brownian motion to be chosen as part of the construction. Here the explicit formula constructs a strong solution, and the uniqueness argument establishes pathwise uniqueness.

Worked Applications

Worked Example: A Drift-Only Equation

Solve $dX_t=(X_t+2)\,dt$ with $X_0=1$. Here $a_t=1$, $f_t=2$, and $c_t=g_t=0$. Thus $\Phi_t=e^t$, $\Psi_t=e^{-t}$, and the stochastic integral vanishes. The formula gives

X_t=e^t\left(1+\int_0^t2e^{-s}\,ds\right) =e^t\left(1+2(1-e^{-t})\right)=3e^t-2.

Check: $X_0=3-2=1$, and differentiating gives $dX_t/dt=3e^t=(3e^t-2)+2=X_t+2$, as required.

Worked Example: Additive Noise with No Multiplicative Noise

Solve $dX_t=2\,dt+3\,dB_t$ with $X_0=-1$. The coefficients are $a_t=c_t=0$, $f_t=2$, $g_t=3$, so $\Phi_t=\Psi_t=1$. The correction $c_tg_t$ is zero, and the formula gives

X_t=-1+\int_0^t2\,ds+\int_0^t3\,dB_s=-1+2t+3B_t.

The integrand $3$ is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^T9\,ds=9T<\infty$; its Itô integral $3B_t$ is a square-integrable martingale. Check: the initial value is $-1$, and differentiating the displayed integral form gives exactly $2\,dt+3\,dB_t$.

Worked Example: A Nonzero Drift Correction

Solve $dX_t=0\,dt+(2X_t+1)\,dB_t$ with $X_0=0$. Here $a_t=f_t=0$, $c_t=2$, and $g_t=1$. The fundamental solution and reciprocal are $\Phi_t=e^{2B_t-2t}$ and $\Psi_t=e^{-2B_t+2t}$. Since $f_t-c_tg_t=-2$, the formula is

X_t=\Phi_t\left[-2t+\int_0^t e^{-2B_s+2s}\,dB_s\right].

The integral is an Itô integral: its integrand is progressively measurable, and $\mathbb{E}\int_0^T e^{-4B_s+4s}\,ds<\infty$ because each Gaussian exponential moment is finite on the finite horizon. To check the drift explicitly, write $Y_t=-2t+\int_0^t\Psi_s\,dB_s$. Then $d\Phi_t=2\Phi_t\,dB_t$, $dY_t=-2\,dt+\Psi_t\,dB_t$, and $d[\Phi,Y]_t=2\,dt$. The product drift is $\Phi_t(-2\Psi_t)dt+2dt=-2dt+2dt=0$; the diffusion is $2\Phi_tY_t\,dB_t+\Phi_t\Psi_t\,dB_t=(2X_t+1)dB_t$. Also $X_0=0$. This verifies both coefficients and the initial condition.

Why the Correction Cannot Be Dropped

A common error is to use the ordinary product rule and write $d(\Phi Y)=Y\,d\Phi+\Phi\,dY$, omitting $d[\Phi,Y]$. In the last example, this would give drift $-2\,dt$, not zero, and would therefore fail to solve the stated SDE. The correct Itô product rule includes $d[\Phi,Y]_t=2\,dt$, which cancels that drift. In differential shorthand, the calculation uses $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

The correction is tied to Itô integration. For continuous semimartingales $U,V$, the Stratonovich product rule has the ordinary form $d(UV)=U\circ dV+V\circ dU$; converting a Stratonovich integral to an Itô integral introduces its own correction. The formula proved here is specifically for the Itô SDE in the definition, and its $-c_tg_t$ term compensates for the cross-variation in the Itô product calculation.

Term	Role in the formula	Source
$\Phi_t$	Homogeneous fundamental solution	Drift $a_tX_t$ and diffusion $c_tX_t$
$\Psi_t(f_t-c_tg_t)\,dt$	Drift inside the bracket	Forcing drift and product-rule correction
$\Psi_tg_t\,dB_t$	Stochastic forcing inside the bracket	Additive diffusion $g_t$

Check Your Understanding

Use the assumptions and formula above to work through these questions.

For $dX_t=(3X_t+1)\,dt$, $X_0=2$, identify $\Phi_t$ and obtain $X_t$ from variation of constants.
In the equation $dX_t=(a_tX_t+f_t)\,dt+(c_tX_t+g_t)\,dB_t$, explain which product-rule term necessitates the correction in the drift integral.
For constant $c=1$, $g=2$, and $f=0$, compute $f-cg$. What sign must this term have in the bracket?
State the integrability condition required for the Itô integral in the solution formula and explain why bounded deterministic coefficients on a finite horizon suffice.
Distinguish pathwise uniqueness from uniqueness in law for the solution constructed here.

Term	Role in the formula	Source
\(\Phi_t\)	Homogeneous fundamental solution	Drift \(a_tX_t\) and diffusion \(c_tX_t\)
\(\Psi_t(f_t-c_tg_t)\,dt\)	Drift inside the bracket	Forcing drift and product-rule correction
\(\Psi_tg_t\,dB_t\)	Stochastic forcing inside the bracket	Additive diffusion \(g_t\)