From a Diffusion to a Backward Equation
A diffusion changes through both a systematic trend and random fluctuations. A partial differential equation describing a quantity evaluated at time \(t\) and state \(x\) must account for both effects. Its spatial operator is the diffusion generator: the drift contributes a first derivative, while the diffusion contributes a second derivative. A terminal-value problem prescribes the quantity at a future time \(T\) and asks for an equation governing its values at earlier times.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a one-dimensional diffusion satisfying \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\). Here \(\mu\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. For definiteness, standard sufficient assumptions for a unique strong solution are that the coefficients are measurable in time, globally Lipschitz in the state variable, and of linear growth. A strong solution is adapted to the given filtration and driven by its given Brownian motion; a weak solution allows the probability space and Brownian motion to be chosen as part of the solution. Under the stated Lipschitz and growth assumptions, the standard existence-and-uniqueness theorem gives a strong solution, and hence pathwise uniqueness. We do not need that theorem’s proof here.
All stochastic integrals below are Itô integrals. Their integrands are progressively measurable and belong to \(\mathcal{L}^2\) on the time interval in use; that is, \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). When a calculation is localized to a bounded region, the stopped integrand is bounded on a finite time interval and therefore satisfies this condition. The differential rules behind the generator are \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\).
The second derivative term is not an adjustment to the drift coefficient. It records the effect of quadratic variation: in Itô’s formula, the second derivative is multiplied by one half of the squared diffusion coefficient. In particular, a diffusion with zero drift can still have a nonzero generator.
The Backward Cauchy Problem
Let \(g(x)\) be terminal data, \(c(t,x)\) a potential or killing rate, and \(f(t,x)\) a source term. The backward Cauchy problem on \([0,T]\times\mathbb{R}\) is
Here \(\mathcal{L}_t u(t,x)\) means that the generator acts on the state variable \(x\), with \(t\) held fixed. A classical solution has one continuous time derivative and two continuous state derivatives on the time-space region, with the terminal condition attained pointwise. Additional growth and coefficient assumptions may be needed for existence of such a solution; writing down the problem alone is not an existence theorem.
The equation’s signs are part of its definition. With this convention, \(f\) enters with a plus sign on the left. The potential enters as \(-cu\). If \(c\ge0\), it is often interpreted as a nonnegative rate that attenuates the quantity \(u\); the PDE itself can also be formulated for \(c\) of either sign. The source \(f\) is an inhomogeneous input. It is distinct from terminal data: \(g\) is specified at \(T\), whereas \(f\) acts throughout the time interval.
Why This Is the Generator
The generator describes the first-order change in the expected value of a smooth test function over a short time interval. The following proposition proves that interpretation under boundedness assumptions that allow the limiting calculation to be made directly.
Proof. Apply Itô’s formula to \(\varphi(X_r^{s,x})\). Its first derivative is \(\varphi'(X_r^{s,x})\), its second derivative is \(\varphi''(X_r^{s,x})\), and the second-order Itô correction is \(\tfrac12\varphi''(X_r^{s,x})\sigma(r,X_r^{s,x})^2\,dr\). Thus \(\varphi(X_{s+h}^{s,x})-\varphi(x)=\int_s^{s+h}\mathcal{L}_r\varphi(X_r^{s,x})\,dr+\int_s^{s+h}\sigma(r,X_r^{s,x})\varphi'(X_r^{s,x})\,dB_r\). The stochastic integrand is progressively measurable and bounded, so it is in \(\mathcal{L}^2\). The Itô integral has expectation zero: this holds for simple adapted integrands by the zero conditional mean of Brownian increments, and extends to the integral by its \(L^2\) construction. Taking expectations therefore leaves the time integral. Bounded coefficients imply \(\mathbb{E}|X_r^{s,x}-x|^2\le 2\|\mu\|_\infty^2(r-s)^2+2\|\sigma\|_\infty^2(r-s)\), so \(X_r^{s,x}\) converges to \(x\) in probability as \(r\downarrow s\). Continuity of the coefficients and derivatives, together with boundedness of the integrand \(\mathcal{L}_r\varphi(X_r^{s,x})\), gives \(\mathbb{E}[\mathcal{L}_r\varphi(X_r^{s,x})]\to\mathcal{L}_s\varphi(x)\). Averaging this expectation over \([s,s+h]\) and dividing by \(h\) proves the asserted limit. \(\square\)
The time-dependent form of the operator also follows directly from Itô’s formula. This identity explains why the backward equation combines a time derivative with the generator.
Proof. For the function \((t,x)\mapsto u(t,x)\), the partial derivatives are \(u_t(t,x)\), \(u_x(t,x)\), and \(u_{xx}(t,x)\). Itô’s formula gives \(du(t,X_t)=u_t(t,X_t)\,dt+u_x(t,X_t)\,dX_t+\tfrac12u_{xx}(t,X_t)\,d[X]_t\). The diffusion has quadratic variation \(d[X]_t=\sigma(t,X_t)^2dt\). Substituting \(dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dB_t\) yields the drift \((u_t+\mu u_x+\tfrac12\sigma^2u_{xx})(t,X_t)dt\) and the stochastic term \(\sigma(t,X_t)u_x(t,X_t)dB_t\), as claimed. On a bounded stopped region the stochastic integrand is progressively measurable and bounded, hence in \(\mathcal{L}^2\). \(\square\)
Consequently, when \(u\) solves the stated backward equation, the drift in this formula is \((c u-f)(t,X_t)\,dt\). This is an algebraic consequence of the PDE, not a representation theorem: it does not by itself assert that a particular expectation solves the PDE or that the PDE has a unique solution.
Worked PDE Calculations
Worked Example: Constant Drift and Volatility with Quadratic Terminal Data
Take \(dX_t=\tfrac12\,dt+2\,dB_t\), set \(c=f=0\), and prescribe \(g(x)=x^2\). Find and check a classical solution.
The coefficients are \(\mu=\tfrac12\) and \(\sigma=2\), so \(\mathcal{L}\varphi=\tfrac12\varphi'+2\varphi''\), since \(\tfrac12\sigma^2=\tfrac12\cdot4=2\). Set \(\tau=T-t\) and try \(u(t,x)=(x+\tfrac12\tau)^2+4\tau\). Its derivatives are \(u_t=-(x+\tfrac12\tau)-4\), \(u_x=2(x+\tfrac12\tau)\), and \(u_{xx}=2\). Therefore
At \(t=T\), \(\tau=0\), so \(u(T,x)=x^2=g(x)\). As an independent algebraic check, expanding the candidate gives \(u=x^2+x\tau+\tfrac14\tau^2+4\tau\). Then \(u_t=-x-\tfrac12\tau-4\), \(u_x=2x+\tau\), and \(u_{xx}=2\), which again gives \(u_t+\tfrac12u_x+2u_{xx}=0\). Both the drift and the diffusion term are needed for the cancellation.
Worked Example: An Ornstein–Uhlenbeck Generator and Linear Terminal Data
Consider \(dX_t=-X_t\,dt+dB_t\), with \(c=f=0\) and \(g(x)=x\). Check the candidate \(u(t,x)=e^{-(T-t)}x\).
The drift is \(-x\) and the diffusion coefficient is \(1\), giving \(\mathcal{L}\varphi=-x\varphi'+\tfrac12\varphi''\). Writing \(\tau=T-t\), the candidate has \(u_t=e^{-\tau}x\), \(u_x=e^{-\tau}\), and \(u_{xx}=0\). Hence \(u_t+\mathcal{L}u=e^{-\tau}x-xe^{-\tau}+\tfrac12(0)=0\). At \(t=T\), \(e^0x=x\), so the terminal condition holds. As a check on the role of the second derivative, there is no diffusion contribution for this linear candidate because \(u_{xx}=0\); the drift term exactly cancels the time derivative.
Worked Example: Constant Source and Potential
Use standard Brownian motion, so \(\mathcal{L}=\tfrac12\partial_{xx}\). Let \(c=2\), \(f=3\), and \(g(x)=0\). Since the data do not depend on \(x\), look for a spatially constant solution \(u(t,x)=y(t)\). The PDE reduces to \(y'(t)-2y(t)+3=0\), with \(y(T)=0\). Test \(y(t)=\tfrac32(1-e^{-2(T-t)})\).
With \(\tau=T-t\), its time derivative is \(y'(t)=-3e^{-2\tau}\). Substitution gives \(y'-2y+3=-3e^{-2\tau}-3(1-e^{-2\tau})+3=0\). Also \(y(T)=\tfrac32(1-1)=0\). Thus the candidate satisfies both the equation and terminal data.
Independent check. Multiplying \(y'-2y=-3\) by the integrating factor \(e^{-2t}\) gives \((e^{-2t}y)'=-3e^{-2t}\). Integrating from \(t\) to \(T\) and using \(y(T)=0\) yields \(-e^{-2t}y(t)=-\tfrac32(e^{-2t}-e^{-2T})\), hence \(y(t)=\tfrac32(1-e^{-2(T-t)})\), the same expression.
Sign check. If the source sign were reversed, the equation would be \(y'-2y-3=0\). Substituting the stated candidate gives \(-3e^{-2\tau}-3(1-e^{-2\tau})-3=-6\), not zero. In particular, the residual is \(-6\) at \(t=T\) as well; reversing the source sign does not leave the candidate as a solution.
What the Terms Do—and a Common Pitfall
| Term | Role in the equation | Effect in the examples |
|---|---|---|
| \(\mu\,\partial_x u\) | Contribution from the drift of the diffusion | Produces the first-derivative term. |
| \(\tfrac12\sigma^2\,\partial_{xx}u\) | Contribution from quadratic variation | Produces the second-derivative term. |
| \(-cu\) | Potential or killing term | Changes the homogeneous part of the equation. |
| \(+f\) | Source term under this sign convention | Makes the equation inhomogeneous. |
| \(u(T,x)=g(x)\) | Prescribed terminal data | Fixes the value at the final time. |
A frequent error is to write down the correct generator but then change the source sign without changing the candidate or rechecking the equation. The constant-source calculation shows why substitution matters: the residual for the reversed sign is exactly \(-6\), not a time-dependent expression that might vanish at some selected times. Another common error is to replace the generator by only its drift term. That loses the second-order contribution even when the drift is zero; for Brownian motion, \(\mathcal{L}=\tfrac12\partial_{xx}\), not the zero operator.
The backward Cauchy problem is useful because it organizes a diffusion’s local dynamics and the terminal information into one equation. The generator is determined by the drift and diffusion coefficients, while the potential and source specify additional terms in the model. The next mathematical task—relating solutions to stochastic expectations and establishing conditions for that representation—is separate from simply formulating and checking this PDE.
Check Your Understanding
For each question, keep the stated sign convention \(\partial_tu+\mathcal{L}_tu-cu+f=0\) and terminal condition \(u(T,x)=g(x)\).
- For \(dX_t=3\,dt+2\,dB_t\), write the generator and identify the coefficient of \(\partial_{xx}\).
- For \(dX_t=-\kappa X_t\,dt+\eta\,dB_t\), with constants \(\kappa,\eta\), write the generator and identify which term comes from quadratic variation.
- If \(u(t,x)=y(t)\) is spatially constant, simplify the backward equation and state the resulting terminal condition.
- For the candidate in the constant-source example, substitute into the equation with source term \(-3\) and compute the residual.
- Explain why the drift term alone cannot be the generator of standard Brownian motion.