advanced33 min studyLesson 678 in Mathematics

Feynman–Kac representations

Represent bounded PDE solutions by weighted paths while checking integrability, killing, and boundary data.

On this page

What you’ll understand

  • Derive a finite-horizon representation with killing and a source term.
  • Identify sufficient boundedness for the martingale verification.
  • Distinguish a verification theorem from an unsupported classical existence claim.

Feynman–Kac turns a linear parabolic equation into an expectation over diffusion paths. The representation becomes trustworthy only after specifying the terminal payoff, killing rate, source, horizon, and any boundary rule. We state a bounded verification theorem first: a sufficiently smooth solution must equal the expectation. Existence and classical smoothness of the function defined by that expectation require further analytic hypotheses and are not automatic consequences of writing down a path average.

A bounded verification setting

Fix a finite terminal time and a nonexplosive diffusion from globally Lipschitz coefficients with linear growth. Let the terminal payoff and source be bounded and continuous, and let the killing rate be bounded, continuous, and nonnegative. Suppose a bounded function, once continuously differentiable in time and twice in space in the interior, solves the stated terminal PDE and is continuous up to its terminal data. These assumptions provide a clean class in which the representation and uniqueness can be verified.

Along a path, multiply the candidate solution by the exponential of minus the accumulated killing rate, then add the accumulated discounted source. Itô's formula shows that the PDE cancels the drift. The resulting process is initially a local martingale. Boundedness of the candidate, source, and finite horizon makes this process bounded, hence a true martingale after localization. Taking expectations and then the terminal limit proves the representation without silently assuming an arbitrary local martingale conserves its mean.

ut+Lu−cu+f=0,u(T,x)=g(x),Dt,s=exp⁡ ⁣(−∫tsc(r,Xr) dr)u_t+Lu-cu+f=0,\qquad u(T,x)=g(x),\qquad D_{t,s}=\exp\!\left(-\int_t^s c(r,X_r)\,dr\right)
The killing rate enters the PDE with a minus sign and the path weight with a negative exponential.↔ Scroll wide equations horizontally.

Terminal payoff and running source

The representation has two terms: a discounted terminal payoff and an integral of discounted running source. Omitting the second term changes an inhomogeneous PDE into a homogeneous one. Since the killing rate is nonnegative, the discount factor is at most one; bounded payoff and source therefore give a finite expectation directly. Other growth conditions are possible, but an unbounded payoff or negative unbounded potential requires a separate integrability argument before the same expression can be asserted.

The verification theorem establishes uniqueness among bounded classical solutions satisfying its terminal condition. To conclude that the expectation itself is classical, one can impose, for example, smooth bounded coefficients and data with bounded derivatives and uniform ellipticity of covariance. Rougher data may yield a viscosity or distributional solution instead. The probabilistic representation does not erase these distinctions, and uniqueness can fail in overly large growth classes even for familiar heat equations.

u(t,x)=Et,x ⁣[Dt,Tg(XT)+∫tTDt,sf(s,Xs) ds]u(t,x)=\mathbb E_{t,x}\!\left[D_{t,T}g(X_T)+\int_t^T D_{t,s}f(s,X_s)\,ds\right]
The first term propagates terminal data and the second accounts for the PDE's running source.↔ Scroll wide equations horizontally.

Stopping adds a boundary decision

On a bounded interval with absorbing boundary, stop the diffusion at its first exit time. A zero Dirichlet boundary condition replaces the terminal payoff by its contribution only on paths surviving to the horizon, and truncates the running-source integral at exit. Compatibility of continuous terminal data with the zero boundary value avoids a corner mismatch. Reflected diffusion would impose a different boundary rule and is not obtained by using the same killed-path indicator.

For nonzero prescribed boundary data, an additional discounted payoff is collected at exit before the terminal horizon. That term cannot be dropped merely because the process has stopped. The stopped Itô argument explains every contribution: what remains at the stopping time must match either terminal data or boundary data. On unbounded domains, controlling the limit of artificial spatial stopping times similarly requires the boundedness or growth assumptions that justify passing expectations to the limit.

A damped trigonometric terminal payoff

Let dXs=2 dWsdX_s=\sqrt2\,dW_s, terminal payoff g(x)=cos⁡(3x)g(x)=\cos(3x), constant killing rate two, and no source. Compute the bounded Feynman–Kac function and verify its terminal PDE.

  1. Over elapsed time h=T−th=T-t, the increment is centered Gaussian with variance 2h2h. Its characteristic function at frequency three is e−9he^{-9h}, so the expected cosine is e−9hcos⁡(3x)e^{-9h}\cos(3x).

  2. The killing weight is e−2he^{-2h}, giving u(t,x)=e−11hcos⁡(3x)u(t,x)=e^{-11h}\cos(3x). It is bounded, smooth, and equals the prescribed payoff at the terminal time.

  3. The generator is ∂xx\partial_{xx}. Thus ut=11uu_t=11u and uxx=−9uu_{xx}=-9u, so ut+uxx−2u=0u_t+u_{xx}-2u=0. The cancellation checks the direction of time and the sign of the killing term.

The result

Diffusive smoothing contributes decay rate nine and killing contributes rate two; the rates add to eleven.

Explore the model

Interactive lab

Interactive lab · heat

Separate diffusion decay from killing

Pause/Reset, then set both mode to 3 and diffusivity to 1. Predict heat amplitude exp(−4.5) after one Step at time 0.5. Multiply that amplitude on paper by exp(−1) for killing rate two, obtaining exp(−5.5). Reset and set both mode to 1 and diffusivity to 1 for the slower diffusive comparison.

Separate diffusion decay from killingThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe panel supplies only the diffusion factor. The extra killing weight is calculated separately; the instrument has no killing-rate control or sampled paths.

Check your understanding

What does the bounded verification theorem establish before additional PDE existence arguments?

Try it yourself

Solve the spatially constant terminal problem ut+12uxx−3u+6=0u_t+\tfrac12u_{xx}-3u+6=0 with u(T,x)=4u(T,x)=4 using Feynman–Kac. Evaluate the answer when T−t=(log⁡2)/3T-t=(\log2)/3.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The expected terminal payoff is deterministic here. Integrate the discounted constant source rather than dropping it.

Compare with the solution

With h=T−th=T-t, the representation gives 4e−3h+6∫0he−3sds=2+2e−3h4e^{-3h}+6\int_0^h e^{-3s}ds=2+2e^{-3h}. It is independent of position, has terminal value four, and satisfies ut−3u+6=0u_t-3u+6=0. At the specified elapsed time the exponential equals one half, giving value three. Keeping only the terminal term would give two there and would fail the inhomogeneous PDE because it omitted the source contribution.

Take it with you

  • Killing contributes a negative PDE term and a discounted path weight.
  • A source produces an accumulated integral in the representation.
  • Verification, classical existence, and boundary behavior have separate hypotheses.

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