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If $q={\sigma }^{2}p$ and $q^{\prime }=2bp$, then $pLu=(q{u}_x{)}^{\prime }/2$. Insert compact cutoffs ${\chi }_R$, integrate the backward equation in time, and integrate by parts twice to move spatial derivatives onto ${\chi }_R$. The error is bounded by $t\Vert u{\Vert }_{\infty }(C{R}^{-2}\Vert q{\Vert }_1+C{R}^{-1}\Vert q^{\prime }{\Vert }_1)/2$. When these norms are finite it vanishes, proving $\int u(t)p=\int fp$ without assuming derivative behavior at infinity.

The coefficient $\sigma$ multiplying the Brownian motion increment in a scalar stochastic differential equation. The local variance rate is ${\sigma }^2$, and the second-order coefficient in the backward Kolmogorov backward equation is ${\sigma }^2/2$. This distinguishes the amplitude from the physical diffusion coefficient convention $D={\sigma }^2/2$.

For the one-dimensional Itô equation $dX=b(X)dt+\sigma (X)dW$, the Infinitesimal generator on smooth test functions is $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. The Kolmogorov backward equation is $u_t=Lu$, while its formal adjoint produces the Fokker-Planck equation.

A Markov diffusion is a continuous-path Markov process whose local evolution is described by a drift vector and diffusion matrix. For sufficiently regular coefficients, its infinitesimal generator is a second-order differential operator and its conditional expectations satisfy a Kolmogorov backward equation.

The transition operators of a time-homogeneous Markov process satisfy $P_{t+s}=P_t{P}_s$ by the Markov property. They preserve constants and positivity. When the generator and function domain permit differentiation, ${\partial }_t{P}_{t}g=L(P_{t}g)$ is the Kolmogorov backward equation. Transition semigroup of a continuous-time Markov chain is the discrete-state special case.

Let $X$ be an adapted Càdlàg process on a finite state space, and let $Q$ be a conservative transition-rate matrix. It solves the martingale problem for $Q$ when, for every real function $f$ on the state space,is a martingale for the specified filtration. On a finite state space all such test functions and their generator values are bounded.

A Continuous-time Markov chain with generator $Q$ has this property by Dynkin's formula. Conversely, fix a terminal time $t$ and put $u(s,x)=(e^{(t-s)Q}f)(x)$ for $0\le s\le t$. The backward equation gives ${\partial }_{s}u+Qu=0$. Applying the martingale identities to its finitely many time-dependent coefficients shows $u(s,X_s)$ is a martingale. ThusTesting state indicators recovers the entire conditional transition law, proving the Markov property and identifying its semigroup. The càdlàg and filtration assumptions make this a path-process characterization, rather than a statement only about separately fixed-time versions.

Write the diffusion process as $d{X}_t=b(X_t)dt+\sigma (X_t)d{B}_t$, with $a=\sigma {\sigma }^{\mathsf{T}}$. Its diffusion generator isThe Itô formula says that $\phi (X_t)-\phi (X_0)-{\int }_0^{t}L\phi (X_s)ds$ is a local martingale, and is a true martingale when the stopped terms are integrable. Taking expectations gives Dynkin formula for a diffusion. This is the bridge from infinitesimal stochastic dynamics to differential equations.

For the transition semigroup $P_{t}g(x)={\mathbb{E}}_{x}g(X_t)$, the Markov property gives $P_{t+s}=P_t{P}_s$. On a smooth generator domain where differentiation is valid, differentiating the semigroup identity yields the Kolmogorov backward equation ${\partial }_t{P}_{t}g=L(P_{t}g)$, with initial value $g$. If a transition density $p(t,x,y)$ exists, integration against smooth compactly supported tests and integration by parts give the Fokker-Planck equationThe backward equation acts on the starting point, while this forward equation describes the evolving probability density.

Here is a full representation proof with a source and potential. Suppose $X$ is nonexplosive, $V\ge 0$ is bounded, $f,g$ are bounded, and $u$ is a bounded classical solution on a finite slab ofFor a process started from $x$ at time $t$, put $D_s=\exp (-{\int }_t^{s}V(X_r)dr)$. The Itô product rule givesTherefore $D_{s}u(s,X_s)+{\int }_t^s{D}_{r}f(r,X_r)dr$ is a local martingale. It is bounded in absolute value by $\Vert u{\Vert }_{\infty }+(T-t)\Vert f{\Vert }_{\infty }$, so it is a true martingale, without requiring a global derivative bound. Its endpoint expectations give the parabolic Feynman-Kac formula with a sourceThis proves uniqueness in the bounded classical class. Setting $V=f=0$ recovers the backward heat or diffusion representation; the minus sign in the killing term produces the negative exponential.

For an elliptic problem, stop at the first exit $\tau$ from a bounded regular domain $D$. Suppose ${\mathbb{E}}_x\tau <\infty$, $u\in C^2(D)\cap C(\bar{D})$ is bounded and solves $(L-V)u=-f$ with boundary data $g$. The same calculation, stopped at $t\wedge \tau$, gives a martingale dominated by $\Vert u{\Vert }_{\infty }+\Vert f{\Vert }_{\infty }\tau$. Continuity puts $X_{\tau }$ on the boundary. Taking the limit and using dominated convergence gives the elliptic Feynman-Kac formulaFor $V=f=0$, harmonic functions for $L$ give stopped martingales and are averages of their exit-boundary values. For Brownian motion $L=\frac{1}{2}\Delta$, this is the probabilistic solution of the Dirichlet problem. Suitable smooth coefficients and uniform ellipticity give familiar settings where the required diffusion, exit-time integrability and classical solutions exist. The representation itself follows from the stated regularity and integrability conditions, rather than claiming classical solvability for arbitrary coefficients.

Define the diffusion generator $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. Fix a horizon $t$ and start the strong solution from the deterministic state $x$. The Itô formula applied to the time-reversed test function givesThe drift vanishes by the Kolmogorov backward equation. This is initially a local martingale; localization on compact state/time sets justifies the stochastic integral without a global derivative bound. Since $u$ itself is bounded, this local martingale is a true martingale on $[0,t]$. Its two endpoint expectations giveThis is the bounded backward-equation stochastic representation. Starting the strong solution at deterministic $x$ is the precise meaning of the conditional notation at $X_0=x$. Also $f=u(0,\cdot )$ is bounded, even though boundedness was not separately imposed on $f$.

The killed transition operator isThe infinitesimal generator of the diffusion isThe Markov property and the Chapman-Kolmogorov equation give, for each fixed $y$,Dividing by $h$, letting $h\downarrow 0$, and using the generator definition together with the assumed $C^{1,2}$ regularity gives the pointwise Kolmogorov backward equation

The backward generator of the drift--diffusion is $\mathcal{L}=\alpha {\partial }_y+D{\partial }_y^2$. The survival probability satisfies the Kolmogorov backward equationwithThe target is absorbing, while reflection gives the Neumann boundary condition at $L$.

The adsorption mechanism is uniform along the two vertical sides, and $X$ evolves independently of $Y$. Consequently the mean first-passage time depends only on the initial $x$-coordinate. Its Kolmogorov backward equation for $a_1=1$ is

The forward partially absorbing boundary condition for a diffusion is $\mathbf{J}\cdot \mathbf{n}=\kappa p$. The boundary term in the adjoint relation isbecause the $X$ diffusion coefficient is one. It vanishes for every admissible $p$ precisely whenFor $\kappa =1$, the two Robin boundary conditions are therefore

The general solution of the ordinary differential equation isThe right condition gives $C=2$, and the left gives $D=-2/e$. At the prescribed initial position,

With an absorbing boundary condition for a diffusion on each vertical side, the splitting probability $u(x)$ of hitting the left side before the right is harmonic for the Kolmogorov backward equation:For $a_1\ne 0$,The particle starts at $x=0$, soThe continuous limit at zero is $g(0)=1/2$, as required by reflection symmetry. A large positive drift drives the particle toward the right and gives $g(a_1)\to 0$; a large negative drift drives it toward the left and gives $g(a_1)\to 1$.

Fix $t$ and apply Itô formula to $M_s=U(t-s,X_s)$ for $0\le s\le t$. Its drift isby the Kolmogorov backward equation. HenceLocalization makes this a martingale, and boundedness of $U$ permits passage to the limit. Conditioning the identity $M_t=U(0,X_t)$ on $X_0$ givesThis is the required special case of the Feynman-Kac formula, proved directly.

For the stochastic differential equation $d{X}_s=b(s,X_s)ds+\Sigma (s,X_s)d{B}_s$, the diffusion generator on smooth functions isOne form of the Feynman-Kac formula is the following. Suppose the stochastic differential equation is well posed and nonexplosive, $c\ge 0$ is bounded, the terminal data $g$ and source $h$ are bounded, and $F$ is a classical solution bounded on finite time intervals ofWith the regularity needed for the Itô formula, the parabolic Feynman-Kac formula with a source statesHere ${\mathbb{E}}_{t,x}$ means expectation for the diffusion process started at $X_t=x$. Under hypotheses ensuring existence of a classical solution, this representation also identifies that solution uniquely in the stated bounded class. In the remaining parts, $c=h=0$, so $F(t,x)={\mathbb{E}}_{t,x}g(X_T)$ solves the terminal-value Kolmogorov backward equation $F_t+{\mathcal{L}}_{t}F=0$. Complex terminal data are allowed by applying the real formula to their real part and imaginary part separately.

At $\alpha =0$ the transverse coordinates are independent Brownian motions with diffusion coefficient $D=1$. The cylinder is infinitely long, so the axial coordinate cannot affect its first contact time. By rotational invariance of Brownian motion, the mean first-passage time $u$ depends only on the transverse radius $\rho =\sqrt{x^2+y^2}$.

The Kolmogorov backward equation for a mean first-passage time is $Lu=-1$. Here $L={\partial }_x^2+{\partial }_y^2$, so the radial boundary value problem isThe absorbing boundary condition for a diffusion is imposed on the circular cross-section. At $\rho =0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating givesSmoothness at the axis excludes the logarithm term, and $u(R)=0$ fixes $C_0=R^2/4$. ConsequentlyThis also verifies the finiteness assumption behind the Kolmogorov backward equation. Let $T$ be the first contact time and apply the stopped Itô formula to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped stochastic integral a martingale, givingBy monotone convergence, $T$ has finite expected value; continuity and dominated convergence then make the final term vanish as $t\to \infty$. Hence $u({\hat{r}}_0)=\mathbb{E}T$. In particular, the mean first-passage time is $R^2/4$ on the axis and tends to zero at the adsorbing wall.

A Splitting probability is the Probability that a stochastic process reaches one target before competing targets. For a diffusion it is harmonic for the Kolmogorov backward equation, with boundary values one on the selected target and zero on the others.