Exit-Time Representations for Dirichlet Problems

From a Dirichlet Problem to an Exit-Time Expectation

A Dirichlet problem specifies a differential equation inside a domain and the values of its solution on the boundary. For a diffusion, the first time the process reaches that boundary supplies a natural random endpoint: the boundary value is then evaluated at the exit location. A potential $c$ discounts contributions before exit, while a source term $f$ accumulates as a running reward. The representation below makes these roles precise.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider a one-dimensional diffusion, started at $X_0=x\in D$, satisfying $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$, where $D$ is a bounded open interval. Assume a continuous solution exists and that the coefficients are locally bounded. Global Lipschitz and linear growth conditions, namely $|\mu(x)-\mu(y)|+|\sigma(x)-\sigma(y)|\le L|x-y|$ and $|\mu(x)|+|\sigma(x)|\le L(1+|x|)$ for a constant $L$, are one standard sufficient set of conditions for a unique strong solution on the whole line. The representation itself applies to any such diffusion for which the conditions stated below hold. A strong solution is built with the given Brownian motion; a weak solution specifies a probability space and Brownian motion as part of the solution. Pathwise uniqueness and uniqueness in law are different assertions. The formula is under $\mathbb{P}$; it does not invoke a risk-neutral measure $\mathbb{Q}$.

Definition: The first exit time is $\tau=\inf\{t\ge0:X_t\notin D\}$, with $\inf\varnothing=\infty$. The generator is $\mathcal{L}v(x)=\mu(x)v'(x)+\tfrac12\sigma(x)^2v''(x)$. The drift is $\mu$, and the diffusion coefficient is $\sigma$. For a potential $c$, set $A_t=\exp(-\int_0^t c(X_s)\,ds)$.

Suppose $c$ is bounded, continuous, and nonnegative on $\overline D$; $f$ is bounded and continuous there; and $g$ is continuous on the boundary. Assume $\mathbb{E}_x[\tau]<\infty$, so $\tau<\infty$ almost surely. The classical solution considered here satisfies $u\in C^2(D)\cap C(\overline D)$, has boundary values $u=g$, and solves $\mathcal{L}u-cu+f=0$ in $D$. Continuity on the compact set $\overline D$ makes $u$ bounded, but does not imply that its derivatives remain bounded near the boundary.

Theorem: Under these assumptions, for every $x\in D$, $$ u(x)=\mathbb{E}_x\left[A_\tau g(X_\tau)+\int_0^\tau A_s f(X_s)\,ds\right]. $$ The expectation is finite: $0<A_t\le1$, $g$ is bounded, and the absolute value of the running integral is at most $\|f\|_\infty\tau$.

Why Localization Is Needed

The stochastic integral in the proof is an Ito integral, not a Stratonovich integral. Ito’s formula accounts for Brownian quadratic variation through the $\tfrac12\sigma^2u''\,dt$ term; in differential notation $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. For a progressively measurable integrand $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, the Ito integral is a square-integrable martingale. A merely local martingale need not be a true martingale, so its expectation cannot be set to zero without an integrability argument.

Proof. Choose an increasing sequence of relatively compact interior sets $D_n=\{y\in D:\operatorname{dist}(y,D^c)>1/n\}$, omitting initial indices for which $x\notin D_n$, and let $\tau_n=\inf\{t\ge0:X_t\notin D_n\}$. For a continuous path, $\tau_n$ increases to $\tau$ almost surely. On the closure of each $D_n$, $u'$ and $u''$ are bounded. The coefficients are bounded there as well. Fix $T<\infty$ and put $\rho_n=\tau_n\wedge T$. Apply Ito’s formula to $u(X_t)$, then the product rule to $A_tu(X_t)$. Explicitly, the derivatives in the spatial Ito formula are $u'(y)$ and $u''(y)$, and its correction term is $\tfrac12u''(X_t)\sigma(X_t)^2\,dt$: $$ du(X_t)=\left(\mu(X_t)u'(X_t)+\tfrac12\sigma(X_t)^2u''(X_t)\right)dt +\sigma(X_t)u'(X_t)\,dB_t. $$ Since $dA_t=-A_tc(X_t)\,dt$, the product rule and the PDE give $$ d(A_tu(X_t))=-A_tf(X_t)\,dt+A_t\sigma(X_t)u'(X_t)\,dB_t. $$ The stochastic integrand stopped at $\rho_n$ is progressively measurable and bounded on $[0,T]$, hence belongs to $\mathcal{L}^2$: its expected squared integral is at most $T$ times the square of that bound. Its Ito integral is therefore a square-integrable martingale, with conditional expectation property and mean zero. Integrating to $\rho_n$ and taking expectations yields $$ u(x)=\mathbb{E}_x[A_{\rho_n}u(X_{\rho_n})] +\mathbb{E}_x\left[\int_0^{\rho_n}A_sf(X_s)\,ds\right]. $$ For fixed $T$, let $n$ increase. Continuity of paths and of $u$ on $\overline D$ gives convergence of the endpoint term to $A_{\tau\wedge T}u(X_{\tau\wedge T})$. The endpoint terms are bounded by $\|u\|_\infty$, and the running integrals by $T\|f\|_\infty$, so dominated convergence passes both expectations to the limit. Now let $T$ increase to infinity. Since $\mathbb{E}_x[\tau]<\infty$, $\tau<\infty$ almost surely, $X_\tau\in\partial D$, and $u(X_\tau)=g(X_\tau)$. The running integrals are dominated in absolute value by $\|f\|_\infty\tau$, which is integrable. Dominated convergence gives the stated formula. In particular, no global martingale claim was made for the unlocalized integral, whose integrand may fail to be square-integrable near the boundary. $\square$

Worked Brownian Interval Problems

Worked Example: Boundary Data Without Discounting

Let $D=(-1,1)$, $X_t=x+B_t$, $c=0$, and $f=0$. Set $g(-1)=0$ and $g(1)=1$. Find the solution and check it.

Here $\mathcal{L}u=\tfrac12u''$, so the equation is $u''=0$. Integrating twice gives $u(x)=a+bx$. The boundary conditions yield $a-b=0$ and $a+b=1$, hence $a=b=\tfrac12$ and $u(x)=\tfrac{x+1}{2}$. Direct substitution gives $u''=0$, $u(-1)=0$, and $u(1)=1$.

The exit-time formula says $u(x)=\mathbb{P}_x(X_\tau=1)$. This agrees with the independently obtained linear function: the probability of exiting at the right endpoint is $(x+1)/2$. At $x=0$, the two exit sides have equal probability, and the formula gives $u(0)=1/2$.

Worked Example: Discounted Boundary Data

Keep $D=(-1,1)$ and $X_t=x+B_t$, take $c=\tfrac12$, $f=0$, and prescribe $g(-1)=0$, $g(1)=1$. Solve the equation and verify both boundary values.

The equation $\tfrac12u''-\tfrac12u=0$ is equivalent to $u''-u=0$. Write $u(x)=A\sinh x+C\cosh x$. The left boundary condition gives $-A\sinh1+C\cosh1=0$, so $C=A\tanh1$. The right condition gives $A\sinh1+C\cosh1=1$, hence $2A\sinh1=1$. Therefore $A=1/(2\sinh1)$ and $C=1/(2\cosh1)$, and equivalently $$ u(x)=\frac{\sinh(x+1)}{\sinh2}. $$ At $x=-1$ the numerator is $\sinh0=0$; at $x=1$ numerator and denominator are both $\sinh2$, so the boundary values are correct. Also $u''=u$, which verifies $\tfrac12u''-\tfrac12u=0$. At zero, $u(0)=\sinh1/\sinh2=1/(2\cosh1)$. The representation identifies this value as $\mathbb{E}_0[e^{-\tau/2}\mathbf{1}_{\{X_\tau=1\}}]$, with the indicator recording the right exit.

Worked Example: A Running Reward Equal to One

For $D=(-1,1)$, $X_t=x+B_t$, $c=0$, and $g(-1)=g(1)=0$, let $f=1$. Find the solution and independently check its value at $x=0$.

The equation is $\tfrac12u''+1=0$, or $u''=-2$. Integrating gives $u(x)=-x^2+a+bx$. The two zero boundary values imply $-1+a-b=0$ and $-1+a+b=0$, so $a=1$ and $b=0$. Thus $u(x)=1-x^2$. Substitution gives $\tfrac12(-2)+1=0$, and $u(-1)=u(1)=0$.

The representation predicts $u(0)=\mathbb{E}_0[\tau]=1$. To check independently, apply Ito’s formula to $X_t^2$: its derivatives are $2X_t$ and $2$, so $d(X_t^2)=2X_t\,dB_t+dt$. Stop at $\tau\wedge T$. The stopped stochastic integrand is bounded because $|X_{t\wedge\tau}|\le1$, so its integral is square-integrable and has mean zero. Consequently $\mathbb{E}_0[X_{\tau\wedge T}^2]=\mathbb{E}_0[\tau\wedge T]$. As $T$ increases, $X_{\tau\wedge T}^2$ converges to $X_\tau^2=1$ and is bounded by $1$, while $\tau\wedge T$ increases to $\tau$. Dominated convergence on the left and monotone convergence on the right give $\mathbb{E}_0[\tau]=1$, as required.

Uniqueness and Scope

Proposition: Under the theorem’s assumptions, there is at most one bounded function in $C^2(D)\cap C(\overline D)$ solving $\mathcal{L}u-cu+f=0$ with boundary data $g$.

Proof. Let $u_1$ and $u_2$ be two such solutions. Their difference $v=u_1-u_2$ is bounded, belongs to $C^2(D)\cap C(\overline D)$, has zero boundary values, and satisfies $\mathcal{L}v-cv=0$. Apply the proved representation with running term zero and boundary data zero. For every $x\in D$, it gives $v(x)=\mathbb{E}_x[A_\tau\,0]=0$. Thus $u_1=u_2$ throughout $D$, and continuity gives equality on its boundary as well. $\square$

The finite-mean-exit assumption is a convenient sufficient condition for the running-reward term when $f$ is bounded. More general domains or potentials may require different integrability hypotheses; without suitable conditions, a formal PDE solution need not yield a finite expectation. The theorem also does not turn every stochastic integral in the argument into a true martingale: localization is essential when derivatives are only controlled on compact subsets of the domain. Finally, the expectation and diffusion law in the formula are those specified under $\mathbb{P}$; changing to $\mathbb{Q}$ would require a separate change-of-measure argument.

Check Your Understanding

Use the hypotheses and the signs in the equation $\mathcal{L}u-cu+f=0$ to check each answer.

For $D=(-a,a)$, Brownian motion, $c=f=0$, and boundary values $g(-a)=0$, $g(a)=1$, find $u(x)$.
Explain why continuity of $u$ on $\overline D$ does not guarantee that $\sigma u'$ is bounded near the boundary.
For the discounted boundary example, verify $u(0)=1/(2\cosh1)$ from $u(0)=\sinh1/\sinh2$.
What integrability condition in the tutorial ensures that the running-reward expectation is finite when $f$ is bounded?
Why does the proof localize the stochastic integral before taking expectations?

Exit-Time Representations for Dirichlet Problems

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