Worked Problems on Explosion and Boundary Behavior

A Framework for Mixed Boundary Problems

Explosion and boundary classification problems often require several steps: calculate scale and speed, inspect improper integrals at each endpoint, and, when useful, compare the diffusion with a process whose behavior is easier to control. The conclusions depend on the endpoint being tested and on the precise integrability conditions; a large drift alone does not settle whether explosion occurs.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider a one-dimensional diffusion on an open interval $I=(\ell,r)$, satisfying $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$. Assume $\mu$ and $\sigma$ are locally Lipschitz on $I$, and $\sigma(x)\ne0$ there. These assumptions give a unique strong solution up to its lifetime, the first time it leaves every compact subinterval of $I$. A strong solution is constructed on the given probability space using the given Brownian motion; uniqueness in law, which concerns only the distribution of solutions, is a different assertion. Stochastic integrals are Ito integrals: for a progressively measurable integrand $H_s$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, $\int_0^T H_s\,dB_s$ is defined in the Ito sense. For locally bounded coefficients, the corresponding integrals stopped on compact subintervals meet this integrability condition.

Definition: Fix an interior point $c\in I$. A scale density and speed density are, respectively, $$ s'(x)=\exp\!\left(-\int_c^x\frac{2\mu(y)}{\sigma^2(y)}\,dy\right), \qquad m(x)=\frac{2}{\sigma^2(x)s'(x)}. $$ A scale function is $S(x)=\int_c^x s'(y)\,dy$. Multiplying $s'$ by a positive constant, or adding a constant to $S$, does not change the boundary classification.

The scale density is chosen so that scale functions are harmonic for the diffusion generator $\mathcal{L}f=\mu f'+\tfrac12\sigma^2f''$. The following identity is useful both as a check on calculations and as the reason for the scale-speed formulas.

Proposition: For every twice continuously differentiable $f$ on $I$, $$ \mathcal{L}f(x)=\frac{1}{m(x)}\left(\frac{f'(x)}{s'(x)}\right)'. $$

Proof. From the definition of $s'$, $(s')'(x)/s'(x)=-2\mu(x)/\sigma^2(x)$. Therefore $$ \frac{1}{m(x)}\left(\frac{f'}{s'}\right)' =\frac{\sigma^2s'}{2}\left(\frac{f''}{s'}-\frac{f'(s')'}{(s')^2}\right) =\frac{\sigma^2}{2}f''+\mu f' =\mathcal{L}f. $$ This also shows $\mathcal{L}S=0$, since $S'=s'$ and $(S'/s')'=1'=0$. $\square$

Feller Tests at an Endpoint

For a right endpoint $r$, take any interior $c$ and form the two boundary integrals $$ \Sigma_r=\int_c^r (S(r)-S(x))m(x)\,dx,\qquad N_r=\int_c^r (S(x)-S(c))m(x)\,dx. $$ If $S(r)=\infty$, the first integral is interpreted as infinite. Feller’s boundary classification says: the endpoint is regular if $\Sigma_r<\infty$ and $N_r<\infty$; exit if $\Sigma_r<\infty$ and $N_r=\infty$; entrance if $\Sigma_r=\infty$ and $N_r<\infty$; and natural if both are infinite. In particular, finite $\Sigma_r$ is the test for accessibility from the interior, and thus for the possibility of reaching that endpoint in finite time. A corresponding pair of integrals, with the orientation reversed, applies at a left endpoint. These are standard Feller classification results; their general proof is not repeated here.

For an endpoint at infinity, accessibility means that the solution can reach arbitrarily large levels in finite time, which is finite-time explosion in that direction. It is not enough that $S(r)$ be finite: the speed-weighted integral $\Sigma_r$ must also be finite. The choice of interior reference point does not affect whether these integrals converge.

Worked Example: A Superlinear Drift and an Exit at Infinity

Consider $dX_t=X_t^2\,dt+dB_t$ on $I=\mathbb{R}$. Find the scale and speed densities and classify both infinite endpoints.

Here $\mu(x)=x^2$ and $\sigma(x)=1$, so $2\mu/\sigma^2=2x^2$. Taking $c=0$ gives $$ s'(x)=\exp\!\left(-\int_0^x2y^2\,dy\right)=e^{-2x^3/3}, \qquad m(x)=2e^{2x^3/3}. $$ At $+\infty$, $S(\infty)=\int_0^\infty e^{-2y^3/3}\,dy<\infty$. The accessibility integral has integrand $$ (S(\infty)-S(x))m(x)=2e^{2x^3/3}\int_x^\infty e^{-2y^3/3}\,dy. $$ For $x>0$, l’Hôpital’s rule gives $$ \lim_{x\to\infty} \frac{\int_x^\infty e^{-2y^3/3}\,dy}{e^{-2x^3/3}/(2x^2)} =1, $$ because the derivatives of the numerator and denominator are both $-e^{-2x^3/3}$ times a factor tending to $1$. Thus the displayed integrand is asymptotic to $1/x^2$, whose integral over $[1,\infty)$ is finite. Hence $\Sigma_{+\infty}<\infty$. Meanwhile $S(x)-S(0)$ tends to a positive finite constant while $m(x)=2e^{2x^3/3}$ grows without bound, so $N_{+\infty}=\infty$. The right endpoint is exit: it is accessible and explosion toward $+\infty$ is possible.

At $-\infty$, $S(-\infty)=-\infty$, since $e^{-2x^3/3}$ grows without bound there. For the entrance test, the integrand is $(S(0)-S(x))m(x)=2e^{2x^3/3}\int_x^0e^{-2y^3/3}\,dy$. As $x\to-\infty$, the integral is asymptotic to $e^{-2x^3/3}/(2x^2)$, by the same derivative-ratio calculation. The product is therefore asymptotic to $1/x^2$, which is integrable near $-\infty$. The left endpoint is entrance. This asymmetric classification reflects the strong positive drift: it pushes the process away from the left endpoint and toward the right.

Check: the right accessibility integral has a tail proportional to $1/x^2$, so its integral is finite; the right entrance integral diverges because its integrand grows exponentially. These are the two conditions required for an exit classification.

A Pair of Inaccessible Boundaries

Worked Example: Logistic Diffusion on the Positive Half-Line

For $dX_t=X_t(1-X_t)\,dt+X_t\,dB_t$, with $X_0>0$, compute scale and speed and classify 0 and $+\infty$.

On $I=(0,\infty)$, $\mu(x)=x(1-x)$ and $\sigma(x)=x$, so $$ \frac{2\mu(x)}{\sigma^2(x)}=2\left(\frac1x-1\right), \qquad s'(x)=x^{-2}e^{2x}, \qquad m(x)=2e^{-2x}. $$ At zero, $S(0)=-\infty$, because $\int_0^c x^{-2}e^{2x}\,dx=\infty$. The lower-endpoint entrance integral is $\int_0^c(S(c)-S(x))m(x)\,dx$. Near zero, $S(c)-S(x)$ is asymptotic to $1/x$, while $m(x)$ tends to $2$, so this integral diverges like $\int_0^c2/x\,dx$. The other lower-endpoint integral is infinite because the scale range to zero is infinite. Thus zero is natural.

At $+\infty$, $S(+\infty)=\infty$ because $s'(x)=x^{-2}e^{2x}$ grows exponentially. For the other test, $S(x)$ is asymptotic to $e^{2x}/(2x^2)$, and hence $$ (S(x)-S(c))m(x)\ \text{is asymptotic to}\ \frac{1}{x^2}. $$ The integral of this quantity over $[c,\infty)$ is finite. Therefore $+\infty$ is entrance, not accessible; the diffusion cannot explode to $+\infty$ in finite time.

Positivity is consistent with the boundary calculation. Applying Ito’s formula to $f(x)=\log x$, whose derivatives are $f'(x)=1/x$ and $f''(x)=-1/x^2$, gives $$ d\log X_t =\left(f'(X_t)\mu(X_t)+\tfrac12 f''(X_t)\sigma^2(X_t)\right)dt +f'(X_t)\sigma(X_t)\,dB_t =\left(\tfrac12-X_t\right)dt+dB_t. $$ Up to any compact exit time, integration yields $\log X_t=\log X_0+B_t+t/2-\int_0^tX_s\,ds$. A continuous path cannot reach zero at a finite time while this logarithm remains finite; the boundary test independently confirms that zero is not accessible.

Check: near zero, the lower entrance integrand is proportional to $1/x$, so it diverges. At infinity, the corresponding upper test integrand is proportional to $1/x^2$, so it converges. The two different powers lead to natural and entrance boundaries, respectively.

Comparison as a Hitting-Time Bound

The scalar comparison theorem applies to one-dimensional SDEs driven by the same Brownian motion, with ordered initial conditions, ordered drift coefficients, and the same diffusion coefficient (under suitable regularity and uniqueness conditions). It yields an almost-sure pathwise order up to the relevant lifetimes. The next example makes the comparison and its hitting-time consequence explicit, so no additional comparison estimate is needed.

Worked Example: Bounding a Level-Hitting Time by Comparison

Let $X_0=Y_0=x$, where $x<R$, and let $$ dX_t=(X_t+1)\,dt+dB_t,\qquad dY_t=Y_t\,dt+dB_t. $$ Compare the first times the processes reach $R$.

Both drift coefficients are globally Lipschitz and satisfy linear growth: $|x+1|\le 1+|x|$, and $|x|\le1+|x|$; the common diffusion coefficient is the constant $1$. Thus both equations have unique global strong solutions. Their drift difference is $1$, so the scalar comparison theorem gives $X_t\ge Y_t$ when they start equal and use the same Brownian motion. In fact, subtracting the equations cancels the Brownian terms and gives $d(X_t-Y_t)=(X_t-Y_t+1)\,dt$, with initial difference zero. Solving this ordinary differential equation gives $X_t-Y_t=e^t-1\ge0$.

Define $\tau_R^X=\inf\{t\ge0:X_t\ge R\}$ and $\tau_R^Y=\inf\{t\ge0:Y_t\ge R\}$, with the infimum of the empty set equal to infinity. If $\tau_R^Y<\infty$, continuity gives $Y_{\tau_R^Y}=R$, and then $X_{\tau_R^Y}=R+e^{\tau_R^Y}-1\ge R$. Since $X_0<R$ and $X$ is continuous, $X$ must have reached $R$ no later. If $\tau_R^Y=\infty$, the inequality $\tau_R^X\le\tau_R^Y$ holds in the extended-real sense as well. Therefore $\tau_R^X\le\tau_R^Y$ almost surely.

To check the comparison by an independent computation, solve for $Y$ using the integrating factor $e^{-t}$. Ito’s formula for $f(t,y)=e^{-t}y$ has $f_t=-e^{-t}y$, $f_y=e^{-t}$, $f_{yy}=0$, so its Ito correction $\tfrac12 f_{yy}(dY_t)^2$ is zero. Substitution gives $d(e^{-t}Y_t)=e^{-t}\,dB_t$, and hence $Y_t=e^t\left(x+\int_0^t e^{-s}\,dB_s\right)$. Adding $e^t-1$ gives the corresponding $X_t$, and direct substitution recovers $dX_t=(X_t+1)\,dt+dB_t$. The comparison bounds the hitting time; it does not assert that either process reaches $R$ almost surely.

A Common Scale-Function Mistake

A frequent error is to omit the minus sign in the exponential defining the scale density. For the drift $\mu(x)=x^2$ and diffusion coefficient $\sigma(x)=1$, the incorrect calculation $s'(x)=\exp(\int_0^x2y^2\,dy)=e^{2x^3/3}$ would make the scale integral at $+\infty$ diverge. That could wrongly suggest the endpoint is inaccessible. The correction follows directly from $\mathcal{L}S=0$: solving $\mu S'+\tfrac12\sigma^2S''=0$ gives $S''/S'=-2\mu/\sigma^2$, hence $s'(x)=e^{-2x^3/3}$, as used in the first example. Always check the sign by substituting the proposed scale density into the generator identity.

Boundary tests and comparison answer different questions. Scale and speed classify endpoint accessibility from the coefficients of one diffusion; comparison transfers pathwise order between two solutions driven by the same noise. In applications, specify the endpoint or level, verify the test’s integrability conditions, and state whether a conclusion concerns possible access, almost-sure access, or only a bound conditional on a comparison process reaching its level.

Check Your Understanding

For each question, specify the endpoint or threshold and use strict inequalities in defining any hitting event or tail event.

For $dX_t=a\,dt+\sigma\,dB_t$ on $\mathbb{R}$, with $\sigma>0$, write the scale and speed densities. What feature of the scale function rules out an accessible infinite endpoint?
In the logistic example, which asymptotic integrand determines the lower-boundary test at zero, and does its integral converge?
For $dX_t=X_t^2\,dt+dB_t$, why does finiteness of $S(+\infty)$ alone not establish accessibility?
In the comparison example, what property of the common Brownian motion is used when subtracting the two SDEs?
State the conditions on $\tau_R^X$ and $\tau_R^Y$ under which the pathwise argument establishes $\tau_R^X\le\tau_R^Y$.