Uniqueness of Bounded Solutions by Stochastic Representation

Why a Stochastic Representation Gives Uniqueness

A terminal-value problem specifies a function at the final time and asks for a solution of a backward parabolic equation at earlier times. A stochastic representation expresses the solution at $(t,x)$ as an expectation over the diffusion started from $x$ at time $t$. Since that expectation is determined by the terminal data and the diffusion law, the representation also gives uniqueness—provided the solution and the stochastic argument satisfy the conditions that justify it.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix a finite horizon $T>0$. Consider the scalar diffusion started at $x$ at time $t$:

dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s,\qquad X_t=x,\qquad t\le s\le T.

Here $\mu$ is the drift and $\sigma$ is the diffusion coefficient. Assume they are continuous in time, globally Lipschitz in the spatial variable uniformly over $[0,T]$, and have linear growth: for some constant $C$, $|\mu(s,y)|+|\sigma(s,y)|\le C(1+|y|)$. These hypotheses give a nonexplosive strong solution: a process adapted to the given Brownian filtration that satisfies the SDE, with pathwise uniqueness. A weak solution instead allows the probability space and Brownian motion to be chosen as part of the solution; uniqueness in law concerns the distribution of the solution. Those SDE notions are distinct from uniqueness of the PDE solution proved here.

For an integrand $H_s$, the Ito integral $\int_t^T H_s\,dB_s$ is used when $H$ is adapted and progressively measurable and $\mathbb{E}\int_t^T H_s^2\,ds<\infty$; these processes form the class $\mathcal{L}^2$ on $[t,T]$. The proof will first stop the diffusion so that the integrand lies in this class, rather than assuming global bounds on derivatives of the PDE solution.

Definition: A bounded classical solution of the terminal-value problem is a bounded function $u:[0,T]\times\mathbb{R}\to\mathbb{R}$, continuous on its domain and of class $C^{1,2}$ (one continuous time derivative and two continuous space derivatives), satisfying $$ u_t(t,y)+\mu(t,y)u_x(t,y)+\tfrac12\sigma(t,y)^2u_{xx}(t,y)=0,\qquad u(T,y)=g(y). $$ The terminal function $g$ is bounded and continuous. The PDE is required for $0\le t<T$; the stated regularity up to $T$ is a convenient sufficient condition for the argument below.

Representation and Uniqueness

Theorem: Under the assumptions above, every bounded classical solution $u$ satisfies $$ u(t,x)=\mathbb{E}\big[g(X_T)\big],\qquad 0\le t\le T,\ x\in\mathbb{R}, $$ where $X$ solves the SDE from $(t,x)$. In particular, there is at most one bounded classical solution for a given terminal function $g$.

Proof. Fix $t<T$ and $x$, and let $X$ be the strong solution started from $x$ at time $t$. For integers $n>|x|$, define the exit time $\tau_n=\inf\{s\ge t:|X_s|\ge n\}$, with the infimum of the empty set taken to be $+\infty$. Apply Ito's formula to $s\mapsto u(s,X_s)$, stopped at $T\wedge\tau_n$. In this formula the time derivative is $u_t$, the first spatial derivative is $u_x$, and the second spatial derivative contributes the correction $\tfrac12u_{xx}\sigma^2\,ds$:

u(T\wedge\tau_n,X_{T\wedge\tau_n}) =u(t,x)+\int_t^{T\wedge\tau_n} \left(u_t+\mu u_x+\tfrac12\sigma^2u_{xx}\right)(s,X_s)\,ds +\int_t^{T\wedge\tau_n}u_x(s,X_s)\sigma(s,X_s)\,dB_s.

The PDE makes the ordinary time integral zero. To justify taking expectations, write the stochastic integral with integrand $H_s=\mathbf{1}_{\{s\le\tau_n\}}u_x(s,X_s)\sigma(s,X_s)$ on $[t,T]$. On the stopped region, the spatial argument lies in $[-n,n]$; continuity of $u_x,\sigma$ on the compact set $[t,T]\times[-n,n]$ bounds their product. Thus $H$ is progressively measurable and $\mathbb{E}\int_t^T H_s^2\,ds<\infty$. The Ito integral is square-integrable and has conditional mean-zero increments: this holds for adapted step integrands by conditioning on Brownian increments, and for $H$ by the $L^2$ construction of the Ito integral. It is therefore a true martingale, not merely a local martingale, and its expectation is zero. Consequently, $$ u(t,x)=\mathbb{E}\big[u(T\wedge\tau_n,X_{T\wedge\tau_n})\big]. $$

The solution $X$ has continuous, nonexplosive paths on the finite interval $[t,T]$. For almost every path, its maximum absolute value on that interval is finite, so for all sufficiently large $n$, $\tau_n>T$. Hence $u(T\wedge\tau_n,X_{T\wedge\tau_n})\to u(T,X_T)=g(X_T)$ almost surely. Boundedness of $u$ supplies a deterministic integrable bound, so dominated convergence gives $u(t,x)=\mathbb{E}[g(X_T)]$. At $t=T$ this identity is the terminal condition itself.

For uniqueness, let $u$ and $v$ be two bounded classical solutions with the same terminal function. The representation just proved gives $u(t,x)=\mathbb{E}[g(X_T)]=v(t,x)$ for every $(t,x)$. Thus $u=v$. The argument also shows exactly where boundedness is used: it permits passage from the stopped terminal values to the unstopped terminal value by dominated convergence. $\square$

The stopped stochastic integral is a true martingale because it is square-integrable. Without stopping, the product $u_x(s,X_s)\sigma(s,X_s)$ need not be square-integrable; an Ito integral defined only locally would then give a local martingale, whose expectation need not be preserved. Localization followed by bounded convergence avoids that unjustified step. No risk-neutral measure is involved: all expectations here are under $\mathbb{P}$.

Worked Representations

Worked Example: The Heat Equation with Cosine Data

Take $\mu=0$, $\sigma=1$, terminal time $T=1$, and terminal function $g(x)=\cos x$. The diffusion started at $(t,x)$ is $X_s=x+B_s-B_t$. Its terminal increment is normal with mean zero and variance $1-t$. The normal characteristic function gives $$ \mathbb{E}[\cos(X_1)]=e^{-(1-t)/2}\cos x. $$ Thus the represented solution is $u(t,x)=e^{-(1-t)/2}\cos x$, which is bounded.

Verify the PDE, including the Ito correction. Its derivatives are $u_t=\tfrac12e^{-(1-t)/2}\cos x$, $u_x=-e^{-(1-t)/2}\sin x$, and $u_{xx}=-e^{-(1-t)/2}\cos x$. Therefore $u_t+\tfrac12u_{xx}=0$, and $u(1,x)=\cos x$. At $(0,0)$, the representation gives $e^{-1/2}$, also obtained directly as $\mathbb{E}[\cos(B_1)]=e^{-1/2}$. This check confirms both the terminal value and the expectation formula.

Worked Example: Drifted Brownian Motion with Sine Data

Let $dX_s=ds+2\,dB_s$, $T=1$, and $g(x)=\sin(2x)$. Starting from $(t,x)$, $X_1=x+(1-t)+2(B_1-B_t)$. The normal increment has variance $4(1-t)$. Using $\mathbb{E}[\sin(a+Z)]=e^{-\operatorname{Var}(Z)/2}\sin a$ for a centered normal $Z$, the representation is $$ u(t,x)=e^{-8(1-t)}\sin\big(2(x+1-t)\big). $$ At $(0,0)$ this is $e^{-8}\sin 2$, and direct substitution into the normal expectation gives $e^{-8}\sin(2\cdot1)$, the same value.

Here the PDE is $u_t+u_x+2u_{xx}=0$. Put $\tau=1-t$ and $a=e^{-8\tau}$. Direct differentiation gives $u_t=8a\sin(2(x+\tau))-2a\cos(2(x+\tau))$, $u_x=2a\cos(2(x+\tau))$, and $u_{xx}=-4a\sin(2(x+\tau))$. The diffusion correction is $\tfrac12(2^2)u_{xx}=2u_{xx}=-8a\sin(2(x+\tau))$. Adding the three PDE terms gives zero. At $t=1$, the formula reduces to $\sin(2x)$, as required.

Worked Example: Ornstein–Uhlenbeck Diffusion with Cosine Data

Let $dX_s=-X_s\,ds+\sqrt{2}\,dB_s$, with terminal time $T$ and $g(x)=\cos x$. This drift and diffusion satisfy the stated Lipschitz and linear growth assumptions. From $(t,x)$, the solution has the form $$ X_T=e^{-(T-t)}x+\sqrt{2}\int_t^T e^{-(T-s)}\,dB_s. $$ The stochastic integral is an Ito integral with deterministic, hence progressively measurable, integrand; its squared-integrand integral is $2\int_t^T e^{-2(T-s)}ds=1-e^{-2(T-t)}<\infty$. It is centered normal with that variance. Thus the representation is $$ u(t,x)=\exp\!\left(-\frac{1-e^{-2(T-t)}}{2}\right)\cos\!\left(e^{-(T-t)}x\right). $$ For $t=T$, the exponent is zero and the cosine argument is $x$, so the terminal value is exactly $\cos x$.

For an independent PDE check, set $\tau=T-t$, $a=\exp(-(1-e^{-2\tau})/2)$, and $q=e^{-\tau}$. Then $u=a\cos(qx)$, $a_t=ae^{-2\tau}$, and $q_t=q$. Consequently $u_t=ae^{-2\tau}\cos(qx)-aqx\sin(qx)$, $u_x=-aq\sin(qx)$, and $u_{xx}=-aq^2\cos(qx)$. The PDE is $u_t-xu_x+u_{xx}=0$; the sine terms cancel, and the cosine coefficient is $a(e^{-2\tau}-q^2)=0$, since $q^2=e^{-2\tau}$. This verifies the candidate and the representation independently.

What the Assumptions Prevent

Common mistake: dropping the Ito correction. In the first example, one might use an ordinary chain-rule calculation and claim that $u(t,x)=\cos x$ solves the heat equation because it does not depend on time. But $u_t=0$ and $u_{xx}=-\cos x$, so the PDE residual is $u_t+\tfrac12u_{xx}=-\tfrac12\cos x$, not zero. The missing term comes from quadratic variation: $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$. The correction is essential to the representation and the verification.

Another error is to conclude that every local martingale has constant expectation. That is not generally true. In the proof, the stopped integrand is in $\mathcal{L}^2$, which makes the stopped stochastic integral a true martingale; boundedness of $u$ then justifies the limit as the stopping levels increase. The theorem proves uniqueness within the stated bounded classical class. It does not assert uniqueness for solutions outside that class, or for functions lacking the regularity needed to apply Ito's formula.

Check Your Understanding

Use the representation argument and its hypotheses to answer these questions.

Which term in the Ito formula for $u(s,X_s)$ disappears when $u$ satisfies the backward PDE?
Why is the stochastic integral stopped at $\tau_n$ square-integrable, even if $u_x$ is not globally bounded?
Where is boundedness of the classical solution used when $n$ tends to infinity?
For $dX_s=\mu\,ds+\sigma\,dB_s$ and $g(x)=\cos(kx)$, use the normal characteristic function to write the terminal expectation from $(t,x)$.
Explain why uniqueness of the PDE solution is not the same statement as pathwise uniqueness of the SDE.

Uniqueness of Bounded Solutions by Stochastic Representation

What You'll Learn