Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 30 / 6

The differential operator without its potential is the diffusion generator of the d-dimensional Ornstein-Uhlenbeck processIts explicit solution $X_s=e^{-\lambda s}x+{\int }_0^s{e}^{-\lambda (s-r)}d{B}_r$ defines its path law ${\mathbb{P}}^x$ on $C([0,t];{\mathbb{R}}^d)$. For the potential $V(x)={\lambda }^2|x{|}^2/2$ and initial function one, the Feynman-Kac formula gives the nonnegative path integralThe sign in the exponential is positive, because the PDE contains $+Vu$. One can first stop the diffusion in bounded balls, where the usual localized Feynman-Kac argument applies, and then examine finiteness of the resulting expectation. The critical quadratic potential for the Ornstein-Uhlenbeck generator below supplies that finiteness check.

Apply Q5(b) in each independent coordinate, with $\mu =\lambda$, to change from the Ornstein-Uhlenbeck path law to d-dimensional Wiener measure ${\mathbb{Q}}^x$. Its density isMultiplication by the positive potential factor cancels the second term exactly. ThereforeThe integral in its exponent is an Itô integral under Wiener measure, not an ordinary pathwise line integral. The Itô formula for $|\omega {|}^2$ givesHence an equivalent endpoint-only Wiener integral is

Under Wiener measure, the endpoint has normal distribution $N(x,t{I}_d)$. For $t>0$ and $1+\lambda t>0$, the Gaussian integral isComplete the square:Evaluating the centered Gaussian gives $(1+\lambda t{)}^{-d/2}\exp [-\lambda |x{|}^2/(2(1+\lambda t))]$. With the preceding prefactor, the critical quadratic potential for the Ornstein-Uhlenbeck generator solution isThe initial value is one. As a direct PDE check, write $u=a(t)e^{b(t)|x{|}^2/2}$, whereSince $\nabla u=bxu$ and $\Delta u=(db+b^2|x{|}^2)u$, these identities give precisely the required time derivative and spatial operator.

For $\lambda \ge 0$, this is finite for every $t\ge 0$; at $\lambda =0$ it reduces to $u=1$. If the parameter is allowed to be negative, the expectation is finite only for $t<-1/\lambda$. At and beyond that time the quadratic coefficient in the endpoint Gaussian integral is nonnegative, so the integral diverges. A nonnegative classical solution on the entire half-line cannot then exist: stopping the Feynman-Kac martingale in balls and discarding its nonnegative boundary term bounds any such solution below by the truncated path expectations; their limit is infinite. Thus the printed global positive-solution premise implicitly requires the nonnegative parameter convention.