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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.

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.