= Solution
Under <Wiener measure>, the endpoint has <normal distribution> $N(x,tI_d)$. For $t>0$ and $1+\lambda t>0$, the <Gaussian integral> is
$$
\mathbb E_{\mathbb Q^x}e^{-\lambda|\omega_t|^2/2}
=(2\pi t)^{-d/2}\int_{\mathbb R^d}
\exp\!\left(-\frac{|z-x|^2}{2t}-\frac\lambda2|z|^2\right)dz.
$$
Complete the square:
$$
-\frac{|z-x|^2}{2t}-\frac\lambda2|z|^2
=-\frac{1+\lambda t}{2t}\left|z-\frac{x}{1+\lambda t}\right|^2
-\frac{\lambda|x|^2}{2(1+\lambda t)}.
$$
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 is
$$
\boxed{u(t,x)=\frac{e^{\lambda dt/2}}{(1+\lambda t)^{d/2}}
\exp\!\left(\frac{\lambda^2t|x|^2}{2(1+\lambda t)}\right).}
$$
The initial value is one. As a direct PDE check, write $u=a(t)e^{b(t)|x|^2/2}$, where
$$
b(t)=\frac{\lambda^2t}{1+\lambda t},\qquad
\frac{a'(t)}{a(t)}=\frac d2b(t),\qquad
b'(t)=(b(t)-\lambda)^2.
$$
Since $\nabla u=bxu$ and $\Delta u=(db+b^2|x|^2)u$, these identities give precisely the required time derivative and spatial operator.
For $\lambda\ge0$, this is finite for every $t\ge0$; 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.
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