On the Relation Between the Girsanov Transform and the Kolmogorov Equations for SPDEs

Abstract

The Girsanov transform and Kolmogorov equations are two useful methods for studying SPDEs. It is shown that, under suitable conditions, the series expansion obtained from the Girsanov transform coincides with the one generated by an iteration scheme for Kolmogorov equations. We also apply the iteration approach to extend the well posedness theory for Kolmogorov equations beyond the boundedness condition on the nonlinear term.

Appendix:: Some Moment Estimates on Gaussian Measures

We have the following estimate on the moments of $\mu = N_{Q_{\infty }} = N_{0, Q_{\infty }}$.

Lemma 4.1

For any $n\in \mathbb N$,

$$ {\int}_{H} |x|^{2n} \mathrm{d}\mu(x) \leq 2^{n} (n!) (\text{Tr} Q_{\infty})^{n}. $$

(1)

Proof

To save notation we write Q instead of $Q_{\infty }$ in the proof below. We use the following fact (see [8, Proposition 1.13]):

$$F(s):= {\int}_{H} e^{s|x|^{2}} \mathrm{d}\mu(x)= \big[\det(1-2s Q) \big]^{-1/2}, \quad s< 1/(2\lambda_{1}), $$

where λ₁ > 0 is the biggest eigenvalue of Q. It is clear that

$$ F^{(n)}(0)= {\int}_{H} |x|^{2n} \mathrm{d}\mu(x). $$

(2)

One has the useful identity (see [8, Example 1.15]):

$$F^{\prime}(s)= F(s) \text{Tr}\big(Q(1-2s Q)^{-1}\big). $$

Then by the combinatorial formula,

$$F^{(n+1)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \big[F(s) \text{Tr}\big(Q(1-2s Q)^{-1}\big) \big] = {\sum}_{k=0}^{n} \binom n k F^{(n-k)}(s) \frac{\mathrm{d}^{k}}{\mathrm{d} s^{k}} \text{Tr}\big(Q(1-2s Q)^{-1}\big),$$

where $\binom n k$ are the combinatorial numbers. Regarding Q as a diagonal matrix and using induction, it is easy to show that

$$\frac{\mathrm{d}^{k}}{\mathrm{d} s^{k}} \text{Tr}\big(Q(1-2s Q)^{-1}\big) = 2^{k} (k!) \text{Tr}\big(Q^{k+1}(1-2s Q)^{-(k+1)}\big). $$

Therefore,

$$F^{(n+1)}(0) = {\sum}_{k=0}^{n} 2^{k} (k!) \binom n k F^{(n-k)}(0) \text{Tr}\big(Q^{k+1} \big).$$

Using the very rough inequality

$$ \text{Tr}\big(Q^{k+1} \big) \leq (\text{Tr} Q)^{k+1}, $$

(3)

we obtain

$$ F^{(n+1)}(0) \leq {\sum}_{k=0}^{n} 2^{k} (k!) \binom n k F^{(n-k)}(0) (\text{Tr} Q)^{k+1}. $$

(4)

Now we prove the assertion by induction. Assume that the estimate Eq. 1 holds for k ≤ n; we want to prove it for k = n + 1. Taking into account Eq. 2, this follows immediately from Eq. 4. Indeed, by the induction hypothesis,

$$\aligned F^{(n+1)}(0) &\leq {\sum}_{k=0}^{n} 2^{k} (k!) \binom n k 2^{n-k} ((n-k)!) (\text{Tr} Q)^{n-k} (\text{Tr} Q)^{k+1} \\ &= 2^{n} (\text{Tr} Q)^{n+1} {\sum}_{k=0}^{n} (n!) \leq 2^{n+1} ((n+1)!) (\text{Tr} Q)^{n+1} . \endaligned $$

The proof is complete. □

Remark 4.2

The estimate Eq. 3 looks very rough, but in a sense it is also sharp. For instance, assume that Q is a diagonal matrix and that the entries on the diagonal are decreasing. Let Q₂₂ ↓ 0 (thus Q_ii ↓ 0 for all i > 2), then $\text {Tr}\big (Q^{k+1} \big ) \to Q_{11}^{k+1}$ and $(\text {Tr} Q)^{k+1} \to Q_{11}^{k+1}$.

Corollary 4.3

There exists a constant C > 0 such that for any p > 1,

$$\bigg({\int}_{H} |x|^{p} \mathrm{d}\mu(x)\bigg)^{1/p} \leq C(\text{Tr} Q_{\infty})^{1/2} \sqrt{p}.$$

Proof

By Lemma 4.1 and the Stirling formula, for any $n\in \mathbb N$,

$$\bigg({\int}_{H} |x|^{2n} \mathrm{d}\mu(x) \bigg)^{1/(2n)} \leq (2 \text{Tr} Q_{\infty})^{1/2} (n!)^{1/(2n)} \leq C (\text{Tr} Q_{\infty} )^{1/2} \sqrt{n} . $$

As a result,

$$\bigg({\int}_{H} |x|^{n} \mathrm{d}\mu(x) \bigg)^{1/n} \leq C (\text{Tr} Q_{\infty} )^{1/2} \sqrt{n} . $$

Hence, for any p > 1, denoting by ⌈p⌉ the smallest integer which is greater than p,

$$\bigg({\int}_{H} |x|^{p} \mathrm{d}\mu(x)\bigg)^{1/p} \leq \bigg({\int}_{H} |x|^{\lceil p \rceil} \mathrm{d}\mu(x)\bigg)^{1/\lceil p \rceil} \leq C (\text{Tr} Q_{\infty} )^{1/2} \sqrt{\lceil p \rceil} $$

which implies the desired estimate with the constant $\sqrt {2} C$. □