Peano-Baker series and ordered exponential

A general solution of the system of the following first order linear ODE

$$\dot{x}(t) = A(t) x(t)$$

is given by the Peano-Baker series. Many books on ODEs or control theory cover it.

Recently, I encountered the (time) ordered exponential. Reading the definition on Wikipedia, I think the Peano-Baker series and the ordered exponential are the same. Is this right? If not, what are the differences?

• Along the same lines, the Peano-Baker series seems to be the same as the Dyson series in mathematical physics (up to a factor of $i$ in the evolution equation.) – Semiclassical Apr 24 '17 at 15:18
• On Wikipedia, the time ordered exponential is defined by the Peano-Baker series. In general, I expect that there may be cases where the Peano-Baker series does not converge, but it still makes sense to speak of the time ordered exponential. The situation is similar for the (ordinary) exponential function (think of the Schrödinger equation with an unbounded time-independent Hamiltonian). Before anyone can give a careful answer to your question, you have to provide a definition of the time ordered exponential (perhaps from a more authoritative source than wikipedia). – Jonas Dahlbæk May 5 '17 at 20:44
• @JonasDahlbæk The definition given in the Wikipedia page is all I have. I'm looking for textbooks which cover the ordered exponential. I found some physics books on quantum mechanics, but I'm not so familiar with quantum mechanics. If you know any reference for non-physicist, please inform me. – leeto May 6 '17 at 8:36
• @leeto According to that definition, it is correct to say that the time ordered exponential is placeholder notation for the Peano-Baker series. You may, however, find Chapter X.12 in Methods of Modern Mathematical Physics by Reed and Simon interesting. There, time-dependent Hamiltonians are discussed, as well as the relation to the Dyson series. I would argue that one should distinguish between the Peano-Baker series and the solution to the ODE $\dot x(t)=A(t)x(t)$; maybe you would be interested in an answer of that sort? – Jonas Dahlbæk May 7 '17 at 12:01
• @JonasDahlbæk Of course, I'm interested in ! – leeto May 7 '17 at 12:16

Let $\mathcal{H}$ be a Hilbert space and let $(A(t))_{0\leqslant t\leqslant 1}$ be a family of operators in $\mathcal{H}$. Consider the differential equation $$x'(t)=A(t)x(t). \tag{1}$$ If, say $A:[0,1]\rightarrow B(\mathcal{H})$ is a continuous map into the bounded operators on $\mathcal{H}$, then there is a unique solution $x:[0,1]\rightarrow B(\mathcal{H})$ to equation (1), and that unique solution is given by the Peano-Baker series $$x(t) = \sum_{n=0}^\infty \int_{t>s_1>\ldots >s_n>0} A(s_1)\cdots A(s_n) ds_n\cdots ds_1.$$ Furthermore, if $A(t)=A$ for some fixed bounded operator, then the Peano-Baker series reduces to the exponential function series $$x(t) = \sum_{n=0}^\infty \frac{t^nA^n}{n!}=e^{tA} \tag{2}.$$
Now, consider the heat equation, which is equation (1) with $A(t)=\Delta$, where $\Delta$ denotes the Laplacian. Here, $\Delta$ is no longer bounded, but rather an unbounded, self-adjoint, non-positive operator in $\mathcal H=L^2(\mathbb{R}^d)$. Then it is still correct that $x(t)=e^{t\Delta}$ solves equation (1), but the exponential is now understood in the sense of functional calculus of a self-adjoint operator. Since $\Delta\leqslant 0$, one finds $\lVert x(t)\rVert \leqslant 1$, i.e. $x(t)$ is still bounded for all $t$. It is then natural to ask if we also have the series expansion (2), corresponding to the Peano-Baker series.
However, it is clear that we cannot have convergence of the series with respect to the operator norm, since $\Delta$ is unbounded. Nonetheless, it is reasonable to expect $$x(t)\psi=\sum_{n=0}^\infty \frac{t^n\Delta^n\psi}{n!} \tag{3}$$ for some suitable subspace of functions $\psi\in L^2(\mathbb{R}^d)$. For instance, any function $\psi$ which satisfies $\hat\psi(k)=0$ if $\lvert k\rvert > M$ (where $\hat\psi$ denotes the Fourier transform of $\psi$) will also satisfy $$\lVert \Delta^n \psi\rVert^2=\int \lvert k\rvert ^{4n}\lvert \hat\psi(k)\rvert^2 \, dk \leqslant M^{4n} \int \lvert \hat\psi(k)\rvert^2 \, dk = M^{4n} \lVert \psi\rVert^2,$$ and therefore $$\sum_{n=0}^\infty \frac{t^n \lVert \Delta^n \psi\rVert}{n!} \leqslant \sum_{n=0}^\infty \frac{t^n M^{2n}\lVert \psi\rVert}{n!} = e^{t M^2}\lVert \psi\rVert < \infty.$$ It follows that the series $\sum_{n=0}^\infty \frac{t^n\Delta^n\psi}{n!}$ converges! With some additional work, one can also show that the identity (3) in fact holds true for such $\psi$.
On the other hand, consider the Gaussian function $\psi(x)=(2\pi)^{-1/4}e^{-\lvert x \rvert^2/4}$. Recalling that the moments of the normal distribution are well known, we find $$\lVert \Delta^n \psi\rVert^2 = \int \lvert k\rvert^{4n}\lvert\psi(k)\rvert^2\, dk = (4n-1)!!= \frac{(4n)!}{2^{2n}(2n)!}.$$ In particular, by Stirling's approximation, $$\frac{t^n \lVert \Delta^n \psi\rVert }{n!} = \frac{t^n}{2^n}\sqrt{\frac{(4n)!}{(2n)!(n!)^2}} \approx \frac{2^{1/4}}{\sqrt{2\pi}} \frac{t^n 4^{n}}{\sqrt{n}}.$$ Thus, when $t>1/4$, the series (3) does not converge with $\psi(x)=(2\pi)^{-1/4}e^{-\lvert x \rvert^2/4}$.