Exact Splitting Methods for Semigroups Generated by Inhomogeneous Quadratic Differential Operators

Abstract

We introduce some general tools to design exact splitting methods to compute numerically semigroups generated by inhomogeneous quadratic differential operators. More precisely, we factorize these semigroups as products of semigroups that can be approximated efficiently, using, for example, pseudo-spectral methods. We highlight the efficiency of these new methods on the examples of the magnetic linear Schrödinger equations with quadratic potentials, some transport equations and some Fokker–Planck equations.

Appendix

1.1 Proof of Lemma 2

Here, the proof relies essentially on computations by block requiring to introduce a more convenient basis than the canonical basis of $\mathbb {C}^{2n+2}$ denoted by $e_1,\dots ,e_{2n+2}$. This basis, denoted $\mathscr {B}$, is just a permutation of the canonical basis and is defined by

$$\begin{aligned} \mathscr {B} = (e_1,\dots ,e_n,e_{n+2},\dots ,e_{2n},e_{2n+1},e_{n+1},e_{2n+2}). \end{aligned}$$

(37)

In this basis the matrix of $J_{2n+2}$ is

$$\begin{aligned} \mathrm {mat}_{\mathscr {B}} \ J_{2n+2} = \begin{pmatrix} J_{2n} &{} \\ &{} J_2 \end{pmatrix} \end{aligned}$$

and the matrix of $\mathbb {P}p_j$ is

Consequently, in this basis the matrix of the Hamiltonian map is

Here, it is really relevant to observe the double triangular structure of this matrix. The four blocks on the top left corner define an upper triangular matrix by blocks, whereas, considering these four blocks as a single block, the global matrix is lower triangular by blocks.

Now observe, through the power expansion series, that if $\varPsi$ is an entire function and

$$\begin{aligned} M = \begin{pmatrix} A &{} B \\ &{} 0 \end{pmatrix} \end{aligned}$$

is an upper triangular matrix by blocks, then

$$\begin{aligned} \varPsi (M) = \begin{pmatrix} \varPsi (A) &{} \left( \frac{\varPsi (z)-\varPsi (0)}{z} \right) (A) B \\ &{} \varPsi (0) \end{pmatrix}. \end{aligned}$$

Consequently, we have

where

At the end of the proof, we will check that $\widetilde{\kappa _j} = \kappa _j$, so for the moment assume that this relation holds.

Thus, realizing a product by block we get by a straightforward induction

Identifying the blocks (1, 1), (3, 1), (3, 2) with those of $\mathrm {mat}_{\mathscr {B}} \ \varPhi _1^{\mathbb {P}p_{m+1}}$, we get the system (24). Conversely, we have to check that if (24) is satisfied, then the blocks (1, 2) are the same. Indeed, consider a complex symplectic matrix $M\in \mathrm {Sp}_{2n}(\mathbb {C})$ with a block structure of the form

$$\begin{aligned} \mathrm {mat}_{\mathscr {B}} \ M = \begin{pmatrix} A &{} B \\ &{} 1 \\ C &{} d &{} 1 \end{pmatrix}. \end{aligned}$$

Note that since M is symplectic, M is invertible and consequently A is also invertible. Since M is symplectic, then is satisfies

(38)

However, due to the double triangular nature of M, its invert can be computed easily and we have

$$\begin{aligned} \mathrm {mat}_{\mathscr {B}} \ M^{-1} = \begin{pmatrix} A^{-1} &{} -A^{-1}B \\ &{} 1 \\ -CA^{-1} &{} CA^{-1}B-d &{} 1 \end{pmatrix}. \end{aligned}$$

Consequently, a straightforward block product leads to

Thus, since M is symplectic, we have

A fortiori, if two symplectic matrices have this block structure and the same top left corner blocks, if their blocks (3, 1) are equal, then their blocks (1, 2) are equal. Consequently, applying this to the symplectic matrices $\varPhi _1^{\mathbb {P}p_{1}} \dots \varPhi _1^{\mathbb {P}p_{m}}$ and $\varPhi _1^{\mathbb {P}p_{m+1}}$, we deduce that if the system (24) is satisfied, then we have the factorization (23).

Finally, we just have to check that $\widetilde{\kappa _j} = \kappa _j$. For this computation, we omit the indices j since they are clearly irrelevant. First, we split the even indices from the odd indices in the power expansion of $\widetilde{\kappa }$ :

Observing that

$$\begin{aligned} (J_{2n}Q )^{2k}J_{2n} = (J_{2n} Q)^{k} J_{2n} (Q J_{2n} )^{k}, \end{aligned}$$

since $J_{2n}$ is skew-symmetric, we have

Consequently, $\varSigma _{even}$ vanishes. Similarly, since we have

we get

1.2 Proof of Theorem 1

Before proving Theorem 1, let us to prove some preparatory lemmas.

Lemma 5

If L is a bounded operator on $L^2(\mathbb {R}^n)$ such that there exists $T\in \mathrm {Sp}_{2n}(\mathbb {R})$ satisfying $L=\pm \mathscr {K}(T)$, then L can be computed by an exact splitting.

Proof

Let G be the group generated by $\varPhi _t^{i x_j^2},\varPhi _t^{i \xi _j^2},\varPhi _t^{i x_j \xi _k}$ for $t\in \mathbb {R}, j,k\in \llbracket 1,n \rrbracket$. Applying Theorem 3, if we prove that $G= \mathrm {Sp}_{2n}(\mathbb {R})$, we deduce that L is a product of operators of the form $e^{it x_j^2 },e^{it \partial _{x_j}^2},e^{t x_j \partial _{x_k}}$ (up to the sign). Thus, since Proposition 6 states that dilatations (i.e. operators of the form $e^{t x_k \partial _k}$) can be factorized similarly, we would deduce that L can be computed by an exact splitting.

Consequently, we aim at proving that $G= \mathrm {Sp}_{2n}(\mathbb {R})$. First, let us prove that G contains a neighborhood of the identity in $\mathrm {Sp}_{2n}(\mathbb {R})$.

Indeed, consider the map $\varPsi : \mathscr {N} \rightarrow \mathfrak {sp}_{2n}(\mathbb {R})$, where $\mathscr {N}$ is a neighborhood of the origin in $\mathfrak {sp}_{2n}(\mathbb {R})$, defined for $Q\in S_{2n}(\mathbb {R})$ such that $J_{2n}Q \in \mathscr {N}$ by

$$\begin{aligned}&\varPsi (J_{2n}Q) = \nonumber \\&\quad \log \left( \prod _{j=1}^{n} \varPhi ^{ i A_{j,j} x_j^2}_{1/2} \prod _{1\le j< k \le n} \varPhi ^{ i A_{j,k} x_j x_k}_{1} \prod _{j=1}^n \prod _{k=1}^{n} \varPhi ^{ i C_{j,k} x_j \xi _k}_{1} \prod _{1 \le j < k \le n} \varPhi ^{ i B_{j,k} \xi _j \xi _k}_{1} \prod _{j=1}^{n} \varPhi ^{ i B_{j,j} \xi _j^2}_{1/2} \right) ,\nonumber \\ \end{aligned}$$

(39)

where the natural block decomposition of Q is

with $A,B\in S_{n}(\mathbb {R})$ and $C\in M_n(\mathbb {R})$. Note that to prove that $\varPsi$ takes its values in $\mathfrak {sp}_{2n}(\mathbb {R})$, it is enough to apply the Baker–Campbell–Hausdorff formula.

Since the differential of the exponential in the origin and the differential of the logarithm in the identity are equal to the identity, we deduce by composition that the differential of $\varPsi$ in the origin is also the identity. Thus, since $\varPsi$ vanishes in the origin, we deduce of the Inverse Function Theorem that $\varPsi$ defines a local homeomorphism around the origin. Furthermore, we recall⁷ that the exponential is an homeomorphism between a neighborhood of the origin in $\mathfrak {sp}_{2n}(\mathbb {R})$ and a neighborhood of the identity in $\mathrm {Sp}_{2n}(\mathbb {R})$. Consequently, we deduce that each matrix in $\mathrm {Sp}_{2n}(\mathbb {R})$ close enough to the identity can be written as a product of the form of the product in the logarithm in (39). A fortiori, we have proved that G contains a neighborhood of the identity in $\mathrm {Sp}_{2n}(\mathbb {R})$. Let V denote this neighborhood.

Since $\mathrm {Sp}_{2n}(\mathbb {R})$ is connected (see, e.g., the subsection 4.4 of [4]), to prove that $G= \mathrm {Sp}_{2n}(\mathbb {R})$, we just have to prove that G is closed and open in $\mathrm {Sp}_{2n}(\mathbb {R})$. Indeed, if $g\in G$, then gV is a neighborhood of g in $\mathrm {Sp}_{2n}(\mathbb {R})$ and since V is included in G, then gV is also included in G. Thus, G is open in $\mathrm {Sp}_{2n}(\mathbb {R})$. Conversely, if $g\notin G$, then gV is also a neighborhood of g in $\mathrm {Sp}_{2n}(\mathbb {R})$, but since G is a group we have $gV \cap G = \emptyset$. Consequently, the complementary of G in $\mathrm {Sp}_{2n}(\mathbb {R})$ is open, i.e., G is closed in $\mathrm {Sp}_{2n}(\mathbb {R})$, which conclude the proof. $\square$

Lemma 6

If $\ell :\mathbb {R}^{2n}\rightarrow \mathbb {R}$ is a real linear form, then $\exp (i \ell ^w)$ can be computed by an exact splitting.

Proof

Let $L\in \mathbb {R}^{2n}$ be the matrix of $\ell$ and let $c=c_1$ where

$$\begin{aligned} c_t = \frac{t}{2} \sum _{j=1}^{n} L_j L_{j+n}. \end{aligned}$$

Applying Lemma 2, we have

$$\begin{aligned} \forall t\in \mathbb {R}, \ \ \varPhi ^{-i \mathbb {P} c_t }_t \prod _{j=1}^n \varPhi ^{-i L_j \mathbb {P} x_j }_t \prod _{j=1}^n \varPhi ^{-i L_{j+n} \mathbb {P} \xi _j }_t = \varPhi ^{-i \mathbb {P} \ell }_t. \end{aligned}$$

Consequently, applying Proposition 4 at $t=1$, we get

$$\begin{aligned} e^{i \ell ^w} = e^{i c_t} \prod _{j=1}^n e^{i L_j x_j} \prod _{j=1}^n e^{ L_j \partial _{x_j} }. \end{aligned}$$

$\square$

Lemma 7

If $p:\mathbb {R}^{2n}\rightarrow \mathbb {R}$ is a real-valued polynomial of degree 2 or less, then there exists a real linear form $\ell :\mathbb {R}^{2n}\rightarrow \mathbb {R}$ and $c\in \mathbb {R}$, such that

$$\begin{aligned} e^{ip^w} =e^{ic} e^{iq^w} e^{i\ell ^w}, \end{aligned}$$

(40)

where q is the quadratic part of p.

Proof

Let l be the linear part of p. Considering the natural action of the entire functions on $M_{2n}(\mathbb {C})$, let $\ell =\ell _1$ and $c=c_1$ where

with $Q\in S_{2n}(\mathbb {R})$, the matrix of q and L the matrix of l. Applying Lemma 2, we have

$$\begin{aligned} \forall t\in \mathbb {R}, \ \ \varPhi ^{-i \mathbb {P} c_t }_t \varPhi ^{-i \mathbb {P} q}_t \varPhi ^{-i \mathbb {P} \ell _t }_t = \varPhi ^{-i \mathbb {P} p}_t. \end{aligned}$$

Consequently, applying Proposition 4 at $t=1$, we get (40). $\square$

Lemma 8

If $p:\mathbb {R}^{2n}\rightarrow \mathbb {R}$ is a real-valued polynomial of degree 2 or less bounded below, then there exists a real-valued linear form $\ell :\mathbb {R}^{2n}\rightarrow \mathbb {R}$ and $c\in \mathbb {R}$ such that

$$\begin{aligned} e^{-p^w} = e^{-c} e^{-i\ell ^w} e^{-q^w} e^{i\ell ^w}, \end{aligned}$$

(41)

where q is the quadratic part of p.

Proof

Applying Lemma 3, we get $Y\in \mathbb {R}^{2n}$ such that $p=p_1$ where

$$\begin{aligned} p_t = q(\cdot \ - \ tY)+p(Y). \end{aligned}$$

and $X=(x_1,\dots ,x_n,\xi _1,\dots ,\xi _n)$.

Let and $c=p(Y)$. Let $\mathscr {B}$ be the basis introduce in the proof of Lemma 2 and defined by (37). Observing that

$$\begin{aligned} \mathrm {mat}_{\mathscr {B}} J_{2n+2}\nabla \mathbb {P}\ell = \begin{pmatrix} x_{n+1} Y \\ 0 \\ \ell \end{pmatrix}, \end{aligned}$$

we deduce that

$$\begin{aligned} (\mathrm {mat}_{\mathscr {B}} \varPhi ^{-i \mathbb {P}\ell }_t) \begin{pmatrix} X \\ x_{n+1} \\ \xi _{n+1}\end{pmatrix} = \begin{pmatrix} X - t x_{n+1} Y \\ x_{n+1} \\ \xi _{n+1} - t \ell \end{pmatrix}. \end{aligned}$$

Consequently, we have

$$\begin{aligned} \mathbb {P}(q + c ) \circ \varPhi ^{-i \mathbb {P}\ell }_t = \mathbb {P}p_t. \end{aligned}$$

However, $\varPhi ^{-i \mathbb {P}\ell }_t$ is a symplectic map, so we have

$$\begin{aligned} \varPhi ^{i \mathbb {P}\ell }_t \varPhi _t^{ \mathbb {P}(q + c)}\varPhi ^{-i \mathbb {P}\ell }_t = \varPhi _t^{ \mathbb {P}(q + c)\circ \varPhi ^{-i \mathbb {P}\ell }_t} = \varPhi _t^{ \mathbb {P}p_t}. \end{aligned}$$

Now observing that the Hamiltonian $\mathbb {P} c$ commutes (i.e., for the canonical Poisson bracket) with all the other Hamiltonians, we deduce of the Noether theorem (or of Lemma 2) that

$$\begin{aligned} \forall t\in \mathbb {R}, \ \varPhi _t^{ \mathbb {P}c} \varPhi ^{i \mathbb {P}\ell }_t \varPhi _t^{ \mathbb {P}q }\varPhi ^{-i \mathbb {P}\ell }_t = \varPhi _t^{ \mathbb {P}p_t}. \end{aligned}$$

Consequently, applying Proposition 4 at $t=1$, we get (41). $\square$

Lemma 9

If $q:\mathbb {R}^{2n}\rightarrow \mathbb {R}$ is a nonnegative real quadratic form, then $\exp (-q^w)$ can be computed by an exact splitting.

Proof

Applying Theorem 21.5.3 of [21], we get a symplectic change of coordinates $T\in \mathrm {Sp}_{2n}(\mathbb {R})$ such that there exists $m\in \llbracket 1,n \rrbracket$ and $0<\lambda _1\le \dots \le \lambda _m$ some real numbers such that

$$\begin{aligned} q\circ T = \sum _{j=1}^m \lambda _j(x_j^2 + \xi _j^2) + \sum _{j=m+1}^n x_j^2. \end{aligned}$$

Consequently, since T is symplectic, applying Noether theorem, we have

$$\begin{aligned} \varPhi ^q_1 = T \varPhi ^{q\circ T}_1 T^{-1} = T \left( \prod _{j=1}^m \varPhi ^{x_j^2 + \xi _j^2}_{\lambda _j} \prod _{j=m+1}^n \varPhi ^{x_j^2}_1 \right) T^{-1}. \end{aligned}$$

Applying Theorem 3, we get, at the level of the Fourier Integral Operators,

$$\begin{aligned} \pm e^{-tq^w}= \mathscr {K}(T) \left( \prod _{j=1}^m e^{-\lambda _j (x_j^2-\partial _{x_j}^2) } \prod _{j=m+1}^n e^{-x_j^2} \right) \mathscr {K}(T^{-1}). \end{aligned}$$

Recalling that, as a consequence of the formula (2), the semigroups $e^{\lambda _j (x_j^2-\partial _{x_j}^2) }$ can be computed by exact splittings, we deduce of Lemma 5 that the same applies for $e^{-tq^w}$.

$\square$

Now, we can prove Theorem 1.

Proof of Theorem 1

Let p be a polynomial of degree 2 or less on $\mathbb {C}^{2n}$ whose real part is bounded below on $\mathbb {R}^{2n}$. We aim at proving that $e^{-p^w}$ can be computed by an exact splitting in the sense of Definition 1.

Applying Corollary 1, we get a constant $c\in \mathbb {R}$ such that $\mathbb {P}(p-c)\ge 0$. Then, applying Theorem 2.1 and Lemma 3.10 of [2], we get $t_0>0$ and two real quadratic forms $a_t,b_t:\mathbb {R}^{2n+2} \rightarrow \mathbb {R}$ depending analytically on $t\in (-t_0,t_0)$, $a_t$ being nonnegative, and such that if $|t| < t_0$, then

$$\begin{aligned} \varPhi _t^{a_t} \varPhi _t^{i b_t} = \varPhi _t^{\mathbb {P}(p-c)}. \end{aligned}$$

Furthermore, it follows of the Baker–Campbell–Hausdorff formulas and formulas of (3.22) and (3.41) of [2] defining $a_t,b_t$ that these quadratic forms belong to the complex Lie algebra generated by $\mathbb {P}(p-c)$ and $\mathbb {P}(\overline{p}-c)$. Observing that for all polynomials $p_1,p_2$ of degree 2 or less on $\mathbb {C}^{2n}$ we have

$$\begin{aligned} \{\mathbb {P}p_1,\mathbb {P}p_2 \} = \mathbb {P}\{p_1,p_2\}, \end{aligned}$$

the image of $\mathbb {P}$ is a Lie algebra. Consequently, $a_t,b_t$ belong to the image of $\mathbb {P}$, i.e., there exist two real polynomials of degree 2 or less on $\mathbb {R}^{2n}$ depending analytically on t and denoted $p_{t}^{(r)}$ and $p_{t}^{(i)}$ such that

$$\begin{aligned} a_t = \mathbb {P}p_{t}^{(r)} \ \mathrm {and} \ b_t = \mathbb {P}p_{t}^{(i)}. \end{aligned}$$

Note that since $p_{t}^{(r)}$ is the restriction of $a_t$ on the affine subspace $\{(x_{n+1},\xi _{n+1})=(1,0)\}$, it is also nonnegative.

Now applying Proposition 4, we deduce that if $0\le t<t_0$ we have

$$\begin{aligned} e^{-t p^w} = e^{-tc} e^{-t (p_{t}^{(r)})^w} e^{-it (p_{t}^{(i)})^w}. \end{aligned}$$

(42)

Let $t_\star \in (0,t_0)$ be such that there exists $n\in \mathbb {N}$ satisfying $t_\star =n^{-1}$. Since $(e^{-t p^w})_{t\ge 0}$ is a semigroup, we have

$$\begin{aligned} (e^{-t_{\star } p^w})^n = e^{-nt_{\star } p^w} = e^{-p^w}. \end{aligned}$$

Consequently, if $e^{-t_{\star } p^w}$ can be computed by an exact splitting, then the same applies for $e^{-p^w}$. Furthermore, from the factorization (42), we deduce that if $e^{-t_{\star } (p_{t_{\star }}^{(r)})^w}$ and $e^{-it_{\star } (p_{t_{\star }}^{(i)})^w}$ can be computed by an exact splitting, then the same applies for $e^{-t_{\star } p^w}$. Consequently, we just have to focus on these two semigroups.

On the one hand, applying Lemmas 7, 6, 5 and Theorem 3 (to justify that the semigroup generated by a quadratic differential operator is a Fourier Integral Operator), we deduce that $\exp (-it_{\star } (p_{t_\star }^{(i)})^w)$ can be computed by an exact splitting.

On the other hand, applying Lemmas 8, 6 and 9, we deduce that $e^{-t_{\star } (p_{t_{\star }}^{(r)})^w}$ can be computed by an exact splitting. $\square$

1.3 Proof of Lemma 1

We aim at proving that if $\mathfrak {R}q\ge 0$ on $\mathbb {R}^{2n}$, then $\varPhi _t^q\equiv e^{-2itJ_{2n}Q} \in \mathrm {Sp}_{2n}^+(\mathbb {C})$, where Q is the matrix of q. First, we recall that from Proposition 3 we know that $\varPhi _t^q$ is a symplectic transformation (i.e., $\varPhi _t^q \in \mathrm {Sp}_{2n}(\mathbb {C})$).

So we aim at proving that $\varPhi _t^q$ is nonnegative, i.e.,

(43)

Since $\varPhi _0^q=I_{2n}$, we have

Since $\mathfrak {R}Q$ is a real symmetric nonnegative matrix, is a Hermitian nonnegative matrix, and thus, is also a Hermitian nonnegative matrix which proves (43).