2 The main results

Let X be a Banach space and assume that $A(t) = \sum _{n \in {\mathbb N}-1} A_nt^n, R(t) = \sum _{n \in {\mathbb N}-1} R_nt^n$ are real analytic. Let ${\boldsymbol {1}} = (1,1,\ldots ) \in ({\mathbb N}-1)^{\infty }$ . Let ${\mathbf {p}} = (p_1,p_2,\ldots ),\, \mathbf{t} = (t_1,t_2,\ldots ) \in ({\mathbb N}-1)^{\infty }$ denote the respective power and index vectors. Define the operator product

whenever $\langle {\mathbf {p}}, {\boldsymbol {1}} \rangle < \infty $ . If $p_j = 0$ or $t_j = 0$ then $R_{t_j}^{p_j} = I$ and we can use a reduced product with finite length $\langle {\mathbf {p}}, {\boldsymbol {1}} \rangle $ . In general, we can write $R_{\mathbf {t}}^{\mathbf {p}} = R_{t_1}^{p_1}R_{t_2}^{p_2} \cdots R_{t_n}^{p_n}$ for some ${n \leq \langle {\mathbf {p}}, {\boldsymbol {1}} \rangle }$ where ${\mathbf {p}},\, \mathbf{t} \in {\mathbb N}^{\infty }$ . The order of terms in the product is important. If ${X = {\mathbb R}}$ then $a(t) = \sum _{n \in {\mathbb N}-1} a_nt^n$ , $r(t) = \sum _{n \in {\mathbb N}-1} r_nt^n$ , where $a_n, r_n \in {\mathbb R}$ . Let ${\mathbf {n}} = (1,2,\ldots ) \in ({\mathbb N}-1)^{\infty }$ and define the real number product

whenever $\langle {\mathbf {p}}, {\boldsymbol {1}} \rangle < \infty $ . In the scalar case the order of terms in the real number product is no longer important. If $p_j = 0$ then $r_j^{p_j} = 1$ and we can use the reduced product

$$ \begin{align*} r_{\mathbf{n}}^{\mathbf{p}} = r_{j(1),j(2),\ldots, j(n)}^{p_{j(1)}, p_{j(2)}, \ldots,p_{j(n)}} = r_{j(1)}^{p_{j(1)}} r_{j(2)}^{p_{j(2)}} \cdots r_{j(n)}^{p_{j(n)}} \end{align*} $$

for some $n \leq \langle {\mathbf {p}}, {\boldsymbol {1}} \rangle $ where $j(1) < j(2) < \cdots < j(n)$ are the indices for the nonzero powers.

2.1 The general case

In the general case

(2.1) $$ \begin{align} A_q = \sum_{\langle \mathbf{p}, \mathbf{t} \rangle = q+1} (-1)^{\langle \mathbf{p}, \boldsymbol{1} \rangle -1} t_1 R_{\mathbf{t}}^{\mathbf{p}} \end{align} $$

for each $q \in {\mathbb N}$ , where $\langle {\mathbf {p}}, {\boldsymbol {1}} \rangle = p_1 + p_2 + \cdots $ is the power sum (the product length) and $\langle {\mathbf {p}},\, \mathbf{t} \rangle = p_1t_1 + p_2t_2 + \cdots $ is the power-weighted index sum.

2.2 The scalar case

In the scalar case the general formula can be replaced by the collective formula

(2.2) $$ \begin{align} a_q = (q+1) \cdot \sum_{k=1}^{q+1} (-1)^{k-1} k^{-1} \bigg[\!\sum_{\mathbf{p} \in {\mathcal P}_{k, q}} \rho_{\mathbf{p}} r_{\mathbf{n}}^{\mathbf{p}} \bigg] \end{align} $$

for each $q \in {\mathbb N}$ , where

for each $k = 1,\ldots ,q+1$ and

for each ${\mathbf {p}} \in {\mathcal P}_{k,q}$ and $\langle {\mathbf {p}}, {\mathbf {n}} \rangle = p_1 + 2p_2 + \cdots $ is the power-weighted ordered index sum.

3 The intuitive argument

We simply use successive applications of the basic recursion.

3.1 The general case

The initial condition shows that $R_0 = I$ is the identity operator. Therefore, we can rearrange the general recursion to show that $A_0 = R_1$ and

$$ \begin{align*} A_q = (q+1)R_{q+1} - \sum_{j = 0}^{q-1} A_jR_{q-j} \end{align*} $$

for each $q \in {\mathbb N}$ . By applying this recursion for $q = 1,2,3,4,\ldots ,$ we obtain

$$ \begin{align*} A_1 & = 2R_2^1 - R_1^2, \\ A_2 & = 3R_3^1 - (R_{1,2}^{1,1} + 2R_{2,1}^{1,1} ) + R_1^3, \\ A_3 & = 4R_4^1 - ( [R_{1,3}^{1,1} + 3R_{3,1}^{1,1} ] + 2R_2^2 ) + [R_{1,2}^{2,1} + R_{1,2,1}^{1,1,1} + 2R_{2,1}^{1,2}] - R_1^4, \\ A_4 & = 5R_5^1 - ( [R_{1,4}^{1,1} + 4R_{4,1}^{1,1}] + [2R_{2,3}^{1,1} + 3R_{3,2}^{1,1} ] ) \nonumber \\ & \quad + ( [R_{1,3}^{2,1} +R_{1,3,1}^{1,1,1} + 3R_{3,1}^{1,2} ] + [R_{1,2}^{1,2} + 2R_{2,1,2}^{1,1,1} + 2R_{2,1}^{2,1} ] ) \nonumber \\ & \quad - [ R_{1,2}^{3,1} + R_{1,2,1}^{2,1,1} + R_{1,2,1}^{1,1,2} + 2R_{2,1}^{1,3} ] + R_1^5, \end{align*} $$

and so on. In order to apply the proposed formula (2.1) to find $A_q$ we must first solve the equation $\langle {\mathbf {p}},\, \mathbf{t} \rangle = q+1$ . When $q = 4$ we must solve the equation $\langle {\mathbf {p}},\, \mathbf{t} \rangle = 5$ with $n \leq \langle {\mathbf {p}}, {\boldsymbol {1}} \rangle \leq 5$ . We can now use (2.1) to write down the relevant terms:

$$ \begin{align*} \langle {\mathbf{p}}, {\boldsymbol{1}} \rangle & = 1: 5R_5^1, \\ \langle {\mathbf{p}}, {\boldsymbol{1}} \rangle & = 2: (-1)R_{1,4}^{1,1}, \ (-1)4R_{4,1}^{1,1}, \ (-1)2R_{2,3}^{1,1}, \ (-1)3R_{3,2}^{1,1}, \\ \langle {\mathbf{p}}, {\boldsymbol{1}} \rangle & = 3: R_{1,3}^{2,1}, \ R_{1,3,1}^{1,1,1}, \ 3R_{3,1}^{1,2}, \ R_{1,2}^{1,2}, \ 2R_{2,1,2}^{1,1,1}, \ 2R_{2,1}^{2,1}, \\ \langle {\mathbf{p}}, {\boldsymbol{1}} \rangle & = 4: (-1)R_{1,2}^{3,1}, \ (-1)R_{1,2,1}^{2,1,1}, \ (-1)R_{1,2,1}^{1,1,2}, \ (-1)2R_{2,1}^{1,3}, \\ \langle {\mathbf{p}}, {\boldsymbol{1}} \rangle & = 5: R_1^5. \end{align*} $$

This confirms the validity of (2.1) when $q = 4$ .

3.2 The scalar case

The initial condition shows that $r_0 = 1$ . Hence, we can rearrange the general recursion to show that $a_0 = r_1$ and

$$ \begin{align*} a_q = (q+1)r_{q+1} - \sum_{j = 0}^{q-1} a_jr_{q-j} \end{align*} $$

for each $q \in {\mathbb N}$ . By applying this recursion for $q = 1,2,3,4,5,\ldots ,$ we obtain

$$ \begin{align*} a_1 & = 2r_2^1 - r_1^2, \\ a_2 & = 3r_3^1 - 3r_{1,2}^{1,1} + r_1^3, \\ a_3 & = 4r_4^1 - (4r_{1,3}^{1,1} + 2r_2^2 ) + 4r_1^2r_2 - r_1^4, \\ a_4 & = 5r_5^1 - ( 5r_{1,4}^{1,1} + 5r_{2,3}^{1,1} ) + ( 5r_{1,3}^{2,1} + 5r_{1,2}^{1,2} ) - 5r_{1,2}^{3,1} + r_1^5, \\ a_5 & = 6r_6 - (6r_1r_5 + 6r_2r_4 + 3r_3^2) + (6r_1^2r_4 + 12 r_1r_2r_3 + 2r_2^3) \nonumber \\ & \quad - (6r_1^3r_3 + 9r_1^2r_2^2) + 6r_1^4r_2 - r_1^6, \end{align*} $$

and so on. In order to apply the proposed formula (2.2) to find $a_q$ we must first find all ${\mathbf {p}} \in {\mathcal P}_{q,k}$ for each $k = 1,\ldots ,q+1$ . Thus, we must solve the simultaneous equations $\langle {\mathbf {p}}, {\mathbf {n}} \rangle = q+1$ and $\langle {\mathbf {p}}, {\boldsymbol {1}} \rangle = k$ for each $k = 1,\ldots ,q+1$ . When $q = 5$ we obtain the following solutions:

$$ \begin{align*} k & = 1: {\mathbf{p}} = (0,0,0,0,0,1), \\ k & = 2: {\mathbf{p}} = (1,0,0,0,1), \ (0,1,0,1), \ (0,0,2), \\ k & = 3: {\mathbf{p}} = (2,0,0,1), \ (1,1,1), \ (0,3,0), \\ k & = 4: {\mathbf{p}} = (3,0,1), \ (2,2), \\ k & = 5: {\mathbf{p}} = (4,1), \\ k & = 6: {\mathbf{p}} = (6). \end{align*} $$

We can now use (2.2) to write down the corresponding terms:

$$ \begin{align*} k &= 1: \binom{1}{1}r_6 = 6 r_6^1; \\ k &= 2: (-1) 3 \binom{2}{1\ 1} r_1r_5 = (-1)6r_{1,5}^{1,1}, \quad (-1) 3 \binom{2}{1\ 1} r_2r_4 = (-1)6r_{2,4}^{1,1}, \\ &\hspace{5.2cm} (-1) 3 \binom{2}{2} r_3^2 = (-1)3r_3^2; \\ k & = 3: 2 \binom{3}{2\ 1} r_1^2r_4 = 6r_{1,4}^{2,1}, \quad 2 \binom{3}{1\ 1\ 1} r_1r_2r_3 = 12r_{1,2,3}^{1,1,1}, \quad 2 \binom{3}{3}r_2^3 = 2r_2^3; \\ k & = 4: 6 (-1) 4^{-1} \binom{4}{3\ 1} r_1^3r_3 = (-1) 6r_{1,3}^{3,1}, \quad 6 (-1) 4^{-1} \binom{4}{2\ 2} r_1^2r_2^2 = (-1) 9r_{1,2}^{2,2}; \\ k & = 5: 6 \cdot 5^{-1} \binom{5}{4\ 1} r_1^4r_2 = 6r_{1,2}^{4,1}; \\ k & = 6: 6 (-1) 6^{-1} \binom{6}{6} r_1^6 = (-1)r_1^6. \end{align*} $$

This confirms the validity of (2.2) when $q = 5$ .

4 Literature review

We refer readers to our recent paper [Reference Bassom, Howlett and Taylor5] for a brief historical review of the extensive literature on this topic. We will simply summarise that review here. Suffice it to say that Kato and Tanabe [Reference Kato and Tanabe11, Reference Tanabe19] and Sobolevski [Reference Sobolevski18] found conditions on the evolutionary kernel $A(t)$ that guarantee existence of an evolutionary operator $R(t)$ satisfying (1.1) subject to the initial condition $R(0) = I$ (see [Reference Yosida21, Section XIV.5, pages 440–445] for an extended discussion and also [Reference Acquistapace and Terrini2, Reference Di Giorgio, Lunardi and Schnaubelt8, Reference Giscard, Lui, Thwaite and Jaksch10, Reference Komatsu12–Reference Latushkin, Pogan and Schnaubelt14, Reference Whittlesey20]). If $A(t) \in {\mathcal B}(X)$ is real analytic at $t=0$ , the Peano–Baker series [Reference Baake and Schlägel3, Reference Baker4, Reference Campbell7, Reference Peano17] exactly reproduces the first $n+1$ terms of the Maclaurin series for $R(t)$ after n iterations but does not reveal an explicit formula. The Magnus expansion [Reference Blanes, Casas, Oteo and Ros6, Reference Magnus15] seeks an infinite series solution in the form of an operator exponential. Explicit formulas for the exponents are notoriously complicated [Reference Blanes, Casas, Oteo and Ros6, Section 2.2, page 20] so recursive calculation is usually preferred. See also fundamental work on the so-called CBHD exponential theorem of Campbell, Baker, Hausdorff and Dynkin [Reference Achilles and Bonfiglioli1, Reference Dynkin9].

The current paper addresses a question posed in a recent paper by Bassom et al. [Reference Bassom, Howlett and Taylor5] and a conditional answer to that question in a subsequent paper by Neto [Reference Neto16].

In [Reference Bassom, Howlett and Taylor5] the authors assumed that the exponential operator and the exponential kernel are both real analytic at $t = 0$ . They justified this assumption by proving that the evolutionary operator is real analytic at $t = 0$ if and only if the evolutionary kernel is real analytic at $t = 0$ . To begin they noted two well-known general results.

They then proved two specific results.

We refer readers to the original paper [Reference Bassom, Howlett and Taylor5] for the proofs of these two crucial results.

In [Reference Neto16] Neto used the $\Omega $ -calculus and an elementary direct argument to find an explicit Maclaurin series solution to the nonhomogeneous evolution equation (1.5). He then used an ingenious role reversal argument to show that in a Hilbert space setting, the adjoint $A(t)^*$ of the original kernel is the evolutionary operator for a nonhomogeneous equation (1.6) with kernel $[R(t)^*]^{-1}dR(t)^*/dt$ and forcing function $[R(t)^*]^{-1}d^2R(t)^*/dt^2$ . Thus, he was able to apply the extended results to find an explicit expression for the coefficients $A_n$ in terms of the coefficients $R_n$ . The role reversal argument is not used here. Consequently, the methodology in [Reference Neto16] is not directly relevant to the current paper.

5 Justification of the main results

We will prove the general formula (2.1) for $A_q$ by induction.

Proof. Formula (2.1) is true for $q = 0$ . Assume (2.1) is true for $r < q \in {\mathbb N}$ . The recursive formula (1.2) shows that

(5.1) $$ \begin{align} A_q = (q+1)R_{q+1} - \sum_{j=0}^{q-1} A_j R_{q-j}. \end{align} $$

for each $q \in {\mathbb N}$ . We argue that each individual product term in $A_q$ is generated by precisely one term in the recursive relationship. Suppose $A_q$ contains the term $c_{\mathbf {p}, \mathbf {t}}R_{t_1,\ldots ,t_r}^{p_1,\ldots ,p_r}$ for some $c_{\mathbf {p}, \mathbf {t}} \in {\mathbb R}$ where ${\mathbf {p}} \in {\mathbb N}^r$ for some $r \leq q+1$ . The assumption that ${\mathbf {p}} \in {\mathbb N}^r$ simply means that trivial terms with $p_j = 0$ are omitted from the product. If $r = q+1$ , the only possible term is $(q+1)R_{q+1}^1$ . Thus, $c_{\boldsymbol {1}, q+1} = q+1$ . If $r < q+1$ , the term $c_{\mathbf {p}, \mathbf {t}} R_{t_1,\ldots ,t_r}^{p_1,\ldots ,p_r}$ must be generated by an individual product within the collective product $A_{q+1 - t_r} R_{t_r}$ . The inductive hypothesis shows that the specific product is

$$ \begin{align*} (-1) \cdot (-1)^{p_1 + \cdots + (p_r-1) - 1} t_1R_{t_1,\ldots,t_r}^{p_1,\ldots,p_r-1} \cdot R_{t_r} = (-1)^{\langle \mathbf{p}, \boldsymbol{1} \rangle - 1} t_1 R_{t_1,\ldots,t_r}^{p_1,\ldots,p_r} \end{align*} $$

where $p_1t_1 + \cdots + (p_r-1)t_r = q+1-t_r$ . Therefore, $c_{\mathbf {p}, \mathbf {t}} = (-1)^{\langle \mathbf {p}, \boldsymbol {1} \rangle - 1}t_1$ and $\langle {\mathbf {p}},\, \mathbf{t} \rangle = q+1$ . It is also clear that every term of the form

$$ \begin{align*} (-1)^{\langle \mathbf{p}, \boldsymbol{1} \rangle - 1} t_1 R_{t_1,\ldots,t_r}^{p_1,\ldots,p_r} \end{align*} $$

with $\langle {\mathbf {p}},\, \mathbf{t} \rangle = q+1$ is generated in this way. Thus, (2.1) is also true for $r = q$ .

Remark 5.1. It is interesting to follow the generational history of the archetypal term

$$ \begin{align*} (-1)^{\langle \mathbf{p}, \boldsymbol{1} \rangle - 1} t_1 R_{\mathbf{t}}^{\mathbf{p}} = (-1)^{\langle \mathbf{p}, \boldsymbol{1} \rangle - 1} t_1 R_{t_1,\ldots,t_n}^{p_1,\ldots,p_n} \end{align*} $$

where $\langle {\mathbf {p}},\, \mathbf{t} \rangle = q+1$ . If each repeated factor $R_t^p$ with $p> 1$ is written as the product of p identical terms $R_t \cdots R_t$ , then we may assume without loss of generality that the archetypal term can be rewritten in the form

$$ \begin{align*} (-1)^{\langle \boldsymbol{1}_m, \boldsymbol{1} \rangle -1}R_{\mathbf{u}}^{\boldsymbol{1}} = (-1)^{m-1}R_{u_1,\ldots,u_m}^{1,\ldots,1}, \end{align*} $$

where $\langle {\boldsymbol {1}}, {\mathbf {u}} \rangle = q+1$ and where $\{{\mathbf {e}}_j\}_{j \in {\mathbb N}}$ are the standard unit vectors in ${\mathbb R}^{\infty }$ . The ancestral term arises at stage $u_1$ in the form $u_1R_{u_1}^1$ as the first term in the recursion

$$ \begin{align*} A_{u_1-1} = u_1R_{u_1} -\sum_{j=0}^{u_1-1} A_jR_{u_1-1-j}. \end{align*} $$

At stage $u_1+u_2$ the term $(-1)R_{u_1,u_2}^{1,1}$ is generated from the parent term $u_1R_{u_1}^1$ as part of the product $(-1)A_{u_1-1}R_{u_2}$ in the recursion

$$ \begin{align*} A_{u_1+u_2-1} = (u_1+u_2)R_{u_1 + u_2} - \sum_{j=0}^{u_1+u_2-1} A_j R_{u_1+u_2-1-j}. \end{align*} $$

In general, at stage $u_1 + \cdots + u_r$ , the term $(-1)^{r-1}u_1R_{u_1,\ldots ,u_r}^{1,\ldots ,1}$ is generated from the parent term $(-1)^{r-2}u_1R_{u_1,\ldots ,u_{r-1}}^{1,\ldots ,1}$ as part of the product $A_{u_1 + \cdots + u_{r-1} - 1}R_{u_r}$ in the recursion

$$ \begin{align*} A_{u_1 + \cdots +u_r - 1} = (u_1+\cdots+u_r)R_{u_1 + \cdots + u_r} - \sum_{j=0}^{u_1 + \cdots +u_{r-1} - 1} A_j R_{u_1 + \cdots + u_r - 1 - j}. \end{align*} $$

The process terminates with the term $(-1)^m u_1R_{u_1,\ldots ,u_m}^{1,\ldots ,1}$ when $u_1+ \cdots + u_m = q+1$ .

We will also prove the scalar formula (2.2) for $a_q$ by induction.

Proof. Formula (2.2) is true for $q = 0$ . Assume (2.2) is true for $r < q \in {\mathbb N}$ . The recursive formula (1.2) shows that

$$ \begin{align*} a_q = (q+1)r_{q+1} - \sum_{j=0}^{q-1} a_j r_{q-j} \end{align*} $$

for each $q \in {\mathbb N}$ . Within the expansion (2.2), suppose there is a particular vector ${\mathbf {p}}$ with $\langle {\mathbf {p}}, {\boldsymbol {1}} \rangle = k$ and $\langle {\mathbf {p}}, {\mathbf {n}} \rangle = q+1$ . Within ${\mathbf {p}}$ consider a particular index j with $p_j \geq 1$ . Now consider a reduced vector $\widehat {\mathbf {p}}_j$ which is identical in structure to ${\mathbf {p}}$ except that $p_j$ is replaced by $p_j-1$ . Therefore,

$$ \begin{align*} \langle\, \widehat{\mathbf{p}}_j, {\boldsymbol{1}} \rangle = k-1 \quad\text{and }\quad \langle\, \widehat{\mathbf{p}}_j, {\mathbf{n}} \rangle = p_1 + 2p_2 + \cdots +j(p_j-1) + \cdots = q + 1 - j. \end{align*} $$

We propose that $\widehat {\mathbf {p}}_j$ appears in the expansion of exactly one $a_s$ with $s < q$ . The inductive hypothesis tells us that the parent term we wish to identify takes the form

$$ \begin{align*} (s+1)(-1)^{\ell-1} \ell^{-1} \binom{\ell}{p_1\ p_2\ \cdots\ p_j-1\ \cdots} r_{\mathbf{n}}^{\widehat{\mathbf{p}}_j} \end{align*} $$

where $\langle \widehat {\mathbf {p}}_j, {\boldsymbol {1}} \rangle = \ell $ and $\langle \widehat {\mathbf {p}}_j, {\mathbf {n}} \rangle = s+1$ . It follows that $\ell = k-1$ and $s + 1 = q + 1 - j$ . Thus, the parent term is

$$ \begin{align*} (q+1-j) (-1)^{k-2} (k-1)^{-1} \binom{k-1}{p_1\ \cdots p_j-1\ \cdots} r_{\mathbf{n}}^{\widehat{\mathbf{p}}_j}. \end{align*} $$

The scalar recursion (5.1) shows that the coefficient of ${\mathbf {p}}$ arising from $\widehat {\mathbf {p}}_j$ is

$$ \begin{align*} & (-1)(q+1-j) (-1)^{k-2} (k-1)^{-1} \binom{k-1}{p_1\ \cdots\ p_j-1\ \cdots} \\ & \quad = (q+1-j)(-1)^{k-1} (k-1)^{-1}k^{-1} p_j \binom{k}{p_1\ p_2\ \cdots}. \end{align*} $$

This is the contribution to the coefficient from the parent term when we reduce the power $p_j$ to $p_j-1$ . The total contribution over all possible parent terms is obtained by summing over all possible values of j. Hence, the coefficient for the product $r_{\mathbf {n}}^{\mathbf {p}}$ in the expansion for $a_q$ is given by

$$ \begin{align*} & \sum_{\mathbf{p} \in {\mathcal P}_{k,q},\, p_j \neq 0} (q+1-j)(-1)^{k-1} (k-1)^{-1}k^{-1} p_j \binom{k}{p_1\ p_2\ \cdots} \\ & \quad = (-1)^{k-1} (k-1)^{-1}k^{-1} \binom{k}{p_1\ p_2\ \cdots} \sum_{\mathbf{p} \in {\mathcal P}_{k,q}} [ (q+1)p_j - jp_j ] \\ & \quad = (-1)^{k-1} (k-1)^{-1}k^{-1} \binom{k}{p_1\ p_2\ \cdots} [ (q+1)k - (q+1)] \\ & \quad = (q+1) (-1)^{k-1} k^{-1} \binom{k}{p_1\ p_2\ \cdots} \end{align*} $$

as required. This completes the proof.