On Expansion of a Solution of General Non-autonomous Polynomial Differential Equation

We give a recursive formula for an expansion of a solution of a general non-autonomous polynomial differential equation. The formula is given on the algebraic level with a use of shuffle product. This approach minimizes the number of integrations on each order of expansion. Using combinatorics of trees we estimate the radius of convergence of the expansion.


Introduction
Consider a non-autonomous polynomial differential equation, known also as a generalized Abel differential equatioṅ x(t) = u 0 (t) + u 1 (t)x(t) + · · · + u i (t)x i (t) + · · · + u n (t)x n (t), (1.1) with a solution x : [0, T ] → R on a small segment of the reals. In this class of differential equation there are: for n = 1 the linear equation with well known formula for the general solution; for n = 2 the Riccati equation well known both for theoretical [17,18,7,3] and practical (see [4] and references therein at the beginning of section 4) reasons; for n = 3 the Abel differential equation of the first kind studied theoretically [15] and for practical reasons ( [14,12,20] and references in [15,6]); for n > 3 the generalized Abel differential equations [2,1]. Assuming X i = x i ∂ ∂x , for i = 0, . . . , n, are differential vector fields on R one can, following Fliess [9] (see also [13,10] and [11] with references therein), expand the solution of the equation in terms of iterated integrals t 0 t k 0 · · · t 2 0 u i k (t k ) · · · u i 1 (t 1 ) dt 1 . . . dt k−1 dt k and iterated differential operators X i 1 · · · X i k acting on the identity function h(x) = x and evaluated at the zero point. In this approach one does not use a specific form of the vector fields, i.e. the fact that they are of polynomial type. In this article we show another approach to expanding a solution of the above equation in terms of iterated integrals with a use of an important feature of Chen's iterated integral mapping that it is a shuffle algebra homomorphism (see the comment after formula (2.4)). In fact with the use of Chen's mapping (2.4) we will be able to consider a purely algebraic problem instead of the analytic one. More precisely, assuming that the solution of (1.1) can be expanded in terms of iterated integrals we show that an algebraic equation must be satisfied in the space of non-commutative series on n + 1 letters. It will be easy to show existence of the solution of the algebraic equation by a recursive formula of its homogeneous parts. The Chen's mapping gives us the expansion of the solution of initial problem as we state in Theorem 1. This is done in section 2. Then, in section 3, by counting elements of a class of trees in two different ways we show convergence of the defined expansion of x(t) for small timesthis is stated in Theorem 2 (in section 2).
As an application of this general approach we consider, in section 4, the cases of the linear equation (i.e., n = 1), the Riccati equation (n = 2) and the one where there are only two non-vanishing summands. In the first case we reestablish a well known formula for the general solution, and in the second case we deduce convergence of the series defining coordinates of the second kind connected with a(1)-type involutive distribution [16].
Finally, in section 5, we compare the Chen-Fliess approach with the one given in this article. It occurs that in the letter case the number of integrals to compute grows significantly slower with the order of approximation, than in the first case.

Existence and convergence of an expansion
Let n ∈ N 1 , T > 0, and let u 0 , . . . , u n : [0, T ] → R be measurable and bounded (by a constant M ∈ R) functions. Consider a non-autonomous polynomial differential equatioṅ Two comments are in order. Firstly, the Newton symbols occurring in the above formula are for convenience reasons -without these constants it would be harder to estimate the radius of convergence of a defined series. Secondly, we assume the initial value equals zero. This is without loss of generality in a sense that with the linear change of variables x → x − x 0 we can transform the equation with the initial value equals x 0 to another equation with different u i 's.
Integrating both sides of the equation we get an integral equation By Caratheory's theorem for a small ǫ > 0 there exists an absolutely continues solution x : [0, ǫ] → R of the initial equation (2.1) in a sense thatx satisfies the integral equation (2.2) for t ∈ [0, ǫ]. We want to express the solutionx by means of iterated integrals of products of u i 's. In order to do this we introduce some algebraic tools.
To each function u i we assign a formal variable a i , which we call a letter. The set of all letters A = {a 0 , . . . , a n } is called an alphabet. Juxtapositioning letters we can obtain words of an arbitrary length k ∈ N; the set of such words is denoted by contains only one -empty -word. The set of all words is denoted by A * = ∞ k=0 A * k . The juxtaposition gives rise to an associative, noncommutative product on the set of words A * × A * ∋ (v, w) → v · w = vw ∈ A * called the concatenation product; then the set A * with the concatenation product and the neutral element 1 ∈ A * is a free monoid generated by A. Taking R-linear combination of words and bilinearly extending the concatenation product we get the R-algebras R A of noncommutative polynomials on A and R A of noncommutative series on A. In both algebras we can consider the bilinear product ¡ : R A ⊗ R A → R A -the shuffle product -defined recursively for words by 1 ¡ w = w ¡ 1 = w for any w ∈ A * , and for all b, c ∈ A and v, w ∈ A * . It is easy to see that the shuffle product is commutative, thus with 1 as the neutral element it gives rise to an additional commutative algebra structure on R A and R A . We will use both -concatenation and shuffle -products in our considerations. It is important to indicate the priority of the shuffle product over the concatenation product in all formulas of this article, so that v ¡ w · a always means (v ¡ w) · a.
On R A we introduce a natural scalar product (·|·) : R A × R A → R, which for elements v, w ∈ A * × A * is given by For S ∈ R A , let S k ∈ R A be the k-degree homogenous part of S, i.e., Clearly, S = ∞ k=0 S k . Define the linear homomorphism Υ t : R A → R by Υ t (1) = 1, and Equivalently, the homomorphism can be defined recursively by for any v ∈ A * and a i ∈ A. Since u i 's are bounded the definition is correct for all t ≥ 0. One can check that Υ t is in fact a shuffle algebra homomorphism, i.e., Υ t (v ¡w) = Υ t (v) Υ t (w) (see [8,19]) which is a crucial feature in what follows. For a general series S ∈ R A the homomorphism Υ t is obviously not well defined since Υ t (S) can be divergent. We restrict the definition of Υ t to series S ∈ R A and times t ≥ 0 for which the series is convergent.
Coming back to the initial problem, assume that there exists a seriesŜ ∈ R A such that the solutionx of (2.1) satisfiesx(t) = Υ t (Ŝ) for t ∈ [0, ǫ] (in particular Υ t (Ŝ) is convergent). Using the recursive definition of Υ t (2.4) and the fact that Υ t is a shuffle algebra homomorphism, we get from (2.2) that Υ t (Ŝ) = Υ t (a 0 + n 1Ŝ · a 1 + n 2Ŝ ¡Ŝ · a 2 + . . . +Ŝ ¡n · a n ), where we abbreviate n i = n i , andŜ ¡n is defined recursively in a natural way, i.e., S ¡0 = 1 andŜ ¡n =Ŝ ¡Ŝ ¡(n−1) . Now the point is that we can forget, for a moment, about the homomorphism and consider only the algebraic equation S = a 0 + n 1Ŝ · a 1 + n 2Ŝ ¡Ŝ · a 2 + . . . +Ŝ ¡n · a n . Proof. The equation under consideration must be satisfied for each homogeneous part, so we can split it into the following series of equationŝ and for arbitrary k ∈ NŜ where the second sum is taken over multi-indices l = (l 1 , . . . , l i ) in We see that the homogeneous parts of the seriesŜ are defined recursively, therefore the series is defined uniquely.
Observe that from the recursive definition of Υ t and a property Υ t (v¡w) Moreover, any permutation of (l 1 , . . . , l i ) gives the same expression under the integral. For l ∈ M (i) denote by R(l) the number of such permutations, i.e., where the second sum is taken over Let us state it in the following theorem.
is a formal solution of the differential equation (2.1). Remark 2.2. It is worth noticing that for a fixed k ≥ 1, the number of integrals in formula (2.8) is the cardinality of n i=1 M ≤ (i). This is less than the cardinality of ∞ i=1 M ≤ (i), which is the number of partitions of k. The first ten of these numbers are 1, 2, 3, 5, 7, 11, 15, 22, 30, 42. It means that the number of integrals that we have to perform to computê x k+1 grows quite slowly. In section 5 we show that this growth is much slower than the growth of the number of non-zero components in the Chen-Fliess expansion.
There remains the problem under what assumption the solution for the algebraic equation is in the domain of the homomorphism Υ t : In order to solve it, we need to compute the number of words (with multiplicities) in each homogeneous part ofŜ. So for S ∈ R A let us introduce the following definition Proposition 2.3. IfŜ k is the k-degree homogeneous part of the solutionŜ of the algebraic equation (2.5), then for k ≥ 1, #Ŝ k = ((n − 1)(k − 1) + 1) · ((n − 1)(k − 2) + 1) · · · n (in particular #Ŝ 1 = 1) and #Ŝ 0 = 0.
The proposition will be proved in section 3. Now we state the theorem about convergence of the expansion.
is the solution of the differential equation (2.1) on the same segment.
For n = 0 the statement is obvious.

Counting trees
In this section we prove Proposition 2.3. In order to do this we consider certain classes of trees. It occurs that the number of trees in these particular classes equal #Ŝ k on the one hand and ((n − 1)(k − 1) + 1) · ((n − 1)(k − 2) + 1) · · · n on the other hand.
For k, n ∈ N let T n k denote the set of planar, rooted, full n-ary and increasingly labeled trees on k vertices. Recall that a tree is rooted if there exists a distinguished vertex called the root; is full n-ary if each vertex has exactly none or n children; is on k vertices if the number of vertices with n children (parent vertices) is equal k; is increasingly labeled if the parent vertices are labeled by natural numbers from 1 to k, and the labels increase along each branch starting at the root (in particular the root is labeled by "1"). A leaf of a tree is a non-parent vertex, i.e., a vertex without children. It is important to note that the number of leafs in each tree in T n k is constant and equals (n − 1)k + 1. Indeed, using induction on k we see that for k = 0 the only tree in T n 0 has 0 children, so the root is the only leaf; each tree T n k can be obtained from a tree t ∈ T n k−1 by adding n leafs to a certain leaf of t, so the number of leafs increases by (n − 1). Now we count the cardinality of T n k in two different ways.
We proceed by induction on k ∈ N. For k = 0 there is only one tree, so the statement is correct. Assume #T n k = ((n − 1)(k − 1) + 1) · ((n − 1)(k − 2) + 1) · · · n. Each tree in T n k+1 comes from the unique tree t in T n k by adding label "k + 1" and k vertices to a leaf of t. Since the number of leafs of t is equal (n − 1)k + 1 we obtain the result.
Proof. First of all observe that for k ∈ N each tree in T n k+1 is uniquely given by n trees t 1 ∈ T n l 1 , . . . , t n ∈ T n ln such that l 1 + · · · + l n = k, and a partition of the set {2, . . . , k + 1} into n disjoint sets I 1 , . . . , I n of the cardinality #I i = l i for i = 1, . . . , n (we do not assume where I(l) is the set of all partitions of the set {2, . . . , k + 1} into n disjoint sets I 1 , . . . , I n s.t. #I i = l i for i = 1, . . . , n. Indeed, the root of a given tree t ∈ T n k+1 has n child vertices v 1 , . . . , v n . Each v i is the root of a certain maximal subtreet i of t. We assume thatt i has l i parent vertices, which are labeled by some numbers 2 ≤ a 1 i < · · · < a l i i ≤ k + 1. Obviously, l 1 + · · ·+ l n = k. Changing the label "a j i " into a label "j" we obtain a tree t i ∈ T n l i . Defining I i = a 1 i , . . . , a l i i for i = 1, . . . , n, we have a partition of {2, . . . , k + 1} into n disjoint sets, i.e., {2, . . . , k + 1} = I 1 ∪ · · · ∪ I n . It is clear how to invert this procedure in order to get its uniqueness.
Using the above correspondence it is easy to establish the formula in the lemma since there are k! l 1 !···ln! possible partitions of the set {2, . . . , k + 1} into n disjoint parts I 1 , . . . , I n such that #I i = l i ∈ N, i.e., #I(l) = k! l 1 !···ln! . We are now ready to prove Proposition 2.3.
Clearly, there are n i different multi-indices (l 1 , . . . , l n ) satisfying this condition, and this is the reason for the Newton symbol to disappear in formula (3.2).
Remark 3.3. The above proof can be simplified for n = 0, 1 when #Ŝ k ≤ 1, but also for n = 2. In this case the recursive formula (2.6) gives Assuming the inductive hypothesis #Ŝ l = l! for l ≤ k we obtain

Examples
In this section we discuss the three simplest cases when n = 0, 1, 2, and the case where only u 0 and u n are not vanishing. We will need one additional intuitive notation. Namely, for S ∈ R A such that (S|1) = 0 we define the shuffle exponent where we recall that S ¡0 = 1 and S ¡k = S ¡ S ¡(k−1) .

(4.1)
This expression looks nice, but there is a problem since a 0 factor is on the left hand side and therefore the expression will not simplify if we apply Υ t to it. In order to obtain the solution in a common form, we prove the following lemma.
Proof. Observe first that for k, l ∈ N a k 1 ¡ (a l 1 a 0 ) = a l 1 a 0 a k 1 + l + 1 1 a l+1 1 a 0 a k−1 1 + · · · + l + k k a l+k 1 a 0 . Indeed, for k = 0 the formula is correct. Using the inductive hypothesis for each m ≤ k, and the defining formula for shuffle product (2.3) we get Since a k+1 , we obtain formula (4.2).
Using the above proved formula we see that ∞ k,l=0 where in the last line we change a method of summation by putting k ′ = k − m and l ′ = l + m. Since the expression in the squared brackets equals 0 l ′ , the sum over l ′ reduces to the one summand with l ′ = 0, and therefore ∞ k,l=0 This ends the proof.
From the lemma it follows that Since Υ t : R A → R is a shuffle-algebra homomorphism, it follows that Υ t (exp ¡ (S)) = exp(Υ t (S)) for all series S such that (S|1) = 0, and thereforê which is the standard formula. For n = 2 the equation under consideration iṡ that is a general Riccati equation. In this case, the seriesŜ is the unique solution of and thereforeŜ = kŜ k , whereŜ k are given by the recurrencê Let us mention that the Riccati equation is a Lie-Scheffers system of the type a 1 (see [7,3] and [17,18]). More precisely, if we take vector fields on R, then they satisfy the following commutation relations It means the vector fields span a simple Lie algebra of the type a 1 (isomorphic to sl(2, R)), and thus (4.3) -equivalent toẋ(t) = u i (t)X i -is a Lie-Scheffers system of this type. The solution in terms of iterated integrals of u i 's for such a system was given in [16]. Let us recall the main theorems of this article.
Theorem 3 (Theorem 1 in [16]). Let X a , X b , X c ∈ Γ(M) be smooth tangent vector fields on a manifold M satisfying [X a , is of the form and S a 1 ∈ R A is the unique solution of the algebraic equation Theorem 4 (Theorem 2 in [16]). For fixed measurable functions u a , u b , u c : [0, T ] → R the function Ξ a : [0, T ] → R, defined in Theorem 3 by Ξ a (t) = Υ t (a · exp ¡ (2S a 1 )), is (locally) the solution of the Riccati equation: Observe that taking X a = X 0 , X b = X 1 , X c = −X 2 (and therefore c = −a 2 ), and u a = u 0 , u b = u 1 , u c = u 2 , and Ξ a (t) = x(t) the system (4.3) can be put into the context of the above theorems in the following way. From Theorem 4 we conclude that the solution of (4.3) is x(t) = Υ t (Ŝ), whereŜ = a 0 · exp ¡ (2S a 1 ), and S a 1 ∈ R A is, by Theorem 3, the unique solution of the algebraic equation  In the discussed article, there was also given a recursive formula for S a 1 , but observe that in fact the algebraic equation (4.4) forŜ is simpler than the equation (4.8) for S a 1 . In consequence, it is reasonable to invert this statement saying that the series S a 1 is given by (4.9), whereŜ is the solution of (4.4). Now using Theorem 2 we get the following corollary about the a 1 -type Lie-Scheffers system considered in Theorem 3. Proof. By Theorem 1 the function Υ t (Ŝ) is well defined for 0 ≤ t < min {T, 1/M }. The above observations (in particular formula (4.9)) implies that Υ t (S a 1 ) is also well defined in this interval. Finally, by formulas (4.7) each function Ξ d (t) : Let us observe that in each of the discussed cases the solution is of the formŜ = a 0 · exp ¡ (L), where L ∈ R A such that (L|1) = 0, and the series L in case n reduces to L in case n − 1 if taking u n ≡ 0. Indeed, L = 0 for n = 0, L = a 1 for n = 1, and L = S a 1 for n = 2 reduces, by (4.8), to a 1 for u 2 ≡ 0. This observation suggests a question: if the same holds for all n ∈ N? Since Riccati equation is essentially the only differential equation on a real line which is connected with the action of a group (namely the special linear group SL(2)) [5,4] one could anticipate that a generalization is impossible. Nevertheless the problem is open.
Another example we are going to consider is the one where there are only two nonvanishing summands, i.e.,ẋ(t) = n m u(t)x m (t) + u n (t)x n (t), x(0) = 0, and 0 ≤ m < n are fixed. The case m = 0 has the trivial solution x(t) ≡ 0, so in fact we consideṙ with n ≥ 1 fixed.
Let us write the first few components of the expansion given in the above proposition.

Comparison with the Chen-Fliess approach
In this section we compare the number of non-zero iterated integrals in two approaches: the one given in this article, and the Chen-Fliess one. Recall that in the letter approach [9] we assume we have a differential equatioṅ x(t) = u 0 (t) X 0 (x(t)) + u 1 (t) X 1 (x(t)) + · · · + u n (t) X n (x(t)), where X i (x) = x i ∂ ∂x , for i = 0, . . . , n, are differential vector fields on R. The solution is given by where for v = a i 1 · · · a i k ∈ A * we define X v (x)(0) := X i 1 · · · X i k (x)(0) as a composition of vector fields acting on the function h(x) = x and evaluated at the initial value x 0 = 0. Since X i (x)(0) = 0 only for i = 0 the sum can be significantly reduced. Our aim is to eliminate all unnecessary summands. Since the second derivative ∂ 2 ∂x 2 x = 0 we need to compute X v (x)(0) modulo the second and higher derivatives.
On the k-th step of approximation there are #M 0 (k) non-trivial integrals to compute. If we assume n = ∞ one can compute that these are Catalan numbers, i.e., #M 0 (k) = 1 k+1 2k k . The first ten of these numbers are 1, 2, 5, 14, 42, 132, 429, 1430, 4862. We see that this growth is much faster than the growth 1, 2, 3, 5, 7, 11, 15, 22, 30, 42 of integrals needed to compute the k-th step of the expansion in our approach, as we mentioned in Remark 2.2.

Concluding remarks
In the article we formulated a scheme for expanding a solution of a general nonautonomous polynomial differential equation. The time dependent homogeneous parts of the expansion were expressed in terms of iterated integrals. The formula for each of this part was given recursively by (2.6). The advantage of our approach is that it is made on algebraic level. We use the shuffle product which is an algebraic analogue of multiplication of iterated integrals. Therefore, the algebraic formula can be easily transformed into the analytic one giving the expansion of the solution of the initial problem, as we stated in Theorem 1.
Finally, there is some work to be done. One way of a development is to write an explicit formula for the algebraic seriesŜ preferably with the use of shuffle product. It would also be important to find a deeper algebraic structure of this solution. Another way is to rewrite the scheme for systems of non-autonomous polynomial differential equations and estimate the radius of convergence in this case. This is important for example to integrate higher order Lie-Sheffers systems.