Characterizing the Codimension of Zero Singularities for Time-Delay Systems

Abstract

The analysis of time-delay systems mainly relies on detecting and understanding the spectral values bifurcations when crossing the imaginary axis. This paper deals with the zero singularity, essentially when the zero spectral value is multiple. The simplest case in such a configuration is characterized by an algebraic multiplicity two and a geometric multiplicity one, known as the Bogdanov-Takens singularity. Moreover, in some cases the codimension of the zero spectral value exceeds the number of the coupled scalar-differential equations. Nevertheless, to the best of the author’s knowledge, the bounds of such a multiplicity have not been deeply investigated in the literature. It is worth mentioning that the knowledge of such an information is crucial for nonlinear analysis purposes since the dimension of the projected state on the center manifold is none other than the sum of the dimensions of the generalized eigenspaces associated with spectral values with zero real parts. Motivated by a control-oriented problems, this paper provides an answer to this question for time-delay systems, taking into account the parameters’ algebraic constraints that may occur in applications. We emphasize the link between such a problem and the incidence matrices associated with the Birkhoff interpolation problem. In this context, symbolic algorithms for LU-factorization for functional confluent Vandermonde as well as some classes of bivariate functional Birkhoff matrices are also proposed.

Appendix

In this section, we first summarize the main notations in Table 1. Then, for the sake of self-containment, we report some results selected from the literature. Finally, some useful auxiliary lemmas are presented and proved. The proofs of Theorems 4.4 and 4.6 are provided.

Table 1 Table of the main notations

Here, we report some useful results from the mentioned literature. The main theorem from [16] emphasizes the link between $\mathbf{card}(\chi_{+})$ and $\mathbf{card}(\chi_{0})$, both take into account the multiplicity.

Theorem 8.1

(Hassard [16, p. 223])

Consider the quasipolynomial function $\Delta$ defined by (4). Let $\rho_{1},\ldots,\rho_{r}$ be the positive roots of $\mathcal{R}(y)= \Re(i^{n} \Delta(i y))$, counted by their multiplicities and ordered so that $0<\rho_{1}\leq\cdots\leq\rho_{r}$. For each $j=1,\ldots,r$ such that $\Delta(i \rho_{j})=0$, assume that the multiplicity of $i\rho_{j}$ as a zero of $\Delta(\lambda,\tau)$ is the same as the multiplicity of $\rho_{j}$ as a root of $\mathcal{R}(y)$. Then $\mathbf{card}(\chi_{+})$ is given by the formula:

$$ \mathbf{card}(\chi_{+})=\frac{n-\mathbf{card}(\chi_{0})}{2}+ \frac{(-1)^{r}}{2}\operatorname{sgn}\mathcal{I}^{(\mu)}(0)+\sum _{j=1} ^{r} \operatorname{sgn}\mathcal{I}( \rho_{j}), $$

(41)

where $\mu$ designate the multiplicity of the zero spectral value of $\Delta(\lambda,\tau)=0$ and $\mathcal{I}(y)=\Im(i^{-n}\Delta(iy))$. Furthermore, $\mathbf{card}( \chi_{+})$ is odd (respectively, even) if $\Delta^{(\mu)}(0)<0$ $(\Delta^{(\mu)}(0)>0)$. If $\mathcal{R}(y)=0$ has no positive zeros, set $r=0$ and omit the summation term in the expression of $\mathbf{card}(\chi_{+})$. If $\lambda=0$ is not a root of the characteristic equation, set $\mu=0$ and interpret $\mathcal{I}^{(0)}(0)$ as $\mathcal{I}(0)$ and $\Delta^{(0)}(0)$ as $\Delta(0)$.

The following result from [15] gives a valuable information allowing to have a first estimation on the bound for the codimension of the zero spectral value.

Proposition 8.2

(Pólya-Szegö [15, p. 144])

Let $\tau_{1}, \ldots, \tau_{N}$ denote real numbers such that

$$ \tau_{1}< \tau_{2}< \cdots < \tau_{N}, $$

and $d_{1}, \ldots, d_{N}$ positive integers satisfying

$$ d_{1}\geq1, d_{2}\geq1 \ldots d_{N}\geq1, \qquad d_{1}+d_{2}+ \cdots+d_{N}=D+N. $$

Let $f_{i,j}(s)$ stands for the function $f_{i,j}(s)=s^{j-1} e^{\tau _{i} s}$, for $1\leq j\leq d_{i}$ and $1\leq i\leq N$.

Let $\sharp$ be the number of zeros of the function

$$ f(s)=\sum_{1\leq i\leq N, 1\leq j\leq d_{i}}c_{i,j} f_{i,j}(s), $$

that are contained in the horizontal strip $\alpha\leq\mathcal{I}(z) \leq\beta$.

Assuming that

$$ \sum_{1\leq k\leq d_{1}}|c_{1,k}|>0, \ldots, \sum _{1\leq k\leq d _{N}}|c_{N,k}|>0, $$

then

$$ \frac{ ( \tau_{N}-\tau_{1} ) ( \beta-\alpha ) }{2 \pi}-D+1\leq\sharp\leq\frac{ ( \tau_{N}-\tau_{1} ) ( \beta-\alpha ) }{2 \pi}+D+N-1. $$

Setting $\alpha=\beta=0$, the above proposition allows to $\sharp_{\mathit{PS}}\leq D+N-1$ where $D$ stands for the sum of the degrees of the polynomials involved in the quasipolynomial function $f$ and $N$ designate the associated number of polynomials. This gives a sharp bound in the case of complete polynomials.

In the sequel, we present some useful lemmas as well as the proofs of the claimed theorems.

Lemma 1

Zero is a root of $\Delta^{(k)}(\lambda)$ for $k\geq0$ if, and only if, the coefficients of $P_{M^{j}}$ for $0\leq j\leq\tilde{N}_{N,n}$ satisfy the following assertion

$$ a_{0,k}=-\sum_{i\in S_{N,n}} \Biggl[ a_{{i,k}}+\sum_{l=0}^{k-1} { \frac{a_{ {i,l}}{\sigma_{{i}}}^{k-l}}{ ( k-l ) !}} \Biggr]. $$

(A.1)

Proof

We define the family $\nabla_{k}$ for all $k\geq0$ by

$$ \nabla_{k}(\lambda)=\sum _{i=0}^{\tilde{N}_{N,n}}{\frac{d ^{k}}{d{\lambda}^{k}}}P_{{M^{i}}} ( \lambda ) +\sum_{j=0} ^{k-1} \Biggl( {k \choose j}\sum_{i=1}^{\tilde{N}_{N,n}}{ \sigma_{{i}}} ^{k-j}{\frac{d^{j}}{d{\lambda}^{j}}} {P}_{{M^{i}}} ( \lambda ) \Biggr), $$

(A.2)

here, $M^{0}\triangleq0$ and ${\frac{d^{0}}{d{\lambda}^{0}}}f( \lambda)\triangleq f(\lambda)$. Obviously, the defined family $\nabla_{k}$ is polynomial since $P_{i}$ and their derivatives are polynomials. Moreover, zero is a root of $\Delta^{(k)}(\lambda)$ for $k\geq0$ if, and only if, zero is a root of $\nabla_{k}(\lambda)$. This can be proved by induction. More precisely, differentiating $k$ times $\Delta(\lambda,\tau)$ the following recursive formula is obtained:

$$ \Delta^{(k)}(\lambda)=\sum _{i=0}^{\tilde{N}_{N,n}}{\frac{d^{k}}{d {\lambda}^{k}}}P_{{M^{i}}} ( \lambda ) e^{\sigma_{i}\lambda } +\sum_{j=0}^{k-1} \Biggl( {k\choose j}\sum_{i=1}^{\tilde{N}_{N,n}} { \sigma_{{i}}}^{k-j}{\frac{d^{j}}{d{\lambda}^{j}}} { P}_{{M^{i}}} ( \lambda ) e^{\sigma_{i}\lambda} \Biggr). $$

Since only the zero root is of interest, we can set $e^{\sigma_{i} \lambda}=1$ which define the polynomial functions $\nabla_{k}$. Moreover, careful inspection of the obtained quantities presented in (A.2) and substituting ${\frac{d^{k}}{d{\lambda}^{k}}}P_{{i}}(0)=k! a_{i,k}$ leads to the formula (A.1). □

Here, we prove the results given in Sect. 4.2.1, that is, we consider the incidence vector:

$$ \mathcal{V}=(\underbrace{x_{1},\ldots,x_{1}}_{d_{1}}, \underbrace{ \star,\ldots,\star}_{d_{*}},x_{2}). $$

The right hand side of the last equality from (23) defining $U_{i,d_{1}+1}$ for $2\leq i\leq d_{1}+1$ can be also written as follows.

Lemma 2

For $2\leq i\leq d_{1}+1$ the following equality is satisfied:

$$ \varUpsilon_{i,d_{1}+1}-(i-1) \int_{0}^{x_{1}} U_{i-1,d_{1}+1}(y,x_{2})dy= \sum_{k=0}^{i-1}{i-1\choose k}(-1)^{i-1-k}x_{1}^{i-1-k} \varUpsilon_{k+1,d_{1}+1}. $$

Proof of Lemma 2

First, one has $U_{2,d_{1}+1}= \varUpsilon_{2,d_{1}+1}-x_{1} \varUpsilon_{1,d_{1}+1}=\varUpsilon_{2,d_{1}+1}- \int_{0}^{x_{1}}U_{1,d_{1}+1}(y, x_{2})dy$ since $U_{1,d1+1}= \varUpsilon_{1,d_{1}+1}(x_{2})$.

Now, let assume that for $2\leq i\leq p$ where $p< d_{1}+1$ the following equality is satisfied:

$$ \sum_{l=0}^{i-1}{i-1\choose l} (-1)^{i-1-l} x_{1}^{i-1-l} \varUpsilon_{l+1,d_{1}+1}= \varUpsilon_{i,d_{1}+1}-(i-1) \int_{0}^{x_{1}}U _{i-1,d_{1}+1}(y,x_{2}) dy. $$

One has to show that for $i=p+1$:

$$ \sum_{l=0}^{p}{p\choose l} (-1)^{p-l} x_{1}^{p-l} \varUpsilon_{l+1,d_{1}+1}= \varUpsilon_{p+1,d_{1}+1}-(p) \int_{0}^{x_{1}}U _{p,d_{1}+1}(y,x_{2}) dy. $$

Indeed,

$$ \textstyle\begin{cases} \displaystyle-\int_{0}^{x_{1}}p U_{p,d_{1}+1}(y,x_{2}) dy=- \int_{0}^{x_{1}}p \sum_{l=0}^{p-1}{p-1 \choose l} (-1)^{p-1-l} s^{p-1-l} \varUpsilon_{l+1,d_{1}+1} ds, \\ \displaystyle\phantom{-\int_{0}^{x_{1}}p U_{p,d_{1}+1}(y,x_{2}) dy}=-\sum_{l=0}^{p-1}\frac{p!}{l! (p-l-1)!} (-1)^{p-1-l} \varUpsilon_{l+1,d_{1}+1} \int_{0}^{x_{1}}s^{p-1-l}ds, \\ \displaystyle\phantom{-\int_{0}^{x_{1}}p U_{p,d_{1}+1}(y,x_{2}) dy}=\sum_{l=0}^{p-1}{p\choose l} (-1)^{p-l} x_{1}^{p-l} \varUpsilon_{l+1,d_{1}+1}. \end{cases} $$

 □

Proof of Theorem 4.4

The only difference between algorithms (23) and (20) lies in definition of the last column of the matrix $U$. Thus, one has to show that for any $2\leq i\leq d_{1}+1$ the following equality holds $\varUpsilon_{i,d_{1}+1}=\sum_{k=1}^{i}L_{i,k}U _{k,d_{1}+1}$. By definition, one has:

$$ \textstyle\begin{cases} \displaystyle\varUpsilon_{2,d_{1}+1} =\sum_{k=1}^{2} L_{2,k} U_{k,d_{1}+1} \\ \phantom{\varUpsilon_{2,d_{1}+1}}=L_{2,1} U_{1,d_{1}+1}+L_{2,2} U_{2,d_{1}+1} \\ \phantom{\varUpsilon_{2,d_{1}+1}}= x_{1} \varUpsilon_{1,d_{1}+1}+U_{2,d_{1}+1}. \end{cases} $$

(42)

Now, let assume that for $2\leq i\leq p$ where $p< d_{1}+1$ the following equality is satisfied:

$$ U_{i,d_{1}+1}=\varUpsilon_{i,d_{1}+1}-(i-1) \int_{0}^{x_{1}}U_{i-1,d_{1}+1}(y,x _{2}) dy, $$

or equivalently, from Lemma 2

$$ U_{i,d_{1}+1}=\sum_{l=0}^{i-1}{i-1 \choose l} (-1)^{i-1-l} x _{1}^{i-1-l} \varUpsilon_{l+1,d_{1}+1}. $$

It stills to show that the last equality from (23) holds for $U_{p+1,d_{1}+1}$ when $p< d_{1}+1$. Indeed, by definition

$$ U_{p+1,d_{1}+1}=\varUpsilon_{p+1,d_{1}+1}-\sum_{k=1}^{p}L_{p+1,k}U_{k,d _{1}+1}. $$

Moreover (for same arguments as the ones given in the proof of Lemma 6 presented in the sequel), one has $L_{p+1,k}= \frac{1}{k-1}\frac{\partial L_{p+1,k-1}}{\partial x_{1}}$. Thus, $L_{p+1,k}=\frac{1}{(k-1)!}\frac{\partial^{k-1} L_{p+1,1}}{\partial x _{1}^{k-1}}=\frac{1}{(k-1)!}\frac{\partial^{k-1}x_{1}^{p} }{\partial x_{1}^{k-1}}=\frac{p! x_{1}^{p-k+1}}{(p-k+1)! (k-1)!}$. So that, one has:

$$ L_{p+1,k}={p\choose k-1} x_{1}^{p-(k-1)}. $$

(43)

Now, by definition of $U_{p+1,d_{1}+1}$ and using (43) as well as the recurrence assumption, we obtain

$$ \textstyle\begin{cases} \displaystyle U_{p+1,d_{1}+1}=\varUpsilon_{p+1,d_{1}+1}- \sum_{\l=1}^{p}L_{p+1,\l}U _{\l,d_{1}+1} \\ \displaystyle\phantom{U_{p+1,d_{1}+1}} =\varUpsilon_{p+1,d_{1}+1}-\sum_{\l=1}^{p} \sum_{l=0}^{k-1} {\l-1\choose l} {p\choose \l-1} (-1)^{\l-l-1} x_{1}^{\l-l-1} x _{1}^{p-(\l-1)} \varUpsilon_{l+1,d_{1}+1} \\ \displaystyle \phantom{U_{p+1,d_{1}+1}} =\varUpsilon_{p+1,d_{1}+1}-\sum_{\l=1}^{p} \sum_{l=0}^{\l-1} {\l-1\choose l} {p\choose \l-1} (-1)^{\l-1-l} x_{1}^{p-l} \varUpsilon_{l+1,d_{1}+1} \end{cases} $$

Thus, one has to prove that

$$\begin{aligned} \sum_{k=0}^{p-1}{p\choose k} (-1)^{p-k} x_{1}^{p-k} \varUpsilon_{k+1,d_{1}+1}=- \sum_{\l=1}^{p}\sum _{l=0}^{\l-1} {\l-1\choose l} {p\choose \l-1} (-1)^{\l-1-l} x_{1}^{p-l} \varUpsilon_{l+1,d_{1}+1}. \end{aligned}$$

(44)

Recall that, the two side expressions of (44) are polynomials in $x_{1}$ and $x_{2}$. The only quantities depending in $x_{2}$ are $(\varUpsilon_{k,d_{1}+1})_{1\leq k\leq p}$. Since, $\deg( \varUpsilon_{k,d_{1}+1})\neq\deg(\varUpsilon_{k',d_{1}+1})$ for $k\neq k'$, it will be enough to we examine the equality of coefficients of the two side expressions in $\varUpsilon_{m+1,d_{1}+1}$ for arbitrarily chosen $0\leq m\leq p-1$. So that, let $m=k_{0}$ for which corresponds $m=l_{0}$ in the right hand side quantity from (44). Then consider the coefficient of $x_{1}^{p-m} \varUpsilon_{m+1,d_{1}+1}$ from the two sides of (44). Now, one easily check that $\sum_{\l=m}^{p}{\l-1\choose m} {p\choose \l-1} ( -1 ) ^{\l-m}= ( -1 ) ^{p-m}{p\choose m}$ is always satisfied, which ends the proof. □

In what follow, we propose some lemmas exhibiting some interesting properties of functional Birkhoff matrices. Those will be useful for the analytical proof of Theorem 4.6.

Lemma 3

Equation (30) is equivalent to:

$$ U_{i,j}=\sum_{l=0}^{i-1}{i-1 \choose l} (-1)^{l} x_{1}^{l} \varUpsilon_{i-l,j} \quad\textit{for } j=d_{1}+d_{2}^{-}+1 \textit{ and } 2 \leq i\leq d_{1}+1. $$

(45)

Proof of Lemma 3

The equality (45) follows directly by induction. First, one checks that

$$ \varUpsilon_{2,d_{1}+d_{2}^{-}+1}=U_{2,d_{1}+d_{2}^{-}+1}+x_{1} \varUpsilon_{1,d_{1}+d_{2}^{-}+1}. $$

Indeed,

$$ \textstyle\begin{cases} \displaystyle\varUpsilon_{2,d_{1}+d_{2}^{-}+1}=\sum _{k=1}^{2} L_{2,k} U_{k,d_{1}+d _{2}^{-}+1} \\ \displaystyle\phantom{\varUpsilon_{2,d_{1}+d_{2}^{-}+1}} =L_{2,1} U_{1,d_{1}+d_{2}^{-}+1}+L_{2,2} U_{2,d_{1}+d_{2}^{-}+1} \\ \displaystyle\phantom{\varUpsilon_{2,d_{1}+d_{2}^{-}+1}} = x_{1} \varUpsilon_{1,d_{1}+d_{2}^{-}+1}+U_{2,d_{1}+d_{2}^{-}+1}, \end{cases} $$

(46)

since $L_{2,2}=1$. Now, let assume that

$$\begin{aligned} U_{i,j}=\sum_{l=0}^{i-1}{i-1 \choose l} (-1)^{l} x_{1}^{l} \varUpsilon_{i-l,j} \quad\text{for } j=d_{1}+d_{2}^{-}+1 \mbox{ and } 2 \leq i\leq p \mbox{ and }p< d_{1}+1. \end{aligned}$$

(47)

From Eq. (30) one has

$$ U_{p+1,d_{1}+d_{2}^{-}+1}=\varUpsilon_{p+1,d_{1}+d_{2}^{-}+1}-p \int _{0}^{x_{1}}U_{p,d_{1}+d_{2}^{-}+1}(y,x_{2})dy. $$

Using (47), one has,

$$ \textstyle\begin{cases} \displaystyle U_{p+1,d_{1}+d_{2}^{-}+1}\\ \displaystyle \quad= \varUpsilon_{p+1,d_{1}+d_{2}^{-}+1} \\ \displaystyle\qquad{}-p \int_{0}^{x_{1}} \Biggl( \varUpsilon_{p,d_{1}+d_{2}^{-}+1}(y,x_{2})+ \sum_{l=1}^{p-1}{{p-1}\choose l} (-1)^{l} y^{l} \varUpsilon_{p-l,d_{1}+d_{2}^{-}+1}(y,x_{2}) \Biggr) dy \\ \displaystyle\quad=\varUpsilon_{p+1,d_{1}+d_{2}^{-}+1}-p \varUpsilon_{p,d_{1}+d_{2}^{-}+1}x _{1}+ \sum_{l=1}^{p-1}p {{p-1}\choose l} (-1)^{l} \varUpsilon_{p-l,d_{1}+d_{2}^{-}+1} \int_{0}^{x_{1}}y^{l}dy \\ \displaystyle\quad =\sum_{l=0}^{p}{p\choose l} (-1)^{l} x_{1}^{l} \varUpsilon_{p+1-l,d_{1}+d_{2}^{-}+1}, \end{cases} $$

which ends the proof. □

Lemma 4

$$\begin{aligned} \varUpsilon_{i+1,j}=x_{2} \varUpsilon_{i,j}+\bigl(d_{2}^{-}+d^{*} \bigr) \int_{0} ^{x_{2}}\varUpsilon_{i,j}(y)dy \quad\textit{for } j=d_{1}+d_{2}^{-}+1 \textit{ and } 1\leq i\leq d_{1}+d_{2}^{-}. \end{aligned}$$

(48)

Proof of Lemma 4

Let consider the coalescence [50] confluent Vandermonde matrix $\hat{\varUpsilon}$ which regularize the considered Birkhoff matrix $\varUpsilon$. That is $\hat{\varUpsilon}$ is the rectangular matrix associated with the incidence matrix

$$ \mathcal{V}=(\underbrace{x_{1},\ldots,x_{1}}_{d_{1}}, \underbrace{x _{2},\ldots,x_{2}}_{d_{2}^{-}}, \underbrace{x_{2},\ldots,x_{2}}_{d _{*}},x_{2}). $$

Here, the “stars” ⋆ in (32) are simply replaced by $x_{2}$. Thus, $\varUpsilon$ and $\hat{\varUpsilon}$ have the same number of rows, but the number of columns of $\hat{\varUpsilon}$ exceeds the columns number of $\varUpsilon$ by $d^{*}$. We point out that $\varUpsilon_{i+1,d_{1}+d_{2}^{-}+1}$ is nothing but $\hat{\varUpsilon} _{i+1,d_{1}+d_{2}^{-}+1+d^{*}}$. This means that the term $(d_{2}^{-}+d ^{*}) \int_{0}^{x_{2}}\varUpsilon_{i,j}$ in (48) is exactly $\hat{\varUpsilon}_{i+1,d_{1}+d_{2}^{-}+d^{*}}$. Thus, equality (48) turns to be

$$ \bar{\varUpsilon}_{i+1,j}=x_{2} \bar{\varUpsilon}_{i,j}+ \bar{\varUpsilon} _{i,j-1} \quad\text{for } j=d_{1}+d_{2}^{-}+1+d^{*} \mbox{ and } 1 \leq i\leq d_{1}+d_{2}^{-}. $$

This last equality can be easily proved by using a 2-D recurrence in terms of $\bar{\varUpsilon}$ (regular matrix) as in the proof of Theorem 4.1 to show that it applies even for $d_{1}+2\leq j\leq d _{1}+d_{2}^{-}+1+d^{*}$. □

The following lemma provides an other way defining the components of $U$ given by (30).

Lemma 5

For all $i=1,\ldots,d_{1}$ and $j=d_{1}+d_{2}^{-}+1$ the following equality applies

$$ U_{i+1,j^{*}}=(x_{2}-x_{1}) U_{i,j^{*}}+\bigl(d_{2}^{-}+d^{*}\bigr) \int_{0} ^{x_{2}}U_{i,j^{*}}(y)dy. $$

(49)

Proof of Lemma 5

Let set

$$\begin{aligned} \mathcal{I}_{k}=U_{k+1,j^{*}}+(x_{1}-x_{2}) U_{k,j^{*}}-\bigl(d_{2}^{-}+d ^{*}\bigr) \int_{0}^{x_{2}}U_{k,j^{*}}(y)dy, \end{aligned}$$

where $j^{*}=d_{1}+d_{2}^{-}+1+d^{*}$ and $1\leq k\leq d_{1}+1$.

Substitute Eq. (45) from Lemma 3 in $\mathcal{I}_{k}$, to obtain

$$\begin{aligned} \mathcal{I}_{k} =&\sum_{l=0}^{k}{k \choose l}(-1)^{l}x_{1}^{l} \varUpsilon_{k+1-l,j^{*}} \\ &{}-\sum_{l=0}^{k-1}{k-1\choose l}(-1)^{l}x_{1}^{l} \biggl( (x_{2}-x _{1})\varUpsilon_{k-l,j^{*}}+\bigl(d_{2}^{-}+d^{*} \bigr) \int_{0}^{x_{2}} \varUpsilon_{k-l,j^{*}}(y)dy \biggr). \end{aligned}$$

Using Lemma 4, one obtains

$$\begin{aligned} \mathcal{I}_{k} = &\sum_{l=1}^{k-1}(-1)^{l} x_{1}^{l} \biggl[ \biggl( {k\choose l}- {k-1\choose l} \biggr) \varUpsilon_{k+1-l,j^{*}}+x_{1}{k-1\choose l} \varUpsilon_{k-l,j^{*}} \biggr] \\ &{}+(-1)^{k} x_{1}^{k} \varUpsilon_{1,j^{*}}+x_{1} \varUpsilon_{k,j^{*}} \\ = &\sum_{l=1}^{k-1}(-1)^{l} x_{1}^{l} \biggl( {k-1\choose l-1} \varUpsilon_{k+1-l,j^{*}}+x_{1}{k-1 \choose l}\varUpsilon_{k-l,j^{*}} \biggr) +(-1)^{k} x_{1}^{k} \varUpsilon_{1,j^{*}} +x_{1} \varUpsilon_{k,j^{*}} \end{aligned}$$

which is as expected identically zero, that ends the proof. □

The following lemma provides a differential relation between the coefficients of $L$ matrix.

Lemma 6

For all $1\leq k\leq p$ the following equality holds

$$ \frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}=k L_{d_{1}+p,d _{1}+k+1} $$

(50)

The following result applies when dealing with $\varUpsilon_{i,j}$ and $\varUpsilon_{i,j-1}$ are in the same variable block. We emphasize that such a property is inherited by the expressions of $L$ defined in (28).

Proof of Lemma 6

The proof is 2-D recurrence-based. First, one easily check that for $p=2$ then $k=1$

$$ L_{d_{1}+2,d_{1}+2}=\frac{\partial L_{d_{1}+2,d_{1}+1}}{\partial x _{2}} $$

since by definition of $L$ one has $L_{d_{1}+2,d_{1}+1}=L_{d_{1}+1,d _{1}}+x_{2} L_{d_{1}+1,d_{1}+1}=L_{d_{1}+1,d_{1}}+x_{2}$ and $\frac{\partial L_{d_{1}+1,d_{1}}}{\partial x_{2}}=0$. When assuming that

$$ L_{d_{1}+p,d_{1}+2}=\frac{\partial L_{d_{1}+p,d_{1}+1}}{\partial x _{2}}, $$

and again, using the definition of $L$, one obtains,

$$\begin{aligned} L_{d_{1}+p+1,d_{1}+2} &=L_{d_{1}+p,d_{1}+1}+x_{2}L_{d_{1}+p,d_{1}+2}, \\ L_{d_{1}+p+1,d_{1}+1} &=L_{d_{1}+p,d_{1}}+x_{2}L_{d_{1}+p,d_{1}+1}, \end{aligned}$$

which as expected gives:

$$\begin{aligned} \frac{\partial L_{d_{1}+p+1,d_{1}+1}}{\partial x_{2}} &=L_{d_{1}+p,d _{1}+1}+x_{2}\frac{\partial L_{d_{1}+p,d_{1}+1}}{\partial x_{2}} \\ &=L_{d_{1}+p,d_{1}+1}+x_{2}L_{d_{1}+p,d_{1}+2}=L_{d_{1}+p+1,d_{1}+2}. \end{aligned}$$

Let assume that for any $2< p< d_{2}^{-}+1$ and $k=1,\ldots,p-1$ one has

$$ \frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}=k L_{d_{1}+p,d _{1}+k+1}. $$

One has to prove the following equalities:

$$ \textstyle\begin{cases} \displaystyle\frac{\partial L_{d_{1}+p+1,d_{1}+k}}{\partial x_{2}} =k L_{d_{1}+p+1,d _{1}+k+1}, \\ \displaystyle\frac{\partial L_{d_{1}+p,d_{1}+k+1}}{\partial x_{2}} =(k+1) L_{d _{1}+p,d_{1}+k+2}, \\ \displaystyle\frac{\partial L_{d_{1}+p+1,d_{1}+k+1}}{\partial x_{2}} =(k+1) L_{d _{1}+p+1,d_{1}+k+2}. \end{cases} $$

(51)

Let us consider the first equality of (51), using the definition of $L$ that asserts that

$$\begin{aligned} L_{d_{1}+p+1,d_{1}+k+1} &=L_{d_{1}+p,d_{1}+k}+x_{2}L_{d_{1}+p,d_{1}+k+1} \\ L_{d_{1}+p+1,d_{1}+k} &=L_{d_{1}+p,d_{1}+k-1}+x_{2}L_{d_{1}+p,d_{1}+k}. \end{aligned}$$

Which gives

$$\begin{aligned} \frac{\partial L_{d_{1}+p+1,d_{1}+k}}{\partial x_{2}} &=\frac{\partial L_{d_{1}+p,d_{1}+k-1}}{\partial x_{2}}+x_{2}\frac{\partial L_{d_{1}+p,d _{1}+k}}{\partial x_{2}}+L_{d_{1}+p,d_{1}+k} \\ &=(k-1)L_{d_{1}+p,d_{1}+k}+k L_{d_{1}+p,d_{1}+k+1}+L_{d_{1}+p,d_{1}+k} \\ &=k L_{d_{1}+p+1,d_{1}+k+1}. \end{aligned}$$

By the same way, the remaining two equality from (51) are obtained:

$$\begin{aligned} \frac{\partial L_{d_{1}+p,d_{1}+k+1}}{\partial x_{2}} &=\frac{\partial ( L_{d_{1}+p-1,d_{1}+k}+x_{2}L_{d_{1}+p-1,d_{1}+k+1} ) }{ \partial x_{2}} \\ &=\frac{\partial L_{d_{1}+p-1,d_{1}+k}}{\partial x_{2}}+x_{2}\frac{ \partial L_{d_{1}+p-1,d_{1}+k+1}}{\partial x_{2}}+L_{d_{1}+p-1,d_{1}+k+1} \\ &=kL_{d_{1}+p-1,d_{1}+k+1}+(k+1) x_{2}L_{d_{1}+p-1,d_{1}+k+2}+L_{d _{1}+p-1,d_{1}+k+1} \\ &=(k+1) L_{d_{1}+p,d_{1}+k+2}, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial L_{d_{1}+p+1,d_{1}+k+1}}{\partial x_{2}} &=\frac{\partial ( L_{d_{1}+p,d_{1}+k}+x_{2}L_{d_{1}+p,d_{1}+k+1} ) }{\partial x_{2}} \\ &=\frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}+x_{2}\frac{ \partial L_{d_{1}+p,d_{1}+k+1}}{\partial x_{2}}+L_{d_{1}+p,d_{1}+k+1} \\ &=kL_{d_{1}+p,d_{1}+k+1}+(k+1) x_{2}L_{d_{1}+p,d_{1}+k+2}+L_{d_{1}+p,d _{1}+k+1} \\ &=(k+1) L_{d_{1}+p+1,d_{1}+k+2}, \end{aligned}$$

that ends the proof. □

Proof of Theorem 4.6

The only change occurring in (26)–(31) compared with (20) is the way in defining the column $d_{1}+d_{2}^{-}+1$ of $U$. Moreover, such a column is only involved in computing the column $d_{1}+d_{2}^{-}+1$ of $\varUpsilon$. Thus, it stills to show that the equalities (30) and (31); this will be done by recurrence. Equation (30) follow directly by induction from Lemma 3.

Let us focus now on (31) and denote $j^{*}=d_{1}+d_{2}^{-}+1$. First, let us check that

$$\begin{aligned} U_{d_{1}+2,j^{*}}= &\bigl(d_{2}^{-}+d^{*}\bigr) \int_{0}^{x_{2}}U_{d_{1}+1,j ^{*}}(x_{1},y)dy. \end{aligned}$$

From the one side, using Lemma 4 one has

$$\begin{aligned} \varUpsilon_{d+2,j^{*}} &=x_{2} \varUpsilon_{d+1,j^{*}}+ \bigl(d_{2}^{-}+d^{*}\bigr) \int_{0}^{x_{2}}\varUpsilon_{d+1,j^{*}}(y)dy, \\ &=x_{2} \sum_{k=1}^{d_{1}+1}L_{d_{1}+1,k}U_{k,j^{*}}+ \bigl(d_{2}^{-}+d ^{*}\bigr) \int_{0}^{x_{2}}\sum_{k=1}^{d_{1}+1}L_{d_{1}+1,k}U_{k,j^{*}}(y) dy. \end{aligned}$$

Since by definition one has $L_{d_{1}+1,d_{1}+1}=1$ and $L_{d_{1}+1,k}=L _{d_{1}+1,k}(x_{1})$ for $k\in\{1,\ldots,d_{1}\}$ then

$$\begin{aligned} \varUpsilon_{d+2,j^{*}} = &x_{2}\sum_{k=1}^{d_{1}+1}L_{d_{1}+1,k}U_{k,j ^{*}}+ \bigl(d_{2}^{-}+d^{*}\bigr)\sum _{k=1}^{d_{1}}L_{d_{1}+1,k} \int_{0}^{x_{2}}U _{k,j^{*}}(y)dy \\ &{}+\bigl(d_{2}^{-}+d^{*}\bigr) \int_{0}^{x_{2}}U_{d_{1}+1,j^{*}}(y)dy. \end{aligned}$$

From the other side and by definition of $\varUpsilon$,

$$\begin{aligned} \varUpsilon_{d_{1}+2,j^{*}} &=\sum_{k=1}^{d_{1}+2}L_{d_{1}+2,k}U_{k,j^{*}}=U _{d_{1}+2,j^{*}}+\sum_{k=1}^{d_{1}+1}L_{d_{1}+2,k}U_{k,j^{*}}. \end{aligned}$$

To prove (31) for $i=d_{1}+2$ one has to prove that

$$\begin{aligned} \sum_{k=1}^{d_{1}+1}L_{d_{1}+2,k}U_{k,j^{*}}=x_{2} \sum_{k=1}^{d_{1}+1}L _{d_{1}+1,k}U_{k,j^{*}}+ \bigl(d_{2}^{-}+d^{*}\bigr)\sum _{k=1}^{d_{1}}L_{d_{1}+1,k} \int_{0}^{x_{2}}U_{k,j^{*}}(y)dy, \end{aligned}$$

or equivalently to prove

$$\begin{aligned} \sum_{k=1}^{d_{1}+1} ( L_{d_{1}+2,k}-x_{2} L_{d_{1}+1,k} ) U _{k,j^{*}}-\bigl(d_{2}^{-}+d^{*} \bigr)\sum_{k=1}^{d_{1}}L_{d_{1}+1,k} \int_{0} ^{x_{2}}U_{k,j^{*}}(y)dy=0 . \end{aligned}$$

(52)

Using Eq. (28), one obtain

$$\begin{aligned} &L_{d_{1}+2,k}-x_{2} L_{d_{1}+1,k}=L_{d_{1}+1,k-1}+ ( x_{1}-x_{2} ) L_{d_{1}+1,k}, \quad\text{for } k=1,\ldots,d_{1}, \\ &L_{d_{1}+2,d_{1}+1}-x_{2} L_{d_{1}+1,d_{1}+1}=L_{d_{1}+1,d_{1}}. \end{aligned}$$

Thus, the right hand side of (52) becomes

$$\begin{aligned} &\sum_{k=1}^{d_{1}}L_{d_{1}+1,k}U_{k+1,j^{*}}+(x_{1}-x_{2}) \sum_{k=1} ^{d_{1}}L_{d_{1}+1,k}U_{k,j^{*}} -\bigl(d_{2}^{-}+d^{*}\bigr)\sum _{k=1}^{d_{1}}L_{d_{1}+1,k} \int_{0}^{x_{2}}U _{k,j^{*}}(y)dy \\ &\quad= \sum_{k=1}^{d_{1}}L_{d_{1}+1,k} \biggl( U_{k+1,j^{*}}+(x_{1}-x_{2}) U _{k,j^{*}}- \bigl(d_{2}^{-}+d^{*}\bigr) \int_{0}^{x_{2}}U_{k,j^{*}}(y)dy \biggr). \end{aligned}$$

Lemma 5 asserts that for all $i=1,\ldots,d_{1}$ and $j=d_{1}+d_{2}^{-}+1$ one has

$$\begin{aligned} U_{k+1,j^{*}}+(x_{1}-x_{2}) U_{k,j^{*}}-\bigl(d_{2}^{-}+d^{*}\bigr) \int_{0} ^{x_{2}}U_{k,j^{*}}(y)dy=0 \end{aligned}$$

(53)

which implies that (31) applies for $i=d_{1}+2$.

Let assume now that, (31) is satisfied for $i=d_{1}+2, \ldots,d_{1}+p$ where $1< p< d_{2}^{-}+d^{*}$. It stills to prove that (31) is satisfied for $i=d_{1}+p+1$.

By the same argument as for $i=d_{1}+2$, one has

$$\begin{aligned} \varUpsilon_{d_{1}+p+1,j^{*}} =&x_{2} \varUpsilon_{d+p,j^{*}}+ \bigl(d_{2}^{-}+d ^{*}\bigr) \int_{0}^{x_{2}}\varUpsilon_{d+p,j^{*}}(y)dy, \\ =&x_{2}\sum_{k=1}^{d_{1}+p}L_{d_{1}+p,k}U_{k,j^{*}} \bigl(d_{2}^{-}+d^{*}\bigr) \sum _{k=1}^{d_{1}+p-1} \int_{0}^{x_{2}}L_{d_{1}+p,k}U_{k,j^{*}}(y)dy \\ &{}+\bigl(d_{2}^{-}+d^{*}\bigr) \int_{0}^{x_{2}}U_{d_{1}+p,j^{*}}(y)dy \\ =&x_{2}\sum_{k=1}^{d_{1}+p}L_{d_{1}+p,k}U_{k,j^{*}}+ \bigl(d_{2}^{-}+d^{*}\bigr) \sum _{k=1}^{d_{1}+p-1} \int_{0}^{x_{2}}L_{d_{1}+p,k}U_{k,j^{*}}(y)dy \\ &{}+(p-1) \int_{0}^{x_{2}}U_{d_{1}+p,j^{*}}(y)dy+ \bigl(d_{2}^{-}+d^{*}-p+1\bigr) \int_{0}^{x_{2}}U_{d_{1}+p,j^{*}}(y)dy \end{aligned}$$

From the other side, we obtain

$$\begin{aligned} \varUpsilon_{d_{1}+p+1,j^{*}} &=\sum_{k=1}^{d_{1}+p}L_{d_{1}+p+1,k}U_{k,j ^{*}}+U_{d_{1}+p+1,j^{*}}. \end{aligned}$$

Hence, we have to prove that

$$\begin{aligned} &\sum_{k=1}^{d_{1}+p}L_{d_{1}+p+1,k}U_{k,j^{*}}-x_{2} \sum_{k=1}^{d _{1}+p}L_{d_{1}+p,k}U_{k,j^{*}}- \bigl(d_{2}^{-}+d^{*}\bigr)\sum _{k=1}^{d_{1}+p-1} \int_{0}^{x_{2}}L_{d_{1}+p,k}U_{k,j^{*}}(y)dy \\ &\quad=(p-1) \int_{0}^{x_{2}}U_{d_{1}+p,j^{*}}(y)dy. \end{aligned}$$

Now, using the result from Lemma 5, one has to prove that

$$\begin{aligned} &\sum_{k=d_{1}+2}^{d_{1}+p}L_{d_{1}+p+1,k}U_{k,j^{*}}-x_{2} \sum_{k=d_{1}+2}^{d_{1}+p}L_{d_{1}+p,k}U_{k,j^{*}} \end{aligned}$$

(54)

$$\begin{aligned} &\qquad{}-\bigl(d_{2}^{-}+d^{*}\bigr)\sum _{k=d_{1}+1}^{d_{1}+p-1} \int_{0}^{x_{2}}L_{d _{1}+p,k}U_{k,j^{*}}(y)dy \end{aligned}$$

(55)

$$\begin{aligned} &\quad=(p-1) \int_{0}^{x_{2}}U_{d_{1}+p,j^{*}}(y)dy. \end{aligned}$$

(56)

Using Eq. (28), one obtains

$$\begin{aligned} &L_{d_{1}+p+1,k}-x_{2} L_{d_{1}+p,k}=L_{d_{1}+p,k-1}, \text{ for } k=d_{1}+2,\ldots,d_{1}+p. \end{aligned}$$

Finally, Eq. (54) becomes

$$\begin{aligned} E =&\sum_{k=1}^{p-1}L_{d_{1}+p,d_{1}+k}U_{d_{1}+k+1,j^{*}}- \bigl(d_{2}^{-}+d ^{*}\bigr)\sum _{k=1}^{p-1} \int_{0}^{x_{2}}L_{d_{1}+p,d_{1}+k}U_{d_{1}+k,j ^{*}}(y)dy \\ &{}-(p-1) \int_{0}^{x_{2}}U_{d_{1}+p,j^{*}}(y)dy=0. \end{aligned}$$

(57)

Differentiating $E$ given in (57) with respect to the variable $x_{2}$ one obtains

$$\begin{aligned} \frac{\partial E}{\partial x_{2}} = &\sum_{k=1}^{p-1} \biggl( \frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}U_{d_{1}+k+1,j^{*}}+L_{d_{1}+p,d _{1}+k}\frac{\partial U_{d_{1}+k+1,j^{*}}}{\partial x_{2}} \biggr) \\ &{}-\bigl(d_{2}^{-}+d^{*}\bigr)\sum _{k=1}^{p-1}L_{d_{1}+p,d_{1}+k}U_{d_{1}+k,j ^{*}}-(p-1)U_{d_{1}+p,j^{*}} \\ = &\sum_{k=1}^{p-1} \frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}U_{d_{1}+k+1,j^{*}} \\ &{}+\sum_{k=1}^{p-1}L_{d_{1}+p,d_{1}+k} \biggl( \frac{\partial U_{d_{1}+k+1,j ^{*}}}{\partial x_{2}}-\bigl(d_{2}^{-}+d^{*} \bigr)U_{d_{1}+k,j^{*}} \biggr) -(p-1)U _{d_{1}+p,j^{*}} \end{aligned}$$

By using the recurrence assumption, one obtains,

$$\begin{aligned} \frac{\partial E}{\partial x_{2}} = &\sum_{k=1}^{p-1} \frac{\partial L _{d_{1}+p,d_{1}+k}}{\partial x_{2}}U_{d_{1}+k+1,j^{*}}-(p-1)U_{d_{1}+p,j ^{*}} \\ &{}+\sum_{k=1}^{p-1}L_{d_{1}+p,d_{1}+k} \bigl( \bigl(d_{2}^{-}+d^{*}-(k-1)\bigr)U _{d_{1}+k,j^{*}}-\bigl(d_{2}^{-}+d^{*} \bigr)U_{d_{1}+k,j^{*}} \bigr) \\ = &\sum_{k=1}^{p-1} \frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}U_{d_{1}+k+1,j^{*}}- \sum_{k=2}^{p-1}(k-1) L_{d_{1}+p,d_{1}+k} U_{d_{1}+k,j^{*}}-(p-1)U _{d_{1}+p,j^{*}} \\ = &\sum_{k=1}^{p-2} \biggl( \frac{\partial L_{d_{1}+p,d_{1}+k}}{\partial x_{2}}-k L_{d_{1}+p,d_{1}+1+k} \biggr) U_{d_{1}+1+k,j^{*}} \\ &{}+ \biggl( \frac{\partial L_{d_{1}+p,d_{1}+p-1}}{\partial x_{2}}-(p-1) \biggr) U _{d_{1}+p,j^{*}}\equiv0, \end{aligned}$$

which is as expected zero since Lemma 6 asserts that each factor is identically zero, that ends the proof. □