We devote this section to prove Theorem 1.1. We borrow ideas from [8]. However, the fact that we are dealing with a system instead of a single equation forces us to develop a significantly different proof. We will organize the proof in several lemmas.

Our first lemma proves part (I) of Theorem 1.1.

**Lemma 4.1**. *If* ${\prod}_{l=1}^{k}\left(1+{p}_{l}\right)\ge {\prod}_{l=1}^{k}2{q}_{l}$ (*i*.*e*. det *A* ≥ 0), *every nonnegative solution of (1.1)-(1.3) is global in time*.

**Proof**. It is enough to construct global supersolutions with initial data as large as needed. We achieve this with the aid of the self-similar solutions of exponential form that we found in Theorem 2.3.

First we choose

${\stackrel{\u0303}{q}}_{1}\ge {q}_{1}$ such that

$2{\stackrel{\u0303}{q}}_{1}{\prod}_{l=2}^{k}2{q}_{l}={\prod}_{l=1}^{k}\left(1+{p}_{l}\right)$ and we let

${\u016b}_{i}\left(x,t\right)={e}^{{\alpha}_{i}\left(t+\tau \right)}{f}_{i}\left(x{e}^{-{\beta}_{i}\left(t+\tau \right)}\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,$

where *α*_{1} > 0 is arbitrary and *β*_{1}, *α*_{
i
} , *β*_{
i
} , (*i* = 2, ..., *k*) are given by (2.7). Now we observe that $\left({\u016b}_{1},{\u016b}_{2},\dots ,{\u016b}_{k-1},{\u016b}_{k}\right)$ is a supersolution of (1.1)-(1.3) as long as ${\u016b}_{1}\left(0,t\right)\ge 1$. This can be done by choosing *τ* large enough. This also allows to assume ${\u016b}_{i}\left(x,0\right)\ge {u}_{0i}\left(x\right)\left(i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k\right)$. Then, by the comparison Lemma 3.1, we obtain that every solution is global.

Now we construct subsolutions with finite time blow-up.

**Lemma 4.2**.

*Let* ${\prod}_{l=1}^{k}\left(1+{p}_{l}\right)<{\prod}_{l=1}^{k}2{q}_{l}$ **(i**.

*e*. det

*A* < 0),

*then there exist compactly supported functions g*_{
i
} *(i* = 1, 2, ...,

*k), such that*${\underset{}{u}}_{i}\left(x,t\right)=(T-t{)}^{{\alpha}_{i}}{g}_{i}\left({\xi}_{i}\right),\phantom{\rule{2.77695pt}{0ex}}{\xi}_{i}=x{\left(T-t\right)}^{-{\beta}_{i}},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,$

*is a subsolution of (1.1)-(1.2)*.

**Proof**. To satisfy (3.2) and (3.3), we need that

$\begin{array}{c}{\left({g}_{i}^{{p}_{i}}\right)}^{\u2033}\left({\xi}_{i}\right)\ge -{\alpha}_{i}{g}_{i}\left({\xi}_{i}\right)+{\beta}_{i}{\xi}_{i}{{g}^{\prime}}_{i}\left({\xi}_{i}\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,\\ -{\left({g}_{i}^{{p}_{i}}\right)}^{\prime}\left(0\right)\le {g}_{i+1}^{{q}_{i+1}}\left(0\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{q}_{k+1}={q}_{1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{g}_{k+1}={g}_{1}.\end{array}$

We choose

${g}_{i}\left({\xi}_{i}\right)={A}_{i}{\left({a}_{i}-{\xi}_{i}\right)}_{+}^{1\u2215\left({p}_{i}-1\right)},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k.$

Inserting this in the equation, we get

$\frac{{p}_{i}}{{\left({p}_{i}-1\right)}^{2}}{A}_{i}^{{p}_{i}-1}\ge -{\alpha}_{i}{\left({a}_{i}-{\xi}_{i}\right)}_{+}-\frac{{\beta}_{i}}{{p}_{i}-1}{\xi}_{i}\phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}0\le {\xi}_{i}\le {a}_{i},\phantom{\rule{1em}{0ex}}i=1,2,\dots ,k.$

Hence, it is enough to impose

$\frac{{p}_{i}}{{\left({p}_{i}-1\right)}^{2}}{A}_{i}^{{p}_{i}-1}\ge -{\alpha}_{i}{a}_{i}+\frac{\left|{\beta}_{i}\right|}{{p}_{i}-1}{a}_{i},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,$

that is

${C}_{i}{A}_{i}^{{p}_{i}-1}\ge {a}_{i},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k.$

(4.1)

The boundary conditions impose

$\frac{{p}_{i}}{{p}_{i}-1}{A}_{i}^{{p}_{i}}{a}_{i}^{1\u2215\left({p}_{i}-1\right)}\ge {A}_{i+1}^{{q}_{i+1}}{a}_{i+1}^{{q}_{i+1}\u2215\left({p}_{i+1}-1\right)},\phantom{\rule{1em}{0ex}}i=1,2,\dots ,k,\phantom{\rule{2.77695pt}{0ex}}{A}_{k+1}={A}_{1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{q}_{k+1}={q}_{1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{a}_{k+1}={a}_{1}.$

(4.2)

Let

${b}_{i}={A}_{i}{a}_{i}^{1\u2215\left({p}_{i}-1\right)},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k.$

Then conditions (4.2) become

$\frac{{p}_{i}}{{p}_{i}-1}{A}_{i}^{{p}_{i}-1}{b}_{i}\ge {b}_{i+1}^{{q}_{i+1}},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,\phantom{\rule{2.77695pt}{0ex}}{b}_{k+1}^{{q}_{k+1}}={b}_{1}^{{q}_{1}}.$

(4.3)

We fix *b*_{
i
} = 1 (*i* = 1, 2, ⋯, *k*) and then *A*_{
i
} large enough (and thus *a*_{
i
} small) to satisfy (4.1) and (4.3).

**Corollary 4.1** *Let* ${\prod}_{l=1}^{k}\left(1+{p}_{l}\right)<{\prod}_{l=1}^{k}2{q}_{l}$ **(i.e**. det *A* < 0). *Then there exist solutions of (1.1)-(1.3) that blow up in a finite time*.

**Proof**. We only have to apply Lemma 3.1, to obtain that every solution (*u*_{1}, ⋯, *u*_{
k
} ) that begins above the subsolutions provided by Lemma 4.2 has finite time blow-up.

**Lemma 4.3** *Let*${\prod}_{l=1}^{k}\left(1+{p}_{l}\right)<{\prod}_{l=1}^{k}2{q}_{l}$*(i.e*. det *A* < 0). *If there exists j (* 1 ≤ *j* ≤ *k) such that α*_{
j
} + *β*_{
j
} ≤ 0, *then every nontrivial solution of (1.1)-(1.3) blows up in finite time*.

**Proof**. Without loss of generality, we consider the case *α*_{1} + *β*_{1} ≤ 0.

Assume that there exists a global nonnegative nontrivial solution of (1.1)-(1.3), we make the following change of variables

${\phi}_{i}\left({\xi}_{i},\tau \right)=(1+t{)}^{-{\alpha}_{i}}{u}_{i}\left({\xi}_{i}{\left(1+t\right)}^{{\beta}_{i}},t\right),\phantom{\rule{2.77695pt}{0ex}}\tau =log\left(1+t\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k.$

(4.4)

These functions satisfy

${\phi}_{i\tau}=({\phi}_{i}^{{p}_{i}}{)}_{{\xi}_{i}{\xi}_{i}}+{\beta}_{i}{\xi}_{i}{\phi}_{i{\xi}_{i}}-{\alpha}_{i}{\phi}_{i},\phantom{\rule{1em}{0ex}}i=1,2,\dots ,k,$

(4.5)

$-{\left({\phi}_{i}^{{p}_{i}}\right)}_{{\xi}_{i}}\left(0,\tau \right)={\phi}_{i+1}^{{q}_{i+1}}\left(0,\tau \right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,\phantom{\rule{2.77695pt}{0ex}}{\phi}_{k+1}^{{q}_{k+1}}={\phi}_{1}^{{q}_{1}}.$

(4.6)

As *u*_{
i
} (*x*, *t*) (*i* = 1, 2, ..., *k*) are by hypothesis global, the same is true for *φ*_{
i
} (*i* = 1, 2, ..., *k*,). We will construct a solution $\left({\hat{\phi}}_{1},\dots ,{\hat{\phi}}_{k}\right)$ to system (4.5)-(4.6) increasing with time, with initial data $\left({\hat{\phi}}_{01},\dots ,{\hat{\phi}}_{0k}\right)$ such that ${\hat{\phi}}_{0i}\left({\xi}_{i}\right)\le {u}_{i}\left({\xi}_{i},0\right)\phantom{\rule{2.77695pt}{0ex}}\left(i=1,2,\dots ,k\right)$. We will prove that $\left({\hat{\phi}}_{1},\dots ,{\hat{\phi}}_{k}\right)$ cannot exists globally, thus contradicting the global existence of (*u*_{1}, ⋯, *u*_{
k
} ). In order to achieve our goal, we use an adaptation for systems of the general monotonicity for single quasilinear equation described in [19].

We take initial data

$\left({\hat{\phi}}_{01},\dots ,{\hat{\phi}}_{0k}\right)$ satisfying

${\left({\hat{\phi}}_{0i}^{{p}_{i}}\right)}_{{\xi}_{i}{\xi}_{i}}+{\beta}_{i}{\xi}_{i}{\left({\hat{\phi}}_{0i}\right)}_{{\xi}_{i}}-{\alpha}_{i}{\hat{\phi}}_{0i}\ge 0,\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,$

and the compatibility conditions

$-{\left({\hat{\phi}}_{0i}^{{p}_{i}}\right)}_{{\xi}_{i}}\left(0\right)={\hat{\phi}}_{0i+1}^{{q}_{i+1}}\left(0\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k,\phantom{\rule{2.77695pt}{0ex}}{\hat{\phi}}_{0k+1}^{{q}_{k+1}}={\hat{\phi}}_{01}^{{q}_{1}}.$

Hence, arguing as in Lemma 3.2, we have that ${\hat{\phi}}_{i\tau}\ge 0\phantom{\rule{2.77695pt}{0ex}}\left(i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k\right)$.

Following an idea for scalar equation from [

8], we set

${\hat{\phi}}_{01}\left({\xi}_{1}\right)=h\left({\xi}_{1}+b\right),$

where

*h* is the Barenblatt profile

$h\left({\xi}_{1}\right)={a}_{{p}_{1}}{\left(c-{\xi}_{1}^{2}\right)}_{+}^{1\u2215\left({p}_{1}-1\right)}.$

Then we have

$\begin{array}{cc}\hfill {\left({\hat{\phi}}_{01}^{{p}_{1}}\right)}_{{\xi}_{1}{\xi}_{1}}+{\beta}_{1}{\xi}_{1}{\left({\hat{\phi}}_{01}\right)}_{{\xi}_{1}}\hfill & \hfill -{\alpha}_{1}{\hat{\phi}}_{01}=-\frac{1}{{p}_{1}+1}b{h}_{{\xi}_{1}}\left({\xi}_{1}+b\right)\hfill \\ \hfill +\left({\beta}_{1}-\frac{1}{{p}_{1}+1}\right){\xi}_{1}{h}_{{\xi}_{1}}\left({\xi}_{1}+b\right)+\left(-{\alpha}_{1}-\frac{1}{{p}_{1}+1}\right)h\left({\xi}_{1}+b\right).\hfill \end{array}$

The last expression is nonnegative if *β*_{1} - 1/(*p*_{1} + 1) ≤ 0 and -*α*_{1} - 1/(*p*_{1} + 1) ≥ 0. But these two conditions are equivalent *α*_{1} + *β*_{1} ≤ 0.

Now we take

${\stackrel{\u0303}{\alpha}}_{i},\phantom{\rule{2.77695pt}{0ex}}{\stackrel{\u0303}{\beta}}_{i}>0$ such that

${\stackrel{\u0303}{\alpha}}_{i}\ge {\alpha}_{i},\phantom{\rule{2.77695pt}{0ex}}{\stackrel{\u0303}{\beta}}_{i}\ge {\beta}_{i}\left(i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k\right)$. We take as

${\hat{\phi}}_{0i}$ a solution to

${\left({\hat{\phi}}_{0i}^{{p}_{i}}\right)}^{\u2033}=-{\stackrel{\u0303}{\beta}}_{i}{\xi}_{i}{\hat{\phi}}_{0i}^{\prime}+{\stackrel{\u0303}{\alpha}}_{i}{\hat{\phi}}_{0i},\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k.$

There is one-parameter family of solution to this equation (see Theorem 2.4), with

${\hat{\phi}}_{0i}\ge 0,\phantom{\rule{2.77695pt}{0ex}}{\hat{{\phi}^{\prime}}}_{0i}\le 0\left(i=2,\cdot \cdot \cdot ,k\right)$. Hence,

${\left({\hat{\phi}}_{0i}^{{p}_{i}}\right)}^{\u2033}\ge -{\beta}_{i}{\xi}_{i}{\hat{\phi}}_{0i}^{\prime}+{\alpha}_{i}{\hat{\phi}}_{0i},\phantom{\rule{1em}{0ex}}i=2,\dots ,k.$

Moreover,

${\hat{\phi}}_{0i}\left(0\right)={U}_{i},\phantom{\rule{2.77695pt}{0ex}}{\left({\hat{\phi}}_{0i}^{{p}_{i}}\right)}^{\prime}\left(0\right)={U}_{i}^{\left({p}_{i}+1\right)\u22152}{V}_{*i},\phantom{\rule{1em}{0ex}}i=2,\dots ,k,$

where *V*_{*i}< 0 is a constant and *U*_{
i
} is the free parameter.

We still have to control the boundary conditions. In order to do this, we choose the constants

*c*,

*b* and

*U*_{
i
} (

*i* = 2, ...,

*k*) conveniently. They have to satisfy

$\begin{array}{c}\frac{2{p}_{1}{a}_{{p}_{1}}}{{p}_{1}-1}b{\left(c-{b}^{2}\right)}^{1\u2215\left({p}_{1}-1\right)}={U}_{2}^{{q}_{2}},\phantom{\rule{2.77695pt}{0ex}}b\in \left(0,{c}^{1\u22152}\right),\\ -{V}_{*i}{U}_{i}^{\left({p}_{i}+1\right)\u22152}={a}_{{p}_{1}}^{{q}_{i+1}}{\left(c-{b}^{2}\right)}^{{q}_{i+1}\u2215\left({p}_{1}-1\right)},\phantom{\rule{2.77695pt}{0ex}}i=2,\dots ,k.\end{array}$

Thus, we choose

$\begin{array}{c}{U}_{2}={c}_{2}{b}^{2{q}_{3}\u2215\left(2{q}_{2}{q}_{3}-{p}_{2}-1\right)},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{U}_{i}={c}_{i}{b}^{2\left({p}_{2}-1\right){q}_{i+1}}\u2215\left(\left({p}_{i}+1\right)\left(2{q}_{2}{q}_{3}-{p}_{2}-1\right)\right),\left(i=3,\dots ,k,\phantom{\rule{2.77695pt}{0ex}}{q}_{k+1}={q}_{1}\right),\\ c={b}^{2}+\gamma {b}^{\left({p}_{1}-1\right)\left({p}_{2}-1\right)\u2215\left(2{q}_{2}{q}_{3}-{p}_{2}-1\right)},\end{array}$

where *c*_{
i
} (*i* = 2, ..., *k*) and *γ* are positive constants. Taking *b* small enough, the initial data $\left({\hat{\phi}}_{01},\dots ,{\hat{\phi}}_{0k}\right)$ is below (*u*_{1}(*ξ*_{1},0), ...,*u*_{
k
} (*ξ*_{
k
} , 0)). This can be done as *u*_{0i}(*i* = 1, 2, ... *k*) can be assumed to be positive at the origin.

To conclude the proof, we will show that $\left({\hat{\phi}}_{1},\dots ,{\hat{\phi}}_{k}\right)$ converge to a self-similar profile that does not exist in this range of parameters.

**Lemma 4.4**.

*There exists j (* 1 ≤

*j* ≤

*k) such that*$\underset{\tau \to \infty}{lim}{\hat{\phi}}_{j}\left({\xi}_{j},\tau \right)={\stackrel{\u0303}{\phi}}_{j}\left({\xi}_{j}\right)<\infty ,\phantom{\rule{2.77695pt}{0ex}}\forall {\xi}_{j}>0.$

(4.7)

**Proof**. It is clear that

${\hat{\phi}}_{i{\xi}_{i}}\le 0\phantom{\rule{2.77695pt}{0ex}}\left(i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k\right)$. Let us suppose that

${\hat{\phi}}_{i}\left({\xi}_{i},\tau \right)\to \infty \phantom{\rule{2.77695pt}{0ex}}uniformly\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}\left(0,{\xi}_{i0}\right),\phantom{\rule{1em}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\dots ,k.$

In the original variables

$\left({\hat{u}}_{1},\dots ,{\hat{u}}_{k}\right)$, we have that for any

*M* > 0 there is a value such that

${\left(1+{t}_{0}\right)}^{{\alpha}_{i}}M\le {\hat{u}}_{i}\left(x,{t}_{0}\right)\phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}0<x{\left(1+{t}_{0}\right)}^{-{\beta}_{i}}<{\xi}_{i0},\phantom{\rule{2.77695pt}{0ex}}i=1,\phantom{\rule{2.77695pt}{0ex}}2,\cdot \cdot \cdot ,k.$

(4.8)

Now we will check that, under these conditions, we can put one of the blowing up subsolutions constructed in Lemma 4.2 below these data. This would lead to a contradiction, as

$\left({\hat{u}}_{1},\dots ,{\hat{u}}_{k}\right)$ is global. In order to do this, we need

$\begin{array}{c}{\left(1+{t}_{0}\right)}^{{a}_{1}}M\ge {A}_{1}{a}_{1}^{1\u2215\left({p}_{1}-1\right)}{T}^{{\alpha}_{1}},\\ {\xi}_{10}{\left(1+{t}_{0}\right)}^{{\beta}_{1}}\ge {a}_{1}{T}^{{\beta}_{1}}.\end{array}$

(4.9)

The first equation says that the height at

*x* = 0 of

${\hat{u}}_{1}$ is bigger than that of

${\underset{}{u}}_{1}$, and the second says that the support of

${\hat{u}}_{1}$ is bigger than the support of

${\underset{}{u}}_{1}$. Imposing analogous conditions for

${\hat{u}}_{i}$ and

${\underset{}{u}}_{i}\left(i=2,\dots ,k\right)$ we get

$\begin{array}{c}{\left(1+{t}_{0}\right)}^{{a}_{i}}M\ge {A}_{i}{a}_{i}^{1\u2215\left({p}_{i}-1\right)}{T}^{{\alpha}_{i}},\\ {\xi}_{i0}{\left(1+{t}_{0}\right)}^{{\beta}_{i}}\ge {a}_{i}{T}^{{\beta}_{i}}.\end{array}$

(4.10)

Taking *T* = 1 + *t*_{0}, then *a*_{
i
} small enough and *A*_{
i
} large enough (*i* = 1, 2, ..., *k*), and then *M* large, then the 2*k* conditions (4.9)-(4.10) are fulfilled.

Let us remark this parametric evolution comparison method to prove global non-existence for arbitrary data first introduced in [20], for scalar quasilinear heat equation.

**End of the proof of Lemma 4.3**. Let us assume that (4.7) holds. Using standard arguments, see [

8], we may pass to the limit to obtain that

${\left({\stackrel{\u0303}{\phi}}_{1}^{{p}_{1}}\right)}_{{\xi}_{1}{\xi}_{1}}+{\beta}_{1}{\xi}_{1}{\stackrel{\u0303}{\phi}}_{1{\xi}_{1}}-{\alpha}_{1}{\stackrel{\u0303}{\phi}}_{1}=0.$

(4.11)

Let

$z={\stackrel{\u0303}{\phi}}_{1}^{{p}_{1}}$, then

${z}_{{\xi}_{1}{\xi}_{1}}+\frac{{\beta}_{1}}{{p}_{1}}{\xi}_{1}{z}^{\left(1-{p}_{1}\right)\u2215{p}_{1}}{z}_{{\xi}_{1}}\le 0.$

Hence, in (0,

*ξ*_{10}),

*z* ≥

*c* > 0,

${z}_{{\xi}_{1}{\xi}_{1}}\le C{z}_{{\xi}_{1}}.$

We conclude that

*z* and therefore

${\stackrel{\u0303}{\phi}}_{1}$ cannot be unbounded at

*ξ*_{1} = 0. In particular,

$0<{\stackrel{\u0303}{\phi}}_{1}\left(0\right)\le C.$ Then, considering the regularity of

${\stackrel{\u0303}{\phi}}_{1}$ in the region where

${\stackrel{\u0303}{\phi}}_{1}>0$, we can pass to the limit in the boundary condition for

${\left({\hat{\phi}}_{1}^{{p}_{1}}\right)}_{{\xi}_{1}}$ to obtain that

$-{\left({\stackrel{\u0303}{\phi}}_{1}^{{p}_{1}}\right)}_{{\xi}_{1}}\left(0\right)={\stackrel{\u0303}{\phi}}_{2}^{{q}_{2}}\left(0\right).$

(4.12)

However, as *α*_{1} + *β*_{1} ≤ 0, problem (4.11)-(4.12) does not have a nontrivial solution, see Theorem 2.4.

If (4.7) holds for some

*j* > 1, we can proceed as before to obtain that

${\stackrel{\u0303}{\phi}}_{j}\left(0\right)<\infty $. Thus, we can pass to the limit in the boundary condition for

${\hat{\phi}}_{j}$, obtaining

$-{\left({\stackrel{\u0303}{\phi}}_{j}^{{p}_{j}}\right)}_{{\xi}_{j}}\left(0\right)={\stackrel{\u0303}{\phi}}_{j+1}^{{q}_{j+1}}\left(0\right).$

As ${\stackrel{\u0303}{\phi}}_{j+1}\left(0\right)\ge {\stackrel{\u0303}{\phi}}_{j+1}\left({\xi}_{j+1}\right)$, this implies that ${\stackrel{\u0303}{\phi}}_{j+1}$ is finite for every *ξ*_{j+1}≥ 0. We get the same contradiction as before.