Positive blow-up solutions of nonlinear models from real world dynamics

where $k\in (1,\mathrm{\infty })$.

Models in the form of (1) arise in many applications. Among others, they occur in the study of phase transitions of Van der Waals fluids [1–3], in population genetics, where they characterize the spatial distribution of the genetic composition of a population [4, 5], in the homogeneous nucleation theory [6], in relativistic cosmology for particles which can be treated as domains in the universe [7], and in the nonlinear field theory, in particular, in context of bubbles generated by scalar fields of the Higgs type in Minkowski spaces [8].

Here, we assume that h is positive and satisfies the Carathéodory conditions on $[0,T]\times [0,\mathrm{\infty })$. We define a positive solution of (1) as a function v which satisfies (1) for a.e. $t\in [0,T]$, is positive on $(0,T)$, and has absolutely continuous first derivative on each compact subinterval in $(0,T]$.

In this case, we speak about a positive solution of problem ( 1 ), ( 2 ). Let us denote by $\mathcal{Z}$ the set of all positive solutions to (1), (2). Moreover, let $\mathcal{R}=\{v\in \mathcal{Z}:t^{k}v\text{ is bounded on }(0,T]\}$ and ${\mathcal{R}}_c=\{v\in \mathcal{R}:v^{\prime }(T)=-c\}$ for $c\ge 0$. Since $v^{\prime }(T)\le 0$ for each $v\in \mathcal{Z}$, it is obvious that $\mathcal{R}={\bigcup }_{c\ge 0}{\mathcal{R}}_c$.

If we denote such v by $v_{\alpha }$ and define $w_{\alpha }(t)=t^{k-1}v(t)$ for $t\in (0,T]$, and $w_{\alpha }(0)=\alpha$, then we find that the graphs of these functions do not intersect, cf. Theorem 8, and that for each ${\alpha }^{\ast }>{\alpha }_0$, the set $\{w_{\alpha }:{\alpha }_0\le \alpha \le {\alpha }^{\ast }\}$ is compact in $C[0,T]$; see Theorem 9.

The study of a structure of positive solutions to other types of ordinary differential equations can be found for example in [17–19].

where $a<-1$. For $k:=-a$ and $v(t):=t^{-k}u(t)$, (3) becomes a special case of (1) and therefore, results obtained for (3) apply for (1). For the further analysis, we assume that f satisfies the following conditions:

(H₁) $f\in Car([0,T]\times [0,\mathrm{\infty }))$,

(H₂) $0<f(t,x)$ for a.e. $t\in [0,T]$ and all $x\in [0,\mathrm{\infty })$,

holds. We call a function $u:[0,T]\to \mathbb{R}$ a positive solution of the Dirichlet problem (3), (4) if $u\in A{C}^1[0,T]$, $u>0$ on $(0,T)$, u satisfies the boundary conditions (4), and (3) holds for a.e. $t\in [0,T]$. Clearly, for each positive solution u of (3), (4), there exists $c\ge 0$ such that (5) is satisfied.

We now denote by $\mathcal{S}$ the set of all positive solutions of problem (3), (4), and ${\mathcal{S}}_c=\{u\in \mathcal{S}:u^{\prime }(T)=-c\}$, where $c\ge 0$.

In the following lemma we cite those results from [20] which will be used in the analysis of problem (1), (2).

in the set $C^1[0,T]$.

We now formulate new results which complete those from [20]. We first establish a relation between ${\mathcal{S}}_0$ and the set $\{(t,x)\in {\mathbb{R}}^2:0\le t\le T,u_{0,min}(t)\le x\le u_{0,max}(t)\}$ if its interior is nonempty. This question was a short time ago an open problem [[20], Remark 4.4]. We note that the relation between ${\mathcal{S}}_c$ and the set $\{(t,x)\in {\mathbb{R}}^2:0\le t\le T,u_{c,min}(t)\le x\le u_{c,max}(t)\}$ with $c>0$ having nonempty interior is described in Lemma 1(d).

Theorem 1 Let (H₁)-(H₃) hold. Let us assume that there exists $t_0\in (0,T)$ such that $u_{0,min}(t_0)<u_{0,max}(t_0)$. Then for any $Q\in (u_{0,min}(t_0),u_{0,max}(t_0))$ there exists $u\in {\mathcal{S}}_0$ satisfying $u(t_0)=Q$.

Proof Step 1. Auxiliary Dirichlet problem.

and taking the limit $t\to T$ we obtain^a $u_{0,min}^{\prime }(T)\ge v^{\prime }(T)\ge u_{0,max}^{\prime }(T)$, which together with $u_{0,min}^{\prime }(T)=u_{0,max}^{\prime }(T)=0$ gives $v^{\prime }(T)=0$.

Consequently, $|v^{\prime }(t)|<S$ for $t\in [t_0,T]$, and so $\chi (v^{\prime })$ = 1 on this interval. Thus, v is a solution of problem (3), (7).

Step 2. Continuation of the solution v.

and either $u(\nu )=u_{0,max}(\nu )$ or $u(\nu )=u_{0,min}(\nu )$. Assume that $u(\nu )=u_{0,max}(\nu )$.

which is not possible. The case $u(\nu )=u_{0,min}(\nu )$ can be discussed analogously. Hence, (14) holds.

holds. This is a contradiction.

Therefore $\gamma =0$, and then (14) yields ${lim}_{t\to 0^{+}}u(t)=0$. Consequently $J=[0,T]$, and the assertion of the theorem follows from (13). □

In the next corollary we extend the statement (d) from Lemma 1 to a large set of A values.

Corollary 1 For each $t_0\in (0,T)$ and each $A\ge u_{0,min}(t_0)$ there exists $u\in \mathcal{S}$ satisfying $u(t_0)=A$.

Proof The result follows immediately from Lemma 1(d) and Theorem 1. □

By Lemma 1(b), (c) we know that functions from $\mathcal{S}$ are uniquely determined by the values −c of their derivatives at the right end point $t=T$ only in the case that ${\mathcal{S}}_c$ is a singleton set for each $c\ge 0$. Since we cannot uniquely determine all functions from $\mathcal{S}$ via their derivatives at $t=T$, see Lemma 1(e), we discuss their derivatives at the singular point $t=0$.

and (15) follows. □

Proof The result follows from Lemma 2, since $u\in {\mathcal{S}}_{-u^{\prime }(T)}$. □

Corollary 3 Let $u,v\in \mathcal{S}$, $u\ne v$. Then $u^{\prime }(0)\ne v^{\prime }(0)$.

which together with Corollary 2 gives $u^{\prime }(0)<v^{\prime }(0)$. □

In particular, $u_{0,min}^{\prime }(0)>0$.

Proof Inequality (16) follows from Lemma 2 and the fact that $f(t,u(t))>0$ for a.e. $t\in [0,T]$ by (H₂). □

Combining the above two equalities yields the result. □

Lemma 3 Let (H₁)-(H₃) hold. Let $c\ge 0$ and ${\mathcal{S}}_c$ be not a singleton set. Then for each $\alpha \in [u_{c,min}^{\prime }(0),u_{c,max}^{\prime }(0)]$ there exists a unique $u\in {\mathcal{S}}_c$ such that $u^{\prime }(0)=\alpha$. Consequently, $[u_{c,min}^{\prime }(0),u_{c,max}^{\prime }(0)]=\{u^{\prime }(0):u\in {\mathcal{S}}_c\}$.

Since ${\mathcal{S}}_c$ is a compact set in $C^1[0,T]$ by Lemma 1(f), the set $I=\{u^{\prime }(0):u\in {\mathcal{S}}_c\}\subset [u_{c,min}^{\prime }(0),u_{c,max}^{\prime }(0)]$ is closed. In fact, let $\{{\beta }_n\}\subset I$, ${\beta }_n=u_n^{\prime }(0)$, where $u_n\in {\mathcal{S}}_c$, and let ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\beta$. Then there exist a subsequence $\{{\beta }_{k_n}\}$ of $\{{\beta }_n\}$ and $u\in {\mathcal{S}}_c$ such that ${lim}_{n\to \mathrm{\infty }}{u}_{k_n}=u$ in $C^1[0,T]$. In particular, ${\beta }_{k_n}=u_{k_n}^{\prime }(0)\to u^{\prime }(0)$ as $n\to \mathrm{\infty }$. Therefore, $\beta =u^{\prime }(0)\in I$.

It remains to prove that $I=[u_{c,min}^{\prime }(0),u_{c,max}^{\prime }(0)]$. Assume that the equality does not hold. Then, from the structure of bounded and closed subsets of ℝ the existence of an open interval $(\alpha ,\beta )\subset [u_{c,min}^{\prime }(0),u_{c,max}^{\prime }(0)]\setminus I$, $\alpha ,\beta \in I$, follows. Let $\alpha =u_{\alpha }^{\prime }(0)$ and $\beta =u_{\beta }^{\prime }(0)$, where $u_{\alpha },u_{\beta }\in {\mathcal{S}}_c$. Due to Lemma 1(c), there exists $t_0\in (0,T)$ such that $u_{\alpha }<u_{\beta }$ on $(0,t_0]$, $u_{\alpha }\le u_{\beta }$ on $(t_0,T]$. Choose $\tau \in (u_{\alpha }(t_0),u_{\beta }(t_0))$. By Corollary 1, there exists $v\in {\mathcal{S}}_c$ such that $v(t_0)=\tau$. Then $u_{\alpha }<v<u_{\beta }$ on $(0,t_0]$ and $u_{\alpha }\le v\le u_{\beta }$ on $(t_0,T]$. Consequently, $u_{\alpha }^{\prime }(0)<v^{\prime }(0)<u_{\beta }^{\prime }(0)$ on $(t_0,T]$, that is, $v^{\prime }(0)\in (\alpha ,\beta )$, which is not possible. □

Lemma 3 implies that functions from $\mathcal{S}$ can be uniquely determined by the values of their derivatives at the singular point $t=0$; see Theorem 2.

Theorem 2 Let (H₁)-(H₃) hold. Then there exists a unique $u\in \mathcal{S}$ satisfying $u^{\prime }(0)=\alpha$ if and only if $\alpha \in J$.

for each $K\ge 0$, where we set $u_{c,min}=u=u_{c,max}$ if ${\mathcal{S}}_c$ is a singleton set and ${\mathcal{S}}_c=\{u\}$. In view of Corollary 4, $u^{\prime }(0)>c/|a+1|$ for $u\in {\mathcal{S}}_c$. Consequently, (18) follows.

Let us now choose $\alpha \in J$. Then there exists $u\in \mathcal{S}$ satisfying $u^{\prime }(0)=\alpha$. The uniqueness of u follows from Corollary 3. □

For $\alpha \in J$, we denote by $u_{\alpha }$ the unique element of $\mathcal{S}$ satisfying $u_{\alpha }^{\prime }(0)=\alpha$.

Theorem 3 Let (H₁)-(H₃) hold. Assume that $\{{\alpha }_n\}\subset J$ is a convergent sequence and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=\alpha$. Then ${lim}_{n\to \mathrm{\infty }}{u}_{{\alpha }_n}=u_{\alpha }$ in $C^1[0,T]$.

Proof Let $K=sup\{-u_{{\alpha }_n}^{\prime }(T):n\in \mathbb{N}\}$. Then $K<\mathrm{\infty }$ because ${\alpha }_n=u_{{\alpha }_n}^{\prime }(0)>-u_{{\alpha }_n}^{\prime }(T)/|a+1|$ for $n\in \mathbb{N}$ by Corollary 4. Since $\{u_{{\alpha }_n}\}\subset {\bigcup }_{0\le c\le K}{\mathcal{S}}_c$ and ${\bigcup }_{0\le c\le K}{\mathcal{S}}_c$ is compact in $C^1[0,T]$ by Lemma 1(f), the sequence $\{u_{{\alpha }_n}\}$ is relatively compact in $C^1[0,T]$. Let $\{u_{{\alpha }_{{\ell }_n}}\}$ be a subsequence of $\{u_{{\alpha }_n}\}$ which is convergent in $C^1[0,T]$, and let ${lim}_{n\to \mathrm{\infty }}{u}_{{\alpha }_{{\ell }_n}}=u$. Then $u\in {\mathcal{S}}_{c_0}$ for a $c_0\in [0,K]$ and $u^{\prime }(0)={lim}_{n\to \mathrm{\infty }}{u}_{{\alpha }_{{\ell }_n}}^{\prime }(0)={lim}_{n\to \mathrm{\infty }}{\alpha }_{{\ell }_n}=\alpha$. Therefore, $u=u_{\alpha }$ and hence any subsequence $\{u_{{\alpha }_{{\ell }_n}}\}$ of $\{u_{{\alpha }_n}\}$ converging in $C^1[0,T]$ has the same limit $u_{\alpha }$. Consequently, $\{u_{{\alpha }_n}\}$ is convergent in $C^1[0,T]$ and $u_{\alpha }$ is its limit. □

This section contains the main results of the paper. First, we present a lemma which describes how positive solutions of (1) may behave at the singular point $t=0$.