1 Introduction

Coagulation and fragmentation events play a crucial role in the dynamics of cluster growth. Coagulation describes the phenomenon where a system of cluster combines together to form a larger one. While fragmentation describe the opposite case. Generally, the coagulation process is always a nonlinear event, whereas, fragmentation depends on the basis of the breakage behavior: (i) linear fragmentation (ii) collisional or nonlinear fragmentation. The linear fragmentation is ruled by the properties of the particle or by external forces and only smaller particles can be produced in this type of breakage process. However, collisional fragmentation occurs due to the collision between two particles. The process allows transfer of mass between the pair of particles, so that it can produce a particle larger than the colliding particles. Before going further it is instructive to discuss about the mathematical model considered in this work. Here we consider a closed system of particles undergoing binary collision such that there is no restriction on the number of daughter particle due to the collision. Moreover, the total volume of the daughter particles is equal to the sum of the volume of the colliding particles. For this case there are three following possible outcomes:

  • when one particle is produced, then coagulation takes place;

  • when two particles are produced, in this case, volume is shared among them;

  • and when three or more than three particles are produced, in this case breakage event takes place.

If the distribution of particles of volume x at time \(t \ge 0\) is given by f(x, t), then the mathematical formulation of the continuous coagulation due to collisional fragmentation model reads as;

$$\begin{aligned} \begin{aligned} \frac{\partial f(x,t)}{\partial t}&= \frac{1}{2} \int _0^x P(x-y,y)K(x-y,y)f(x-y,t)f(y,t)\mathrm {d}y\\&\quad - f(x,t) \int _0^\infty K(x,y)f(y,t)\mathrm {d}y \\&\quad + \frac{1}{2}\int _x^\infty \int _0^y b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,t)f(z,t)\mathrm {d}z \mathrm {d}y \end{aligned} \end{aligned}$$
(1.1)

here \(P_1(x,y) = 1-P(x,y)\) and supported with the initial data,

$$\begin{aligned} f(x,0)=f_{0}(x)\left( \ge 0\right) ,\quad \text{ for } \text{ all } \quad x>0. \end{aligned}$$
(1.2)

The function K(x, y) be the collision kernel which describe the rate at which particle of volumes x and y are colliding. P(x, y) describe the probability that the two colliding particles aggregate to form a single particle. Therefore it is clear that P(x, y)K(x, y) represent the coagulation kernel and \((1-P)(x,y)K(x,y)\) be the breakage kernel. In addition, it is clear that both the collision kernel K(x, y) and collision probability P(x, y) are symmetric, i.e. \(K(x,y)=K(y,x)\) and \(P(x,y)=P(y,x)\) with \(0 \le P(x,y) \le 1\) for all \((x,y) \in [0,\infty )^2:=\bar{\mathbb {R}}_+\). The term b(x|y, z) is the distribution function describing the expected number of particles of volume x produced from the breakage event arising from the collision of particles of volume y and z. This distribution function b has the following properties,

  1. (i)

    b(x|y, z) is nonnegative and symmetric with respect to y and z, that is \(b(x|y,z)=b(x|z,y)\ge 0\),

  2. (ii)

    the total number of particles resulting from the collisional breakage event is given by

    $$\begin{aligned} \int _0^{y+z} b(x|y,z)\mathrm {d}x = \bar{N}(\ge 2), \quad \text{ and }\quad b(x|y,z)=0\quad x>y+z; \end{aligned}$$
    (1.3)
  3. (iii)

    the necessary condition for mass conservation during collisional breakage events is

    $$\begin{aligned} \int _0^{y+z} xb(x|y,z)\mathrm {d}x =y+z \quad \text {for all } y>0,z>0. \end{aligned}$$
    (1.4)

From the condition (1.4) the total volume \(y+z\) of particles remains conserved during the collisional breakage of particles of volume y and z.

The first term in the right hand side of (1.1) represent the birth particle of volume x due to coagulation among the particles of volume \(x-y\) and y. The second term represent the disappearance of x volume particles from the system, And the last term represent the birth particle of volume x due to coagulation among the particles of volume \(y-z\) and z. The factor 1/2 multiplied with the first and third term to avoid the double counting of particles formation.

We also denote the \(r^{th}-\)moment of the number density f(x, t) by \(\mathcal {M}_r(t)\) which is defined by

$$\begin{aligned} \mathcal {M}_r(t) := \int _0^\infty x^r f\left( x,t\right) \mathrm {d}x ~ \text { where }r\ge 0. \end{aligned}$$

The first moment represent the total mass of the particles where the zeroth moment represent the total number of particles. It is worth to mention that due to collision breakage \(\mathcal {M}_0(t)\) increases where \(\mathcal {M}_0(t)\) decreases for the coagulation event. However some time due to rapid growth coagulation or breakage rate the gelation or shattering transition may appear in the system.

Coagulation with collision breakage has significant affect in the different fields of science and engineering studies such as communication system [1, 2], raindrop formation or deformation [3,4,5], several types of milling process [6], granulation and polymarization or depolymarization process [7,8,9,10] etc. From the theoretical perspective, the existence and uniqueness of solution to the classical coagulation fragmentation model have been studied by applying various mathematical technique [11,12,13]. Beside these analytic technique, there are some semi-analytic and numerical techniques are described in the articles [14,15,16]. However, there are very less number of articles available in the literature which are studied about the mathematical aspect of coagulation with collision breakage model. The model considered here was first described in the articles [17, 18]. The article [19] studies about the existence of mass conserving weak solution to the discrete form of the Eqs. (1.1)–(1.2) by the help of weak compactness technique. Recently, the existence and uniqueness result for the pure binary collisional breakage models in article [20]. The recent works [21, 22] studied the mathematical aspect of continuous coagulation with collisional breakage model, which is considered in the present work. In this regards, the article [22] deals with the weak solution to the considering model while article [21] discusses about the existence of strong solution with the help of some strong compactness theorem.

The focus of articles [21, 22] is to explore the existence result by using unbounded collisional kernel. However, the authors directly stated the existence of solution, when the kernel is compactly supported. Hence, the detailed derivation of the existence theorem for the compact supported kernels to the coagulation with collision breakage model are not available in the literature. Therefore, the present work can be assumed as a prequel to the articles [21, 22] that proves the existence theorem for unbounded or singular collisional kernel. The derivation of the main theorem of the present work, is based on the method used in article [23] that examines the existence to the coagulation with binary fragmentation for compactly supported kernel. Furthermore, large time behavior of the existing solution is investigated for a wide class of unbounded breakage function satisfying “power-law” rates [22].

The organization of the paper is in the following manner. In Sect. 2, we will state and prove the main existence theorem of mass conserving solution for the model considered in (1.1)–(1.2). Firstly, we will define our main functional space in the Sect. 2.1, which forms a Banach space and then in the Sect. 2.2 we will give the statement and proof for the existence of solution, when the collisional kernel is compactly supported. In the Sect. 2.3, we will show that the existing solution follows the mass conservation law. In Sect. 3, we will prove the uniqueness of the existing solution without any additional assumption. In Sect. 4 we investigate the behavior of the existing solution for a long time interval. Finally, future scopes of the present work are discussed in Sect. 5.

2 Existence of solution

2.1 Functional space

For a fixed \(T(>0)\), we consider the strip

$$\begin{aligned} \mathcal {X} := \left\{ (x,t) : 0<x<\infty , 0\le t\le T \right\} , \end{aligned}$$

and define \(\Omega _{r,\sigma } (T)\) to be the space of all continuous functions f with the bounded norm

$$\begin{aligned} \left\| f \right\| _\Omega := \sup _{0\le t\le T} \int _0^\infty \left( x^{r} + \frac{1}{x^{2\sigma }}\right) \left| f(x,t)\right| \mathrm {d}x, \quad r\ge 1,~ 0\le \sigma <1. \end{aligned}$$
(2.1)

Clearly, \(\Omega _{r,\sigma }(T)\) form a Banach space [12, 22]. Furthermore, let \(\Omega _{r,\sigma }^+(T)\) denotes the space of all non-negative functions from \(\Omega _{r,\sigma } (T)\).

The motivation behind using this space is because the earlier authors [21, 22] directly stated the existence of solution for compact supported kernel but not provide the detail derivation. In the subsequent section, we present the detail derivation of existence and uniqueness theorem of mass conserving solution for the coagulation with collisional fragmentation model (1.1)–(1.2) under the assumption that the collisional kernel have compact support.

2.2 Main theorem

Theorem 2.1

(Existence theorem for compactly supported kernels) Let the functions K(x, y), P(x, y) and b(x|y, z) are non-negative and continuous \(\mathbb {R}_+ \times \mathbb {R}_+\), \(\bar{\mathbb {R}}_+ \times \bar{\mathbb {R}}_+\) and \(\mathbb {R}_+ \times \mathbb {R}_+ \times \mathbb {R}_+\) respectively and in addition, the collision kernel K has compact support. If the initial data \(f_0(x)\) belongs to \(\Omega _{r,\sigma }^+(0)\cap \Omega _{0,0}^+(0)\) and the breakage function satisfy the following conditions,

$$\begin{aligned} \displaystyle {\int _0^{y+z} x^{-2\sigma }b(x|y,z)\mathrm {d}x} \le \nu (2\sigma )\left( y+z\right) ^{-2\sigma } \end{aligned}$$
(2.2)

where \(\nu (\ge 1)\) is a positive constant. Then the IVP (1.1)–(1.2) has at least one solution in \(\Omega _{r,\sigma }^+(T)\).

Let us take the following example of ‘power law kernels’ for the breakage function b;

$$\begin{aligned} b(x|y,z) = \gamma \frac{x^\beta }{\left( y+z\right) ^{\beta +1}} ~\text { where } \gamma >1 \text { and } -1< \beta \le 0. \end{aligned}$$

Now, the assumption (2.2) will be valid for \(0<2\sigma<1+\beta <\gamma\). Then,

$$\begin{aligned} \int _0^{y+z} x^{-2\sigma }b(x|y,z)\mathrm {d}x=\gamma \int _0^{y+z} \frac{x^{\beta -2\sigma }}{\left( y+z\right) ^{\beta +1}}= \frac{\gamma }{\beta +1-2\sigma }:=\nu (2\sigma ) \end{aligned}$$

For the above choice of \(\sigma\), we have \(\nu (2\sigma )>1\). Therefore, (2.2) is a realistic assumption on b(x|y, z) which satisfied by the power law kernels.

Proof

We prove the theorem in the following steps;

  • local existence of the solution, that is, there exists a \(\tau > 0\) such that the IVP (1.1)–(1.2) has at least one solution \(f \in \Omega _{r,\sigma }^+(\tau )\),

  • nonnegativity of the local solution,

  • global existence of the unique solution to the space \(\Omega _{r,\sigma }^+(T)\).

Existence of local solution Let there exists a fixed \(\mathcal {M}(>0)\), such that for each \(t\in [0,T]\) the kernels K(x, y) have compact supports in the intervals \(\left[ \frac{1}{\mathcal {M}},\mathcal {M}\right] \times \left[ \frac{1}{\mathcal {M}},\mathcal {M}\right]\). In accordance with Eq. (1.1), the truncated problem can be written as

$$\begin{aligned} f(x,t)&= f_0(x) + \int _0^t \left[ \frac{1}{2}\int _0^x P(x-y,y)K(x-y,y)f(x-y,s)f(y,s)\mathrm {d}y\right. \nonumber \\&\quad \left. - f(x,s)\int _0^\infty K(x,y)f(y,s)\mathrm {d}y \right. \nonumber \\&\quad \left. + \frac{1}{2}\int _x^\infty \int _0^y b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\mathrm {d}z \mathrm {d}y \right] \mathrm {d}s. \end{aligned}$$
(2.3)

Hence the solution to (1.1)–(1.2) for \(x>2\mathcal {M}\) takes the value

$$\begin{aligned} f(x,t) = f_0(x). \end{aligned}$$
(2.4)

The relation (2.4) gives an estimate of the solution function beyond the right hand side of the compact domain where, the tails of the solution f(x, t) actually does not change with time and coincides with the tails of the initial data \(f_0(x)\). We now proceed to prove the local existence of unique solution for \(0< x \le 2\mathcal {M}\).

In this regard, we define the integral operator \(\mathcal {G}\) as follows;

$$\begin{aligned} \mathcal {G}(f)(x,t) := \text{ right } \text{ hand } \text{ side } \text{ of } \text{ Eq. } (2.3). \end{aligned}$$
(2.5)

Since K has compact supports and \(f_0\) is a nonnegative continuous function therefore, the integral operator \(\mathcal {G}\) is well-defined on \(\Omega _{r,\sigma }(\tau )\). We prove this result by using the contraction mapping principle. Let us first prove that for small \(\tau > 0\) there exist a closed ball in \(\Omega _{r,\sigma }(\tau )\) which is invariant relatively to the mapping \(\mathcal {G}\). Let \(L_0 (>0)\) be a constant such that

$$\begin{aligned} \Vert f\Vert ^{(\tau )}_{\Omega } := \sup _{0 \le t \le \tau } \int _{0}^{\infty }\left( x^r + \frac{1}{x^{2\sigma }}\right) |f(x,t)|\mathrm {d}x \le L_0. \end{aligned}$$
(2.6)

Multiplying both sides of Eq. (2.3) with the weight \(\left( x^r + \frac{1}{x^{2\sigma }}\right)\) and integrating with respect to x, we get

$$\begin{aligned} \Vert \mathcal {G}(f)\Vert _{\Omega }^{(\tau )}&\le \Vert f_0\Vert _{\Omega }^{(\tau )}\nonumber \\&\quad + \int _{0}^{t}\left[ \frac{1}{2} \int _{0}^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) \int _0^x P(x-y,y)K(x-y,y)f(x-y,s)f(y,s) \mathrm {d}y\mathrm {d}x \right. \nonumber \\&\quad - \int _{0}^{\infty }\int _0^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) K(x,y)f(x,s)f(y,s) \mathrm {d}y\mathrm {d}x + \frac{1}{2}\int _{0}^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) \nonumber \\&\quad \times \left. \left[ \int _{x}^\infty \int _{0}^{y} b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\mathrm {d}z \mathrm {d}y \right] \mathrm {d}x\right] \mathrm {d}s. \end{aligned}$$
(2.7)

Now, for the third integral of right hand side we first change the order of the integration with respct to y and x then apply the Fubini’s theorem on the integral with respect to z and x

$$\begin{aligned}&\frac{1}{2}\int _{0}^{\infty }\int _{x}^\infty \int _{0}^{y} \left( x^r + \frac{1}{x^{2\sigma }}\right) \left[ b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\right] \mathrm {d}z \mathrm {d}y\mathrm {d}x \\&\le \frac{1}{2}\int _{0}^{\infty }\int _{0}^{y}\int _{0}^{y} \left( x^r + \frac{1}{x^{2\sigma }}\right) \left[ b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\right] \mathrm {d}z \mathrm {d}x\mathrm {d}y \\&= \frac{1}{2}\int _{0}^{\infty }\int _{0}^{y}\int _{0}^{y} x^r\left[ b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\right] \mathrm {d}x \mathrm {d}z\mathrm {d}y \\&\quad + \frac{1}{2}\int _{0}^{\infty }\int _{0}^{y}\int _{0}^{y} x^{-2\sigma }\left[ b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\right] \mathrm {d}x \mathrm {d}z\mathrm {d}y. \end{aligned}$$

By using fact that \(P_1 \le 1\) and the conditions (1.3), (2.2) for the first and second integral respectively, we have

$$\begin{aligned} \le \frac{\nu _0}{2} \int _{0}^{\infty }\int _{0}^{y} \left( y^r + \frac{1}{y^{2\sigma }}\right) K(y-z,z)f(y-z,s)f(z,s) \mathrm {d}z\mathrm {d}y. \end{aligned}$$

Where \(\nu _0=\max \{\bar{N}, \nu \}\). Now first change the order of the integral. Then put \(\bar{y} = y-z\), \(\bar{z} = z\) and again replace \(\bar{y}\) by y and \(\bar{z}\) by z to get

$$\begin{aligned}&= \frac{\nu _0}{2} \int _{0}^{\infty }\int _{0}^{\infty } \left( (y+z)^r+(y+z)^{-2\sigma }\right) K(y,z)f(y,s)f(z,s)\mathrm {d}y \mathrm {d}z. \end{aligned}$$

Since K has compact support, so their supremum exist. Let \(\kappa _0 = \underset{\frac{1}{\mathcal {M}}\le x,y \le \mathcal {M}}{\sup } K(x,y)\). Therefore rearranging the terms in the above relation and using the inequalities

$$\begin{aligned} (x+y)^r \le 2^r\left( x^r+y^r\right) ,\quad (x+y)^{-2\sigma }\le x^{-2\sigma } + y^{-2\sigma }, \end{aligned}$$

we can find the following estimate

$$\begin{aligned}&\le 2^{r} \kappa _0 \nu _0 \int _{0}^{\infty }\int _{0}^{\infty } (y^r+y^{-2\sigma })(z^r+z^{-2\sigma })f(y,s)f(z,s)\mathrm {d}y \mathrm {d}z. \end{aligned}$$

Using this estimation and apply similar operation for the first integral of (2.7) we get

$$\begin{aligned} \Vert \mathcal {G}(f)\Vert _{\Omega }^{(\tau )} \le&\Vert f_0\Vert _{\Omega }^{(\tau )} + \int _{0}^{t}2^{r} \kappa _0 (\nu +1)\int _{0}^{\infty }\int _{0}^{\infty } (y^r+y^{-2\sigma })(z^r+z^{-2\sigma })f(y,s)f(z,s) \mathrm {d}y \mathrm {d}z. \end{aligned}$$
(2.8)

Further let \(\zeta _1 := \max \{\Vert f_0\Vert _{\Omega }^{(\tau )}, 2^r\kappa _0(\nu +1)\}\) then the inequality (2.8) reduces to

$$\begin{aligned} \Vert \mathcal {G}(f)\Vert _{\Omega }^{(\tau )} \le \zeta _1\left( 1 + \tau L_0^2\right) . \end{aligned}$$

Hence, \(\Vert \mathcal {G}(f)\Vert _{\Omega }^{(\tau )} \le L_0\), if \(\zeta _1\left( 1+ \tau L_0^2\right) \le L_0\). This inequality hold if \(\tau < \frac{1}{4\zeta _1^2}\) and

$$\begin{aligned} \frac{1 - \sqrt{1 - 4\zeta _1^2\tau }}{2\zeta _1\tau } \le L_0 \le \frac{1 + \sqrt{1 - 4\zeta _1^2\tau }}{2\zeta _1\tau }. \end{aligned}$$
(2.9)

Now, we proceed to show that the mapping \(\mathcal {G}\) is contracting. From (2.3) we have

$$\begin{aligned} \Vert \mathcal {G}(f)-\mathcal {G}(g)\Vert _{\Omega }^{(\tau )}&\le \int _{0}^{t}\left[ \frac{1}{2} \int _{0}^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) \int _0^x P(x-y,y)K(x-y,y)\left| A(x-y,y,s)\right| \mathrm {d}y\mathrm {d}x \right. \nonumber \\&\quad +\int _{0}^{\infty }\int _0^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) K(x,y)\left| A(x,y,s) \right| \mathrm {d}y\mathrm {d}x + \frac{1}{2}\int _{0}^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) \nonumber \\&\quad \times \left. \left[ \int _{x}^\infty \int _{0}^{y} b(x|y-z,z)P_1(y-z,z)K(y-z,z)\left| A(y-z,z,s)\right| \mathrm {d}z \mathrm {d}y \right] \mathrm {d}x\right] \mathrm {d}s. \end{aligned}$$
(2.10)

where, \(\displaystyle A(x,y,s) = f(x,s)f(y,s) - g(x,s)g(y,s)\). Now, for the third integral of right hand side we first change the order of the integration with respct to y and x then apply the Fubini’s theorem on the integral with respect to z and x;

$$\begin{aligned}&\frac{1}{2}\int _{0}^{\infty }\int _{x}^\infty \int _{0}^{y} \left( x^r + \frac{1}{x^{2\sigma }}\right) b(x|y-z,z)P_1(y-z,z)K(y-z,z)|A(y-z,z,s)|\mathrm {d}z \mathrm {d}y\mathrm {d}x \\&\quad \le \frac{1}{2}\int _{0}^{\infty }\int _{0}^{y}\int _{0}^{y} \left( x^r + \frac{1}{x^{2\sigma }}\right) b(x|y-z,z)P_1(y-z,z)K(y-z,z)|A(y-z,z,s)|\mathrm {d}z \mathrm {d}x\mathrm {d}y \\&\quad = \frac{1}{2}\int _{0}^{\infty }\int _{0}^{y}\int _{0}^{y} x^r b(x|y-z,z)P_1(y-z,z)K(y-z,z)|A(y-z,z,s)|\mathrm {d}x \mathrm {d}z\mathrm {d}y \\&\qquad + \frac{1}{2}\int _{0}^{\infty }\int _{0}^{y}\int _{0}^{y} x^{-2\sigma } b(x|y-z,z)P_1(y-z,z)K(y-z,z)|A(y-z,z,s)|\mathrm {d}x \mathrm {d}z\mathrm {d}y. \end{aligned}$$

By using fact that \(P_1 \le 1\) and the conditions (1.3), (2.2) for the first and second integral respectively, we have

$$\begin{aligned} \le \frac{\nu _0}{2} \int _{0}^{\infty }\int _{0}^{y} \left( y^r + \frac{1}{y^{2\sigma }}\right) K(y-z,z)|A(y-z,z,s)| \mathrm {d}z\mathrm {d}y. \end{aligned}$$

Now first change the order of the integral. Then put \(\bar{y} = y-z\), \(\bar{z} = z\) and again replace \(\bar{y}\) by y and \(\bar{z}\) by z to get

$$\begin{aligned}&= \frac{\nu _0}{2} \int _{0}^{\infty }\int _{0}^{\infty } \{(y+z)^r+(y+z)^{-2\sigma }\}K(y,z)|A(y,z,s)|\mathrm {d}y \mathrm {d}z \\&\quad \le \frac{\nu _0}{2} \int _{0}^{\infty }\int _{0}^{\infty } \{2^r(y^r+z^r)+(y^{-2\sigma }+z^{-2\sigma })\}K(y,z)|f(z,s)\left( f(y,s)-g(y,s)\right) \\&\qquad +g(y,s)\left( f(z,s)-g(z,s)\right) |\mathrm {d}y \mathrm {d}z \\&\quad \le 2^{r} \kappa _0 \nu _0 \int _{0}^{\infty }\int _{0}^{\infty } (y^r+y^{-2\sigma })(z^r+z^{-2\sigma })|f(z,s)\left( f(y,s)-g(y,s)\right) \\&\qquad +g(y,s)\left( f(z,s)-g(z,s)\right) |\mathrm {d}y \mathrm {d}z\\&\quad \le 2^{r} \kappa _0 \nu _0 \left[ \Vert f-g\Vert _{\Omega }^{(\tau )} \Vert f\Vert _{\Omega }^{(\tau )} + \Vert f-g\Vert _{\Omega }^{(\tau )} \Vert g\Vert _{\Omega }^{(\tau )}\right] = 2^{r} \kappa _0 \nu \Vert f-g\Vert _{\Omega }^{(\tau )} \left[ \Vert f\Vert _{\Omega }^{(\tau )}+\Vert g\Vert _{\Omega }^{(\tau )}\right] . \end{aligned}$$

Using this estimation and apply similar operation for the first integral of (2.8) we get

$$\begin{aligned} \Vert \mathcal {G}(f)-\mathcal {G}(g)\Vert _{\Omega }^{(\tau )}&\le (2^{r}+1) \kappa _0 \nu \int _{0}^{\tau }\left[ \Vert f-g\Vert _{\Omega }^{(\tau )} \Vert f\Vert _{\Omega }^{(\tau )} + \Vert f-g\Vert _{\Omega }^{(\tau )} \Vert g\Vert _{\Omega }^{(\tau )} \right] \mathrm {d}s\nonumber \\&\quad +2^{r} \kappa _0 \nu \int _{0}^{\tau }\Vert f\nonumber \\&\quad -g\Vert _{\Omega }^{(\tau )} \left[ \Vert f\Vert _{\Omega }^{(\tau )}+\Vert g\Vert _{\Omega }^{(\tau )}\right] \mathrm {d}s \le 2(2^{r+1}+1) L_0 \kappa _0 \nu \tau \Vert f-g\Vert _{\Omega }^{(\tau )}. \end{aligned}$$
(2.11)

Further let \(\zeta _2 := 2(2^{r+1}+1) \kappa _0 \nu\) then the inequality (2.9) reduces to

$$\begin{aligned} \Vert \mathcal {G}(f) - \mathcal {G}(g)\Vert _\Omega ^{(\tau )} \le \tau \zeta _2 L_0\Vert f-g\Vert _\Omega ^{(\tau )}. \end{aligned}$$
(2.12)

So, the mapping \(\mathcal {G}\) is contractive in \(\Omega _{r,\sigma }^+(\tau )\) for \(\tau < \left[ \zeta _2 L_0 \right] ^{-1}\). Using this result together with the inequalities (2.7), there exist an invariant ball of radius \(L_0\) for sufficiently small \(\tau > 0\). In that ball \(\mathcal {G}\) is contractive. Consequently, that ball contains a fixed point of \(\mathcal {G}\).

Nonnegativity: Case I: If \(f_0 >0\), then we assume \(\left( x_0,t_0\right)\) be the point such that

$$\begin{aligned} f\left( x_0, t_0\right) = 0 \quad \text {and}\quad f(x,t)\ne 0 \quad \text {for all}\quad 0< x< \max \{x_0, \mathcal {M}\},\quad 0\le t < t_0. \end{aligned}$$
(2.13)

Since, f is continuous, so this is possible only if \(f(x,t)> 0\) for all \(0 < x \le \max \{x_0, \mathcal {M}\}\), \(0\le t < t_0\). Otherwise, \(\left( x_0,t_0\right)\) fails to satisfy (2.13) for which \(f\left( x_0, t_0\right) = 0\). Again since, the solution is continuous and satisfy (2.1) it must be continuously differentiable w.r.t. t. Therefore,

$$\begin{aligned} \left. \frac{\partial f(x,t)}{\partial t}\right| _{\left( x_0,t_0\right) }&= \frac{1}{2} \int _0^{x_0} P(x_0-y,y)K(x_0-y,y)f(x_0-y,t_0) f(y,t_0) \mathrm {d}y\nonumber \\&\quad + \int _{x_0}^\mathcal {M}\int _{0}^{y} b(x_0|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\mathrm {d}z\mathrm {d}y. \end{aligned}$$
(2.14)

  • If \(x_0 \le \mathcal {M}\), then \(f(x,t)> 0\) for all \(0 < x \le \mathcal {M}\) and \(0\le t < t_0\). The positivity of right hand side of (2.14) implies \(\displaystyle \left. \frac{\partial f(x,t)}{\partial t}\right| _{\left( x_0,t_0\right) } > 0\).

  • If \(x_0 > \mathcal {M}\), then \(\displaystyle \int _{x_0}^\mathcal {M}\int _{0}^{y} b(x_0|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\mathrm {d}z\mathrm {d}y = -\int _\mathcal {M}^{x_0}\int _{0}^{y} b(x_0|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,s)f(z,s)\mathrm {d}z\mathrm {d}y = 0\) by the virtue of (1.3), we have, \(\displaystyle \left. \frac{\partial f(x,t)}{\partial t}\right| _{\left( x_0,t_0\right) } > 0\).

Moreover, positivity of the time derivative proves that there exist a point \(\left( x_0, t\right)\), with \(t < t_0\) such that \(f\left( x_0, t\right) <0\). This contradicts the assumption that \(\left( x_0, t_0\right)\) is the point with the property (2.13). Hence, no such point \(\left( x_0, t_0\right)\) exists. Consequently, f(x, t) is strictly positive provided the initial data is strictly positive.

Case II: Let \(f_0\) is not strictly positive. Then we construct the sequence of positive function \(\{f_0^n\}\) which satisfies the condition of Theorem 2.1 and converges to \(f_0\) uniformly in \(\Omega _{r,\sigma }(\tau )\). We already proved that the family of operators \(\displaystyle \mathcal {G}_n : \Omega _{r,\sigma }(\tau ) \rightarrow \Omega _{r,\sigma }(\tau )\) defined as

$$\begin{aligned} \mathcal {G}_n(f)(x,t)&= f_0^n(x) + \int _0^t \left[ \frac{1}{2}\int _0^x P(x-y,y)K(x-y,y)f(x-y,s)f(y,s)\mathrm {d}y\right. \\&\quad \left. -f(x,s)\int _0^\infty K(x,y)f(y,s)\mathrm {d}y \right. \nonumber \\&\quad \left. + \frac{1}{2}\int _x^\infty \int _0^y b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,t)f(z,t)\mathrm {d}z \mathrm {d}y \right] \mathrm {d}s. \end{aligned}$$

is a contraction mapping. Therefore as \(n \rightarrow \infty\), we have

$$\begin{aligned} \sup _{\Vert f\Vert ^{(\tau )}_\Omega \le L} \left\| \mathcal {G}_n(f) - \mathcal {G}(f)\right\| _\Omega ^{(\tau )} \le \int _{0}^{\infty } \left( x^r + \frac{1}{x^{2\sigma }}\right) \left| f_0^n(x) - f_0(x)\right| \mathrm {d}x \rightarrow 0. \end{aligned}$$

Since the mapping is contractive in \(\Omega _{r,\sigma }(\tau )\), therefore

$$\begin{aligned} \Vert f^n - f\Vert _\Omega ^{(\tau )} = \Vert \mathcal {G}_n(f^n) - \mathcal {G}(f)\Vert _\Omega ^{(\tau )}&\le \Vert \mathcal {G}_n(f^n) - \mathcal {G}(f^n)\Vert _\Omega ^{(\tau )} + \Vert \mathcal {G}(f^n) - \mathcal {G}(f)\Vert _\Omega ^{(\tau )} \\&\le \Vert \mathcal {G}_n(f^n) - \mathcal {G}(f^n)\Vert _\Omega ^{(\tau )} + \bar{\zeta }\Vert f^n - f\Vert _\Omega ^{(\tau )}. \end{aligned}$$

Which implies

$$\begin{aligned} \left( 1 - \bar{\zeta }\right) \Vert f^n - f\Vert _\Omega ^{(\tau )} = \Vert \mathcal {G}_n(f^n) - \mathcal {G}(f^n)\Vert _\Omega ^{(\tau )} \rightarrow 0\quad \text{ whenever }\quad n \rightarrow \infty . \end{aligned}$$

This proves that the solution f is nonnegative for a nonnegative initial data.

Global existence of unique solution: We first observe that the boundedness of moments

$$\begin{aligned} \mathcal {M}_k(t) = \int _{0}^{\infty } x^k f(x,t)\mathrm {d}x; \quad \text{ where }\quad 0\le k \le r \quad \text{ and }\quad k=-2\sigma , \end{aligned}$$

for compactly supported kernels. Simple calculations will lead us to the following results;

$$\begin{aligned} \mathcal {M}_1(t) \le \bar{m}_1,\quad \mathcal {M}_{-2\sigma }(t) \le \bar{m}_{-2\sigma },\quad \mathcal {M}_0(t) \le \bar{m}_0,\quad \mathcal {M}_2(t) \le \bar{m}_2, \end{aligned}$$

and so on. Here terms \(\bar{m}_k\), \(k=-2\sigma ,0,1,\ldots ,r\) are all constants [for detailed calculation please see Appendix-1]. Moreover, it can be noted that for \(k=2,3,\ldots ,r\) the boundedness of \((k+1)\)-th moment depends upon the boundedness of \(k-\)th moment. Thus all the above results together implies

$$\begin{aligned} \Vert f\Vert _\Omega \le \bar{m}_r + \bar{m}_{-2\sigma }. \end{aligned}$$

Therefore, the solution of IVP (1.1)–(1.2) is bounded in the norm \(\Vert .\Vert _\Omega\). Taking into account nonnegativity of the local solution, we prolong it for all \(0\le t \le T\). Recalling Theorem 2.2 of [23], the global existence of the solution belonging to \(\Omega _{r,\sigma }^+(T)\) can easily be proved. \(\square\)

2.3 Conservation of mass

For the mass conservation law we multiplying with the weight x and integrating, Eq. (1.1) is written as

$$\begin{aligned} \frac{\mathrm {d}\mathcal {M}_1 (t)}{\mathrm {d}t}&= \frac{\mathrm {d}}{\mathrm {d}t} \int _{0}^{\infty }xf(x,t)\mathrm {d}x \nonumber \\&= \frac{1}{2} \int _{0}^{\infty }\int _0^x xP(x-y,y)K(x-y,y)f(x-y,t)f(y,t)\mathrm {d}y\mathrm {d}x\nonumber \\&\quad - \int _{0}^{\infty } \int _0^\infty xK(x,y)f(x,t) f(y,t)\mathrm {d}y \mathrm {d}x\nonumber \\&\quad + \underbrace{\frac{1}{2}\int _0^\infty \int _x^\infty \int _0^y xb(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,t)f(z,t)\mathrm {d}z \mathrm {d}y \mathrm {d}x}_{J_0}. \end{aligned}$$
(2.15)

Now, for the integral \(J_0\) we first change the order of the integration with respect to y and x. Then apply the Fubini’s theorem and (1.4) we get

$$\begin{aligned} J_0&= \frac{1}{2}\int _0^\infty \int _x^\infty \int _0^y xb(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,t)f(z,t)\mathrm {d}z \mathrm {d}y \mathrm {d}x \\&=\frac{1}{2}\int _0^\infty \int _0^y \int _0^y xb(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,t)f(z,t)\mathrm {d}x \mathrm {d}z \mathrm {d}y \\&=\frac{1}{2}\int _0^\infty \int _0^y y \left( 1-P(y-z,z)\right) K(y-z,z)f(y-z,t)f(z,t) \mathrm {d}z \mathrm {d}y. \end{aligned}$$

Using this in (2.15) we get

$$\begin{aligned} \frac{\mathrm {d}\mathcal {M}_{1} (t)}{\mathrm {d}t} = 0. \end{aligned}$$

Hence, we can say that the mass conservation law obeyed unconditionally for all \(n\ge 1\) and \(0\le t\le T\).

3 Uniqueness

Theorem 3.1

(Uniqueness theorem) Let \(\mathcal {M}(0)<\infty\) and all the assumptions of Theorem 2.1 hold true. Then the IVP (1.1)–(1.2) has a unique solution \(f \in \Omega _{r,\sigma }^+(T)\).

Proof

If possible let f and g be two solution of the IVP (1.1)–(1.2) with the same initial data \(f_0\). Let \(\vartheta (x,t) := f(x,t)-g(x,t)\) and we construct an auxiliary function

$$\begin{aligned} \mathcal {F}(t) := \int _0^\infty \left| \vartheta (x,t)\right| {d}x. \end{aligned}$$

Then the solutions f and g both satisfy the relation (2.3), and we get the difference

$$\begin{aligned} \mathcal {F}(t)&\le \int _{0}^{t}\left[ \underbrace{\frac{1}{2} \int _{0}^{\infty } \int _0^x P(x-y,y)K(x-y,y)\left| A(x-y,y,s)\right| \mathrm {d}y\mathrm {d}x}_{J_1}\right. \nonumber \\&\quad \left. +\underbrace{\int _{0}^{\infty }\int _0^{\infty } K(x,y)\left| A(x,y,s) \right| \mathrm {d}y\mathrm {d}x}_{J_2} \right. \nonumber \\&\quad \left. + \underbrace{\frac{1}{2}\int _{0}^{\infty } \int _{x}^\infty \int _{0}^{y} b(x|y-z,z)P_1(y-z,z)K(y-z,z)\left| A(y-z,z,s)\right| \mathrm {d}z \mathrm {d}y\mathrm {d}x}_{J_3} \right] \mathrm {d}s. \end{aligned}$$
(3.1)

Now by changing the order of the integration followed by applying Fubini’s theorem and the condition (2.2) the integral \(J_3\) can be reduced to;

$$\begin{aligned} \displaystyle J_3 \le \frac{\kappa _0\bar{N}}{2}\left( \Vert f\Vert _\Omega +\Vert g\Vert _\Omega \right) \mathcal {F}(s). \end{aligned}$$
(3.2)

Using similar operation on the integrals \(J_1\) and \(J_2\) we get the estimation;

$$\begin{aligned} J_1+J_2 \le \frac{3\kappa _0}{2} \left( \Vert f\Vert _\Omega +\Vert g\Vert _\Omega \right) \mathcal {F}(s). \end{aligned}$$
(3.3)

Using (3.2) and (3.3) on (3.1) we get;

$$\begin{aligned} \mathcal {F}(t) \le \displaystyle \frac{\kappa _0}{2}\left( 3+\bar{N}\right) \left( \Vert f\Vert _\Omega +\Vert g\Vert _\Omega \right) \int _{0}^{t}\mathcal {F}(s)\displaystyle . \end{aligned}$$
(3.4)

Since, f and g both are belongs to the space \(\Omega _{r,\sigma } (T)\), so both the norm \(\Vert f\Vert _\Omega\) and \(\Vert g\Vert _\Omega\) are uniformly bounded with respect to t on \(\left[ 0,T\right]\). Then by applying Grownwall’s inequality on (3.4)

$$\begin{aligned} \mathcal {F}(t) = 0,\quad \text{ which } \text{ implies, }\quad f(x,t) = g(x,t) \quad \text{ for } \text{ all }\quad 0\le t\le T. \end{aligned}$$

Hence, the uniqueness of the solution is obtained. \(\square\)

4 Long term behavior of the solution

By the physical point of view we can expect that, under mild conditions, the number of particles (represented by zeroth moment: \(\mathcal {M}_0\)) in the system may increase or decrease with time. This change of particles may result to mass loss from the system, that is shattering or gelation onsets after a certain time. Therefore, we plan to investigate the large-time behavior of the obtained solution for the following particular breakage distribution function;

$$\begin{aligned} b(x|y,z)=\left( \theta +2\right) \frac{x^\theta }{\left( y+z\right) ^{\theta +1}} ~\text { where } -1< \theta \le 0. \end{aligned}$$
(4.1)

One can easily verify that, for a suitable choice of \(\sigma\), the breakage distribution function b satisfy the condition (2.2). The possibility of shattering condition can be observed by the following proposition.

Proposition 1

Let K(x, y), P(x, y) and b(x|y, z) all are satisfies the condition of theorem () together with the breakage distribution function b satisfy (4.1) and if f(x, t) is a solution of (1.1) with the initial data \(f(x,0)=f_0(x) \in \Omega _{r,\sigma }^+(0)\cap \Omega _{0,0}^+(0)\), defined on \(\left[ 0,\infty \right)\). Then for \(P \left( x,y\right) \ge -\theta\) and \(s,t \in \mathbb {R}_+\) with \(s\le t\),

$$\begin{aligned} \mathcal {M}_0\left( t\right) \le \mathcal {M}_0\left( s\right) . \end{aligned}$$

Moreover, if there exist \(\mu \in \left( 0,1\right]\) such that \(P\left( x,y\right) +\theta >\mu\) then for any \(x \in \mathbb {R}_+\), \(f\left( x,t\right) \rightarrow 0\) as \(t\rightarrow \infty\).

Proof

By successive integration of (1.1) with respect to x and t over the range of interval \(0 \text {to }\infty\) and \(s \text { to }t\) respectively, we can obtain;

$$\begin{aligned} \mathcal {M}_0\left( t\right) - \mathcal {M}_0\left( s\right)&= \int _{s}^{t}\left[ \underbrace{\frac{1}{2} \int _{0}^{\infty }\int _0^x \nonumber P(x-y,y)K(x-y,y)f(x-y,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x }_{J_4}\right. \\ \nonumber&\quad + \underbrace{\frac{1}{2} \int _{0}^{\infty }\int _x^\infty \int _0^y b(x|y-z,z)P_1(y-z,z)K(y-z,z)f(y-z,\xi )f(z,\xi )\mathrm {d}z \mathrm {d}y\mathrm {d}x}_{J_5}\\&\quad \left. - \underbrace{\int _{0}^{\infty }\int _0^\infty K(x,y)f(x,\xi )f(y,\xi )\mathrm {d}y\mathrm {d}x}_{J_6}\right] \mathrm {d}\xi . \end{aligned}$$
(4.2)

Now for the term \(J_4\) we first change the order of the integral. Then put \(\bar{x} = x-y\), \(\bar{y} = y\) and again replace \(\bar{x}\) by x and \(\bar{y}\) by y to get

$$\begin{aligned} J_4= \frac{1}{2} \int _{0}^{\infty }\int _0^\infty P(x,y)K(x,y)f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x. \end{aligned}$$
(4.3)

By using the condition (4.1) and applying Fubini theorem on the term \(J_5\) we can obtain;

$$\begin{aligned} J_5 \nonumber&= \frac{\theta +2}{2\left( \theta +1\right) }\int _0^\infty \int _0^yP_1(y-z,z)K(y-z,z)f(y-z,\xi )f(z,\xi )\mathrm {d}z \mathrm {d}y\\&= \frac{\theta +2}{2\left( \theta +1\right) }\int _0^\infty \int _0^\infty \left[ 1-P(x,y)\right] K(x,y)f(y,\xi )f(x,\xi )\mathrm {d}y \mathrm {d}x. \end{aligned}$$
(4.4)

Using the simplification (4.3), (4.4) in (4.2) we obtain

$$\begin{aligned} \mathcal {M}_0\left( t\right) - \mathcal {M}_0\left( s\right)&= -\frac{1}{2\left( \theta +1\right) } \int _{s}^{t}\int _{0}^{\infty }\int _0^\infty \left[ \theta +P(x,y)\right] K(x,y)f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x \mathrm {d}\xi \nonumber \\&\le 0. \end{aligned}$$
(4.5)

From which we can conclude that \(\mathcal {M}_0\left( t\right)\) is a monotonic non increasing function. Also it bounded below (by zero), so the limit \(t \rightarrow \infty\) exist and let the value of the limit is \(\mathcal {M}^*\). Now consider the case \(t=s+1\) on (4.5) and then taking limit \(s \rightarrow \infty\) on the moment equation

$$\begin{aligned}&\mathcal {M}_0\left( s+1\right) - \mathcal {M}_0\left( s\right)&=\nonumber \\&\quad =-\frac{1}{2\left( \theta +1\right) } \int _{s}^{s+1}\int _{0}^{\infty }\int _0^\infty \left[ \theta +P(x,y)\right] K(x,y)f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x \mathrm {d}\xi \nonumber \\&\quad \Bigg \downarrow {s\rightarrow \infty } \nonumber \\&\quad 0= \lim \limits _{s\rightarrow \infty }\int _{s}^{s+1}\int _{0}^{\infty }\int _0^\infty \left[ \theta +P(x,y)\right] K(x,y)f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x \mathrm {d}\xi . \end{aligned}$$
(4.6)

Now we need to prove that for any \(x \in \mathbb {R}_+\), the function \(f\left( x,t\right) \rightarrow 0\) as \(t\rightarrow \infty\), if not let there exist a fixed \(\bar{x}\in \mathbb {R}_+\) such that, \(f\left( \bar{x},t\right) \rightarrow L\) as \(t\rightarrow \infty\), where \(L>0\). Then from the definition of limit at infinity, for an arbitrary \(\epsilon >0\), there exist \(t_0>0\) such that \(f(\bar{x},t) > L-\epsilon\) for all \(t\ge t_0\). Again by the continuity of the solution f we can say \(f\left( x,t\right) \ge L-\epsilon\) for all \(x\in N_\delta (\bar{x})\). Also K has compact support, so their infimum exist. Let \(\kappa _1 = \underset{\frac{1}{\mathcal {M}}\le x,y \le \mathcal {M}}{\inf } K(x,y)\), then right hand side of (4.6) can be estimated as;

$$\begin{aligned}&\lim \limits _{s\rightarrow \infty }\int _{s}^{s+1}\int _{0}^{\infty }\int _0^\infty \left[ \theta +P(x,y)\right] K(x,y) f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x \mathrm {d}\xi \\&\quad> \mu \kappa _1 \lim \limits _{s\rightarrow \infty }\int _{s}^{s+1}\int _{0}^{\infty }\int _0^\infty f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x \mathrm {d}\xi \\&\quad \ge \mu \kappa _1 \lim \limits _{s\rightarrow \infty }\int _{s}^{s+1}\int _{ N_\delta (\bar{x}) } \int _{ N_\delta (\bar{x}) } f(x,\xi )f(y,\xi )\mathrm {d}y \mathrm {d}x \mathrm {d}\xi \\&\quad \ge 4\mu \delta ^2 \kappa _1(L-\epsilon )^2\\&\quad >0 \qquad \text { for an arbitrary }\epsilon , \end{aligned}$$

which contradicts Eq. (4.6), and hence it proves that \(L=0\). \(\square\)

5 Conclusion

In this work, we have thoroughly investigated the mathematical aspect for the existence and uniqueness of mass conserving solution to the coagulation with collision induced fragmentation model. Here the collisional kernel includes the class of compactly supported function. At first, we have shown the existence of the non-negative solution locally for this compactly supported kernels, then we have extended it globally in the space \(\Omega _{r,\sigma }\left( T\right)\) of the IVP (1.1)–(1.2). We have also established that the solution satisfies the mass conserving law and uniqueness property without any additional restriction. The study is completed by the investigation of large time dynamics of the existing solution. A mathematical treatment of the considered model equipped with singular collision kernel can be considered in future studies.