Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 50, pp. 1–19.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

MATHEMATICAL METHODS FOR THE RANDOMIZED

NON-AUTONOMOUS BERTALANFFY MODEL

JULIA CALATAYUD, TOMÁS CARABALLO,

JUAN CARLOS CORTÉS, MARC JORNET

Abstract. In this article we analyze the randomized non-autonomous Berta-

lanffy model

x′(t, ω) = a(t, ω)x(t, ω) + b(t, ω)x(t, ω)2/3, x(t0, ω) = x0(ω),

where a(t, ω) and b(t, ω) are stochastic processes and x0(ω) is a random vari-
able, all of them defined in an underlying complete probability space. Under

certain assumptions on a, b and x0, we obtain a solution stochastic process,

x(t, ω), both in the sample path and in the mean square senses. By using the
random variable transformation technique and Karhunen-Loève expansions,

we construct a sequence of probability density functions that under certain

conditions converge pointwise or uniformly to the density function of x(t, ω),
fx(t)(x). This permits approximating the expectation and the variance of

x(t, ω). At the end, numerical experiments are carried out to put in practice

our theoretical findings.

1. Introduction and motivation

Bertalanffy model [2] is a biological ordinary differential equation model that
describes the relationship between the metabolism and the growth of an organism.
The metabolism is divided into anabolism (synthesis) and catabolism (destruction).
The model assumes that the body weight W (t), at the time instant t, of an animal
is the result of the counteraction of the processes of anabolism and catabolism,

W ′(t) = ηWm(t)− κWn(t),

where η and κ are the constants of anabolism and catabolism, respectively; both
constants are proportional to some power of the body weight at the time t. This
model follows the law of allometry: the rate of change of weight depends on the
constants of anabolism and catabolism via a power of the weight.

The surface rule states that the dependence of anabolism on body weight takes
the power m = 2/3 [1, 26] (however, other exponents have been suggested, see [18]).
Bertalanffy justified, by using physiological facts and mathematical considerations,
that the rate of catabolism should have the power n = 1. Bertalanffy model thus

2010 Mathematics Subject Classification. 34F05, 60H35, 60H10, 65C30.
Key words and phrases. Random non-autonomous Bertalanffy model;

random differential equation; random variable transformation technique;

Karhunen-Loève expansion; probability density function.
c©2020 Texas State University.

Submitted July 20, 2019. Published May 26, 2020.

1



2 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

becomes

W ′(t) = ηW 2/3(t)− κW (t).

In this article, we want to perform a mathematical study of the randomized
non-autonomous Bertalanffy model,

x′(t, ω) = a(t, ω)x(t, ω) + b(t, ω)x(t, ω)2/3, t ∈ [t0, T ], ω ∈ Ω,

x(t0, ω) = x0(ω), ω ∈ Ω.
(1.1)

In this setting, we consider an underlying complete probability space (Ω,F ,P). The
set Ω contains the outcomes, that will be generically denoted by ω, the set F is
the σ-algebra of events and P is the probability measure. In (1.1), we are also
considering the stochastic processes

a = {a(t, ω) : t ∈ [t0, T ], ω ∈ Ω}, b = {b(t, ω) : t ∈ [t0, T ], ω ∈ Ω}

and the random variable x0(ω), all of them defined in the probability space (Ω,F ,P).
The solution to the stochastic system, x(t, ω), becomes a stochastic process.

The mathematical term that encompasses equations such as (1.1) is random
differential equation (RDE) [20, Ch. 8], [22, 23]. RDEs are differential equations
in which randomness enters via the input data (coefficients and initial conditions).
Any probability distribution for the input data is allowed. RDEs can be interpreted
in a sample path sense or in a random Lp sense. In both senses, one of the main
goals is to compute or approximate the solution stochastic process x(t, ω) and to
calculate its main statistical information, say the expectation, E[x(t, ω)], and the
variance, V[x(t, ω)] (uncertainty quantification). Both statistics can be computed
or approximated if one is able to compute or approximate the probability density
function of the solution process, fx(t)(x), since all one-dimensional moments and
probabilities can be determined:

E[x(t, ω)k] =

∫
R
xkfx(t)(x) dx,

P(a1 ≤ x(t, ω) ≤ a2) =

∫ a2

a1

fx(t)(x) dx.

This target has been achieved in [5, 6, 7, 8, 10, 12, 13], by using the Random
Variable Transformation technique [15, Lemma 4.12], although the inputs are ran-
dom variables (time invariant) rather than stochastic processes. The random au-
tonomous Bertalanffy model has been studied, with applications to fish weight
growth modelling, in [5]. The recent paper [4], that has become the main reference
for our article, studied the random non-autonomous inhomogeneous linear differ-
ential equation, by focusing on approximating the probability density function of
the solution stochastic process. Its main contribution has been using Karhunen-
Loève expansions [15, Ch. 5] to construct the approximating sequence of density
functions. Other mathematical studies of the Bertalanffy model, in the sense of
stochastic differential equations (SDEs) of Itô type [16], have been carried out in
[17, 21]. In SDEs of Itô type, the uncertainty is introduced in the deterministic dif-
ferential equation by means of stochastic perturbations driven by the white noise
process.

The goal of this paper is to perform a mathematical study of the random non-
autonomous Bertalanffy model (1.1). It is proved that, under certain conditions,
there is a solution stochastic process in the sample path and the mean square



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 3

senses. This is done by relating the random non-autonomous Bertalanffy model
to the random non-autonomous linear differential equation from [4], via the usual
change of variables performed in Bernoulli differential equations. The Random
Variable Transformation technique permits relating the probability density func-
tions of the solution processes to the random non-autonomous Bertalanffy model
and the random non-autonomous linear differential equation. The Karhunen-Loève
expansions of the coefficient stochastic processes in the RDE (1.1) allow us to obtain
an approximating sequence of probability density functions for the density fx(t)(x)
of the solution process x(t, ω). A numerical and computational treatment of the
theoretical results obtained will be performed at the end.

2. Solving the random non-autonomous Bertalanffy model

There are different ways of interpreting the random differential equation (1.1).
One way, which strongly uses the deterministic theory on differential equations, is
the sample path interpretation [20, p. 440], [23, p. 2, SP problem]: fixed ω ∈ Ω, the
random problem (1.1) becomes a deterministic problem, so one looks for stochastic
processes x(t, ω) with absolutely continuous sample paths that solve (1.1). Another
way consists in using Lp(Ω) random calculus (see [22, Ch. 4] and [15, Section 5.5]
for an introduction). We will study solutions to (1.1) in the mean square sense,
according to the definition in [22, p. 118] or [23, p. 3, Lp problem].

Consider the stochastic process

x(t, ω) =
(
x0(ω)1/3e

1
3

∫ t
t0
a(s,ω) ds

+
1

3

∫ t

t0

b(s, ω)e
1
3

∫ t
s
a(r,ω) dr ds

)3

. (2.1)

This stochastic process comes from randomizing the deterministic solution to the
deterministic non-autonomous Bertalanffy model. The integrals in (2.1) are consid-
ered in the Lebesgue sense, for each ω ∈ Ω fixed (sample path Lebesgue integral),
or in the mean square sense.

2.1. Solution stochastic process with absolutely continuous sample paths.
By the classical theory on deterministic differential equations [11, p. 28–30], the
following theorem holds.

Theorem 2.1 (Sample path solution to the random Bertalanffy model). Suppose
that the processes a(·, ω), b(·, ω) ∈ L1([t0, T ]), for a.e. ω ∈ Ω. Then the stochastic
process x(t, ω) given by (2.1) satisfies that, for a.e. ω ∈ Ω, x(·, ω) is absolutely
continuous on [t0, T ] and satisfies (1.1) for a.e. t ∈ [t0, T ].

If a(·, ω) and b(·, ω) are continuous on [t0, T ], then x(·, ω) is in C1([t0, T ]) and
satisfies (1.1) for all t ∈ [t0, T ].

2.2. Solution stochastic process in the mean square sense. Consider the
random linear differential equation

y′(t, ω) =
1

3
a(t, ω)y(t, ω) +

1

3
b(t, ω), t ∈ [t0, T ], ω ∈ Ω,

y(t0, ω) = x0(ω)1/3, ω ∈ Ω.
(2.2)

Notice that, if a stochastic process y(t, ω) is a solution to (2.2) in the L6(Ω)
sense, then the stochastic process defined as x(t, ω) = y(t, ω)3 is a solution to (1.1)
in the L2(Ω) sense. Indeed, if y(t, ω) is differentiable in the L6(Ω) sense, then x(t, ω)



4 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

is differentiable in the L2(Ω) sense, by [25, Lemma 3.14], and by the product rule,
x(t, ω) solves (1.1) in the L2(Ω) sense.

Let y(t, ω) be a stochastic process solution to (2.2) in the L6(Ω) sense. By
[20, Thm. 8–20] and [23, Thm. 3(a)], there exists a stochastic process ϕ(t, ω),
measurable on [t0, T ] × Ω, equivalent to y(t, ω) (meaning that ϕ(t, ·) = y(t, ·) a.s.,
for all t ∈ [t0, T ]), such that ϕ(·, ω) is absolutely continuous on [t0, T ] and solves
(2.2) in the sample path sense, for each ω ∈ Ω. By [4, Thm. 1.3], up to equivalence,

y(t, ω) = x0(ω)1/3e
1
3

∫ t
t0
a(s,ω) ds

+
1

3

∫ t

t0

b(s, ω)e
1
3

∫ t
s
a(r,ω) dr ds (2.3)

Thus, (2.3) is the unique candidate, up to equivalence, that solves (2.2) in the
L6(Ω) sense. We are going to show that, under some conditions on the stochastic
processes a(t, ω) and b(t, ω), the stochastic process y(t, ω) given by (2.3) is indeed
a solution to (2.2) in the L6(Ω) sense. From this fact, (2.1) will solve the random
Bertalanffy model (1.1) in the mean square sense.

Theorem 2.2 (Mean square solution to the random Bertalanffy model). Suppose

(i) x0, a and b are independent.
(ii) x0 ∈ L4(Ω), and a and b are L12(Ω) continuous.
(iii) There exist r > 12 and δ > 0 such that

sup
s,s∗∈[−δ,δ]

E
[
er

∫ t+s∗
x+s

a(u) du
]
<∞,

for each t0 ≤ x ≤ t ≤ T .

Then the stochastic process y(t, ω) defined by (2.3) (with mean square Riemann
integrals) is differentiable in the L6(Ω) sense and satisfies the random problem (2.2).
As a consequence, the stochastic process x(t, ω) defined by (2.1) is differentiable in
the mean square sense and satisfies the random Bertalanffy model (1.1).

The proof of the above theorem is straightforward from [9]. Denote y0(ω) =
x0(ω)1/3, which belongs to L12(Ω). We apply the theory from [9, Section 3] adapted
to L6(Ω) calculus.

Example 2.3 (Applications of Theorem 2.2). Let us see some examples of processes
a(t, ω) for which the hypotheses of Theorem 2.2 fulfill.
• Let [t0, T ] = [0, 1] and a(t, ω) be a standard Brownian motion on [0, 1] [15,

p. 185–186]. We have a(t+ h, ω)− a(t, ω) ∼ Normal(0, h), so

‖a(t+ h, ω)− a(t, ω)‖L12(Ω) = ‖
√
hZ‖L12(Ω) =

√
h ‖Z‖L12(Ω) → 0 as h→ 0,

being Z ∼ Normal(0, 1). This shows that a(t, ω) is continuous in the L12(Ω) sense.
Another way of checking the L12(Ω)-continuity of the Brownian motion uses [25,
Lemma 3.11]: a(t, ω) is continuous in the L12(Ω) sense if and only if the function
E[a(t1, ω) · · · a(t12, ω)] defined on R12 is continuous on the diagonal (t, . . . , t) ∈
[t0, T ]12. This is clear by [22, p. 28].

On the other hand, since a(t, ω) is a Gaussian process, its mean square integral
is normally distributed [22, Thm. 4.6.4]. Bearing in mind the moment generating
function of a normal distribution,

E
[
er

∫ t+s∗
x+s

a(u,ω) du
]
<∞.

• If a(t, ω) is a standard Brownian bridge on [0, 1], [15, p. 193–195], the same
holds, as a(t, ω) is a Gaussian process with stationary increments.



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 5

• An example of a non-Gaussian process is given by a(t, ω) = t U(ω), t ∈ [t0, T ],
being U any bounded random variable. Indeed, a(t, ω) is L12(Ω)-continuous, since

‖a(t+ h, ω)− a(t, ω)‖L12(Ω) = |h|‖U‖L12(Ω)
h→0−→ 0.

On the other hand,

E
[
er

∫ t+s∗
x+s

a(u,ω) du
]
≤ E

[
er‖U‖L∞(Ω)

∫ t+s∗
x+s

u du
]

= er‖U‖L∞(Ω)
(t+s∗)2−(x+s)2

2 <∞,

so the hypotheses of Theorem 2.2 hold for a(t, ω).

3. Obtaining the probability density function of the solution
stochastic process

The aim of this section is at approximating the probability density function of the
solution stochastic process to the random Bertalanffy model, x(t, ω) given by (2.1).
To achieve this goal, we will use existing results on the random linear differential
equation from [4], together with a version of the Random Variable Transformation
technique and the Karhunen-Loève expansions of both processes a(t, ω) and b(t, ω)
from (1.1).

Lemma 3.1 (Random Variable Transformation technique [14, p. 115]). Let X be
an absolutely continuous random variable with density fX and with support DX

contained in an open set D ⊆ R. Let g : D → R be such that D = ∪ni=1Di and
gi = g|Di is injective and C1(Di) with non-vanishing derivative. Then the random
variable Y = g(X) is absolutely continuous, with density function given by

fY (y) =

{∑
i:y∈g(Di) fX(g−1

i (y))|dg
−1
i (y)

dy |, y ∈ g(D),

0, y /∈ g(D).

Lemma 3.2 (Karhunen-Loève Theorem [15, Thm. 5.28]). Consider a stochastic
process {X(t, ω) : t ∈ T , ω ∈ Ω} in L2(T × Ω). Then

X(t, ω) = µ(t) +

∞∑
j=1

√
νj φj(t)ξj(ω), (3.1)

where the sum converges in L2(T ×Ω), µ(t) = E[X(t)], {φj}∞j=1 is an orthonormal

basis of L2(T ), {(νj , φj)}∞j=1 is the set of pairs of (nonnegative) eigenvalues and
eigenvectors of the operator

C : L2(T )→ L2(T ), Cf(t) =

∫
T

Cov[X(t), X(s)]f(s) ds, (3.2)

and {ξj}∞j=1 is a sequence of random variables with zero expectation, unit variance
and pairwise uncorrelated. In (3.2), Cov[·, ·] stands for the covariance operator.
Moreover, if {X(t) : t ∈ T } is a Gaussian process, then {ξj}∞j=1 are independent
and Gaussian.

3.1. Main results. Let a(t, ω) and b(t, ω) be stochastic processes in L2([t0, T ]×Ω).
According to Lemma 3.2, we can expand both a(t, ω) and b(t, ω) via a Karhunen-
Loève expansion:

a(t, ω) = µa(t) +

∞∑
j=1

√
νj φj(t)ξj(ω), b(t, ω) = µb(t) +

∞∑
j=1

√
γj ψj(t)ηj(ω), (3.3)



6 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

respectively. The summation symbol in both expansions will be always written
up to ∞, although it could be possible that their corresponding covariance integral
operators C have only a finite number of nonzero eigenvalues, so that the summation
symbol finishes at an index J <∞. As the more complex case arises when the sum
arrives at infinity, we will always write the Karhunen-Loève expansions of a(t, ω)
and b(t, ω) up to infinity.

Notice that, from a, b ∈ L2([t0, T ] × Ω), we have a(·, ω), b(·, ω) ∈ L1([t0, T ]),
therefore the process x(t, ω) has absolutely continuous sample paths and solves the
random Bertalanffy model (1.1), by Theorem 2.1. Under the stricter assumptions
of Theorem 2.2, the process x(t, ω) will be a mean square solution too.

For convenience, let us consider the truncation of the Karhunen-Loève expansions
(3.3) of the stochastic processes a and b of common order N

aN (t, ω) = µa(t) +

N∑
j=1

√
νj φj(t)ξj(ω), bN (t, ω) = µb(t) +

N∑
j=1

√
γj ψj(t)ηj(ω).

This gives a truncation of the solution stochastic process x(t, ω) given by (2.1),

xN (t, ω) =

(
x0(ω)1/3e

1
3

∫ t
t0
aN (s,ω) ds

+
1

3

∫ t

t0

bN (s, ω)e
1
3

∫ t
s
aN (r,ω) dr ds

)3

, (3.4)

We also obtain a truncation of the solution stochastic process y(t, ω) to the random
linear differential equation, given by (2.3):

yN (t, ω) = x0(ω)1/3e
1
3

∫ t
t0
aN (s,ω) ds

+
1

3

∫ t

t0

bN (s, ω)e
1
3

∫ t
s
aN (r,ω) dr ds. (3.5)

Naturally, the relation between both truncations (3.4) and (3.5) is that xN (t, ω) =
yN (t, ω)3.

We denote, as in [4], the following vectors in bold letters, ξN = (ξ1, . . . , ξN ) and
ηM = (η1, . . . , ηM ), understanding this as a random vector or as a deterministic
real vector, depending on the context. Denote

Ka(t, ξN ) =

∫ t

t0

(
µa(s) +

N∑
j=1

√
νj φj(s)ξj

)
ds,

Sb(s,ηN ) = µb(s) +

N∑
i=1

√
γi ψi(s)ηi.

Suppose that x0 and (ξ1, . . . , ξN , η1, . . . , ηN ) are absolutely continuous and in-
dependent, for each N ≥ 1. Let y0(ω) = x0(ω)1/3 be the initial condition of the
random linear differential equation (2.2). By Lemma 3.1 applied with the transfor-
mation mapping g(x) = x1/3 on D = R\{0}, with domain partition D = D1 ∪D2,
being D1 = (0,∞) and D2 = (−∞, 0), we have y0(ω) is absolutely continuous, with
density function

fy0
(y) = fx0

(y3)3y2, (3.6)

for y ∈ R.
By using the version of the Random Variable Transformation technique of [4,

Lemma 2.1], in [4, Expression (10)] it was obtained the probability density function
of yN (t, ω),

fyN (t)(y)



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 7∫
R2N

fy0

(
y e−

1
3Ka(t,ξN ) − 1

3

∫ t

t0

Sb(s,ηN )e−
1
3Ka(s,ξN ) ds

)
× e−

1
3Ka(t,ξN )fξN ,ηN (ξN ,ηN ) dξN dηN

= E
[
fy0

(
y e−

1
3Ka(t,ξN ) − 1

3

∫ t

t0

Sb(s,ηN )e−
1
3Ka(s,ξN ) ds

)
e−

1
3Ka(t,ξN )

]
= E

[
fx0

({
y e−

1
3Ka(t,ξN ) − 1

3

∫ t

t0

Sb(s,ηN )e−
1
3Ka(s,ξN ) ds

}3)
× 3
{
y e−

1
3Ka(t,ξN ) − 1

3

∫ t

t0

Sb(s,ηN )e−
1
3Ka(s,ξN ) ds

}2

e−
1
3Ka(t,ξN )

]
,

for y ∈ R.
Since xN (t, ω) = yN (t, ω)3, by Lemma 3.1 with the transformation mapping

g(x) = x3 on D = R\{0}, with domain partition D = D1 ∪D2, being D1 = (0,∞)
and D2 = (−∞, 0), we have that xN (t, ω) is an absolutely continuous random
variable for each t ∈ [t0, T ], with density function

fxN (t)(x)

= fyN (t)

(
x1/3

) 1

3x2/3

=
1

x2/3
E
[
fx0

({
x1/3 e−

1
3Ka(t,ξN ) − 1

3

∫ t

t0

Sb(s,ηN )e−
1
3Ka(s,ξN ) ds

}3)
{
x1/3 e−

1
3Ka(t,ξN ) − 1

3

∫ t

t0

Sb(s,ηN )e−
1
3Ka(s,ξN ) ds

}2

e−
1
3Ka(t,ξN )

]
,

(3.7)

for 0 6= x ∈ R. Density functions are defined up to sets of Lebesgue measure 0, so
the fact that fxN (t)(x) is not defined at x = 0 is not a problem.

Under some assumptions, for instance, by taking into account [4, Thm. 2.9,
Thm. 2.12], the random variable y(t, ω) is absolutely continuous for each t ∈ [t0, T ],
with density function

fy(t)(y) = lim
N→∞

fyN (t)(y), (3.8)

for all y ∈ R. Bearing in mind that x(t, ω) = y(t, ω)3, by Lemma 3.1 with the trans-
formation mapping g(x) = x3 on D = R\{0}, we get that x(t, ω) is an absolutely
continuous random variable, for each t ∈ [t0, T ], with density function

fx(t)(x) = fy(t)

(
x1/3

) 1

3x2/3
, (3.9)

for 0 6= x ∈ R. By combining (3.7), (3.8) and (3.9),

fx(t)(x) = fy(t)

(
x1/3

) 1

3x2/3
= lim
N→∞

fyN (t)

(
x1/3

) 1

3x2/3
= lim
N→∞

fxN (t)(x),

for all 0 6= x ∈ R.
The goal is to find out under which conditions on the stochastic processes a(t, ω)

and b(t, ω) and on the random variable x0(ω) from (1.1), the solution stochastic
process x(t, ω) given by (2.1) is an absolutely continuous random variable, for each
t ∈ [t0, T ], with density function satisfying

fx(t)(x) = lim
N→∞

fxN (t)(x), (3.10)



8 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

for each 0 6= x ∈ R. For this purpose, we will use results on the random linear
differential equation (2.2) [4, Thm. 2.9, Thm. 2.12], that establish under which
conditions the limit (3.8) is justified.

In the next theorem we use the following assumptions:

Theorem 3.3. Assume the following hypotheses:

(1) a, b ∈ L2([t0, T ]× Ω);
(2) x0 and (ξ1, . . . , ξN , η1, . . . , ηN ) are absolutely continuous and independent,

N ≥ 1;
(3) the density function of x0, fx0

, is continuous on R and fx0
(x) ≤ C

|x|2/3 , for

x 6= 0;
(4) ‖e− 1

3Ka(t,ξN )‖L2(Ω) ≤ C for all N ≥ 1 and all t ∈ [t0, T ].

Then, for all 0 6= x ∈ R and t ∈ [t0, T ], the sequence {fxN (t)(x)}∞N=1 given by (3.7)
converges to the density fx(t)(x) of the solution process x(t, ω) given by (2.1).

Proof. Since y0(ω) = x0(ω)1/3, by hypothesis (2), y0 and (ξ1, . . . , ξN , η1, . . . , ηN )
are absolutely continuous and independent, for N ≥ 1. By (3.6) and hypothesis
(3), fy0(y) is continuous on R and

fy0
(y) ≤ C

|y3|2/3
3y2 = 3C,

for y 6= 0, therefore bounded. The hypotheses of [4, Thm. 2.9] are fulfilled for (2.2),
therefore y(t, ω) is an absolutely continuous random variable for each t ∈ [t0, T ],
with density function satisfying (3.8). Then, x(t, ω) is absolutely continuous and
verifies (3.10). �

Theorem 3.4. Assume the following hypotheses:

(1) a, b ∈ L2([t0, T ]× Ω);
(2) x0, η1, (ξ1, . . . , ξN , η2, . . . , ηN ) are absolutely continuous and independent,

N ≥ 1;
(3) the density function of η1, fη1

, is continuous and bounded on R;
(4) ξ1, ξ2, . . . have compact support in [−A,A] (A > 0) and ψ1 > 0 on (t0, T ).

Then, for each 0 6= x ∈ R and t ∈ (t0, T ], the sequence {fxN (t)(x)}∞N=1 given by
(3.7) converges to the density fx(t)(x) of the solution process x(t, ω) given by (2.1).

Proof. By [4, Thm. 2.12], y(t, ω) is an absolutely continuous random variable for
each t ∈ [t0, T ], with density function satisfying (3.8). Then, x(t, ω) is absolutely
continuous and verifies the desired limit (3.10). �

By using the following lemma, we will establish a theorem similar to Theorem 3.3,
but which substitutes the continuity hypothesis in (3) by a.e. continuity. This is
important, as in (3) we will allow densities with some discontinuities in R, such as
the uniform distribution, exponential distribution, etc.

Lemma 3.5. Let U and V be two independent random variables. If U is absolutely
continuous, then U + V is absolutely continuous.

Proof. For any Borel set A, by the convolution formula [3, p. 266] we have P(U+V ∈
A) =

∫
R P(U ∈ A − v)PV (dv), where PV = P ◦ V −1 is the law of V . If A is null,

then A− v is null, so P(U ∈ A− v) = 0. Thus, if A is null, then P(U +V ∈ A) = 0.
By the Radon-Nikodym Theorem [27, Ch. 14], U + V has a density. �



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 9

Theorem 3.6. Assume the following five hypotheses:

(1) a, b ∈ L2([t0, T ]× Ω);
(2) x0, ξ1, . . . , ξN , η1, . . . , ηN are absolutely continuous and independent, N ≥

1;
(3) the density function of x0, fx0

, is a.e. continuous on R and fx0
(x) ≤

C/|x|2/3, for a.e. x 6= 0;

(4) ‖e− 1
3Ka(t,ξN )‖L2(Ω) ≤ C for all N ≥ 1 and all t ∈ [t0, T ];

(5) ψ1(t) 6= 0 for all t ∈ (t0, T ).

Then, for all 0 6= x ∈ R and t ∈ [t0, T ], the sequence {fxN (t)(x)}∞N=1 given by (3.7)
converges to the density fx(t)(x) of the solution process x(t, ω) given by (2.1).

Proof. The proof is analogous to Theorem 3.3, but with a slight modification. We
analyze the proof of [4, Thm. 2.9]. In the notation of [4, Thm. 2.9], xY − Z is
absolutely continuous, by Lemma 3.5, (2) and (5). Then the probability that xY −Z
belongs to the discontinuity set of fx0 is 0. Recalling that in [4, Thm. 2.9] one has
xYN (ω) − ZN (ω) → xY (ω) − Z(ω) a.s. as N → ∞, by the Continuous Mapping
Theorem [24, p. 7, Thm. 2.3] it follows |f0(xYN (ω)−ZN (ω))−f0(xY (ω)−Z(ω))|2 →
0 a.s. as N →∞. With this fact, the proof of [4, Thm. 2.9] is applicable, as we did
in Theorem 3.3. �

Finally, we can establish results on uniform convergence of {fxN (t)(x)}∞N=1, as a
consequence of [4, Thms. 2.4, 2.7].

Theorem 3.7. Assume the following four hypotheses:

(1) a, b ∈ L2([t0, T ]× Ω);
(2) x0 and (ξ1, . . . , ξN , η1, . . . , ηN ) are absolutely continuous and independent,

N ≥ 1;
(3) the function fx0(x3)x2 is Lipschitz on R;
(4) there exist p, q with 2 ≤ p ≤ ∞ and 4 ≤ q ≤ ∞ such that 1/p+ 2/q = 1/2,

‖µb‖Lp(t0,T ) +

∞∑
j=1

√
γj ‖ψj‖Lp(t0,T )‖ηj‖Lp(Ω) <∞,

‖e− 1
3Ka(t,ξN )‖Lq(Ω) ≤ C, for all N ≥ 1, t ∈ [t0, T ].

Then the sequence {fxN (t)(x)}∞N=1 given by (3.7) converges in L∞(J × [t0, T ]) for
every bounded set J ⊆ R\[−δ, δ], for every δ > 0, to the density fx(t)(x) of the
solution process x(t, ω) given by (2.1).

Proof. By (3.6), fy0
is Lipschitz on R. As a consequence of [4, Thm. 2.4], the limit

(3.8) holds in L∞(J × [t0, T ]). Then

‖fxN (t)(x)− fx(t)(x)‖L∞(J×[t0,T ])

= ‖
(
fyN (t)(x

1/3)− fy(t)(x
1/3)

) 1

3x2/3
‖L∞(J×[t0,T ])

≤ 1

3δ2/3
‖fyN (t)(x

1/3)− fy(t)(x
1/3)‖L∞(J×[t0,T ]) → 0 as N →∞

(3.11)

This concludes the proof. �

Theorem 3.8. Assume that

(1) a, b ∈ L2([t0, T ]× Ω), x0 ∈ L2/3(Ω);



10 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

(2) x0, η1, (ξ1, . . . , ξN , η2, . . . , ηN ) are absolutely continuous and independent,
N ≥ 1;

(3) the density function of η1, fη1 , is Lipschitz on R;
(4) ξ1, ξ2, . . . have compact support in [−A,A] (A > 0) and ψ1 > 0 on (t0, T ).

Then, for each fixed t ∈ (t0, T ], the sequence {fxN (t)(x)}∞N=1 given by (3.7) con-
verges in L∞(J) for every bounded set J ⊆ R\[−δ, δ], for every δ > 0, to the density
fx(t)(x) of the solution process x(t, ω) given by (2.1).

Proof. Since y0(ω) = x0(ω)1/3, it follows that E[y2
0 ] = E[x

2/3
0 ] < ∞ by hypothesis

(1). Then, y0 ∈ L2(Ω) and therefore hypothesis (1) of [4, Thm. 2.7] holds. In fact,
the four hypotheses of [4, Thm. 2.7] are fulfilled. Hence, fixed t ∈ (t0, T ], the limit
in (3.8) holds in L∞(J), then (3.11) follows and we are done. �

3.2. Comments on the hypotheses of the theorems. We comment on some
examples where the hypotheses of the previously established theorems hold. We
point out that these comments will be very useful later in the examples exhibited
in Section 4.

(1) In [4, pp. 29, 30], it was proved that, if a(t, ω) is Gaussian or if ξ1, ξ2, . . .
have a common compact support, then given any c ∈ R, there is a constant
C > 0 such that the inequality E[ecKa(t,ξN )] ≤ C holds for all N ≥ 1 and
t ∈ [t0, T ].

(2) The continuity on R is satisfied, for instance, by the density function of
the distributions Normal(µ, σ2), µ ∈ R and σ2 > 0; Beta(α, β), α > 1
and β > 1; Gamma(α, β), α > 1 and β > 0. The a.e. continuity from
Theorem 3.6 is satisfied in more cases: Beta(α, β), α ≥ 1 and β ≥ 1;
Uniform(α, β), α < β; Gamma(α, β), α ≥ 1 and β > 0 (in particular,
Exponential(β)); truncated normal distribution; etc.

(3) In hypothesis (3) of Theorem 3.3, the hypotheses fx0
continuous on R and

fx0
(x) ≤ C/|x|2/3 are independent, merely because there are unbounded

continuous density functions.
(4) Hypothesis (4) of Theorem 3.7 is satisfied, for example, when b(t, ω) is

a standard Brownian motion or a Brownian bridge, as it was seen in [4,
pp. 30–31] with p = 3.

3.3. Approximation of the expectation and variance of the solution sto-
chastic process. We have seen that, under some assumptions, (3.10) holds. We
would like to derive conditions under which the expectation and variance of x(t, ω)
can be approximated:

E[xN (t, ω)] =

∫
R
xfxN (t)(x) dx→

∫
R
xfx(t)(x) dx = E[x(t, ω)] (3.12)

and

V[xN (t, ω)] =

∫
R
x2fxN (t)(x) dx− E[xN (t, ω)]2

→
∫
R
x2fx(t)(x) dx− E[x(t, ω)]2 = V[x(t, ω)].

(3.13)

as N → ∞. From [4, p. 13], yN (t, ω) → y(t, ω) a.s., therefore xN (t, ω) → x(t, ω)
a.s. This implies that xN (t, ω)→ x(t, ω) in probability. By [3, p. 338 Corollary] or



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 11

[25, Thm. 2.4], if we check that

sup
N≥1

E[|xN (t, ω)|2+ε] <∞ (3.14)

for some ε > 0, then (3.12), (3.13) and x(t, ω) ∈ L2(Ω) will follow.
In what follows, we will use a consequence of Jensen’s inequality [3, p. 80]: if

a1, . . . , am ≥ 0 and p ≥ 1, then (a1 + · · · + am)p ≤ mp−1(ap1 + · · · + apm). For ease
of notation, we will denote by Cε any positive constant only depending on ε.

By the triangular, Jensen’s, and Hölder’s inequalities, we have

E[|xN (t)|2+ε]

= E
[∣∣∣x1/3

0 e
1
3Ka(t,ξN ) +

1

3

∫ t

t0

Sb(s,ηN )e
1
3 (Ka(t,ξN )−Ka(s,ξN )) ds

∣∣∣3(2+ε)]
≤ Cε

(
E
[∣∣x1/3

0 e
1
3Ka(t,ξN )

∣∣3(2+ε)]
+ E

[∣∣∣1
3

∫ t

t0

Sb(s,ηN )e
1
3 (Ka(t,ξN )−Ka(s,ξN )) ds

∣∣∣3(2+ε)])
≤ Cε

(
E
[
|x0|2+ε e(2+ε)Ka(t,ξN )

]
+ E

[ ∫ t

t0

|Sb(s,ηN )|3(2+ε)e(2+ε)(Ka(t,ξN )−Ka(s,ξN )) ds
])
.

(3.15)

If ξ1, ξ2, . . . have compact support in a common interval, then there is a constant
Cε > 0 such that e(2+ε)Ka(t,ξN ) ≤ Cε, for all N ≥ 1 and t ∈ [t0, T ]. This is a
consequence of [4, p. 30]. In this case, from (3.15),

E[|xN (t)|2+ε] ≤ Cε
(
E[|x0|2+ε] + E

[ ∫ t

t0

|Sb(s,ηN )|3(2+ε) ds
])
.

Thus, it suffices to have ‖x0‖L2+ε(Ω) <∞ and

‖µb‖L3(2+ε)([t0,T ]) +

∞∑
i=1

√
γi‖ψi‖L3(2+ε)([t0,T ])‖ηi‖L3(2+ε)(Ω) <∞,

to ensure that (3.14) holds.
Otherwise, if ξ1, ξ2, . . . do not have compact support in a common interval, one

continues from (3.15) by applying Hölder’s inequality with exponents r1 = 1+δ > 1
and r2 = (1 + δ)/δ:

E[|xN (t)|2+ε]

≤ Cε
(
E
[
|x0|(2+ε)(1+δ)

] 1
1+δE

[
e(2+ε)r2Ka(t,ξN )

]1/r2
+

∫ t

t0

E
[
|Sb(s,ηN )|3(2+ε)(1+δ)

] 1
1+δE

[
e(2+ε)r2(Ka(t,ξN )−Ka(s,ξN ))

]1/r2
ds
)
.

If a(t, ω) is a Gaussian process, then, by the first paragraph of Subsection 3.2 and
Hölder’s inequality,

E[|xN (t)|2+ε]

≤ Cε,δ
(
E
[
|x0|(2+ε)(1+δ)

] 1
1+δ +

∫ t

t0

E
[
|Sb(s,ηN )|3(2+ε)(1+δ)

] 1
1+δ ds

)



12 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

≤ Cε,δ
(
E
[
|x0|(2+ε)(1+δ)

] 1
1+δ +

(∫ t

t0

E
[
|Sb(s,ηN )|3(2+ε)(1+δ)

]
ds
) 1

1+δ
)
.

Thereby, if ‖x0‖L2+s(Ω) <∞ and

‖µb‖L6+s([t0,T ]) +

∞∑
i=1

√
γi‖ψi‖L6+s([t0,T ])‖ηi‖L6+s(Ω) <∞,

Inequality (3.14) holds. Summarizing, the following theorem has been established.

Theorem 3.9. If a(t, ω) is a Gaussian process or ξ1, ξ2, . . . have a common com-
pact support, if ‖x0‖L2+s(Ω) <∞ and if

‖µb‖L6+s([t0,T ]) +

∞∑
i=1

√
γi‖ψi‖L6+s([t0,T ])‖ηi‖L6+s(Ω) <∞ (3.16)

for some s > 0, then x(t, ω) ∈ L2(Ω) and xN (t, ω) tends in L2(Ω) to x(t, ω), for each
t ∈ [t0, T ]. As a consequence, E[xN (t, ω)]→ E[x(t, ω)] and V[xN (t, ω)]→ V[x(t, ω)]
as N →∞, for each t ∈ [t0, T ].

4. Numerical examples

In this section we show examples where the theoretical findings of this paper
are illustrated. We choose specific stochastic processes a(t, ω) and b(t, ω) (via
their Karhunen-Loève expansions) and an initial condition x0(ω) in the random
Bertalanffy model (1.1), and then we compute the approximating density function
fxN (t)(x) given by (3.7) for different values of N ≥ 1. We also check that it con-
verges to a function, which will be fx(t)(x), as an application of Theorems 3.3, 3.4,
3.6, 3.7 or 3.8. We inform the reader that (1)–(4) listed in Subsection 3.2 will be ex-
tensively used throughout this section to check that the hypotheses of the involved
theorems hold within the context of each example.

To compute fxN (t)(x), we have used the software Mathematica. We have pro-
grammed a Monte Carlo procedure to compute the expectation in (3.7): fixed
N ≥ 1, we obtain M realizations of each random variable ξ1, . . . , ξN , η1, . . . , ηN :

realizations of ξ1: ξ
(1)
1 , . . . , ξ

(M)
1 ,

. . .

realizations of ηN : η
(1)
N , . . . , η

(M)
N .

We denote ξ
(i)
N = (ξ

(i)
1 , . . . , ξ

(i)
N ) and η

(i)
N = (η

(i)
1 , . . . , η

(i)
N ), i = 1, . . . ,M . Then

fxN (t)(x)

≈ 1

x2/3

1

M

M∑
i=1

[
fx0

({
x1/3e−

1
3Ka(t,ξ

(i)
N ) − 1

3

∫ t

t0

Sb(s,η
(i)
N )e−

1
3Ka(s,ξ

(i)
N ) ds

}3)
×
{
x1/3e−

1
3Ka(t,ξ

(i)
N ) − 1

3

∫ t

t0

Sb(s,η
(i)
N )e−

1
3Ka(s,ξ

(i)
N ) ds

}2

e−
1
3Ka(t,ξ

(i)
N )
]
,

(4.1)

with convergence as M → ∞, by the Law of Large Numbers. We take M large
enough in such a way that expression (4.1) coincides for orders of truncation M
and M ′ > M (convergence in (4.1)).



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 13

We perform three examples in what follows. In the first example, both processes
a(t, ω) and b(t, ω) are Gaussian. In the second and third examples, the processes
involved may not be Gaussian.

Example 4.1. Let

a(t, ω) =

∞∑
j=1

√
2

(j − 1
2 )π

sin
(
t(j − 1

2
)π
)
ξj(ω),

where ξ1, ξ2, . . . are independent and Normal(0, 1) random variables, be a standard
Brownian motion on [t0, T ] = [0, 1] [15, Exercise 5.12]. Let

b(t, ω) =

∞∑
i=1

√
2

iπ
sin(tiπ)ηi(ω),

where ξ1, ξ2, . . . , η1, η2, . . . are independent and Normal(0, 1) distributed random
variables, be a standard Brownian bridge on [0, 1] [15, Example 5.30]. The sums
defining a(t, ω) and b(t, ω) converge in L2([0, 1] × Ω). Let x0 ∼ Beta(4, 6). It is
assumed x0, ξ1, ξ2, . . . and η1, η2, . . . to be independent.

Since a(·, ω) and b(·, ω) are continuous on [0, 1], by Theorem 2.1, the stochastic
process x(t, ω) given by (2.1) has C1([t0, T ]) sample paths that satisfy the random
Bertalanffy model (1.1). Moreover, by Theorem 2.2 and the subsequent Exam-
ple 2.3, x(t, ω) is a solution to the random Bertalanffy model (1.1) in the mean
square sense.

The hypotheses of Theorems 3.3 and 3.7 are satisfied. Indeed, fx0 is continuous
on R by (2) in subsection 3.2 and fx0

(x) ≤ C/|x|2/3 holds because fx0
is continuous

with compact support, therefore (3) of Theorem 3.3 fulfills. Hypothesis (4) of The-
orem 3.3 is satisfied because a(t, ω) is Gaussian and (1). Regarding Theorem 3.7,
hypothesis (3) holds, since g(x) := fx0

(x3)x2 = 504x11(1 − x3)51(0,1)(x) is differ-

entiable on R, with g′(x) = 504x10(1 − x3)4(11(1 − x3) − 15x3)1(0,1)(x) bounded
on R. Moreover, (4) is fulfilled by (1) in subsection 3.2 applied to a(t, ω) and (4)
in subsection 3.2 applied to b(t, ω).

Hence, the sequence {fxN (t)(x)} converges in L∞(J × [t0, T ]) for every bounded
set J ⊆ R\[−δ, δ], for every δ > 0, to the density fx(t)(x) of the solution process
x(t, ω).

The assumptions of Theorem 3.9 are fulfilled. Indeed, by taking s = 1 in (3.16),

‖µb‖L7([0,1]) +

∞∑
i=1

√
γi‖ψi‖L7([0,1])‖ηi‖L7(Ω)

= M7

∞∑
i=1

√
2

iπ

(∫ 1

0

| sin(tiπ)|7 dt
)1/7

= M7

∞∑
i=1

1

iπ

2 14
√

2 7
√

1/i
7
√

35π
<∞,

(4.2)

where M7 = ‖Z‖L7(Ω), being Z ∼ Normal(0, 1). Hence, E[xN (t, ω)] → E[x(t, ω)]
and V[xN (t, ω)]→ V[x(t, ω)] as N →∞.



14 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

In Figure 1, we observe the graph of fxN (0.5)(x), x ∈ R, for N = 1, 2, 3, 4, 5, 6.
The theoretical convergence of the sequence {fxN (0.5)(x)}∞N=1 agrees with the nu-
merical results of Figure 1. In Table 1, we have presented the expectation and vari-
ance for N = 1, 2, 3, 4, 5, 6, by using the formulas E[xN (0.5, ω)] =

∫
R xfxN (0.5)(x) dx

and V[xN (0.5, ω)] =
∫
R x

2fxN (0.5)(x) dx− E[xN (0.5, ω)]2.

-1.0 -0.5 0.5 1.0 1.5 2.0
x

0.5

1.0

1.5

2.0

f(x)

N=1

N=2

N=3

N=4

N=5

N=6

Figure 1. Density fxN (0.5)(x) for N = 1, 2, 3, 4, 5, 6. Example 4.1.

Table 1. E[xN (0.5, ω)] and V[xN (0.5, ω)] for N = 1, 2, 3, 4, 5, 6 in
Example 4.1

N 1 2 3 4 5 6
E[xN (0.5, ω)] 0.4109 0.4139 0.4145 0.4145 0.4146 0.4146
V[xN (0.5, ω)] 0.0345 0.0391 0.0396 0.0396 0.0397 0.0397

Example 4.2. We work on [t0, T ] = [0, 1]. Let

a(t, ω) =

∞∑
j=1

√
2

j3
sin(tjπ)ξj(ω), (4.3)

where ξ1, ξ2, . . . are independent with distribution Uniform(−
√

3,
√

3). The sum in
(4.3) converges in L2([0, 1] × Ω): given M < N , by using Pythagoras’s Theorem,
we have

‖
N∑

j=M+1

√
2

j3
sin(tjπ)ξj(ω)‖2L2([0,1]×Ω) =

N∑
j=M+1

2

j6
,

and since
∑∞
j=1 2/j6 <∞, the partial sums {

∑N
j=1(
√

2/j3) sin(tjπ)ξj(ω)}∞N=1 form

a Cauchy sequence in L2([0, 1]× Ω), therefore convergent.
Let

b(t, ω) =

∞∑
i=1

√
2

i4 + 6
sin(tiπ)ηi(ω), (4.4)

where η1, η2, · · · ∼ Normal(0, 1) are independent. The sum defining b(t, ω) in (4.4)
exists in L2([0, 1]× Ω), reasoning as in (4.3).



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 15

Let x0 follow an Exponential(2) distribution. It is assumed x0, ξ1, ξ2, . . . and
η1, η2, . . . to be independent.

By Theorem 2.1, the process x(t, ω) given by (2.1) has absolutely continuous
sample paths that solve the random Bertalanffy model (1.1). In fact, a(t, ω) and
b(t, ω) have continuous sample paths. To prove the continuity for a(t, ω), we bound

√
2

j3
| sin(tjπ)ξj(ω)| ≤

√
6

j3
,

with
∑∞
j=1 1/j3 <∞, and by using Weierstrass M-test for uniform convergence of

series [19, Thm. 7.10], we deduce that the series in (4.3) converges uniformly on
[0, 1], so a(t, ω) has continuous sample paths. To prove the continuity for b(t, ω),
one has to work a bit more, since ηi is not bounded. Notice that

P(|ηi| ≥ i) = 2

∫ ∞
i

1√
2π

e−
x2

2 dx ≤ 2√
2π

∫ ∞
i

x

i
e−

x2

2 dx =
2√
2π i

e−i
2/2 ≤ e−i

2/2;

therefore
∞∑
i=1

P(|ηi| ≥ i) ≤
∞∑
i=1

e−i
2/2 <∞.

By Borel-Cantelli lemma [3, Thm. 4.3], for a.e. ω ∈ Ω, there exists an i0(ω) ≥ 1
such that, for all i ≥ i0(ω), |ηi(ω)| ≤ i. Thus, for all i ≥ i0(ω),

√
2

i4 + 6
| sin(tiπ)ηi(ω)| ≤

√
2 i

i4 + 6
.

Since
∑∞
i=1 i/(i

4 +6) <∞, by Weierstrass M-test for uniform convergence of series,
we derive that the series defining b(t, ω) in (4.4) converges uniformly on [0, 1],
therefore b(t, ω) has continuous sample paths. As a consequence, by Theorem 2.1,
x(t, ω) has C1([0, 1]) sample paths that solve the random Bertalanffy model (1.1).

The process x(t, ω) is a mean square solution to the random Bertalanffy model
(1.1), as the hypotheses of Theorem 2.2 are accomplished. Indeed, a(t, ω) is con-
tinuous in the L12(Ω) sense, since

‖a(t+ h, ω)− a(t, ω)‖L12(Ω) ≤
√

3

∞∑
j=1

√
2

j3
| sin((t+ h)jπ)− sin(tjπ)| → 0

as h → 0, where we have used the fact that |ξj(ω)| ≤
√

3 and, for the last limit,
the Dominated Convergence Theorem [19, result 11.32, p. 321] applied to the last
series. On the other hand,∣∣ ∫ t

0

a(s, ω) ds
∣∣ ≤ ∫ t

0

|a(s, ω)|ds ≤
∞∑
j=1

√
2

j3

√
3 <∞,

so the third hypothesis of Theorem 2.2 holds. The other process, b(t, ω), is L12(Ω)-
continuous:

‖b(t+ h, ω)− b(t, ω)‖L12(Ω) ≤M12

∞∑
i=1

√
2

i4 + 6
| sin((t+ h)jπ)− sin(tjπ)| → 0

as h → 0, where M12 = ‖Z‖L12(Ω), being Z ∼ Normal(0, 1). For the limit, the
Dominated Convergence Theorem has been applied again. Thus, the hypotheses of
Theorem 2.2 hold, as stated, and the process x(t, ω) is a mean square solution.



16 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

Theorems 3.4 and 3.8 are applicable. Indeed, hypothesis (3) of Theorem 3.4
is satisfied, because fη1 is the density function of a normal distribution, which is
continuous and bounded. Assumption (4) of Theorem 3.4 is accomplished, because

the density function of a Uniform(−
√

3,
√

3) distribution has compact support and

ψ1(t) =
√

2 sin(tπ) > 0 on (0, 1). Concerning Theorem 3.8, (3) holds because the
density function of a normal distribution is Lipschitz and (4) is the same as in
Theorem 3.4.

Therefore, for each t ∈ (t0, T ], the sequence {fxN (t)(x)} converges in L∞(J) for
every bounded set J ⊆ R\[−δ, δ], for every δ > 0, to the density fx(t)(x) of the
solution process x(t, ω).

The assumptions of Theorem 3.9 are fulfilled, by doing similar computations to
the ones in (4.2). As a consequence, E[xN (t, ω)] → E[x(t, ω)] and V[xN (t, ω)] →
V[x(t, ω)] as N →∞.

In Figure 2 we show the graph of fxN (0.3)(x) for N = 1, 2, 3, 4, 5, 6. In Table 2,
we approximate the mean and variance of x(t, ω), with orders of truncation N =
1, 2, 3, 4, 5, 6. The numerical results agree with our theoretical findings.

-4 -2 2 4
x

0.5

1.0

1.5

2.0

f(x)

N=1

N=2

N=3

N=4

N=5

N=6

Figure 2. Density fxN (0.3)(x) for N = 1, 2, 3, 4, 5, 6 in Example 4.2.

Table 2. E[xN (0.3, ω)] and V[xN (0.3, ω)] for N = 1, 2, 3, 4, 5, 6 in
Example 4.2

N 1 2 3 4 5 6
E[xN (0.3, ω)] 0.5071 0.5078 0.5076 0.5076 0.5076 0.5076
V[xN (0.3, ω)] 0.2754 0.2769 0.2768 0.2768 0.2768 0.2768

Example 4.3. We work on [t0, T ] = [0, 1]. Let

a(t, ω) =

∞∑
j=1

√
2

j
sin(tjπ)ξj(ω),

where ξ1, ξ2, . . . are independent with distribution Uniform(−
√

3,
√

3). Let

b(t, ω) =

∞∑
i=1

√
2

i
3
2

sin(tiπ)ηi(ω),



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 17

where η1, η2, . . . are independent having distribution Uniform(−
√

3,
√

3). Let the

initial condition be x0 ∼ Uniform(−
√

3,
√

3). It is assumed x0, ξ1, ξ2, . . . and
η1, η2, . . . to be independent. As in Example 4.2, both series converge in L2([0, 1]×
Ω), as a consequence of Pythagoras’s Theorem.

The process x(t, ω) has absolutely continuous sample paths that solve the random
Bertalanffy model, by Theorem 1.1.

Let us see that the hypotheses of Theorem 3.6 are satisfied. The density function
fx0 is a.e. continuous (see (2) in subsection 3.2). The inequality fx0(x) ≤ C/|x|2/3
holds because x0 has compact support, so (3) is satisfied. Hypotheses (4) is a
consequence of (1) in subsection 3.2. Finally, (5) holds because ψ1(s) = sin(sπ) >
0 for s ∈ (0, 1). Thus, by Theorem 3.6, the sequence {fxN (t)(x)}∞N=1 converges
pointwise to fx(t)(x).

The hypotheses of Theorem 3.9 are fulfilled, by doing similar computations to
the ones in (4.2). Hence, E[xN (t, ω)] → E[x(t, ω)] and V[xN (t, ω)] → V[x(t, ω)] as
N →∞.

In Figure 3, we observe that the sequence {fxN (0.3)(x)}∞N=1 seems to converge,
for N = 1− 6. In Table 3, we approximate the expectation and variance of x(t, ω).

-4 -2 2 4
x

0.1

0.2

0.3

0.4

0.5

f(x)

N=1

N=2

N=3

N=4

N=5

N=6

Figure 3. Density fxN (0.3)(x) for N = 1, 2, 3, 4, 5, 6 in Example 4.3.

Table 3. E[xN (0.3, ω)] and V[xN (0.3, ω)] for N = 1, 2, 3, 4, 5, 6 in
Example 4.3

N 1 2 3 4 5 6
E[xN (0.3, ω)] 0.0014 0.0001 0.0001 0.0002 0.0002 0.0002
V[xN (0.3, ω)] 1.1240 1.1914 1.2213 1.2298 1.2299 1.2300

Conclusions. In this paper, we have analyzed the random non-autonomous Berta-
lanffy model: x′(t, ω) = a(t, ω)x(t, ω) + b(t, ω)x(t, ω)2/3, t ∈ [t0, T ], with initial
condition x(t0, ω) = x0(ω). The coefficients are stochastic processes a(t, ω) and
b(t, ω) and the initial condition is a random variable x0(ω) in an underlying com-
plete probability space. Via the usual change of variables for solving determinis-
tic Bernoulli differential equations, we have related the random non-autonomous



18 J. CALATAYUD, T. CARABALLO, J. C. CORTÉS, M. JORNET EJDE-2020/50

Bertalanffy model with a random non-autonomous linear differential equation, and
thus we have obtained a formal solution stochastic process x(t, ω). Theorem 2.1
tells us when x(t, ω) has absolutely continuous sample paths that solve the ran-
dom Bertalanffy model a.e. Theorem 2.2 gives conditions on the moments of a
and b under which x(t, ω) is a mean square solution. By using existing results
on the random non-autonomous linear differential equation, the Random Variable
Transformation technique and Karhunen-Loève expansions, we have constructed a
sequence of density functions that, under certain assumptions, converge pointwise
(Theorems 3.3, 3.4 and 3.6) and uniformly (Theorems 3.7 and 3.8) to the density
function of x(t, ω), fx(t)(x). Results on the approximation of the expectation and
the variance of x(t, ω) have been achieved. Finally, these theoretical findings have
been numerically assessed in the computer.

Acknowledgements. This work was supported by the Spanish Ministerio de
Economı́a, Industria y Competitividad (MINECO), by the Agencia Estatal de
Investigación (AEI) and Fondo Europeo de Desarrollo Regional (FEDER UE)
grant MTM2017–89664–P. Marc Jornet acknowledges the doctorate scholarship
granted by Programa de Ayudas de Investigación y Desarrollo (PAID), Universitat
Politècnica de València.

References

[1] J. R. Banavar, J. Damuth, A. Maritan, A. Rinaldo; Ontogenetic growth. Modelling univer-
sality and scaling, Nature, 420 (2002) 626. doi: 10.1038/420626a.

[2] L. V. Bertalanffy; Quantitative laws in metabolism and growth, The Quarterly Review of

Biology, 32(3) (1957) 217–231.
[3] P. Billingsley; Probability and Measure, 3rd Edition, John Wiley & Sons, 1995.

[4] J. Calatayud, J.-C. Cortés, M. Jornet; Constructing approximate solutions of the random-
ized non-autonomous complete linear differential equation via probability density functions,

Electronic Journal of Differential Equations, 2019 (85) (2019) 1–40.

[5] M.-C. Casabán, J.-C. Cortés, A. Navarro-Quiles, J.-V. Romero, M.-D. Roselló; Computing
probabilistic solutions of the Bernoulli random differential equation, Journal of Computa-

tional and Applied Mathematics, 309 (2017) 396–407.

[6] M.-C. Casabán, J.-C. Cortés, J.-V. Romero, M.-D. Roselló; Probabilistic solution of ran-
dom SI-type epidemiological models using the Random Variable Transformation technique,

Communications in Nonlinear Science and Numerical Simulation, 24 (1-3) (2015) 86–97. doi:

10.1016/j.cnsns.2014.12.016.
[7] M.-C. Casabán, J.-C. Cortés, J.-V. Romero, M.-D. Roselló; Determining the first probability

density function of linear random initial value problems by the random variable transforma-

tion (RVT) technique: a comprehensive study, Abstract and Applied Analysis 2014–ID248512
(2014) 1–25. doi: 10.1155/2013/248512.

[8] M.-C. Casabán, J.-C. Cortés, J.-V. Romero, M.-D. Roselló; Solving Random Homogeneous
Linear Second-Order Differential Equations: A Full Probabilistic Description, Mediterranean

Journal of Mathematics, 13(6) (2016) 3817–3836. doi:10.1016/10.1007/s00009-016-0716-6.
[9] J.-C. Cortés, L. Jódar, M.-D. Roselló, L. Villafuerte; Solving initial and two-point boundary

value linear random differential equations: A mean square approach, Applied Mathematics
and Computation, 219(4) (2012) 2204–2211.

[10] J.-C. Cortés, A. Navarro-Quiles, J.-V. Romero, M.-D. Roselló; Probabilistic solution of ran-
dom autonomous first-order linear systems of ordinary differential equations, Romanian Re-

ports in Physics, 68 (4) (2016) 1397–1406.
[11] J. K. Hale; Ordinary Differential Equations, Malabar: Robert E. Krieger Publishing Com-

pany, Second edition, 1980.
[12] A. Hussein, M. M. Selim; Solution of the stochastic radiative transfer equation with Rayleigh

scattering using RVT technique, Applied Mathematics and Computation 218(13) (2012)
7193–7203. doi: 10.1016/j.amc.2011.12.088.



EJDE-2020/50 RANDOMIZED NON-AUTONOMOUS BERTALANFFY MODEL 19

[13] A. Hussein, M. M. Selim; Solution of the stochastic generalized shallow-water wave

equation using RVT technique, European Physical Journal Plus (2015) 130:249.

doi:10.1140/epjp/i2015-15249-3.
[14] H. Kobayashi, B. L. Mark, W. Turin; Probability, Random Processes, and Statistical Analysis,

Cambridge University Press, 2011.

[15] G. J. Lord, C. E. Powell, T. Shardlow; An Introduction to Computational Stochastic PDEs,
Cambridge Texts in Applied Mathematics, Cambridge University Press, New York, 2014.

[16] B. Øksendal; Stochastic Differential Equations. An Introduction with Applications, Springer-

Verlag, Series: Stochastic Modelling and Applied Probability 23, Heidelberg and New York,
2003.

[17] L. V. Qiming, J. W. Pitchford; Stochastic von Bertalanffy models with applications to fish re-

cruitment, Journal of Theoretical Biology 244 (2007) 640–655. doi: 10.1016/j.jtbi.2006.09.009.
[18] K. Renner-Martin, N. Brunner, M. Kuhleitner, W. G. Nowak, K. Scheicher; On the exponent

in the Von Bertalanffy growth model, PeerJ 6:e4205 (2018). doi:10.7717/peerj.4205.
[19] W. Rudin; Principles of Mathematical Analysis, 3rd Edition, International Series in Pure &

Applied Mathematics, 1976.

[20] T. L. Saaty; Modern Nonlinear Equations, Dover Publications Inc., New York, 2003.
[21] M. A. Shah; Stochastic logistic model for fish growth, Open Journal of Statistics, 4 (2014)

11–18. doi: 10.4236/ojs.2014.41002.

[22] T. T. Soong; Random Differential Equations in Science and Engineering, Academic Press,
New York, 1973.

[23] J. L. Strand; Random Ordinary Differential Equations, Journal of Differential Equations, 7

(1970) 538–553.
[24] A. W. van der Vaart; Asymptotic Statistics, Cambridge University Press, 1998.

[25] L. Villafuerte, C. A. Braumann, J.-C. Cortés, L. Jódar; Random differential operational

calculus: Theory and applications, Computers and Mathematics with Applications, 59 (2010)
115–125.

[26] C. R. White, R. S. Seymour; Mammalian basal metabolic rate is proportional to body mass
2/3, in: Proceedings of the National Academy of Sciences of the United States of America,

100, 4046–4049. doi: 10.1073/pnas.0436428100.

[27] A. C. Zaanen; Introduction to Operator Theory in Riesz Spaces, Springer, Berlin Heidelberg,
1996.

Julia Calatayud

Instituto Universitario de Matemática Multidisciplinar, Universitat Politècnica de

València, Camino de Vera s/n, 46022, Valencia, Spain
Email address: jucagre@doctor.upv.es

Tomás Caraballo
Dpto. Ecuaciones Diferenciales y Análisis Numérico, Universidad de Sevilla, c/ Tarfia

s/n, 41012, Sevilla, Spain

Email address: caraball@us.es

Juan Carlos Cortés
Instituto Universitario de Matemática Multidisciplinar, Universitat Politècnica de
València, Camino de Vera s/n, 46022, Valencia, Spain

Email address: jccortes@imm.upv.es

Marc Jornet
Instituto Universitario de Matemática Multidisciplinar, Universitat Politècnica de

València, Camino de Vera s/n, 46022, Valencia, Spain
Email address: marjorsa@doctor.upv.es


	1. Introduction and motivation
	2. Solving the random non-autonomous Bertalanffy model
	2.1. Solution stochastic process with absolutely continuous sample paths
	2.2. Solution stochastic process in the mean square sense

	3. Obtaining the probability density function of the solution stochastic process
	3.1. Main results
	3.2. Comments on the hypotheses of the theorems
	3.3. Approximation of the expectation and variance of the solution stochastic process

	4. Numerical examples
	Conclusions
	Acknowledgements

	References