Mathematical methods for the randomized non-autonomous Bertalanffy model

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, TOM´ AS CARABALLO, JUAN CARLOS CORT´ ES, MARC JORNET Abstract. In this article we analyze the randomized non-autonomous Bertalanffy model x0(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 variable, all of them defined in an underlying complete probability space. Under certain assumptions on a,band 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-Lo`eve 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, W0(t) = ηW m(t)−κWn(t), where ηand κare 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-Lo`eve 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. CORT´ ES, M. JORNET EJDE-2020/50 becomes W0(t) = ηW 2/3(t)−κW(t). In this article, we want to perform a mathematical study of the randomized non-autonomous Bertalanffy model, x0(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 Fis the σ-algebra of events and Pis 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 Lpsense. 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] = ZR xkfx(t)(x) dx, P(a1≤x(t, ω)≤a2) = Za2 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 random variables (time invariant) rather than stochastic processes. The random autonomous 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 differential equation, by focusing on approximating the probability density function of the solution stochastic process. Its main contribution has been using Karhunen- Lo`eve 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 Itˆo type [16], have been carried out in [17, 21]. In SDEs of Itˆo type, the uncertainty is introduced in the deterministic differential 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 nonautonomous 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 functions of the solution processes to the random non-autonomous Bertalanffy model and the random non-autonomous linear differential equation. The Karhunen-Lo`eve 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, Lpproblem]. Consider the stochastic process x(t, ω) = x0(ω)1/3e1 3Rt t0a(s,ω) ds+1 3Zt t0 b(s, ω)e1 3Rt sa(r,ω) drds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 considered 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 y0(t, ω) = 1 3a(t, ω)y(t, ω) + 1 3b(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)3is 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. CORT´ ES, 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)1/3e1 3Rt t0a(s,ω) ds+1 3Zt t0 b(s, ω)e1 3Rt sa(r,ω) drds(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,aand bare independent. (ii) x0∈L4(Ω), and aand bare L12(Ω) continuous. (iii) There exist r > 12 and δ > 0such that sup s,s∗∈[−δ,δ] EerRt+s∗ x+sa(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 ka(t+h, ω)−a(t, ω)kL12(Ω) =k√h ZkL12(Ω) =√hkZkL12(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erRt+s∗ x+sa(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 Uany bounded random variable. Indeed, a(t, ω) is L12(Ω)-continuous, since ka(t+h, ω)−a(t, ω)kL12(Ω) =|h|kUkL12(Ω) h→0 −→ 0. On the other hand, EerRt+s∗ x+sa(u,ω) du≤EerkUkL∞(Ω) Rt+s∗ x+sudu= erkUkL∞(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-Lo`eve 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 Xbe an absolutely continuous random variable with density fXand with support DX contained in an open set D⊆R. Let g:D→Rbe such that D=∪n i=1Diand gi=g|Diis 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) = (Pi:y∈g(Di)fX(g−1 i(y))|dg−1 i(y) dy|, y ∈g(D), 0, y /∈g(D). Lemma 3.2 (Karhunen-Lo`eve 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) + ∞ X j=1 √νjφj(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) = ZT 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- Lo`eve expansion: a(t, ω) = µa(t) + ∞ X j=1 √νjφj(t)ξj(ω), b(t, ω) = µb(t) + ∞ X j=1 √γjψj(t)ηj(ω),(3.3) 6 J. CALATAYUD, T. CARABALLO, J. C. CORT´ ES, 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 Chave 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-Lo`eve 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-Lo`eve expansions (3.3) of the stochastic processes aand bof common order N aN(t, ω) = µa(t) + N X j=1 √νjφj(t)ξj(ω), bN(t, ω) = µb(t) + N X j=1 √γjψj(t)ηj(ω). This gives a truncation of the solution stochastic process x(t, ω) given by (2.1), xN(t, ω) = x0(ω)1/3e1 3Rt t0aN(s,ω) ds+1 3Zt t0 bN(s, ω)e1 3Rt saN(r,ω) drds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)1/3e1 3Rt t0aN(s,ω) ds+1 3Zt t0 bN(s, ω)e1 3Rt saN(r,ω) drds. (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) = Zt t0µa(s) + N X j=1 √νjφj(s)ξjds, Sb(s, ηN) = µb(s) + N X i=1 √γiψi(s)ηi. Suppose that x0and (ξ1, . . . , ξN, η1, . . . , ηN) are absolutely continuous and independent, for each N≥1. Let y0(ω) = x0(ω)1/3be the initial condition of the random linear differential equation (2.2). By Lemma 3.1 applied with the transformation mapping g(x) = x1/3on 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 ZR2N fy0ye−1 3Ka(t,ξN)−1 3Zt t0 Sb(s, ηN)e−1 3Ka(s,ξN)ds ×e−1 3Ka(t,ξN)fξN,ηN(ξN,ηN) dξNdηN =Ehfy0ye−1 3Ka(t,ξN)−1 3Zt t0 Sb(s, ηN)e−1 3Ka(s,ξN)dse−1 3Ka(t,ξN)i =Ehfx0nye−1 3Ka(t,ξN)−1 3Zt t0 Sb(s, ηN)e−1 3Ka(s,ξN)dso3 ×3nye−1 3Ka(t,ξN)−1 3Zt t0 Sb(s, ηN)e−1 3Ka(s,ξN)dso2e−1 3Ka(t,ξN)i, for y∈R. Since xN(t, ω) = yN(t, ω)3, by Lemma 3.1 with the transformation mapping g(x) = x3on 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 x2/3Ehfx0nx1/3e−1 3Ka(t,ξN)−1 3Zt t0 Sb(s, ηN)e−1 3Ka(s,ξN)dso3 nx1/3e−1 3Ka(t,ξN)−1 3Zt t0 Sb(s, ηN)e−1 3Ka(s,ξN)dso2e−1 3Ka(t,ξN)i, (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 transformation mapping g(x) = x3on 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. CORT´ ES, 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) x0and (ξ1, . . . , ξN, η1, . . . , ηN)are absolutely continuous and independent, N≥1; (3) the density function of x0,fx0, is continuous on Rand fx0(x)≤C |x|2/3, for x6= 0; (4) ke−1 3Ka(t,ξN)kL2(Ω) ≤Cfor all N≥1and all t∈[t0, T ]. Then, for all 06=x∈Rand 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), y0and (ξ1, . . . , ξN, η1, . . . , ηN) are absolutely continuous and independent, for N≥1. By (3.6) and hypothesis (3), fy0(y) is continuous on Rand fy0(y)≤C |y3|2/33y2= 3C, for y6= 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>0on (t0, T). Then, for each 06=x∈Rand 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 Uand Vbe two independent random variables. If Uis absolutely continuous, then U+Vis absolutely continuous. Proof. For any Borel set A, by the convolution formula [3, p. 266] we have P(U+V∈ A) = RRP(U∈A−v)PV(dv), where PV=P◦V−1is the law of V. If Ais null, then A−vis null, so P(U∈A−v) = 0. Thus, if Ais null, then P(U+V∈A) = 0. By the Radon-Nikodym Theorem [27, Ch. 14], U+Vhas 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, . . . , ηNare absolutely continuous and independent, N≥ 1; (3) the density function of x0,fx0, is a.e. continuous on Rand fx0(x)≤ C/|x|2/3, for a.e. x6= 0; (4) ke−1 3Ka(t,ξN)kL2(Ω) ≤Cfor all N≥1and all t∈[t0, T ]; (5) ψ1(t)6= 0 for all t∈(t0, T). Then, for all 06=x∈Rand 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 −Zis absolutely continuous, by Lemma 3.5, (2) and (5). Then the probability that xY −Z belongs to the discontinuity set of fx0is 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) x0and (ξ1, . . . , ξN, η1, . . . , ηN)are absolutely continuous and independent, N≥1; (3) the function fx0(x3)x2is Lipschitz on R; (4) there exist p, q with 2≤p≤ ∞ and 4≤q≤ ∞ such that 1/p + 2/q = 1/2, kµbkLp(t0,T )+ ∞ X j=1 √γjkψjkLp(t0,T )kηjkLp(Ω) <∞, ke−1 3Ka(t,ξN)kLq(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), fy0is Lipschitz on R. As a consequence of [4, Thm. 2.4], the limit (3.8) holds in L∞(J×[t0, T]). Then kfxN(t)(x)−fx(t)(x)kL∞(J×[t0,T ]) =kfyN(t)(x1/3)−fy(t)(x1/3)1 3x2/3kL∞(J×[t0,T ]) ≤1 3δ2/3kfyN(t)(x1/3)−fy(t)(x1/3)kL∞(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); 16 J. CALATAYUD, T. CARABALLO, J. C. CORT´ ES, M. JORNET EJDE-2020/50 Theorems 3.4 and 3.8 are applicable. Indeed, hypothesis (3) of Theorem 3.4 is satisfied, because fη1is 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 N1 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) = ∞ X j=1 √2 jsin(tjπ)ξj(ω), where ξ1, ξ2, . . . are independent with distribution Uniform(−√3,√3). Let b(t, ω) = ∞ X i=1 √2 i3 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 fx0is a.e. continuous (see (2) in subsection 3.2). The inequality fx0(x)≤C/|x|2/3 holds because x0has 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 N1 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 Bertalanffy model: x0(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 complete probability space. Via the usual change of variables for solving deterministic Bernoulli differential equations, we have related the random non-autonomous 18 J. CALATAYUD, T. CARABALLO, J. C. CORT´ ES, 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 random Bertalanffy model a.e. Theorem 2.2 gives conditions on the moments of a and bunder 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-Lo`eve 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 Econom´ıa, Industria y Competitividad (MINECO), by the Agencia Estatal de Investigaci´on (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 Investigaci´on y Desarrollo (PAID), Universitat Polit`ecnica de Val`encia. References [1] J. R. Banavar, J. 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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 Matem´ atica Multidisciplinar, Universitat Polit` ecnica de Val` encia, Camino de Vera s/n, 46022, Valencia, Spain Email address:[email protected] Tom´ as Caraballo Dpto. Ecuaciones Diferenciales y An´ alisis Num´ erico, Universidad de Sevilla, c/ Tarfia s/n, 41012, Sevilla, Spain Email address:[email protected] Juan Carlos Cort´ es Instituto Universitario de Matem´ atica Multidisciplinar, Universitat Polit` ecnica de Val` encia, Camino de Vera s/n, 46022, Valencia, Spain Email address:[email protected] Marc Jornet Instituto Universitario de Matem´ atica Multidisciplinar, Universitat Polit` ecnica de Val` encia, Camino de Vera s/n, 46022, Valencia, Spain Email address:[email protected]