Application of Differential Equations for Population Estimation

Study Case Application of Differential Equations for Population Estimation Michael Kurniawan (2018123456) University of Cincinnati, 2600 Clifton Ave, Cincinnati, OH 45221, United States kurniaml@mail.uc.edu Abstract This paper discusses the theory and application of partial differential equations. The purpose of this study is to examine these concepts and apply them in mathematical computations, thereby supporting aspects of life outside the realm of mathematics. Differential equations have various applications in everyday life, one of which is to estimate the population in a certain area at a given time. Suppose that population changes or increases/decreases are represented by N(t). If it is assumed that P(t) is a function that can be derived over time, then this function is continuous. Furthermore, if dN/dt is the rate of change of the population that is proportional to the total population, then dN/dt=kN or dN/dt-kN=0, where k is a proportional constant. Keywords: Applications of differential equations, population estimation INTRODUCTION A differential equation is a mathematical equation that relates a function of one or more variables to its derivatives of various orders. Differential equations play an important role in engineering, physics, economics, and various other disciplines. These equations arise when deterministic relationships involving continuously varying quantities are modeled through mathematical functions, with the rate of change expressed as derivatives. One application of derivatives in everyday life is to create mathematical models for phenomena involving instantaneous and average rates of change. Differential equations form the basis for the development of science and technology. Examples include: radioactive decay, population growth and decline, heating and cooling, evaporation, MICHAEL KURNIAWAN 1 Study Case electric current in circuits, motion in gravitational fields, market price dynamics, financial growth and stock prices, drug dosage regulation, and cell division. First-order ordinary differential equations are the simplest form because they only contain the first derivative of the unknown function. Although simple, many real phenomena can be modeled using these equations. Models are constructed under the assumption that the observed changes are proportional to the assumed variables or functions. Two quantities P and Q are said to be proportional if one is a constant multiple of the other, for example, P = kQ. If y = f(x) can be differentiated at x₀, then f'(x₀) is the rate of change of y with respect to x at the point x₀. Geometrically, f'(x₀) is the slope of the tangent line to the curve y = f(x) at the point (x₀, f(x₀)). The interpretation of the derivative as the instantaneous rate of change is a fundamental concept in many applications, while the geometric interpretation is useful for numerical solutions. The ratio 𝑓(𝑥) − 𝑓(𝑥0 ) Δ𝑦 = 𝑥 − 𝑥0 Δ𝑥 is the average rate of change of the function y=f(x) when x changes by Δx. A first-order linear differential equation has the form: 𝑎1 (𝑥)𝑦 ′ + 𝑎0 (𝑥)𝑦 = 𝑓(𝑥)(1.1) Assuming that the coefficients a₁(x), a₀(x), and f(x) are continuous on the interval I and a₁(x) ≠ 0 for all x in I. If divided by a_1 (x) and denoted by a(x)=a_0 (x)/a_1 (x), b(x)=f(x)/a_1 (x), then the equivalent equation becomes: 𝑦 ′ + 𝑎(𝑥)𝑦 = 𝑏(𝑥)(1.2) The general solution to this equation can be found explicitly by substitution: 𝑤 = 𝑦𝑒 ∫ 𝑎(𝑥)𝑑𝑥 (1.3) So that equation (1.2) becomes: 𝑤 ′ = 𝑏(𝑥)𝑒 ∫ 𝑎(𝑥)𝑑𝑥 (1.4) Considering the many applications of differential equations, this paper summarizes examples of their use for population estimation. The purpose of this research is to examine these concepts and implement them in mathematical computation, thereby supporting real-life aspects. Exponential Growth Model In 1798, Thomas Malthus introduced the exponential growth model, which is used to estimate the population of living things with a constant growth rate. Let P(t) be the MICHAEL KURNIAWAN 2 Study Case population size at time t. If P(t) is continuous and differentiable, and the population change is proportional to the population size, then: 𝑑𝑃 = 𝑘𝑃(1.5) 𝑑𝑡 This is a separable differential equation. The general solution is: ∫ 𝑑𝑃 = ∫ 𝑘𝑑𝑡 ⇒ ln⁡ 𝑃(𝑡) = 𝑘𝑡 + 𝑐 ⇒ 𝑃(𝑡) = 𝑒 𝑘𝑡+𝑐 (1.6) 𝑃 With the initial condition P(0) = P_0, we obtain c = ln⁡P_0, so the particular solution is: 𝑃(𝑡) = 𝑃0 𝑒 𝑘𝑡 (1.7) If k>0, the population increases exponentially; if k<0, the population decreases. Logistic Growth Model Pierre Verhulst (1838) introduced the logistic model for large populations with limited space. This model takes into account carrying capacity (K), which is the maximum population that the environment can support. The logistic model states that population growth approaches a point of equilibrium, where the number of births equals the number of deaths: 1 𝑑𝑃 𝑃 = 𝑘(1 − )(1.8) 𝑃 𝑑𝑡 𝐾 When multiplied by P, we obtain: 𝑑𝑃 𝑃 = 𝑘𝑃(1 − )(1.9) 𝑑𝑡 𝐾 Interpretation: if P≪K, growth approaches exponential; if P≈K, the growth rate decreases; if P>K, the population declines. The solution to the logistic equation is obtained through: ∫ 𝑑𝑃 𝑃 = ∫ 𝑘𝑑𝑡 ⇒ ln⁡ = 𝑘𝑡 + 𝑐 𝑃(1 − 𝑃/𝐾) 𝐾−𝑃 With the initial condition P(0)=P_0: 𝑐 = ln⁡ 𝑃0 𝐾 ⇒ 𝑃(𝑡) = (1.11) 𝐾 − 𝑃0 −𝑘𝑡 𝐾 − 𝑃0 1 + ( 𝑃 )𝑒 0 Where: P= population size at time t P_0= initial population K= environmental capacity (carrying capacity) k= per capita growth rate MICHAEL KURNIAWAN 3 Study Case t= time Equation (1.11) is a specific solution to the logistic model that can be used to project city populations. The value of K is usually determined through trial and error so that the model results approximate real data. RESEARCH METHODOLOGY This research is a theoretical study of the application of differential equations in everyday life. The methodology used includes collecting theoretical sources from books and journals, which are then analyzed to find common threads between existing concepts. The results of the study are accompanied by examples of real-world applications in everyday life. PROBLEM FORMULATION This study constructs a population growth model for the States of Rhode Island using population data from 2010 to 2015. The models used include exponential and logistic growth. The environmental capacity value K is determined through trial and error, namely by entering the estimated K into the model based on data from the previous year. Furthermore, the model obtained is used to predict the population in 2020. DISCUSSION Table 1. Population of Rhode Island States (2010–2015) The population growth model was constructed using exponential and logistic equations. MICHAEL KURNIAWAN 4 Study Case Solution of the Exponential Population Growth Model The data in Table 1 shows an increase in population between 2010 and 2015. Using the exponential model: 𝑑𝑃 − 𝑘𝑃 = 0 𝑑𝑡 the proportional constant k can be found. Suppose the population in 2010 is P_0, and P(t) is the population after t years from 2010. Then Table 1 can be rewritten as: Table 2 population of Rhode Island States To find k, use the data for 2011 (t=1): 613,388 = 𝑃0 𝑒 𝑘 (4.1) and for 2015 (t=5): 1,531,402 = 𝑃0 𝑒 5𝑘 (4.2) Solving this system yields: k = 0.228735 Thus, the exponential population growth equation becomes: 𝑃(𝑡) = 613,388 × 𝑒 0.228735𝑡 (4.3) For example, the estimated population in 2020 (t=10): 𝑃(10) = 613,388 × 𝑒 0.228735⋅10 ≈ 6,041,138 This means that the estimated population in 2020 using the exponential model is around 6,041,138 people. Solution of the Logistic Population Growth Model For the logistic model, the environmental capacity K=2,000,000 is assumed because the population during 2010–2015 was still below 2 million. With P(0)=P_0, the specific logistic equation becomes: MICHAEL KURNIAWAN 5 Study Case 𝑃(𝑡) = 2,000,000 (4.4) (2.62965)𝑒 −𝑘𝑡 + 1 Using data from 2011 (t=1, P(1)=613,388), we obtain: 𝑘 = 0.15123 Thus, the first logistic model: 𝑃(𝑡) = 2,000,000 (4.5) (2.62965)𝑒 −0.15123𝑡 + 1 Using data from 2012 (t=2, P(2)=897,951), we obtain: 𝑘 = 0.38102 ⇒ 𝑃(𝑡) = 2,000,000 (4.6) (2.62965)𝑒 −0.38102𝑡 + 1 Similar steps are performed for t=3,4,5, resulting in Logistic Models III, IV, and V: 2,000,000 (4.7) (2.62965)𝑒 −0.428423𝑡 + 1 2,000,000 𝑃(𝑡) = (4.8) (2.62965)𝑒 −0.36936𝑡 + 1 2,000,000 𝑃(𝑡) = (4.9) (2.62965)𝑒 −0.430208𝑡 + 1 𝑃(𝑡) = Table 3. Comparison of Census Results and Logistic Model Estimates Based on the population comparison graph (Figure 1), the Logistic Model III most closely approximates the census results, especially in 2015. Therefore, this model is used to predict the 2020 population: 𝑃(10) = 2,000,000 ≈ 1,930,061 (2.62965)𝑒 −0.428423⋅10 + 1 This means that the population of Rhode Island is predicted to be around 1,930,061 in 2020. MICHAEL KURNIAWAN 6 Study Case Figure 1. Population Based on Census Results and Model Results CONCLUSIONS AND RECOMMENDATIONS Differential equations have a variety of applications. In population estimation, two main models are used: exponential and logistic. The logistic model has proven to be more accurate because it takes environmental capacity into account. • Exponential model: 𝑃(𝑡) = 𝑃0 𝑒 𝑘𝑡 • Logistic model 𝑃(𝑡) = 𝑒 −𝑘𝑡 (𝐾/𝑃 −1)+1 𝐾 0 Both models can also be applied in other fields, such as banking to calculate savings balances with compound interest. For population estimates that are closer to reality, it is recommended to use the logistic model, as its results are most consistent with actual census data. MICHAEL KURNIAWAN 7 Study Case REFERENCE MICHAEL KURNIAWAN 8