heating_pulse1: Heating pulse-1
In epcc: Simulating Populations of Ectotherms under Global Warming

Description Usage Arguments Details Value References Examples

This function allows simulating the effect of an environmental warming pulse on the abundance of ectotherm populations. After the pulse, the temperature stabilizes at a specific value (temp_a).

heating_pulse1(
  y_ini = c(N = 400, N = 400, N = 400),
  temp_ini = rep(20, 3),
  temp_cmin = rep(18, 3),
  temp_cmax = c(25, 28, 32),
  ro = rep(0.7, 3),
  lambda = rep(5e-05, 3),
  temp_peak = rep(25, 3),
  time_peak = rep(2060, 3),
  sd = rep(2, 3),
  time_start = 2005,
  time_end = 2100,
  leap = 1/12
)

`y_ini`	Initial population values (must be written with its name: N).
`temp_ini`	Initial temperatures.
`temp_cmin`	Minimum critical temperature.
`temp_cmax`	Maximum critical temperature.
`ro`	Population growth rate at optimum temperature.
`lambda`	Marginal loss by non-thermodependent intraspecific competition.
`temp_peak`	Peak pulse temperature.
`time_peak`	Time when temp_peak is reached.
`sd`	Vector of standard deviations associated with temperature trend.
`time_start`	Start of time sequence.
`time_end`	End of time sequence.
`leap`	Time sequence step.

Three populations and/or scenarios can be simulated simultaneously. The temperature trend corresponds to a heating pulse determined by a Gaussian function, and the characteristics of the pulse are determined by the mean and the standard deviation. In each input vector, the parameters f or the three simulations must be specified (finite numbers for the initial population abundance). The simulations are obtained by a model that incorporates the effects of temperature over time, which leads to a non-autonomous ODE approach. This is function uses the ODE solver implemented in the package deSolve (Soetaert et al., 2010).

(1) A data.frame with columns having the simulated trends.

(2) A two-panel figure in which (a) shows the population abundance curves represented by solid lines and the corresponding carrying capacities are represented by shaded areas. In (b) the temperature trend is shown. The three simultaneous simulations are depicted by different colors, i.e. 1st brown, 2nd green and 3rd blue.

Soetaert, K., Petzoldt, T., & Setzer, R. (2010). Solving Differential Equations in R: Package deSolve. Journal of Statistical Software, 33(9), 1 - 25. doi:http://dx.doi.org/10.18637/jss.v033.i09

#######################################################################
  #Example 1: Different initial population abundances.
#######################################################################

heating_pulse1(y_ini = c(N = 100, N = 200, N = 400),
              temp_ini = rep(26,3),
              temp_cmin = rep(18,3),
              temp_cmax = rep(30,3),
              ro = rep(0.7,3),
              lambda = rep(0.00005,3),
              temp_peak = rep(29,3),
              time_peak = rep(2060,3),
              sd = rep(10,3),
              time_start = 2005,
              time_end = 2100,
              leap = 1/12)

#######################################################################
  #Example 2: Different thermal tolerance ranges.
#######################################################################

temp_cmin3 <- 18
temp_cmin2 <- 10/9*temp_cmin3
temp_cmin1 <- 10/9*temp_cmin2

temp_cmax1 <- 32.4
temp_cmax2 <- 10/9*temp_cmax1
temp_cmax3 <- 10/9*temp_cmax2

heating_pulse1(y_ini = c(N = 100, N = 100, N = 100),
              temp_ini = rep(26,3),
              temp_cmin = c(temp_cmin1,temp_cmin2,temp_cmin3),
              temp_cmax = c(temp_cmax1,temp_cmax2,temp_cmax3),
              ro = rep(0.7,3),
              lambda = rep(0.00005,3),
              temp_peak = rep(31,3),
              time_peak = rep(2060,3),
              sd = rep(10,3),
              time_start = 2005,
              time_end = 2100,
              leap = 1/12)

#######################################################################
  #Example 3: Different relationships between initial environmental
  #           temperature and optimum temperature.
#######################################################################

temp_cmin <- 18
temp_cmax <- 30

# Temperature at which performance is at its maximum value.
temp_op <- (temp_cmax+temp_cmin)/3+sqrt(((temp_cmax+temp_cmin)/3)^2-
           (temp_cmax*temp_cmin)/3)

temp_ini1 <- (temp_cmin+temp_op)/2
temp_ini2 <- temp_op
temp_ini3 <- (temp_op+temp_cmax)/2

heating_pulse1(y_ini = c(N = 100, N = 100, N = 100),
              temp_ini = c(temp_ini1,temp_ini2,temp_ini3),
              temp_cmin = rep(temp_cmin,3),
              temp_cmax = rep(temp_cmax,3),
              ro = rep(0.7,3),
              lambda = rep(0.00005,3),
              temp_peak = rep(29,3),
              time_peak = rep(2060,3),
              sd = rep(10,3),
              time_start = 2005,
              time_end = 2100,
              leap = 1/12)


#######################################################################
  #Example 4: Different peaks temperature.
#######################################################################

temp_peak3 <- 30
temp_peak2 <- 9/10*temp_peak3
temp_peak1 <- 9/10*temp_peak2

heating_pulse1(y_ini = c(N = 100, N = 100, N = 100),
              temp_ini = rep(22,3),
              temp_cmin = rep(18,3),
              temp_cmax = rep(30,3),
              ro = rep(0.7,3),
              lambda = rep(0.00005,3),
              temp_peak = c(temp_peak1,temp_peak2,temp_peak3),
              time_peak = rep(2060,3),
              sd = rep(10,3),
              time_start = 2005,
              time_end = 2100,
              leap = 1/12)

#######################################################################
  #Example 5: Different marginal losses by a non-thermodependent
  #           component of intraspecific competition.
#######################################################################

lambda3 <- 0.01
lambda2 <- 1/2*lambda3
lambda1 <- 1/2*lambda2

heating_pulse1(y_ini = c(N = 100, N = 100, N = 100),
              temp_ini = rep(20,3),
              temp_cmin = rep(18,3),
              temp_cmax = rep(30,3),
              ro = rep(0.7,3),
              lambda = c(lambda1,lambda2,lambda3),
              temp_peak = rep(29,3),
              time_peak = rep(2060,3),
              sd = rep(10,3),
              time_start = 2005,
              time_end = 2100,
              leap = 1/12)