Description Usage Arguments Details Value References Examples
View source: R/decreasing_linear.R
This function simulates the effect of a linear decreasing trend in environmental temperature on the abundance of ectotherm populations.
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y_ini |
Initial population values (must be written with its name: N). |
temp_ini |
Initial temperature. |
temp_cmin |
Minimum critical temperature. |
temp_cmax |
Maximum critical temperature. |
ro |
Population growth rate at optimum temperature. |
m |
Slope of the temperature decreasing trend. |
lambda |
Marginal loss by non-thermodependent intraspecific competition. |
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 is determined by decreasing a linear function. The slope can be modified. In each input vector, the parameters for 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
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#Example 1: Different initial population abundances.
#######################################################################
decreasing_linear(y_ini = c(N = 100, N = 200, N = 400),
temp_ini = rep(30,3),
temp_cmin = rep(18,3),
temp_cmax = rep(35,3),
ro = rep(0.7,3),
m = rep(1/5,3),
lambda = rep(0.00005,3),
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
decreasing_linear(y_ini = c(N=100,N=100,N=100),
temp_ini = rep(32,3),
temp_cmin = c(temp_cmin1,temp_cmin2,temp_cmin3),
temp_cmax = c(temp_cmax1,temp_cmax2,temp_cmax3),
ro = rep(0.7,3),
m = rep(1/5,3),
lambda = rep(0.00005,3),
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 <- 35
# 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
decreasing_linear(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),
m = rep(1/5,3),
lambda = rep(0.00005,3),
time_start = 2005,
time_end = 2100,
leap = 1/12)
#######################################################################
#Example 4: Different marginal losses by a non-thermodependent
# component of intraspecific competition.
#######################################################################
lambda3 <- 0.01
lambda2 <- 1/2*lambda3
lambda1 <- 1/2*lambda2
decreasing_linear(y_ini = c(N = 100, N = 100, N = 100),
temp_ini = rep(30,3),
temp_cmin = rep(18,3),
temp_cmax = rep(35,3),
ro = rep(0.7,3),
m = rep(1/5,3),
lambda = c(lambda1,lambda2,lambda3),
time_start = 2005,
time_end = 2100,
leap = 1/12)
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