Question

我正试图利用离散状态解决两个相互关联的差异方程式系统。

我的第一个模式是:

library(package = "deSolve")

growth <- function(t, state, parms) {
   with(as.list(c(state, parms)), {
   dx1 <- a*x - b*x*x - c*x*y
   dy2 <- d*y - e*y*x - f*y*y
   return(list(c(dx1, dy2)))   
   })
}

parameters <- c(a = 0.5, b = 10, c = 0.7, d = 0.1, e = 1.2, f = 0.5)
init <- c(x = 0.1, y = 0.2)

model1 <- ode(y = init, func = growth, times = seq(from = 0, to = 10, by = 0.1), parms = parameters)

and then i want to use the values of dx1 and dy1 at a time t to modify the values of model2 at time t.

growth2 <- function(t, state, parms) {
   with(as.list(c(state, parms)), {
   dx2 <- a*x - b*x*x - c*x*y*dx1 
   dy2 <- d*y - e*y*x - f*y*y*dy1
   return(list(c(dx2, dy2)))   
   })
}

parameters <- c(a = 0.5, b = 10, c = 0.7, d = 0.1, e = 1.2, f = 0.5)
init <- c(x = 0.1, y = 0.2)

model2 <- ode(y = init, func = growth2, times = seq(from = 0, to = 10, by = 0.1), parms = parameters)

如何在t时使用dx1的数值,在t时解决模型2。

现在,即解决模式2问题,为此,它分了1个步骤,然后利用循环改变公式的价值。它花费了大量时间。由于我的模式非常大,希望缩短模拟时间。

你们会就如何提高其效率提出建议。

Answer 1

我以同样的方式在评论中解释你的问题。我认为,你希望把这一模式作为单一系统。在这方面,我如何使第二个“模型”取决于第一个“模型”。

library(package = "deSolve")

growth <- function(t, state, parms) {
  with(as.list(c(state, parms)), {
    dx1 <- a*x1 - b*x1*x1 - c*x1*y1
    dy1 <- d*y1 - e*y1*x1 - f*y1*y1
    dx2 <- a*x2 - b*x2*x2 - c*x2*y2*dx1 
    dy2 <- d*y2 - e*y2*x2 - f*y2*y2*dy1
    return(list(c(dx1, dy1, dx2, dy2)))   
  })
}

parameters <- c(a = 0.5, b = 10, c = 0.7, d = 0.1, e = 1.2, f = 0.5)
init <- c(x1 = 0.1, y1 = 0.2, x2 = 0.1, y2 = 0.1)

model1 <- ode(y = init, func = growth, times = seq(from = 0, to = 10, by = 0.1), parms = parameters)

If you are looking for something different, can you clarify your question?

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