Question

我在R中使用deSolve解决ODE系统。我打算计算一阶导数的时间平均值作为我的二阶导数。那就是：

dCt1 = (g/V - NCt1)dt
dCt2 = (1/t * Ct1)dt

以下是我的代码：

myfun <- function(tm, state, parms) {
  with(as.list(c(tm, state, parms)),
   {
     dC1 <- (g/V) - (N*Ct1)
     dCt2 <- Ct1 * 1/tm
     res <- c(dCt1, dCt2)
     return(list(res))
    })
}

库（ “deSolve”）

> duration <- seq(0, 4.5, by = 0.5);
> flow  <- 0.5;
> source  <- 128;
> Vol <- 168;

> state = c(Ct1 = 0, Ct2 = 0) # initial conditions
> 
> para = list(V = Vol, g = source, N = flow)
> 
> sol = ode(y = state, times = duration, parms = para, func = myfun,
> method = rkMethod("rk45ck"))

我收到错误消息：

警告消息：在rk（y，times，func，parms，method = method，...）中：在t = 0时，时间步数1超过最大步数

然而，我能够使用另一种方法无误地解决：

myfun <- function(tm, state, parms) {
      with(as.list(c(tm, state, parms)),
       {
         dCt1 <- (g/V) - (N*Ct1)
         dCt2 <- Ct1
         res <- c(dCt1, dCt2)
         return(list(res, dC3 = dC2/tm))
         })
    }


>  sol = ode(y = state, times = duration, parms = para, func = myfun,
>  method = rkMethod("rk45ck"))
>  sol[is.na(sol)] <- 0;
>  head(sol ,10)

> #which gives the following

      time       C1          C2        C3
 [1,]  0.0 0.0000000 0.00000000 0.0000000
 [2,]  0.5 0.3370655 0.08777375 0.1755475
 [3,]  1.0 0.5995724 0.32466476 0.3246648
 [4,]  1.5 0.8040129 0.67768842 0.4517923
 [5,]  2.0 0.9632314 1.12115626 0.5605781
 [6,]  2.5 1.0872308 1.63506211 0.6540248
 [7,]  3.0 1.1838017 2.20382513 0.7346084
 [8,]  3.5 1.2590112 2.81531096 0.8043746
 [9,]  4.0 1.3175844 3.46006934 0.8650173
[10,]  4.5 1.3632012 4.13074041 0.9179423

我担心的是我真的不需要Ct2并且不想输出它。我认为第一种计算方法dCt2 = (1/tm)*dCt1应该有一种解决方法。所有必要的帮助将受到高度赞赏。

谢谢大家

使用deSolve对ODE状态变量进行时间平均？

0 个答案: