Question

大家好我在我的ODE系列中遇到了一个问题。我有两个完全相同的方程式，但它们的答案却截然不同。我的方程式有不同答案的原因吗？这两个方程是控制e和r的方程式。

library(sigmoid)

parameters <- c(
                a = 0.032,
                b = (9 / 140),
                c = (5 / 1400),
                d = (95 / 700),
                k = 1 / 140,
                i = 0.25,
                # r = 0.2,
                n = 6000000,
                x = 0.5 ,
                y = 0.25,
                t = 1 / 180,        # important in looking at the shape
                u = 1 / 180,        # important in looking at the shape
                v = 1 / 180,        # important in looking at the shape
                p = 10,
                s = 100,
                g = 100

                # e = .4,
                #h = 1000
  ) 

  state <- c(
            S = 5989900,
            E = 0,
            I = 0,
            Q = 0,
            D = 100,
            B = 0,
            C = 100,
            Y = 100,
            H = 100,
            R = 10,
            J = 10,
            h = 100,
            e = 0.1,
            r = 0.1
   )

 # Function will transition between 0 and 1 when h and Q are approximately 
 equal
 smooth.transition <- function(h, Q, tune = 0.01){
 sigmoid((h/Q - 1)/tune)
 }
 Q <- 1
 h <- seq(0.001, 5, by = 0.001)

 plot(h/Q, j, type = "l")


 # set up the equations

 equation <- (function(t, state, parameters)
  with(as.list(c(state, parameters)), {
    # rate of change

    dS <- (-(a * S * I) / n) - (((1 / r) * S * D) / n)
    dE <- (a * S * I) / n + (((1 / r) * S * D) / n) - i * E
    j <- smooth.transition(h, Q)
    dI <- i * (j) * E - (e) * I - c * I - d * I
    dQ <- (j) * (e) * I - b * Q - k * Q
    dD <- d * I - r * D
    dB <- b * Q + r * D
    dC <- c * I + k * Q

    dY <- p * (b * Q + r * D)
    dR <- j*(1 - x - y) * (p * (b * Q + r * D))  - j*t * (R)
    de <- j*t * (s / R)
    dJ <- (y) * (p * (b * Q + r * D))  - v * (J)
    dr <- v * (s / J)
    dH <- (x) * (p * (b * Q + r * D)) - u * (H)
    dh <- u * (H / g)


    # return the rate of change
    list(c(dS, dE, dI, dQ, dD, dB, dC, dY, dR, de, dJ, dr, dH, dh))
  }))
#

# solve the equations for certain starting parameters


library(deSolve)
times <- seq(0, 200, by = 1)

out <-
  ode(y = state,
    times = times,
    func = equation,
    parms = parameters,
    maxsteps = 1e5
  )
# , method = "vode"
head(out)
tail(out)

# graph the results

par(oma = c(0, 0, 3, 0))
plot(out, xlab = "Time", ylab = "People")
#plot(out[, "X"], out[, "Z"], pch = ".")
mtext(outer = TRUE, side = 3, "Ebola Model",cex = 1.5
)

我在R，e，J和r的初始迭代中得到的是：

   R: 10, 10.05540, 10.11050
   e: 0.1, 59, 138

   J: 10, 39, 79
   r: 0.1, 0.105, 0.11

J和r表现得像我期望他们行动而R和e不行。任何人都可以在我的编码中看到问题。我认为我的数学很稳定。

Answer 1

您已将参数定义为t，解决方案中的时间也定义为t。

Answer 2

您返回的变化率的顺序与您最初定义状态变量的顺序不同。

CF

 state <- c(
            S = 5989900,
            E = 0,
            I = 0,
            Q = 0,
            D = 100,
            B = 0,
            C = 100,
            Y = 100,
            H = 100,
            R = 10,
            J = 10,
            h = 100,
            e = 0.1,
            r = 0.1
   )

带

list(c(dS, dE, dI, dQ, dD, dB, dC, dY, dR, de, dJ, dr, dH, dh))

返回列表应为：

list(c(dS, dE, dI, dQ, dD, dB, dC, dY, dH, dR, dJ, dh, de, dr))

相同的ODE方程表现不同

2 个答案: