Hello,
I am using R package deSolve to solve a system of two differential equations
for a one-dimensional spatial and time-based problem. There is one ODE and a
second-order PDE. In order to solve with the function ode.1D, I've discretized
the spatial derivative and put both equations in terms of the time derivative
only. However, I'm now stuck with the problem of being unable to impose
boundary conditions on the spatial derivative for flux at the edges of the
system. How can I force a value for the discretized dE/dx part of my equations
at x = 0 and x = 1?
The code I have been using is below. The ode.1D solver will run on it, but the
solutions aren't correct for my system owing to the missing boundary conditions.
Thanks,
Molly C Brockway
Materials Science - PhD Candidate
Metallurgical and Materials Engineering - B.S.
Montana Technological University
```
BVmod1D <- function(time, state, parms, N, dx) {
with(as.list(parms), {
U <- state[1 : N]
E <- state[(N + 1) : (2 * N)]
coef1 <- Sv * io / (Qox - Qred)
coef2 <- Tau * io / Cd
BV <- exp(aa * f * (E - 0.5 * (U) )) - exp(-ac * f * (E - 0.5 * (1 + U)))
dEdx <- diff(c(E, E[N])) / dx # first spatial derivative of E
ddEddx <- diff(c(dEdx, dEdx[N])) / dx # second spatial derivative of E
dU <- coef1 * BV
dE <- (ddEddx - (coef2 * BV)) / Tau
return(list(c(dU, dE)))
})
}
pars <- c(Sv = 1.72e5,
io = 1.71e-6,
Qox = 1.56,
Qred = 0,
Tau = 3.79e-6,
Cd = 7.42e-8,
aa = 0.7674,
ac = 0.2326,
ks = 0.042992,
sig = 1.67e-6,
f = 38.92237)
R <- 1
N <- 50
dx <- R / N
Vo <- 0.5
# Initial and Boundary Conditions
yini <- rep(0, (2 * N))
yini[ 1 : N ] <- 2 * Vo
yini[ (N + 1) : (2 * N)] <- 1
times <- seq(0, 1 , by = 0.002 )
out <- ode.1D(
y = yini,
times = times,
func = BVmod1D,
parms = pars,
nspec = 2,
N = N,
dx = dx
)
```
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