Hi Mahesh,
Here is a solution to your problem. I have used the power-series approach.
You can solve for any positive value of k1 and k2 (except that k2
cannot be
unity). I think this is correct, but you can compare this with a numerical
ODE solver, if you wish.
pseries <- function(u, k2, eps=1.e-08) {
fn <- function(x) x * log(u) - log(eps) - log(x)
N <- ceiling(uniroot(fn, c(1, 1/eps))$root)
n <- seq(1, N+1)
u^(-k2) * sum(u^n / (n - k2))
}
logistic.soln <- function(t, k1, k2, Rm, R0) {
C <- k1 * Rm^k2
y0 <- pseries(R0/Rm, k2)
rhs <- C*t + y0
ff <- function(x, k2) pseries(x, k2) - rhs
ans <- try(uniroot(ff, c(1.e-03, 1-1.e-03), tol=1.e-06, k2=k2)$root,
silent=TRUE)
y <- if (class(ans) !="try-error") ans * Rm else Rm
y
}
t <- seq(0, 10, length=200)
# R0 > 0, Rm > R0, k1 > 0, k2 > 0, k2 != 1
k1 <- 0.5
k2 <- 1.5
Rm <- 2
R0 <- 0.2
system.time(y <- sapply(t, function(t) logistic.soln(t, k1, k2, Rm, R0)))
plot(t, y, type="l")
Hoep this helps,
Ravi.
-------------------------------------------------------
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns
Hopkins University
Ph. (410) 502-2619
email: rvaradhan at jhmi.edu
-----Original Message-----
From: r-help-bounces at r-project.org [ On
Behalf Of Ravi Varadhan
Sent: Wednesday, December 15, 2010 5:08 PM
To: 'Scionforbai'; 'mahesh samtani'
Cc: r-help at r-project.org
Subject: Re: [R] Solution to differential equation
This integral is NOT easy. Your solution is wrong. You cannot integrate
term-by-term when the polynomial is in the *denominator* as it is here.
I am not sure if there is a closed-form solution to this. I remember
you
had posed a problem before that is only slightly different from this.
Previous formulation:
dR/dt = k1*R*(1-(R/Rmax)^k2); R(0) = Ro
Current formulation:
dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
Note that the difference is that the exponent `k2' is in different places.
Initially I thought that this should not make much difference. But
appearances of similarity can be quite deceiving. Your previous formulation
admitted a closed-form solution, which I had shown you before. But this
current formulation does not have a closed-form solution, at least
not to my
abilities.
For the current formulation, you have u^k2 * (1-u) in the denominator
of the
integrand. This cannot be simplified in terms of partial fractions.
In the
case of previous formulation, the denominator was u * (1 - u^k2), which
could be simplified as (1/u) + u^(k2-1)/(1-u^k2). So, we are stuck,
it
seems.
We can expand 1/(1 - u) in a power-series expansion, provided |x| <
1, and
then integrate each term of the expansion. However, this is not very
helpful
as I don't know what this series converges to.
May be I am missing something simple here? Any ideas?
Ravi.
-------------------------------------------------------
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns
Hopkins University
Ph. (410) 502-2619
email: rvaradhan at jhmi.edu
-----Original Message-----
From: r-help-bounces at r-project.org [ On
Behalf Of Scionforbai
Sent: Wednesday, December 15, 2010 12:28 PM
To: mahesh samtani
Cc: r-help at r-project.org
Subject: Re: [R] Solution to differential equation
> I am trying to find the analytical solution to this differential equation
>
> dR/dt = k1*(R^k2)*(1-(R/Rmax)); R(0) = Ro
> If there is an analytial solution to this differential equation
then it
It is a polynomial function of R, so just develop the expression and
when you get the two terms in R (hint: one has exponent k2, the other
k2+1) you can simply apply the basic integration rule for polynomials.
It literally doesn't get easier than that.
the resulting function is:
k1/(k2+1) * R^(k2+1) - K1/[Rmax*(k2+2)] * R^(k2+2) + Ro
scion