# 
# Example of exponential decay - fluorescent light
#  first by the y' = k y
#  then by      y' = IC (1-y) - k y
#
# Written by Doug Hensley
# Consider the question of how a fluorescent keychain is recharged 
# when exposed to light. (A battery, when exposed to current, a 
# capacitor, when exposed to voltage...).
# 
# The keychain has embedded in the plastic certain molecules in 
# which some electrons can take two energy states: ground state, 
# and excited. 
# A photon of sufficiently high frequency (say, UV) can bump one 
# of these electrons to the excited state, where it remains for a time. 
# But like with radioactivity, this state is unstable and sooner or later 
# the electron will drop back to the ground state, emitting a (green) 
# photon.
# 
# The probability that a particular excited electron will drop back to 
# the ground state within 1 second (from whenever you start timing) 
# is independent of how long it has been in an excited state. Let y(t) 
# denote the fraction of the fluorescent molecules with an excited 
# electron. A reasonable model for the decay of fluorescent intensity 
# is 
# y'=-Ky.
# 
# But when the keychain is exposed to light, we have also excitation 
# to think about. The probability that, within an arbitrary 1-second 
# timespan,  a given molecule will have an electron bumped up from 
# its ground state, assuming that ground state is occupied, is 
# proportional to the intensity of (UV) illumination. Let I denote this 
# illumination rate, and C, the constant of proportionality. The rate 
# at which further charging takes place will also depend, 
# (negatively) upon the fraction of molecules which are already in an 
# excited state. 
# Model for this: 
# 
# y'=IC(1-y)-Ky. 
# Problem: My (fully discharged) keychain becomes 10% charged in 
# 1 second of exposure to direct sunlight. It discharges by 3% per 
# second when in the dark. 
# (a) What is the value of K?
# 
# (b) What is the value of IC?
# 
# The decay DE is y'=-Ky, which has the very simple solution 
# y=y_0 *exp(-K*t). Setting t=1 and solving for 3% decay:
> solve(exp(-K)=.97,K);

                                  .03045920748
> K:=";

                                K := .03045920748
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# The next question is modeled by the IVP y'=IC(1-y)-Ky. 
# (Writing this down is actually the hard part).
# 
# Now this is again both FOL and separable. Either way, the 
# solution is y=(IC/(IC+K))*(1-exp(-(IC+K)*t)). The task of (b) 
# reduces to finding IC for which this expression takes the value 
# 1/10 at t=1.
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> IC:=fsolve(1/10=(IC)/(IC+K)*(1-exp(-IC-K)),IC);

                                IC := .1070320451
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# How fully will the keychain be charged after 5 seconds exposure?
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> evalf(subs(t=5,(IC)/(IC+K)*(1-exp(-IC*t-K*t))));

                                   .3870107954
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> res:=dsolve( diff(y(x),x) = IC_u*(1-y(x)) - K_u * y(x) , y(x) ); 

res :=

           IC_u + _C1 exp(- (IC_u + K_u) x) IC_u + _C1 exp(- (IC_u + K_u) x) K_u
    y(x) = ---------------------------------------------------------------------
                                         IC_u + K_u
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> subs( IC_u=IC, K_u=K, rhs(res) );

               .7784643974 + 1.000000000 _C1 exp( - .1374912526 x)
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>