dp/dt = exp(p) means dp exp(-p) = dt. Integration
on both sides gives - exp(-p) = t -c, where c is a constant
so that p(t) = -log(c-t).
Integrating dp/(exp(p)+1) = dt gives p-log(1+exp(p)) = t+c so
that p = t+c + log(1+exp(p)), or exp(p) = exp(t+c) (1+exp(p))
Solving for p gives exp(p(t))=exp(t+c)/(1-exp(t+c))=1/(exp(-t-c)-1) which can also
be rewritten as p(t) = -log(-1+exp(c-t)) for some constant c.
c) To integrate dp/(p(1-p)) = dt we write 1/(p(1-p) = 1/p + 1/(1-p)
and get log(p) + log(1-p) = t+c or p = exp(t+c) (1-p) so that
p = exp(t+c)/(1-exp(t+c)) which can also be written as
p(t) = exp(t)/(exp(t)-C) for some other constant C.
- In biology , p(t) might be the number of sick people as a
function of time at the start of an infectious epidemic. Here, a is proportional to the
number of people that a given sick individual will interact with (when there are a lot of
sick people, this equation will no longer be a good approximation to p(t) since sick
people will by more likely interact with other sick people.)
In Sociology p(t) might be the number of people at Harvard as a function of time
who have heard a particular joke if the joke has't been around too long. In this case,
a represents the average number of friends that a given person has.
Again, if the joke has been around for a while, there will be a greater chance that a
reasonable percentage of friends of any given individual
will have heard the joke.)
In economics , p(t) might represent the number of people as a
function of time (if not too much time has gone by) who are involved in a chain letter
type of scheme to make money (a pyramid scheme, such as that which bankrupted
Albania a few years ago.) Here, a is going to be proportional to the average number of
acquaintances of a given individual.
In environmental science
p(t) is the number of isotopes of a radioacitive container stored
at the Hanford facility ( checkout this for
more information). The number of decays in some time interval is proportional to p(t).
Here a is negative. The time for 1/2 of the isotopes to be decayed is log(2)/a.
a)u(t, x) = e-2t e-4(x-3t) = e-4x+10t.
b)u(t, x) = e-2t e -(x-3t)2.
- The solution that provides the sketches is u(t, x) = 1/(1 + (x - 3t)2).
- The function r(x) is the death rate of people at age x. Thus, it should be roughly
constant until x = 15 and then increase to a peak at x = 19, decrease again until x = 30
and then slowly increase for x larger than 30, maybe more steeply for x larger than 55. The speed is one
since each person gets one year older after one year.
- ut = r u + (-2 m t + x2)/4 m t2) u
uxx = (-2 m t + x2)/4 m t2) u
- a) ut = l u, uxx = (l-r)/m u
b) ut = l u, uxx = (l-r)/m u
c) ut = r u, uxx = 0
d) ut = l u, uxx = (l-r)/m u
e) ut = l u, uxx = (l-r)/m u
- a = u(1, 0)2/u(2, 0)), r = log[ u(1, 0)/(u(2, 0)) ], m = 1/(log[ u(1, 0)/u(1, 2)]).
- In this case, the standard choice takes q=-c u - m ux. Here is how one can
argue for this term: without random motions, the net number of particles which cross x
between time t-dt and t (for very small dt) is going to be roughly the number of
particles with positions between x- c dt and x at time t, which is roughly u(t,x) c dt.
With random motion, this number must be adjusted because some particles with x
coordinate slightly greater than x-c dt at time t-dt will have velocity less than c and will
not make it to x in time dt, while others with x coordinate slightly less than x-c dt will
have velocity greater than c, and so make it past x in the time dt. If the velocities are
assigned randomly, then this correction factor to the term u(t,x - c dt) should be roughly
proportional to ux(t-dt,x-c dt) dt. In particular, when dt is very small, then the
the x-derivative of u at (t-dt, x-c dt) is almost the same as that of u at (x,t). Thus, the
correction to the original c u dt term gives (c u + m ux) dt as the small dt
approximation for the net number of particles which cross x between t-dt and t.
Granted that such is the case, this leads to q = -c u + m ux by the arguments used
previously in the notes.