Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discretization problem help

  1. Feb 8, 2006 #1
    If I have a logged temperature change over time which makes up a plottet graph. I denote this change

    {{dT} \over {dt}}

    People have told me this can be solved using discretization, but I have no idea what that is. Apparently it is something like this:

    {{T_1 - T_2 } \over {\Delta t}}

    but I don't really know what it means. Also, isn't a differential equation supposed to end up as a function of some variable? If so, how can an approximation like this wind up as one? Could someone explain it to me? Thx :smile:
    Last edited: Feb 8, 2006
  2. jcsd
  3. Feb 8, 2006 #2
    Say you have logged time T = [T1,T2,T3....TN] over time values t = [t1,t2,t3,...tN]
    now, if you do a plot of T vs t, you would get a curve. If all goes well, your plot may represent a function, say like newton's law of cooling ..some exponential form may be available. Now, a continuous function like exp is continuously differentiable. Hence, assume you got a solution

    T =f(t),
    at any time t, you can get dT/dt leading to

    dT/dt = f'(t) evaluated at some time t.

    Now, this is straightforward.

    Also, thiink abt a curve made up of discrete points.

    So to get a change or derivative (the slope), we can also do

    dT/dt at t1 = (T2-T1)(t2-t1) like forward differenced

    or dT/dt at t1 = (T1-T2)/(t1-t2) backward differenced.

    However, since these are first order approximations of the original functions, f(t) , your solution of evaluating dT/dt is first order accurate.

    Higher order can be achieved by taking more grid points. Look at finite difference approximations.

    the smaller the difference between t1 and t2, the better approximation of your time derivative.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook