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Why assume a solution in exponential form?

  1. Dec 24, 2008 #1
    Hello Physics Forums,

    I sometimes come across statements like this (from here):

    Now assume a solution of the following form: x(t) = Aeat

    What is the motivation/background to make such an assumption? In this case the equation to be solved is x'' = 100x. Why would you assume the solution can usefully be written in that exponential form, or at all? Why does that type of step seem to be used so often? What is the role of A (a matrix I suppose...?) and a?

    Thanks!
     
  2. jcsd
  3. Dec 24, 2008 #2

    quasar987

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    I guess it comes from intuition. You know that x(t)=e^t equals itself when derived two times. So one could hope that with appropriate constants A and a, one could account for the 100 factor.
     
  4. Dec 24, 2008 #3

    symbolipoint

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    Try examing at a very low level what happens in a quantity which changes according to a set percentage; the rate stays the same, as a percentage, and this rate is applied as a factor, repeatedly. Try this using something common, like bank account interest. Do it step by step for several months, use symbols for numbers, and you will be able to derive an exponential expression for the new account quantity.
     
  5. Dec 26, 2008 #4
    Thanks for the answers.

    quasar987, that makes sense! I lacked that intuition.

    symbolipoint, I played around with your suggestion, and for one thing realised that while an interest rate r gives exponential growth of your cash thanks to dx/dt = (1+r)x (or rather a discrete-time version of that), the big exponential takeoff does not quite happen in the first 50-odd years, where growth seems fairly linear :bugeye:

    But so... after n years you get x(1+r)n so that's how it gets exponential. Not entirely sure how this is related to the x''=100x equation though. I rewrote that one as

    dx/dt = y
    dy/dt = 100x​

    and integrating numerically in Matlab confirm both x and x' being exponential with respect to t. But this doesn't seem quite the same as the interest business, where dx/dt is as stated, and so x'' = 1+r rather than x'' = c*x with c a constant, as for the x'' = 100x case. Maybe it's just the difference between a differential equation and a difference equation?
     
  6. Dec 26, 2008 #5

    jambaugh

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    Functions of the form [itex] f(t) = Ae^{rt}[/itex] are eigen-vectors of the differential operator [itex] \frac{d}{dt}[/itex], where you think of a function as an element of a vector space (you can add functions and multiply by constants thus they are "vectors" in the abstract sense.)

    [edit:] above,... eigen-vector (eigen-function) with eigen-value [itex]r[/itex].
     
  7. Dec 26, 2008 #6

    jgens

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    You could also assume solutions of that form if you've done much work with functions of the form x'' = ax

    Example:

    x'' = ax
    (x')(x'') = a(x)(x')
    (x')^2 = ax^2 + C, assuming C equals 0
    x' = bx
    (1/x') = 1/(bx)
    t = (1/b)ln(x)
    bt = ln(x)
    x = e^(bt)

    Not sure all of that is correct but it conveys the general idea.
     
  8. Dec 26, 2008 #7

    HallsofIvy

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    Notice that "assuming" an exponential solution then leads to many other non-exponential solutions!
     
  9. Dec 27, 2008 #8
    It seems to me that a lot of work in the area of differential equations was developed specifically to solve physics problems, and that because of this, the general solution (the set of all solutions) is almost never required. Even in a simple linear DE: ay'' + by + c = 0, the solution set is quite hard to figure out (and many cases depend on the implicit restrictions placed on y). Constant solutions are usually omitted as trivial.

    If we allow y to be a partial function (which isn't too absurd when we start throwing in negative powers, square roots, and logarithms into our formulas), we could even define solutions in a piece-wise fashion. For example, if y'' = 2y, we have a solution y(x) = e^2x, but we also have silly solutions such as:

    y(x) = e^(2x) if x > 0,
    y(x) = 2e^(2x) if x < 0, and
    y(0) is undefined.

    But of course, functions like this are silly for a physicist's purposes. It's easier to just teach the one useful solution and let the mathematics department deal with the rest.
     
  10. Dec 27, 2008 #9

    arildno

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    It is even easier, and better, to teach the aspiring physicist why it is crucial to formulate his problem in a precise manner. For example, in order to avoid "pathologies" in his solution set.
    Of course, it isn't the "weird", unphysical element in the solution set that is weird or unphysical; rather, it is the physicist who failed to specify the proper "physicality" requirements that his solutions are to have.
     
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