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Unique solutions

  1. Dec 1, 2004 #1
    Is there a general method whereby one can proove that a general solution one has obtained to a D.E. is unique?
     
  2. jcsd
  3. Dec 1, 2004 #2
    I'm not sure I understand what you mean. A "general" solution is not unique; in fact it generates a whole family (typically infinitely many) of solutions.

    Do you mean something like "Once I have a general solution, how can I be sure that every function that satisfies the DE is generated by my general solution?"
     
  4. Dec 2, 2004 #3

    matt grime

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    In generaly the thing you ask for is false, mutatis mutandis. There are sufficient and necessary conditions for for "uniqueness", try googling wolfram unique solution differential equation. One of them is called Lipschitz.
     
  5. Dec 2, 2004 #4

    mathwonk

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    the basic technique for proving uniqueness of solutions, is the mean value theorem. it implies that if f is any function on an interval with f'=0 and f(0)= c then f is the constant function = c.

    for example this proves that the solution f to an equation of form f' = g(x), and f(0) = c, must be unique when g is given.

    i.e. then if f,h are two such solutions we get (f-h)' = 0 and (f-h)(0) = 0, so by MVT f-h is the zero function hence f = h.

    this generalizes to show for example that the solution to f' = rf, f(0) = c is also unique, by transforming it into an equation of the previous kind, namely by showing that

    if f satisifes f' = rf and f(0) = c, then f/e^rx is constant equal to f(0) = c, so f = ce^(rx).

    this generalizes further to prove uniqueness of a whole class of "linear constant coefficient" differential equations such as (D^2-1)f =f'' - f = 0, and f(0) = a, f'(0) = b.

    The idea is to factor the associated equation x^2 - 1 into (x-1)(x+1) hence factoring the differential operator in the equation into D^2 -1 = (D-1)(D+1)f = 0.

    then we know the only solutions to (D+1)f = f' + f = 0, from the discussion above. these are some of the solutions of our equation since if (D+1)f = 0, then also (D-1)(D+1)f = 0. recall all solutions of (D+1)f = 0 have form ce^(-x).

    looking at this factorization further we see that f solves (D-1)(D+1)f = 0 if and only if h = (D+1)f solves

    (D-1)h = 0. since we know all solutions of (D-1)h=0 have form h = ce^(x), we are trying to find all solutions of (D+1)f = h = ce^(x).

    since we know all solutions of (D+1)f = 0, if we can find even one solution of

    (D+1)f = h = ce^(x), we can get all others by adding solutions of

    (D+1)f = 0.

    but (D+1)(ce^x) = ce^x + ce^x = 2ce^x. so f = c/2 e^x solves (D+1)f = ce^x.
    hence all solutions of (D-1)(D+1)f = 0 have form ce^x + de^(-x).

    this is another uniqueness theorem. and applies to any equation of form

    (D^n + a D^(n-1) +.....+ cD + d)f = 0.
     
    Last edited: Dec 2, 2004
  6. Dec 2, 2004 #5
    Thank you mathwonk - very interesting
     
  7. Dec 3, 2004 #6

    matt grime

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    Sorry, but the proof of uniqueness of solutions of DEs is relies on the contraction mapping theorem. Or at least the proofs I know do, should I say.
     
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