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Showing a diff. eq. has no solutions

  1. Jan 3, 2006 #1

    es

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    Unfortunately I have read few proofs and have written fewer. So I am asking for advice on how to write this one so hopefully I can get better at it.

    I am to show that

    Eq1: Abs[ D[y,x] ] + Abs[ y ] + 1 = 0

    has no solutions.

    So this is what I was thinking.

    Let
    f(y) = Abs[ y ]
    g(y) = Abs[ D[y,x] ]
    Thus we can rewrite Eq1 as
    Eq2: f(y) + g(y) + 1 = 0

    However due to the Abs operator the domain of both f and g
    must be greater than or equal to zero therefore there is no y
    in the range of f or g that can satisfy Eq2.

    I guess first off, is this proof even correct?
    Is there anything I should do to improve it?

    Is ok for me to imply and assume the fact that 1 plus a
    non-negative cannot be less than 1 thus the sum cannot be
    zero? I suppose the answer depends on the audience. Here I
    assumed the audience was my college student peers with
    the material we have covered to date, including previous
    classes, therefore the proof of this fact should be pretty easily
    available.

    I guess I am most concerned with where to draw the "it's trivial
    from here" line. I mean, to be honest, this proof in itself seems
    a bit trivial which, ironically, is what makes it hard to write. :)
     
  2. jcsd
  3. Jan 3, 2006 #2

    LeonhardEuler

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    It isn't even really necessary to introduce f and g. I would just move the 1 over to the other side as a -1, then observe that the left hand side is necessarily non-negative, while the right hand side is negative.
     
  4. Jan 3, 2006 #3

    HallsofIvy

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    Actually, the fact that one absolute value is dy/dx and the other is of x is not relevant: call dy/dx, A and y, B: then the euation says:
    |A|= -|B|- 1= -(|B|+1).

    that's obviously impossible because the left side is positive and the right side is negative.
     
  5. Jan 3, 2006 #4

    mathwonk

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    what is becoming extremely clear by now, is that a necessary condition for there to be a function y(x) such that F(y, dy/dx,x) = 0, is that there should at least exist numbers a,b,c such that F(a,b,c) = 0.
    Thus we can make up more "differential equations" with no solutions, such as
    (dy/dx)^2 + y^2 + 1 = 0. can you make up another one?
    so existence theorems usually start off with equations of the form:
    dy/dx = F(x,y), where this is not a problem.
    even ones where we have a(x) dy/dx = F(y,x) lead to difficulties where a(x) = 0, and these are called "singular points".
    The same sort of problem arises in
    implicit differentiation" p[roblems when we are supposed to find the derivative of the function defined by F(x,y) = 0, since if there are no numbers a,b, with F(a,b) = 0, there certainly can be no function y(x) satisfying F(y(x),x) = 0.
    This problem is usually not mentiuonewd in elementary calkculus, but is swept, under the rug, by saying something like: find the derivative y'(x) of the function defined by y^2 + x^2 = 1, near the point (sqrt(1/2), sqrt(1/2)).
    i.e. they pick the F carefully, and then even give us the point.
    Even in advanced calculus where they tell us that an implicit function y(x) defined by F(x,y) = 0, exists locally near (a,b), if the partial Fy(a,b) is not zero, they should emphasize too that it must be true that F(a,b) = 0 as well.
    This part is sometimes given only passing mention, whereas finding such points (a,b) is really the hardest part.
    Similarly, in stating a condition for a d.e. F(x,y,y') = 0 to have a solution, one should begin by specifying a triple (a,b,c) such that F(a,b,c) = 0, and then ask for a function y(x) with y(a) = b, and y'(a) = c, and satisfying
    F(x,y(x), y'(x)) = 0 on some x- interval around a.
    In your case, the original function, as has been pointed out, F(x,y,y')
    was 1 + |y| + |y'| = 0, for which such a triple (a,b,c) does not exist.
    so always ask your self, whether there are any numbers at all making your equation true. If not then there certainly cannot be a whole family of them, i.e. no function.
    thanks for asking this question, as it (and the answers) have helped me, since I have to start teaching o.d.e. next week. It is not exactly my forte, but I am beginning to find it fascinating while preparing.
    As I said elsewhere, perhaps an o.d.e. is a wind blowing, and a solution a flow of liquid blown by that wind. So maybe the creator started the world with an o.d.e.
    or maybe not: the king james text only says the spirit of god moved upon the waters, but older sanskrit writings suggest that the breath (spirit) of god came first, and then the waters, perhaps moving under the impulse of that holy word.
    I guess the solar system is also a diff eq, and the planets and stars moving together merely solutions.
    Wow, people argue about the origins of life on this planet, but what about the origins of planets? I.e. this one planet and its little solar system all flowered during the stable existence of this one set of 9 solutions. what about before this planet arose? and this sun? and this universe? was this the creator's first try? it seems unlikely, and more likely that the beginning of "our world"
    was just another chapter in a much more lengthy succession of creations and destructions, experiments that flourished for a time, solutions of a d.e. that had a finite domain of definition. Indeed the sanskrit sources known to me say "in the beginning god created heaven and earth, as before."
    can't you just picture zeus flinging a handful of planets out into orbit around a sun, and watching them arc back under its pull, and glide wondrously around each other in graceful sync?
    gee, d.e. is cool. all this fun, just from 1 + |y| + |y'| = 0,

    and so is xmas break, time to reflect on the bigger picture, so precious.
    happy new year all.
     
    Last edited: Jan 3, 2006
  6. Jan 17, 2006 #5

    HallsofIvy

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    mathwonk, what have you been smoking and where can I get some?
     
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