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Uniquely Specifying a Function

  1. May 18, 2005 #1
    Many physical process, and other processes that we like to think about, can be described using functions. In general, the function we seek is a complete description of the process, and we must construct this function from our incomplete knowledge of the process.

    Differential Equations are the best example of this construction that I know of. Based on the rate of change for a process, you can determine a function that describes the process at all times. Because we know algebra and calculus, we can also find an explicit formula for the function in many cases. Another advantage of diff-eqs is that the solutions satisfy uniqueness in many cases.

    I believe that mathematical physics needs more tools for uniquely specifying functions given incomplete, semi-unrelated information (such as the rate of change). I am aware that there is a generalized concept of a functional equation, which may involve derivatives, integrals, or inverses etc of an unknown function. I do not understand how these equations could naturally arise (other then just writing one down and then trying to solve it). This is my essential question: What other methods exist for uniquely specifying a function and how are these methods applied?


    Here is a much more specific question: Suppose we have f(a) = 0 and f(b) = 0, and we have that the area under f from a to b is A, and we have that the arc length of f from a to b is C.

    The question is, is f determined uniquely? Intuitively I say yes, but I am asking if it has been proven. Opinions and counterexamples are welcome.

    If f is determined uniquely, can we construct an explicit formula for f? Perhaps this falls under the branch of integral equations (although I doubt it). I also thought it might be a problem in the calculus of variations involving constraints, but can't really say more then that.


    Edit: I have added the condition that the function be strictly positive in the interval a to b, and that uniqueness be relaxed to mean "within a multiplicative constant", as in differential equations.
     
    Last edited: May 18, 2005
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  3. May 18, 2005 #2

    arildno

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    Counter-example:
    [tex]a=0, b=2\pi, f_{1}=\sin(x), f_{2}=-\sin(x), f_{1}\neq{f}_{2}[/tex]
     
  4. May 18, 2005 #3
    Totally unfair of me to throw this in now, but in my mind one of the conditions is that the function is strictly positive. This is to avoid a CE like yours, where the functions are essentially the same (in the sense of differential equation uniqeness). Perhaps, I should even say "uniqueness up to a multiplicative constant".

    I say strictly positive because I get a degenerate feeling from functions which have zero area in the interval.

    Thank you for the reply, and I hope you will accept my attempt to patch the holes in a sinking ship. It is not unusual in mathematics to propose an ad hoc condition to eliminate a trivial (no offense) counter example :smile:
     
    Last edited: May 18, 2005
  5. May 18, 2005 #4

    arildno

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    Allright, how about this one:
    Consider (for simplicity) an equilateral triangle whose base is the interval [0,1] on the x-axis.
    Now, on the interval (0,1/2), let f(x) be some function which is 0 at x=0 and coincident with the vertex at x=1/2, but strictly beneath the leg of the triangle elsewhere.
    Let g(x) lie above the same leg with the same vertical displacement to the leg as f(x) lies below the leg.
    Now, reflect f and g about the line x=1/2, so that you have an [tex]\hat{f}[/tex] lying below the other leg, and, similarly a [tex]\hat{g}[/tex] above the other leg.

    Now, we have two functions on (0,1):
    [tex]h_{1}(x)=f(x),0\leq{x}\leq\frac{1}{2}, h_{1}(x)=\hat{g}(x),\frac{1}{2}\leq{x}\leq{1}[/tex]
    [tex]h_{2}(x)=g(x),0\leq{x}\leq\frac{1}{2}, h_{2}(x)=\hat{f}(x),\frac{1}{2}\leq{x}\leq{1}[/tex]
    Evidently, [tex]h_{1}\neq{h}_{2}[/tex], but your demands ought to be fulfilled..
     
  6. May 18, 2005 #5
    I appreciate your patience arildno, but again I am dissastisfied by your counter example. I could shift your functions by 1/2 to center them around the origin, and then h1(x) = h2(-x). This shows that h1 and h2 are the same "function", aside from coordinate transformations.

    In analysis, we would say that these are different functions. Strictly speaking, this is correct. But in differential geometry, the fundamental statement of uniqueness (for a given curvature and torsion) requires that we ignore reflections and rotations around a particular coordinate axis.

    I should have phrased my entire question in terms of "curves" rather then "functions". In other words, I am interested in uniquesness only up to translations and rotations.

    It also occured to me that you could take any function on an interval, and devide it in to subintervals. Then, randomly arrange these intervals and construct a piecewise function that has the same arc length and area. I don't know how to say "this doesn't count", but it shouldn't count. To this extent, I specify the condition that the function be differentiable at all point between a and b (which is also implied by it having an arc length).
     
  7. May 18, 2005 #6

    Galileo

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    If f is a function satisfying the conditions, then the function g specified by g(x)=f(a+b-x) does too. So unless f has this symmetry wtr (a+b)/2 it is not uniquely determined.

    EDIT: I just read Crosson's post. Nevermind then.
     
  8. May 18, 2005 #7

    arildno

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    So, you have switched your interest into differentiable curves rather than arbitrary functions now?
    Well, differentiable curves are certainly interesting as well..:wink:

    However, one flaw in what you said here:
    "To this extent, I specify the condition that the function be differentiable at all point between a and b (which is also implied by it having an arc length)."

    A curve can certainly be assigned an "arc length" even though it isn't differentiable at all points (i.e, it might fail pointwise, as for the curve generated by |x|). If the curve IS differentiable at all points, then we gain the usual formula.
     
    Last edited: May 18, 2005
  9. May 19, 2005 #8

    matt grime

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    Of course there are counter examples, uncountably many of them I'd guess.

    We may as well assume that the end points are 0 and , and the f(0)=f(1)=0, and that the arc length is 2, so you're saying that any two curves with the same area length must be the same?

    If we drop the unnecessary condition of smooth then it is rather obvious how to make infinitely many curves with the same area.

    I can't think of any nice ones that are differentiable but that is merely because I can't off the top of my head think of that many sufficiently different curves (ie ones you won't automatically dismiss) that I can integrate and whose areas I know, and whose arc lengths I can find using, again, the integral formula.

    one simple way to see it though is to imagine a piece of string between the top points with a fixed length. Let is put it in the shape of a semicircle. Any deformation changes the area, agreed, but fixes the arc length.

    Now, deform it smoothly so that we flatten out both ends; so that in the limit it looks like a spike in the middle. Now do it so we have in the limit two spikes, one at each end. and for the sake of argument do lots of deformations of the semicircle into many spikes. In the limit these all have area zero. So the all started with area pi, ended with area 0 and were deformed smoothly, thus at some point thy must have all had area 1 and were smooth curves; none were the same, or even the same up to translation or anything. Indeed they all had areas anywhere between pi and 0.
     
  10. May 19, 2005 #9
    I'm obviously way beyond myself here but couldnt the same argument be applied towards finding two polygons with the same area and perimeter? Isnt that essentially what were doing here? If so there exists a square with perimeter 4s and area s^2, such that the lengths of a triangle equal 4s and area s^2 if

    4s = x+y+z and
    2s^2 = xy (simplified for a right triangle)

    (x+y+z)^2 = 8xy

    z = sqrt(x^2+y^2)

    (x+y+sqrt(x^2+y^2))^2 = 8xy

    Bah, three equations, four unknowns.

    I'll leave this just incase someone had an opinion.

    *Great thread (and idea) by the way.
     
    Last edited: May 19, 2005
  11. May 19, 2005 #10

    matt grime

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    yes, in polgonal paths it's fine, but the OP has now specified that the curve be smooth, ie have no sharp corners. of course polgonal paths are dense in all paths so it's no surprise it also holds for smooth paths.
     
  12. May 19, 2005 #11
    Thanks Matt, your counter example involving the string convinced me.

    As we have all just seen, it is difficult to uniquely specify a function using global properties (area, arc length) as opposed to local ones (the derivative at a point).

    Does anyone know of a global method for uniquely specifying functions?
     
  13. May 19, 2005 #12

    matt grime

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    Yes, require it to be analytic (or harmonic), then it can be globally defined using local properties. Or require it to satisfy a differential equation that is Lipschitz with sufficient boundary conditions, then it will be uniquely determined, if it exists.
     
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