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Boundary conditions

  1. Nov 30, 2009 #1
    If you have the value of a function of many variables, and its 1st-derivatives, at a single point, and a 2nd-order partial differential equation, then haven't you determined the entire function? You can use a Taylor expansion about that point to build the entire function because you have the value of the function at that point, the value of the 1st derivatives, and the value of higher derivatives (from the differential equation and differentiation of the differential equation).

    Cauchy boundary conditions require the value of a function and the value of the normal derivative to an entire boundary curve of the region of interest. But don't you just need these things at a single point, and not an entire boundary curve?
     
  2. jcsd
  3. Nov 30, 2009 #2
    Consider and arbitrary function [tex]u: \mathbb{R} ^n \rightarrow \mathbb{R}[/tex] such that [tex]u[/tex] is harmonic ie. [tex]\Delta u =0[/tex] in [tex]\mathbb{R} ^n[/tex] (actually you could do this in any domain [tex]\Omega \subset \mathbb{R} ^n[/tex] ) and suppose you want a solution of the problem:

    [tex]\Delta v=0[/tex] on [tex]\mathbb{R} ^n[/tex] , [tex]v(0)=v_0[/tex] and [tex]\frac{ \partial v}{\partial x_i } (0)=v_i[/tex]

    Then it's trivial to see [tex]w:= u+(v_0 -u(0))+\sum_{i=1}^{n} (v_i- \frac{ \partial u}{\partial x_i} (0))x_i[/tex] is harmonic and satisfies the given conditions, but [tex]u[/tex] was arbitrary so there are as many solutions to this as there are harmonic functions.
     
  4. Nov 30, 2009 #3
    But it ought to be true that if you were also given all the 2nd derivatives at the origin, along with the first derivatives and the function value at the origin, then the solution of the 2nd order PDE would be completely determined?

    The question then becomes how do you find the 2nd derivatives? Having the value of the function at all points on a boundary curve, along with the normal first derivative on that curve, can usually give you enough information to determine the 2nd derivatives.

    But what I don't get is why do you need an entire curve for your boundary? You have shown that a single point is not enough to constitute a boundary, but why isn't a tiny curve that passes through that point not enough?

    Also, is it odd that a function with no singularities is completely determined by all the derivatives at a single point via a Taylor expansion? All the derivatives at a point constitute a countable infinite set, but the function itself is uncountably infinite?
     
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