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Classical Wave Equation

  1. Jul 28, 2007 #1
    can someone tell me why this is true

    [tex]\vec{F}=T(\frac{dy(x+\Delta x)}{dx}-\frac{dy(x)}{dx})\cong T(\frac{d^2y(x)}{dx^2})\Delta x[/tex]

    and am i correct in understanding the notation in that that dy(x) simply means the same as dy when it is implied dy is a function of x?
     
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  3. Jul 28, 2007 #2

    AlephZero

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    if [tex]\frac{dy(x+\Delta x)}{dx}[/tex] means [tex]\frac{dy}{dx}[/tex] evaluated at [tex](x+\Delta x)[/tex], your equation follows from Taylor's theorem:

    [tex]f(x+h) = f(x) + h\frac{df(x)}{dx} + \dots[/tex]

    where [tex]f(x) = \frac{dy(x)}{dx}[/tex]
     
  4. Jul 28, 2007 #3
    sorry don't see it? do you mean to expand the two differential functions on the left within the parenthesis in a taylor series and then find their difference? and their difference is supposed to look like the right hand side?

    the relationship is from this page

    http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/AnalyzingWaves.htm
     
  5. Jul 29, 2007 #4

    AlephZero

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    You agree [tex]f(x+h) \cong f(x) + h\frac{df}{dx}(x)[/tex] for any differentiable function f, when h is small?

    So [tex]f(x+h) - f(x) \cong h\frac{df}{dx}(x)[/tex]

    Take [tex]f(x) = \frac{dy}{dx}(x)[/tex]

    [tex]\frac{dy}{dx}(x+h) - \frac{dy}{dx}(x) \cong h\frac{d^2y}{dx^2}(x) [/tex]

    Reformat the notation a bit, and write [tex]\Delta x[/tex] instead of [tex]h[/tex] and the result is your equation.
     
  6. Aug 2, 2007 #5
    so that's a good explanation aleph but i was rusty about taylor's theorem. so i looked that up and that led me to mean value theorem

    this restatement of mean value is for my own edification mostly as i'm sure you know it but ok so we have

    [tex]\frac{f(b)-f(a)}{b-a}=f'(c)[/tex]

    which can be rearranged to be

    [tex]f(b)=f(a)+(b-a)f'(c) \hspace{2mm} \mbox{or} \hspace{2mm} f(x+\Delta{x})=f(x)+\Delta{x}f'(x+\Delta{x})[/tex]

    and then differentiating with respect to x at that same point, with a change of notation

    [tex]\frac{dy(x+\Delta{x})}{dx}=\frac{dy(x)}{dx}+\Delta{x}\frac{d^2y(x+\Delta{x})}{dx^2}[/tex]

    which is not just congruent but exact right? i think this is the relationship they've used in the page i linked to, though i know taylor's theorem is derived from the mean value theorem.

    the problem is why is this mean value relationship exact but the one on the page only congruent? but i think i've discovered the reason for the approximation, i think that it is considered congruent because [itex] \Delta{x}[/tex] is not the infinitesmal while they are looking at an infinitesmal piece of string. of course it is even mentioned that the relationship becomes exact when [itex]x\rightarrow 0[/itex]. am i correct?
     
  7. Aug 2, 2007 #6

    Claude Bile

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    You can think in terms of the definition of a derivative as well - If you divide both sides of the equation by Dx, and let, say g(x) = dy/dx, the LHS becomes the difference equation corresponding to a derivative, i.e.

    [g(x+Dx)-g(x)]/[(x+Dx-x)] = [g(x+Dx)-g(x)]/dx = LHS

    Taking the limit as dx approaches 0;

    [g(x+Dx)-g(x)]/Dx = dg/dx = d^2y/dx^2 = RHS

    By definition of a derivative.

    Essentially this means that as long as Dx is small enough, the LHS and RHS will be approximately equal.

    Claude.
     
  8. Aug 3, 2007 #7

    AlephZero

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    Not quite. What the mean value theorem says is that for any interval [a,b] there is some value of c where a <= c <= b that satisfies [tex]\frac{f(b)-f(a)}{b-a}=f'(c)[/tex].

    But the value of c depends on a, b, and the function f, and the theorem doesn't tell you anything about how to find the value of c, it just says there always is a value.

    You can't make a free choice of a, b, and c to get the equation you said was "not just conjugate but exact".
     
  9. Aug 3, 2007 #8
    yes you're right, that i replaced c with x+[itex]\Delta[/itex]x, don't know why i did that

    all is clear now
     
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