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Periodicity of Fourier's heat theory

  1. Aug 8, 2013 #1
    Can I please get some help in understanding how Fourier developed his idea of heat transfer being a periodic phenomena?
     
  2. jcsd
  3. Aug 9, 2013 #2

    HallsofIvy

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    I don't know what you mean by that. Heat is NOT a "periodic phenomenon" and Fourier never said it was. You may be mistaking Fourier series (infinite sums of sine and cosine) with finite sums. Finite sums of periodic functions are periodic. Infinite sums are not.
     
  4. Aug 9, 2013 #3
    So why did he use sine and cosine functions for heat analysis? Orthogonal reasons?
     
  5. Aug 11, 2013 #4

    pasmith

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    Because sine and cosine are the solutions of the equations for the spatial variation which arise from separation of variables in the heat equation.
     
  6. Aug 13, 2013 #5
    Ok. I'm sorry for being dense, but what do you mean by spatial variation?
     
  7. Aug 13, 2013 #6

    HallsofIvy

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    The wave equation is
    [tex]\frac{\partial^2 \phi}{\partial x^2}= \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}[/tex]

    If you use "separation of variables" you would look for solutions of the form [tex]\phi(x, t)= X(x)T(t)[/tex] (separating x and t into separate functions) so that the equation becomes
    [tex]T\frac{d^2X}{dx^2}= \frac{1}{c^2}X\frac{d^2T}{dt^2}[/tex]

    Divide both sides by XT to get
    [tex]\frac{1}{X}\frac{d^2X}{dx^2}= \frac{1}{c^2T}\frac{d^2T}{dt^2}[/tex]

    Now the left side is a function of x only and the right side is a function of t only. In order to be the same for all x and t, each side must be equal to a constant:
    [tex]\frac{1}{X}\frac{d^2X}{dx^2}= \lambda[/tex]
    [tex]\frac{d^2X}{dx^2}= \lambda X[/tex]
    This is what pasmith is referring to as the "spatial variation".

    IF we are given the heat problem on a finite interval, say T(0)= T(P)= 0, then we can show that the "spatial dependence", the function X(x), must have sine and cosine solutions.

    If, for example, [itex]\lambda= 0[/itex] this becomes just [itex]d^2X/dx^2= 0[/itex] so the general solution is y= Ax+ B which is NOT periodic and so cannot satisfy the boundary conditions.
    If, for example, [itex]\lambda> 0[/itex] the general solution is [itex]y= Ae^{x\sqrt{\lambda}}+ Be^{x\sqrt{\lambda}}[/itex].

    So [itex]\lambda[/itex] must be negative. If we write [itex]\lambda= -\alpha^2[/itex], the equation becomes [itex]d^2X/dx^2= -\alpha^2 X[/itex], which has general solution [tex]Acos(\alpha x)+ B sin(\alpha x)[/tex]. That's where the

    [tex]\frac{1}{c^2T}\frac{d^2T}{dt^2}= \lambda[/tex]
    [tex]\frac{d^2T}{dt^2}= \lambda c^2T[/tex]

    Again, that is for the heat problem on a finite interval. If we had and infinite interval, say the heat equation on [itex]x\in [0, \infty)[/itex], the solutions are Fourier Transforms, not Fourier series, and are NOT periodic at all.
     
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