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Homework Help: Fourier Series - Will someone walk me through this

  1. Apr 13, 2006 #1
    The book for the class that I'm currently taking is "Introduction to Applied Mathematics" by Gilbert Strang. Things have been good with this class until this chapter. If you have used this book, you will understand what I mean when I say it is different. Things have been ok, because I've been understanding it. But I seriously need someone to hand hold me through an example. Here is one of the two worked examples on this material. (pg 278)

    EXAMPLE 2 [itex] u_0(\theta) = \theta,\,\,\,-\pi<\theta<\pi.[/itex]

    The Fourier series for [itex] u_0 [/itex] was calculated in equation (12);
    [tex]\theta = 2\left(\frac{\sin\theta}{1}-\frac{\sin 2\theta}{2}+\frac{\sin 3\theta}{3}- \ldots \right). [/tex]

    Therefore the Fourier series (21) for [itex] u [/itex] is
    [tex] u(r,\theta)=2\left(\frac{r \sin\theta}{1}-r^2 \frac{\sin2\theta}{2}+r^3 \frac{\sin3\theta}{3}-\ldots \right). [/tex]

    On the circle [itex] u [/itex] has a sudden jump at [itex] \theta = \pi [/itex], but inside the circle that jump is gone. The powers [itex] r^k [/itex] damp the high frequencies and Laplace's equation always has smooth solutions.

    If someone would please help me fill in the missing steps, I would be relieved.

    I feel like someone has showed me: [tex] 5x^2 + 30x +2 =0[/tex] and asked me to solve it for [itex] x [/itex] but I've never heard of the quadratic equation, or completing the square :)

    I mean yes it would be nice if I could figure out those "in between steps". But, it is not coming to me... and we are moving quickly into the next chapter. So I need to know this. Thank you.
    Last edited: Apr 13, 2006
  2. jcsd
  3. Apr 13, 2006 #2


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    What you have given here, simply doesn't make sense. Given the Fourier series for [itex]u_0(\theta)[/itex] does not automatically give a Fourier series for [itex]u(r,\theta)[/itex] without some more information- for example the fact that [itex]u(r,\theta)[/itex] satisfies some partial differential equation inside the circle r= 1.

    I suspect, although you didn't say it, that this [itex]u(r,\theta)[/itex] satisfies Laplace's[\b] equation inside the circle. Certainly any (reasonable) function of [itex]\theta[/itex] can be written as a Fourier series. If it is also a function of r, then the coefficients may be functions or r. Assume a solution of the form
    [tex]u(r,\theta)= u_1(r)\frac{sin \theta}{1}- u_2(r)\frac{sin 2\theta}{2}+ u_3(r)\frac{sin 3\theta}{3}- ... [/tex]
    , plug it into the partial differential equation and see what differential equations you get for un(r).
  4. Apr 13, 2006 #3
    Sorry HallsofIvy. This example was under the section of Solution of Laplaces Equation.

    Thank you by the way! It's how you worded it, but something clicked (I think, or hope anyways).

    [tex] \nabla^2 = 0 [/tex]

    Laplaces equation in polar coordinates:
    [tex] \frac{1}{r} \frac{\partial}{\partial r}\left( r \frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}=0 [/tex]

    [tex] u(r,\theta) = a_0 + \sum_{n=1}^{\infty}a_n r^n \cos n\theta + \sum_{n=1}^{\infty} b_n r^n \sin n\theta [/tex]

    When [itex] r=1 [/itex] then [itex] u(r=1,\theta) = u_0(\theta) [/itex]

    [tex] u_0 = \theta [/tex]

    We rewrite this as a fourier series:
    [tex] u_0(\theta) = 2\left( \frac{\sin \theta}{1}-\frac{\sin 2\theta}{2}+\frac{\sin 3 \theta}{3} - \ldots \right) =\ldots[/tex]

    [tex] \ldots = a_0 + (a_1 r \cos \theta + a_2 r^2 \cos 2 \theta + \ldots) + (b_1 r \sin \theta + b_2 r^2 \sin 2\theta+\ldots) [/tex]

    Now [itex] u_0 = u [/itex] when r=1 so:
    [tex] a_n =0 [/tex] and [tex] r=1 [/tex] yields:

    [tex] (b_1 r \sin \theta + b_2 r^2 \sin 2\theta + b_3 r^3 \sin 3 \theta + \ldots) [/tex] which is close to [tex] u_0 [/tex] but not it yet so, we have to solve for [itex] b_n [/itex]

    by inspection:
    [tex] b_n = \frac{2\alpha}{n}\,\,\left|\,\,\alpha = \left\{\begin{array}{c} 1\,\,\,\,n=\{1,3,5,\ldots\} \\ -1\,\,\,\,n=\{2,4,6,\ldots\} \end{array} [/tex]

    [tex] u(r,\theta) = \sum_{n=1}^{\infty} \frac{2 \alpha}{n}(r^n \sin n\theta ) [/tex]

    plugging [itex] u_n(r,\theta) = \frac{2 \alpha}{n}(r^n \sin n\theta ) [/itex] into Laplaces equation:
    [tex] \nabla^2(u_n(r,\theta)) = 0 [/tex]

    Thus satisfying the boundary condition, and Laplaces equation.
    Last edited: Apr 13, 2006
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