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Laplace’s equation inside a semi-infinite strip

  1. Feb 21, 2013 #1


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    Hello everyone,
    can anyone help me with solving Laplace’s equation inside a semi-infinite strip.
    Is there specific steps to follow. I'm gonna give an example and I will be really grateful
    if someone explains to me.

    Solve Laplace’s equation ∇2u = 0 inside a semi-infinite strip (0 < x < ∞, 0 < y < H) with the following boundary condi- tions:
    u(x,0) = 0, u(x,H) = 0, u(0,y) = f(y).

    I miss the classes and I feel I'm lost.
  2. jcsd
  3. Feb 21, 2013 #2


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    Homework Helper

    You will want to solve this using separation of variables. The idea is to take a linear combination of solutions of Laplace's equation of the form [itex]X(x)Y(y)[/itex]. Then
    \nabla^2 (XY) = X''(x)Y(y) + X(x)Y''(y) = 0
    so that
    \frac{X''}{X} + \frac{Y''}{Y} = 0.
    The first term on the left is a function only of x and the second a function only of y. The only way this equation can hold for all x and y is if each term is constant. Hence
    [tex]X'' = CX \\
    Y'' = -CY
    for some real constant [itex]C[/itex] (known as a separation constant). The values of [itex]C[/itex] we need to take depend on the boundary conditions, which are:
    X(0) = 1,\qquad \lim_{x \to \infty} X(x) = 0 \\
    Y(0) = Y(h) = 0
    with [itex]Y(y)[/itex] not identically zero (actually all that's required is [itex]X(0) \neq 0[/itex], but it is convenient to specify [itex]X(0) = 1[/itex]).

    The easiest boundary condition to satisfy is that [itex]X(x) \to 0[/itex] as [itex]x \to \infty[/itex]. We must have [itex]X(x) = e^{-kx}[/itex] for some [itex]k > 0[/itex]. This means that [itex]C = k^2[/itex] so that
    Y'' = -k^2 Y
    subject to [itex]Y(0) = Y(h) = 0[/itex] but with [itex]Y(y)[/itex] not identically zero. That can be done if we take [itex]k = (n\pi)/h[/itex] for some positive integer [itex]n[/itex] with
    Y(y) = B\sin \left(\frac{n\pi y}{h}\right)
    where the constant [itex]B[/itex] cannot be determined from the boundary conditions on [itex]Y[/itex]. But given the next stage of the solution we may as well take [itex]B= 1[/itex].

    Putting this together, we have, for each positive integer [itex]n[/itex], an eigenfunction
    X_n(x) Y_n(y) = \exp\left(-\frac{n\pi x}{h}\right) \sin\left(\frac{n\pi y}{h}\right)
    and the natural thing to do is to take a linear combination of these,
    u(x,y) = \sum_{n=1}^{\infty} a_n \exp\left(-\frac{n\pi x}{h}\right) \sin\left(\frac{n\pi y}{h}\right),[/tex]
    and choose the coefficients [itex]a_n[/itex] to satisfy the boundary condition [itex]u(0,y) = f(y)[/itex]. We then have
    [tex]f(y) = u(0,y) = \sum_{n=1}^{\infty} a_n \sin\left(\frac{n\pi y}{h}\right)
    which is the fourier sine series for [itex]f(y)[/itex] on the interval [itex]0 \leq y \leq h[/itex]. Thus
    a_n = \frac{2}{h} \int_0^h f(y) \sin\left(\frac{n\pi y}{h}\right)\,\mathrm{d}y.[/tex]
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