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Laplace’s equation inside a semiinfinite strip 
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#1
Feb2113, 12:48 AM

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Hello everyone,
can anyone help me with solving Laplace’s equation inside a semiinfinite 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 semiinfinite 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
Feb2113, 03:29 PM

HW Helper
Thanks
P: 1,008

[tex] \nabla^2 (XY) = X''(x)Y(y) + X(x)Y''(y) = 0 [/tex] so that [tex] \frac{X''}{X} + \frac{Y''}{Y} = 0. [/tex] 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 [/tex] 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: [tex] X(0) = 1,\qquad \lim_{x \to \infty} X(x) = 0 \\ Y(0) = Y(h) = 0 [/tex] 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 [tex] Y'' = k^2 Y [/tex] 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 [tex] Y(y) = B\sin \left(\frac{n\pi y}{h}\right) [/tex] 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 [tex] X_n(x) Y_n(y) = \exp\left(\frac{n\pi x}{h}\right) \sin\left(\frac{n\pi y}{h}\right) [/tex] and the natural thing to do is to take a linear combination of these, [tex] 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) [/tex] which is the fourier sine series for [itex]f(y)[/itex] on the interval [itex]0 \leq y \leq h[/itex]. Thus [tex] 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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