Laplace’s equation inside a semi-infinite strip

[Hio]

Hello everyone,
can anyone help me with solving Laplace’s equation inside a semi-infinite strip.
Is there specific steps to follow. I'm going to 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.

 

[pasmith]

Hello everyone,
can anyone help me with solving Laplace’s equation inside a semi-infinite strip.
Is there specific steps to follow. I'm going to 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.

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 $X(x)Y(y)$. 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
$$X'' = CX \\ Y'' = -CY$$
for some real constant $C$ (known as a separation constant). The values of $C$ 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 $Y(y)$ not identically zero (actually all that's required is $X(0) \neq 0$, but it is convenient to specify $X(0) = 1$).

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

Putting this together, we have, for each positive integer $n$, 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,
$$u(x,y) = \sum_{n=1}^{\infty} a_n \exp\left(-\frac{n\pi x}{h}\right) \sin\left(\frac{n\pi y}{h}\right),$$
and choose the coefficients $a_n$ to satisfy the boundary condition $u(0,y) = f(y)$. We then have
$$f(y) = u(0,y) = \sum_{n=1}^{\infty} a_n \sin\left(\frac{n\pi y}{h}\right)$$
which is the fourier sine series for $f(y)$ on the interval $0 \leq y \leq h$. Thus
$$a_n = \frac{2}{h} \int_0^h f(y) \sin\left(\frac{n\pi y}{h}\right)\,\mathrm{d}y.$$