# [SOLVED]Laplace semi-infinite domain

#### dwsmith

##### Well-known member
The domain is $$0\leq x < \infty$$ and $$0\leq y\leq b$$.
\begin{align*}
\phi(x, 0) &= f(x)\\
\phi(x, b) &= 0
\end{align*}
I want to show that
$\phi(x, y) = \frac{1}{\pi}\int_0^{\infty} \int_{-\infty}^{\infty}f(\xi) \frac{\sinh[u(b - y)]}{\sinh(ub)} \cos[u(\xi - x)]d\xi du$
Why isn't the periodic function cosine used on y? How should this be started then?

#### dwsmith

##### Well-known member
Question is at the bottom in red.
Our boundary conditions on $$x$$ are
$\lim_{x\to\pm\infty}\phi(x, y) = 0.$
Let $$\phi(x, y) = \varphi(x)\psi(y)$$.
Then $$\frac{\varphi''}{\varphi} = - \frac{\psi''}{\psi} = -k^2$$.
\begin{align}
\varphi(x) &\sim\left\{\cos(kx), \sin(kx)\right\}\\
\psi(y) &\sim\left\{\cosh(ky), \sinh(ky)\right\}
\end{align}
From the boundary conditions on $$y$$, we see that we need to make a change of
variables.
That is, $$y\to b - y^*$$; however, since the choice of variables are
arbitrary, let the change of variable be $$b - y$$.
So we have is
$\psi(y) \sim\left\{\cosh(k(b - y)), \sinh(k(b - y))\right\}.$
Using the boundary condition $$\phi(x, b) = 0$$, we have that
$\psi(y) \sim\sinh(k(b - y)).$
The general solution is then
\begin{alignat}{2}
\phi(x, y) &= \int_0^{\infty}\left[A(k)\cos(kx) + B(k)\sin(kx)\right]
\sinh(k(b - y))dk\\
\phi(x, 0) &= \int_0^{\infty}\left[A(k)\cos(kx) + B(k)\sin(kx)\right]
\sinh(kb)dk && ={} f(x)
\end{alignat}
Let $$A^*(k) = A(k)\sinh(kb)$$ and $$B^*(k) = B(k)\sinh(kb)$$,
and let's multiple through by $$\cos(k'x)$$.
\begin{alignat}{2}
\phi(x, 0) &= \int_{-\infty}^{\infty}\int_0^{\infty}
\left[A^*(k)\cos(kx)\cos(k'x) + B^*(k)\sin(kx)\cos(k'x)\right]dkdx
&& ={} \int_{-\infty}^{\infty}f(x)\cos(k'x)dx\\
\phi(x, 0) &= \int_0^{\infty}\int_{-\infty}^{\infty}
A^*(k)\cos(kx)\cos(k'x)dxdk && ={} \int_{-\infty}^{\infty}f(x)\cos(k'x)dx
\end{alignat}
Note that $$\cos(kx)\cos(k'x) = \frac{1}{2}\left[\cos(x\Delta k) + \cos[x(k + k')]\right]$$.
From class, we know that the integral of $$\cos[x(k + k')]$$ is zero.
\begin{align}
\int_{-\infty}^{\infty}\cos(kx)\cos(k'x)dx &=
\int_{0}^{\infty}\cos(x\Delta k)dx\\
&= \mathcal{R}e
\left\{\int_0^{\infty}\exp\left[(i\Delta k - \eta)x\right]dx\right\}\\
&= \mathcal{R}e\left\{\left.
\frac{\exp\left[(i\Delta k - \eta)x\right]}
{i\Delta k - \eta}\right|_0^{\infty}\right\}\\
&= \mathcal{R}e\left\{\frac{1}{-i\Delta k + \eta}\right\}\\
&= \frac{\eta}{\eta^2 + (\Delta k)^2}
\end{align}
In class, we have shown that
$$\frac{\eta}{\eta^2 + (\Delta k)^2} = \pi\delta(k - k')$$.
Therefore, we have that
$\int_{-\infty}^{\infty}\cos(kx)\cos(k'x)dx = \pi\delta(k - k').$
Now we have that
$\phi(x, 0) = \pi\int_0^{\infty}A^*(k)\delta(k - k')dk = \int_{-\infty}^{\infty}f(x)\cos(k'x)dx$
The integral $$\int_0^{\infty}A^*(k)\delta(k - k')dk = A^*(k')$$ so
\begin{align}
A^*(k') &= \frac{1}{\pi}\int_{-\infty}^{\infty}f(x)\cos(k'x)dx\\
A(k) &= \frac{1}{\pi\sinh(kb)}\int_{-\infty}^{\infty}f(x')\cos(kx')dx'
\end{align}
Now we can write the general solution \cref{lapgensoln} as
$\phi(x, y) = \frac{1}{\pi}\int_0^{\infty}\int_{-\infty}^{\infty} \frac{\sinh(k(b - y))}{\sinh(kb)}f(x')\cos(kx)\cos(kx')dx'dk.$
Again we have that $$\cos(kx)\cos(kx') = \frac{1}{2}\left[\cos(k(x' - x)) + \cos(k(x' + x))\right]$$ and $$\cos(k(x' + x))$$ integrates out to zero.
Let $$x' = \xi$$ and $$k = u$$.
Then the solution is
$\phi(x, y) = \frac{1}{2\pi}\int_0^{\infty}\int_{-\infty}^{\infty} \frac{\sinh(u(b - y))}{\sinh(ub)}f(\xi)\cos(u(\xi - x))d\xi dk.$
I have an extra factor of 1/2. Where am I missing a factor of 2?

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