A problem from Fourier Analysis An Introduction. Chapter 2, 19:
Solve Laplace's equation $\Delta u=0$ in the semi infinite strip
$$S=\{(x,y):0<x<1,0<y\},$$
subject to the following boundary conditions
$$ u(0,y)=0,\ \ when\ 0\le y,$$
$$ u(1,y)=0, \ \ when\ 0\le y,$$
$$ u(x,0)=f(x)\ \ when\ 0\le x\le 1$$
where $f$ is a given function, with $f(0)=f(1)=0.$
$$f(x)=\sum_{n=1}^{\infty}a_{n}\sin(n\pi x)$$
and expand the general solution in terms of the special solutions give by
$$u_{n}(x,y)=e^{-n\pi y}\sin(n\pi x).$$
Express $u$ as an integral involving $f$, analogous to the Poisson integral formula:
$$u(r,\theta)=(f*P_{r})(\theta)$$
$$=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\varphi)(\sum_{n=-\infty}^{\infty}r^{|n|}e^{-in(\varphi-\theta)})d\varphi.$$
I have no idea how to express u as an interal involving $f$.
Best Answer
Separation of variables gives solutions $$ u(x,y)=\sum_{n=1}^{\infty}A_n\sin(n\pi x)e^{-n\pi y}, $$ assuming some requirement of boundedness near $\infty$, for example. The requirement that $u(x,0)=f(x)$ determines the coefficients $A_n$ through the Fourier sine series:
$$ f(x)=u(x,0) = \sum_{n=1}^{\infty}A_n\sin(n\pi x). $$ The functions $\{ \sin(n\pi x)\}_{n=1}^{\infty}$ automatically form a complete orthogonal basis of $L^2[0,1]$. So the coefficients $A_m$ are uniquely determined by multiplying both sides by $\sin(m\pi x)$, integrating over $[0,1]$ in $x$, and using the orthogonality of the $\sin$ functions in order to obtain $$ A_n = \frac{\int_0^1f(x)\sin(n\pi x)dx}{\int_0^1\sin^2(n\pi x)dx} = 2\int_0^1f(x)\sin(n\pi x)dx $$ So $$ u(x,y)=2\sum_{n=1}^{\infty}\int_0^1f(x')\sin(n\pi x')dx'\sin(n\pi x)e^{-n\pi y} \\ = 2\int_0^1\left(\sum_{n=1}^{\infty}\sin(n\pi x')\sin(n\pi x)e^{-n\pi y}\right)f(x')dx' $$ You can write $$ \sin(n\pi x')\sin(n\pi x)=\frac{1}{2}(\cos(n\pi(x-x'))-\cos(n\pi(x+x'))) \\ = \frac{1}{2}\Re\left(e^{in\pi(x-x')}-e^{in\pi(x+x')}\right). $$ That will give you integrals in terms of $x-x'$ and $x+x'$.