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I'm looking to solve Laplace's equation on the infinite strip $(x,y)\in(0,1)\times(0,\infty)$ with mixed boundary conditions identical to this question:

$$\Delta u=0$$ $$u_x(0,y)=u_x(1,y)=0$$ $$u(x,0)=f(x)$$ The answer provided there is from a separation-of-variables approach, but I am interested in an integral representation if possible (mainly for numerical evaluation because $f$ is given numerically). Several examples where there are Dirichlet conditions on the boundaries (rather than Nuemann conditions) have integral representations such as this question and this question. With Dirchlet conditions, one can employ the conformal mapping $z\mapsto\cos(z)$ (mapping the strip to a half plane) and use the Poisson kernel to get an integral representation. However, the Poisson kernel does not apply for mixed-boundary conditions like those above. I would appreciate any assistance determining if an integral representation exists.

superckl
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2 Answers2

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Starting with the separation of variables solution, there is a way to sum the series $$ u(x,y)=\sum_{n=0}^{\infty}A_n\cos(n\pi x)e^{-n\pi y}. $$ $u(x,0)=f(x)$ determines the coefficients $A_n$: $$ f(x)=\sum_{n=0}^{\infty}A_n\cos(n\pi x) $$ $$ \int_0^1 f(x)\cos(n\pi x)dx = A_n\int_0^1\cos(n\pi x)^2dx $$ $$ A_n = \frac{\int_0^1 f(x)\cos(n\pi x)dx}{\int_0^1\cos^2(n\pi x)dx}. $$ Therefore, $$ u(x,y)=\sum_{n=0}^{\infty}\frac{\int_0^1f(x')\cos(n\pi x')dx'}{\int_0^1\cos^2(n\pi x')dx'}\cos(n\pi x)e^{-n\pi y} \\ = \sum_{n=0}^{\infty}2\int_0^1f(x')\cos(n\pi x')\cos(n\pi x)e^{-n\pi y}dx' \\ = 2\int_0^1f(x')\sum_{n=0}^{\infty}\cos(n\pi x')\cos(n\pi x)e^{-n\pi y}dx' \\ = 2\int_0^1f(x')\sum_{n=0}^{\infty}(\cos(n\pi(x'-x))+\cos(n\pi(x+x'))e^{-n\pi y}dx' \\ = 2\Re\int_0^1f(x')\sum_{n=0}^{\infty}(e^{in\pi(x'-x)}+e^{in\pi(x'+x)})e^{-n\pi y}dx' $$ Now it comes down to summing a geometric series, which I'll leave to you.

Disintegrating By Parts
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The solution form is given by \begin{equation*} u=A_0+\sum_{n=1}^{\infty} A_n\cos(\beta_nx)e^{-\beta_ny}, \end{equation*} where $\beta_n:=n\pi/L$. Imposing the boundary condition on $y=0$ leads to \begin{align*} A_0&=\frac{1}{L}\int_0^L f(\hat{x})\,d\hat{x}, \\ A_n&=\frac{2}{L}\int_0^L f(\hat{x})\cos(\beta_n\hat{x})\,d\hat{x}. \end{align*} On substituting the above expressions into the original solution form, and carrying out the summation, we get \begin{equation*} u=\frac{1}{L}\int_0^L\frac{\sinh\frac{\pi y}{L}\left[\cosh\frac{\pi y}{L}-\cos\frac{\pi x}{L}\cos\frac{\pi\hat{x}}{L}\right]f(\hat{x})\,d\hat{x}} {\left(\cosh\frac{\pi y}{L}-\cos\frac{\pi(x-\hat{x})}{L}\right)\left(\cosh\frac{\pi y}{L}-\cos\frac{\pi(x+\hat{x})}{L}\right)}. \end{equation*} $L=1$ in your case.

Jog
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