# A series question related to solution of Laplace equation

Let $u(x,y)$ be the solution of the Laplace equation $\Delta u=0$ on the unit square $(0,1)\times (0,1)$ with boundary condition: $$u(x,1)=1, u(x,0)=0, u(0,y)=0, u(1,y)=0$$ The series solution is $$\sum_{n=1}^\infty \frac{4\sin((2n-1)\pi x)\mbox{sinh}((2n-1)\pi y)}{(2n-1)\pi\mbox{sinh}((2n-1)\pi)}$$ The value of $u(1/2,1/2)$ is then (use identity for hyperbolic sine) the series: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}2}{(2n-1)\pi\cosh((2n-1)\pi/2)}$$ Numerical evidence suggests that this series sums to 1/4.

But this is not obvious at all. Does anyone know how to sum this series? Or use other methods to get this answer?

This problem was posted in math.stackexchange.

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One can make a symmetry argument that the value must be 1/4... – Mike Jury Apr 12 '13 at 4:13

Let $v,w,z$ be the functions obtained from $u$ by composing with a rotation of angle $\frac\pi4,\frac\pi2,\frac{3\pi}4$ about $(\frac12,\frac12)$. The sum $u+v+w+z$ is a harmonic function, whose value at the boundary is constant equal to $1$. Thus $u+v+w+z\equiv1$. This gives you $4u\left(\frac12,\frac12\right)=1$.