### 16 Reduced Helmholtz PDE in Cartesian coordinates

16.1 Inside square

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#### 16.1 Inside square

problem number 113

Added December 20, 2018.

Solve for $$u\left ( x,y\right )$$

\begin{align*} \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^2} - k u(x,y) & = 0 \end{align*}

With $$k>0$$. It is called reduced Helmholtz, because of the minus sign above. Otherwise, standard Helmholtz has a positive sign.

Boundary conditions

\begin{align*} u(x,0) &= 0 \\ u(x,\pi ) &= 0 \\ u(0,y) &= 1 \\ u(\pi ,y) &=0 \end{align*}

Mathematica

$\text{DSolve}\left [\left \{-k u(x,y)+u^{(0,2)}(x,y)+u^{(2,0)}(x,y)=0,\{u(x,0)=0,u(x,\pi )=0,u(0,y)=1,u(\pi ,y)=0\}\right \},u(x,y),\{x,y\},\text{Assumptions}\to k>0\right ]$

Maple

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }2\,{\frac{\sin \left ( ny \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) \left ( -{{\rm e}^{- \left ( -2\,\pi +x \right ) \sqrt{{n}^{2}+k}}}+{{\rm e}^{\sqrt{{n}^{2}+k}x}} \right ) }{ \left ({{\rm e}^{2\,\sqrt{{n}^{2}+k}\pi }}-1 \right ) \pi \,n}}$