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Helmholtz Differential Equation--Cartesian Coordinates

In 2-D Cartesian Coordinates, attempt Separation of Variables by writing

\begin{displaymath}
F(x, y) = X(x)Y(y),
\end{displaymath} (1)

then the Helmholtz Differential Equation becomes
\begin{displaymath}
{d^2X\over dx^2}Y + {d^2Y\over dy^2}X +k^2 XY= 0.
\end{displaymath} (2)

Dividing both sides by $XY$ gives
\begin{displaymath}
{1\over X}{d^2X\over dx^2}+ {1\over Y}{d^2Y\over dy^2}+k^2= 0.
\end{displaymath} (3)

This leads to the two coupled ordinary differential equations with a separation constant $m^2$,
$\displaystyle {1\over X}{d^2X\over dx^2}$ $\textstyle =$ $\displaystyle m^2$ (4)
$\displaystyle {1\over Y}{d^2Y\over dy^2}$ $\textstyle =$ $\displaystyle -(m^2+k^2),$ (5)

where $X$ and $Y$ could be interchanged depending on the boundary conditions. These have solutions
$\displaystyle X$ $\textstyle =$ $\displaystyle A_me^{mx}+B_me^{-mx}$ (6)
$\displaystyle Y$ $\textstyle =$ $\displaystyle C_me^{i\sqrt{m^2+k^2}\,y}+D_me^{-i\sqrt{m^2+k^2}\,y}$  
  $\textstyle =$ $\displaystyle E_m\sin (\sqrt{m^2+k^2}\,y)+F_m\cos (\sqrt{m^2+k^2}\,y).$  
      (7)

The general solution is then


\begin{displaymath}
F(x, y) = \sum_{m=1}^\infty (A_me^{mx}+B_me^{-mx})[E_m\sin(\sqrt{m^2+k^2}\,y)+F_m\cos(\sqrt{m^2+k^2}\,y)].
\end{displaymath} (8)


In 3-D Cartesian Coordinates, attempt Separation of Variables by writing

\begin{displaymath}
F(x, y, z) = X(x)Y(y)Z(z),
\end{displaymath} (9)

then the Helmholtz Differential Equation becomes
\begin{displaymath}
{d^2X\over dx^2}YZ + {d^2Y\over dy^2}XZ + {d^2Z\over dz^2}XY+k^2XY = 0.
\end{displaymath} (10)

Dividing both sides by $XYZ$ gives
\begin{displaymath}
{1\over X}{d^2X\over dx^2}+ {1\over Y}{d^2Y\over dy^2}+{1\over Z}{d^2Z\over dz^2}+k^2= 0.
\end{displaymath} (11)

This leads to the three coupled differential equations
$\displaystyle {1\over X}{d^2X\over dx^2}$ $\textstyle =$ $\displaystyle l^2$ (12)
$\displaystyle {1\over Y}{d^2Y\over dy^2}$ $\textstyle =$ $\displaystyle m^2$ (13)
$\displaystyle {1\over Z}{d^2Z\over dz^2}$ $\textstyle =$ $\displaystyle -(k^2+l^2+m^2),$ (14)

where $X$, $Y$, and $Z$ could be permuted depending on boundary conditions. The general solution is therefore

$F(x, y, z) = \sum_{l=1}^\infty \sum_{m=1}^\infty (A_le^{lx}+B_le^{-lx})(C_me^{my}+D_me^{-my})$
$\times (E_{lm}e^{-i\sqrt{k^2+l^2+m^2}\,z}+F_{lm}e^{i\sqrt{k^2+l^2+m^2}\,z}).\quad$ (15)


References

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 501-502, 513-514 and 656, 1953.



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© 1996-9 Eric W. Weisstein
1999-05-25