info prev up next book cdrom email home

Biharmonic Equation

The differential equation obtained by applying the Biharmonic Operator and setting to zero.

\begin{displaymath}
\nabla^4\phi = 0.
\end{displaymath} (1)

In Cartesian Coordinates, the biharmonic equation is


$\displaystyle \nabla^4\phi$ $\textstyle =$ $\displaystyle \nabla^2(\nabla^2)\phi$  
  $\textstyle =$ $\displaystyle \left({{\partial^2\over\partial x^2}+{\partial^2\over\partial y^2...
... x^2}+{\partial^2\over \partial y^2}+{\partial^2\over \partial z^2}}\right)\phi$  
  $\textstyle =$ $\displaystyle {\partial^4\phi\over\partial x^4}+{\partial^4\phi\over\partial y^...
...r\partial y^2\partial z^2} +2{\partial^4\phi\over\partial x^2\partial z^2} = 0.$ (2)

In Polar Coordinates (Kaplan 1984, p. 148)
$\displaystyle \nabla^4\phi$ $\textstyle =$ $\displaystyle \phi_{rrrr}+{2\over r^2}\phi_{rr\theta\theta}+{1\over r^4} \phi_{\theta\theta\theta\theta}+{2\over r}\phi_{rrr}$  
  $\textstyle \phantom{=}$ $\displaystyle -{2\over r^3}\phi_{r\theta\theta}-{1\over r^2}\phi_{rr}+{4\over r^4}\phi_{\theta\theta}+{1\over r^3}\phi_r = 0.$ (3)

For a radial function $\phi(r)$, the biharmonic equation becomes
$\displaystyle \nabla^4\phi$ $\textstyle =$ $\displaystyle {1\over r}{d\over dr}\left\{{r{d\over dr}\left[{{1\over r}{d\over dr}\left({r{d\phi\over dr}}\right)}\right]}\right\}$  
  $\textstyle =$ $\displaystyle \phi_{rrrr}+{2\over r}\phi_{rrr}-{1\over r^2}\phi_{rr}+{1\over r^3}\phi_r = 0.$ (4)

Writing the inhomogeneous equation as
\begin{displaymath}
\nabla^4\phi = 64\beta,
\end{displaymath} (5)

we have
\begin{displaymath}
64\beta r\,dr = d\left\{{r{d\over dr}\left[{{1\over r}{d\over dr}\left({r{d\phi\over dr}}\right)}\right]}\right\}
\end{displaymath} (6)


\begin{displaymath}
32\beta r^2+C_1=r{d\over dr}\left[{{1\over r}{d\over dr}\left({r {d\phi\over dr}}\right)}\right]
\end{displaymath} (7)


\begin{displaymath}
\left({32\beta r+{C_1\over r}}\right)\, dr=d\left[{{1\over r}{d\over dr}\left({r{d\phi\over dr}}\right)}\right]
\end{displaymath} (8)


\begin{displaymath}
16\beta r^2+C_1 \ln r+C_2={1\over r}{d\over dr}\left({r{d\phi\over dr}}\right)
\end{displaymath} (9)


\begin{displaymath}
(16\beta r^3+C_1 r\ln r+C_2 r)\, dr=d\left({r {d\phi\over dr}}\right).
\end{displaymath} (10)

Now use
\begin{displaymath}
\int r\ln r\, dr = {\textstyle{1\over 2}}r^2\ln r-{\textstyle{1\over 4}}r^2
\end{displaymath} (11)

to obtain
\begin{displaymath}
4\beta r^4+C_1({\textstyle{1\over 2}}r^2\ln r-{\textstyle{1\over 4}}r^2)+{\textstyle{1\over 2}}C_2r^2+C_3=r{d\phi\over dr}
\end{displaymath} (12)


\begin{displaymath}
\left({4\beta r^3+C_1'r\ln r+C_2'r+{C_3\over r}}\right)\, dr=d\phi
\end{displaymath} (13)


$\displaystyle \phi(r)$ $\textstyle =$ $\displaystyle \beta r^4+C_1'\left({{\textstyle{1\over 2}}r^2\ln r-{\textstyle{1\over 4}}r^2}\right)+{\textstyle{1\over 2}}C_2' r^2+C_3\ln r+C_4\hfill$  
  $\textstyle =$ $\displaystyle \beta r^4+ar^2+b+(cr^2+d)\ln\left({r\over R}\right).$ (14)

The homogeneous biharmonic equation can be separated and solved in 2-D Bipolar Coordinates.


References

Kaplan, W. Advanced Calculus, 4th ed. Reading, MA: Addison-Wesley, 1991.



info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein
1999-05-26