The differential equation obtained by applying the Biharmonic Operator and setting to zero.
![\begin{displaymath}
\nabla^4\phi = 0.
\end{displaymath}](b_1179.gif) |
(1) |
In Cartesian Coordinates, the biharmonic equation is
In Polar Coordinates (Kaplan 1984, p. 148)
For a radial function
, the biharmonic equation becomes
Writing the inhomogeneous equation as
![\begin{displaymath}
\nabla^4\phi = 64\beta,
\end{displaymath}](b_1190.gif) |
(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}](b_1191.gif) |
(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}](b_1192.gif) |
(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}](b_1193.gif) |
(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}](b_1194.gif) |
(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}](b_1195.gif) |
(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}](b_1196.gif) |
(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}](b_1197.gif) |
(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}](b_1198.gif) |
(13) |
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.
© 1996-9 Eric W. Weisstein
1999-05-26