Biharmonic Equation

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

$$\nabla^4\phi = 0.$$ (1)

In Cartesian Coordinates, the biharmonic equation is

$$\nabla^4\phi = \nabla^2(\nabla^2)\phi$$

$$= \left({{\partial^2\over\partial x^2}+{\partial^2\over\partial y^2... ... x^2}+{\partial^2\over \partial y^2}+{\partial^2\over \partial z^2}}\right)\phi$$

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

In Polar Coordinates (Kaplan 1984, p. 148)

$$\nabla^4\phi = \phi_{rrrr}+{2\over r^2}\phi_{rr\theta\theta}+{1\over r^4} \phi_{\theta\theta\theta\theta}+{2\over r}\phi_{rrr}$$

$$\phantom{=} -{2\over r^3}\phi_{r\theta\theta}-{1\over r^2}\phi_{rr}+{4\over r^4}\phi_{\theta\theta}+{1\over r^3}\phi_r = 0.$$ (3)

For a radial function $\phi(r)$, the biharmonic equation becomes

$$\nabla^4\phi = {1\over r}{d\over dr}\left\{{r{d\over dr}\left[{{1\over r}{d\over dr}\left({r{d\phi\over dr}}\right)}\right]}\right\}$$

$$= \phi_{rrrr}+{2\over r}\phi_{rrr}-{1\over r^2}\phi_{rr}+{1\over r^3}\phi_r = 0.$$ (4)

Writing the inhomogeneous equation as

$$\nabla^4\phi = 64\beta,$$ (5)

we have

$$64\beta r\,dr = d\left\{{r{d\over dr}\left[{{1\over r}{d\over dr}\left({r{d\phi\over dr}}\right)}\right]}\right\}$$ (6)

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

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

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

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

Now use

$$\int r\ln r\, dr = {\textstyle{1\over 2}}r^2\ln r-{\textstyle{1\over 4}}r^2$$ (11)

to obtain

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

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

$$\phi(r) = \beta r^4+C_1'\left({{\textstyle{1\over 2}}r^2\ln r-{\textstyle{1\over 4}}r^2}\right)+{\textstyle{1\over 2}}C_2' r^2+C_3\ln r+C_4\hfill$$

$$= \beta r^4+ar^2+b+(cr^2+d)\ln\left({r\over R}\right).$$ (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.