Thin-plate versus biharmonic equation

The biharmonic equation uses the Laplacian operator twice: The biharmonic equation results from minimizing the quadratic form

$$\begin{eqnarray} B(u) &=& u' (\partial_{xx}+\partial_{yy})' (\partial_{xx}+\part... ..._{yy}+ \partial_{yy}'\partial_{xx}+ \partial_{yy}'\partial_{yy} )u\end{eqnarray}$$ (18) (19)

To minimize, simply cancel off u' and set to zero.

The thin plate equation resembles the biharmonic equation but differs in a subtle but important way. The quadratic form minimized for a thin plate is:

$$T(u) \quad =\quad v'v \quad \quad \quad {\rm where \ } v = ... ...{xy} \\ \partial_{yx} \\ \partial_{yy} \end{array} \right] u$$ (20)

or

$$T(u) \quad =\quad u'( \partial_{xx}' \partial_{xx} + \partia... ... \partial_{xy}' \partial_{xy} + \partial_{yx}' \partial_{yx} )u$$ (21)

Again, we find the associated differential equation by canceling off the u'.

What is bothering me is that the dispersion relations look the same but the quadratic forms look different. The difference between the biharmonic quadratic form and the thin plate quadratic form lies in the cross term. Let us form half this difference G=(B-T)/2.

$$G(u) \quad =\quad u_{xx}u_{yy}- u_{xy}u_{xy}$$ (22)

We see the difference has turned out to be the Gaussian curvature. Although the difference is a quadratic form, it is not uniformly positive or negative, as it can have both signs.

By means of rotation of coordinates, we can diagonalize the matrix

$$\left[ \begin{array} {cc} u_{xx} & u_{xy} \\ u_{yx} & u_{... ...[ \begin{array} {cc} K_1 & 0 \\ 0 & K_2 \end{array} \right]$$ (23)

Thus we can think of G=K₁ K₂ as the product of the curvatures while the biharmonic quadratic form is the square of the sum B=(K₁+K₂)². In terms of curvatures, in the rotated coordinates the thin plate operator is

$$\begin{eqnarray} T &=& B -2 G \\ T &=& (K_1+K_2)^2 - 2 K_1 K_2 \\ T &=& K_1^2 +K_2^2\end{eqnarray}$$ (24) (25) (26)

which is the sum of the squares of the curvatures.

The meaning is this: The biharmonic equation zeroes B=(K₁+K₂)² so its solution could be expected to have many places of K₁=-K₂ where the curvature on one axis is the negative of that on the other axis. In other words, solving the biharmonic equation might give us a function containing many saddles. On the other hand, the thin-plate equation T = K₁² +K₂² tries to eliminate both curvatures (not allowing credit for when one cancels the other). However, with respect to optimization, both quadratic forms are equivalent.