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The thin-plate energy functional is defined by : $$ E = \int \kappa_1^2 + \kappa_2^2 $$ while the Willmore energy functional is defined by : $$ W = \int (\kappa_1 - \kappa_2)^2 $$

For a closed surface, if we develop $W$: $$ W = \int \kappa_1^2 + \kappa_2^2 - 2\int K = E -4\pi\chi $$ where $\chi$ is the Euler characteristic of the surface that appears following the Gauss-Bonnet theorem.

In conclusion, for a closed surface, the thin-plate and the Willmore energies differ by $4\pi\chi$.

Is it correct?

If we look at the Euler-Lagrange equations related to these energies we have : $$ \Delta^2 f = 0 $$ for the thin-plate energy, and : $$ \Delta H + 2H(H^2-K) = 0 $$ for the Willmore energy.

But since these energies differ by a topological invariant of the surface, a critical point of one energy would be also critical for the other, which means the Euler-Lagrange equations above are equivalent.

Where am I wrong ?

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  • $\begingroup$ What is $f$? Can't make much progress without knowing that . . . $\endgroup$ Mar 12, 2020 at 19:39
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    $\begingroup$ It's the surface : the map on which E(f) and W(f) are computed. The values k1, k2, H and K are the principal, mean and gaussian curvatures of the surface f. $\endgroup$
    – T.L
    Mar 12, 2020 at 19:49
  • $\begingroup$ So to be more precise, is the surface given as $z = f(x, y)$ or as $f(x, y, z) = 0$? I am inclined towards the latter since $z = f(x, y)$ is topologically restrictive. $\endgroup$ Mar 12, 2020 at 19:51
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    $\begingroup$ It's a 2D surface in 3D : $f(u,v) = [ x(u,v), y(u,v), z(u,v) ]$. But the simpler case $z = f(x,y)$ is also interesting since any 3D surface can be locally seen as the graph of such function. $\endgroup$
    – T.L
    Mar 13, 2020 at 23:16
  • $\begingroup$ You imply differential area $E = \int (\kappa_1^2 + \kappa_2^2)\, dA$ etc? $\endgroup$
    – Narasimham
    Apr 3, 2020 at 16:44

1 Answer 1

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The biharmonic equation $\Delta^2 f = 0$ is NOT the Euler-Lagrange equation related to the energy defined by that first integral you posted.

Rather, it's what you get when you apply a thin plate approximation: $\nabla f \approx 0$. Under that, $\kappa_1^2 + \kappa_2^2 \approx f_{xx}^2 + 2 f_{xy}^2 + f_{yy}^2$. Applying Euler-Lagrange to that last expression leads to the (notably linear) biharmonic equation.

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