Suppose we have a surface $\Omega$ with prescribed principal curvatures, $\kappa_1$, $\kappa_2$, say. An isometric deformation ${\bf r}:\Omega\rightarrow\mathbb{R}^3$ maps the surface into $\mathbb{R}^3$ and changes its curvature into $\kappa_{11}$, $\kappa_{21}$, say. How to express the second fundamental form of the deformed surface in terms of the second gradient of the deformation: $\nabla^2{\bf r}$.

The second fundamental form is defined as ${\bf II}=(\nabla{\bf r})^T\nabla{\bf n}$, ${\bf n}$ is the unit normal to the surface: ${\bf n}=\frac{\partial{\bf r}}{\partial x_1}\wedge\frac{\partial{\bf r}}{\partial x_2}$.

${\bf Note}$: I am very sorry, in the above definition of the second fundamental form I did a mistake, it should be ${\bf II}=(\nabla{\bf r})^T\nabla{\bf n}$ (and not "dot" product).

  • $\begingroup$ Note that you need $\Omega\subset\Bbb R^3$ in the first place in order for principal curvatures to make sense. Presumably your deformation has an extra $t$ parameter, and $t=0$ corresponds to the original position of the surface. Usually, there's a negative sign in your definition of $\mathbf{II}$ and then, using $\nabla\mathbf r\cdot \mathbf n = 0$, you get $\mathbf{II} = \nabla^2\mathbf r \cdot\mathbf n$. This is the usual way the second fundamental form is calculated. $\endgroup$ – Ted Shifrin Aug 10 '15 at 15:30
  • $\begingroup$ Indeed, $\Omega\subset\mathbb{R}$, but I do not quite understand the role and meaning of $t$ parameter. What is also not clear for me is why $\nabla{\bf r}\cdot{\bf n}=0$ and how do you use the $isometry$ of ${\bf r}$. I used the definition of ${\bf II}$ from G. Friesecke, R.D. James, S. Muller, A theorem on geometric rigidity and the derivation of nonlinear plate theory from three-dimensional elasticity. Comm. Pure Appl. Math. 55 (11) (2002) 1461–1506. $\endgroup$ – Mechanical engineer Aug 10 '15 at 17:03
  • $\begingroup$ Ordinarily, a deformation is given by a smooth family of mappings, depending on an auxiliary parameter $t$ (say, time). For example, we can isometrically deform a helicoid to a catenoid (with every surface being minimal, in fact). I honestly don't think having the deformation be through isometries is relevant for the computation you want to do. $\nabla\mathbf r\cdot\mathbf n$ because $\mathbf n$ is normal to the tangent plane of the surface. $\endgroup$ – Ted Shifrin Aug 10 '15 at 17:15
  • $\begingroup$ Dear Ted! Thank you for explanation, but I have something to clarify. As I see, from $\nabla\left(\nabla{\bf r}\cdot{\bf n}\right)=\nabla^2{\bf r}\cdot{\bf n}+\nabla{\bf r}\cdot\nabla{\bf n}=0$ follows that $\nabla{\bf r}\cdot\nabla{\bf n}=-\nabla^2{\bf r}\cdot{\bf n}$, but in the definition above ${\bf II}=(\nabla{\bf r})^T\cdot{\bf n}$. I do not have strong knowledge, that is why I cannot understand is $(\nabla{\bf r})^T\cdot{\bf n}=\nabla{\bf r}\cdot\nabla{\bf n}$? $\endgroup$ – Mechanical engineer Aug 10 '15 at 17:51
  • $\begingroup$ Sorry, ${\bf II}=(\nabla{\bf r})^T\cdot\nabla{\bf n}$. $\endgroup$ – Mechanical engineer Aug 10 '15 at 17:57

So, let me summarize what do you have and what do you need. You have $$ \nabla{\bf r}\cdot{\bf n}=0~~ {\rm and}~~ (\nabla{\bf r})^T\cdot\nabla{\bf r}={\rm Id}, $$ and must apply $\nabla$ to the first equation, so actually, you have $$ \nabla\left(\nabla{\bf r}\cdot{\bf n}\right)=0~~ {\rm and}~~ (\nabla{\bf r})^T\cdot\nabla{\bf r}={\rm Id}. $$ Here $\nabla{\bf r}\in\mathbb{R}^{2\times 3}$ and ${\bf n}\in\mathbb{R}^3$.

You have to clarify what do you understand by dot product of $2\times 3$ matrix and $3$-vector, i.e. $\nabla{\bf r}\cdot{\bf n}$ and calculate $\nabla\left(\nabla{\bf r}\cdot{\bf n}\right)$. It will be expressed in terms of ${\bf II}$ and $\nabla^2{\bf r}$. That would be the answer.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.