# How to express the second fundamental form in terms of deformation second gradient

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).

• 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. – Ted Shifrin Aug 10 '15 at 15:30
• 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. – Mechanical engineer Aug 10 '15 at 17:03
• 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. – Ted Shifrin Aug 10 '15 at 17:15
• 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 $(\nabla{\bf r})^T\cdot{\bf n}=\nabla{\bf r}\cdot\nabla{\bf n}$? – Mechanical engineer Aug 10 '15 at 17:51
• Sorry, ${\bf II}=(\nabla{\bf r})^T\cdot\nabla{\bf n}$. – 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.