Energy and Gradients

This is given in Etienne Vouga’s notes here:

https://www.cs.utexas.edu/users/evouga/uploads/4/5/6/8/45689883/turning.pdf

Bending Energy Hessian

For a minimal example of a 2 segment rod with 3 vertices $x_{i-1}, x_i, x_{i+1}$, bending energy is:

\[E(\theta (v1(x_i, x_{i-1}), v2(x_{i+1}, x_i))) = 0.5*K*(\theta(v1, v2) - \theta_0)^2\]

Since our rods are meant to be as straight as possible, $\theta_0 = 0$. This assumption simplifies the energy and gradients to the following:

\[E = 0.5*K*(\theta(v1, v2))^2\]

This comma notation below means derivative $\frac{d f}{dx} = f_{,x}$ and I didn’t know what the comma notation for 2nd derivative was, so I use 2 commas $f’‘(x) = f_{,x ,x}$. Also I drop the $v1, v2$ since they’re just functions of $x$ :

Jacobians
$\bold{E}{,i-1} = \frac{dE}{d x{i-1}} = K\theta(\cdot)\theta_{,i-1}$
$\bold{E}{,i} = \frac{dE}{d x{i}} = K\theta(\cdot)(\theta_{,i}) = K\theta(\cdot)(-\theta_{,i-1} - \theta_{,i+1})$
$\bold{E}{,i+1} = \frac{dE}{d x{i+1}} = K\theta(\cdot)\theta_{,i+1}$

Then the $9x9$ symmetric hessian matrix is made up of 9 individual $3x3$ blocks which are:

9 x 9	Hessian	Structure
$\bold{E}_{,i-1, i-1}$	$\bold{E}_{,i-1, i}$	$\bold{E}_{,i-1, i+1}$
$\bold{E}_{,i, i-1}$	$\bold{E}_{,i, i}$	$\bold{E}_{,i, i+1}$
$\bold{E}_{,i+1, i-1}$	$\bold{E}_{,i+1, i}$	$\bold{E}_{,i+1, i+1}$

---

First row
$\bold{E}{,i-1, i-1} = K*\theta{,i-1}*\theta_{,i-1}^T + K *\theta *\theta_{,i-1, i-1}$
$\bold{E}{,i-1, i} = K*\theta{,i-1}*\theta_{,i}^T + K *\theta *\theta_{,i-1, i}$
$\bold{E}{,i-1, i+1} = K*\theta{,i-1}\theta_{,i+1}^T + \sout{K\theta *\theta_{,i-1, i+1}} = 0$
$\bold{E}{,i, i-1} = H{,i-1, i}^T$
$\bold{E}{,i, i} = K*\theta{,i}\theta_{,i}^T + K *\theta *\theta_{,i, i} \newline = K(-\theta_{,i-1} - \theta_{,i+1})(-\theta_{,i-1} - \theta_{,i+1})^T + K\theta*(-\theta_{,i-1,i} - \theta_{,i+1,i})$
$\bold{E}{,i, i+1} = K*\theta{,i+1}\theta_{,i}^T + K *\theta \theta_{,i+1, i}$
$\bold{E}{,i+1, i-1} = E{,i-1, i+1}^T$
$\bold{E}{,i+1, i} = E{,i, i+1}^T$
$\bold{E}{,i+1, i+1} = K*\theta{,i+1}\theta_{,i+1}^T + K *\theta \theta_{,i+1, i+1}$

---

Now I will define each individual component and jacobian from above:

Components
$v_1 = x_i - x_{i-1}$
$v_2 = x_{i+1} - x_{i}$
$z = \frac{v_1 \times v_2}{max(|v_1 \times v_2|, \epsilon)}$
$[z]$ cross product matrix of $z$

---

Jacobians of Theta
$\frac{d\theta}{d x_{i-1}} = \frac{-(x_i - x_{i-1}) \times \hat z}{| x_i - x_{i-1} |^2} = \frac{-v_1 \times \hat z}{| v_1 |^2}$
$\frac{d \theta}{d x_{i+1}} = \frac{-(x_{i+1} - x_{i}) \times \hat z}{| x_{i+1} - x_{i} |^2} = \frac{-v_2 \times \hat z}{| v_2 |^2}$
$\frac{d \theta}{d x_{i}} = (-\frac{d \theta}{d x_{i-1}} - \frac{d \theta}{d x_{i+1}})$

---

Hessians of Theta
$\frac{d^2\theta}{d x_{i-1} ^2} = \frac{|v_1 |^2 [z] - 2v_1 (v_1 \times \hat z)^T}{|v_1|^4}$
$\frac{d^2\theta}{d x_{i+1} ^2} = \frac{|v_2 |^2 [z] + 2v_2 (v_2 \times \hat z)^T}{|v_2|^4}$
$\frac{d^2\theta}{d x_{i-1} dx_i} = \frac{|v_1 |^2 [z] + 2v_1 (v_1 \times \hat z)^T}{|v_1|^4}$
$\frac{d^2\theta}{d x_{i+1} dx_i} = \frac{|v_2 |^2 [z] + 2v_2 (v_2 \times \hat z)^T}{|v_2|^4}$
$\frac{d^2\theta}{d x_i dx_i} =-\frac{d^2\theta}{dx_{x-1} dx_i} - \frac{d^2\theta}{dx_{x+1} dx_i}$