Given $$S_\mu(u) := \sum_{i=1}^n \sqrt{u_i^2+\mu^2}-\mu$$ a smoothed $1$-norm. Using Newton's method, calculate $$\min_{u\in \mathbb{R}^n}\frac{1}{2} \|u-u_0\|_2^2 + \alpha S_\mu(\nabla u)$$ where $$(\nabla u)_i = \begin{cases} u_{i+1} - u_i & 1 \leq i < n\\0 & i = n\end{cases}$$
In order to do that, I will need the gradient and Hessian for the expression. I tried representing $\nabla$ as follows
$$\nabla = \begin{pmatrix} -1 & 1 & \dots & \dots & 0 \\ 0 & -1 & 1 & \dots & 0 \\ 0 & 0 & -1 & \dots & 0 \\ \vdots & \vdots & \ddots & \ddots & 0\\ 0 & \dots & 0 & -1 & 1\\ 0 & \dots & 0 & 0 & 0 \end{pmatrix}$$
and I know that the first parts derivative is just $u$ itself but am confused about $\frac{\partial}{\partial u_i} S_\mu(\nabla u)$.
The cost function writes $$ \phi(\mathbf{u}) = \frac12 \|\mathbf{u}-\mathbf{u}_0 \|_2^2 +\alpha \mathbf{1}_N^Tf(\mathbf{Du}) +c $$ where $f(\mathbf{y})$ is applied elementwise on each element of $\mathbf{y}$ and thus is a column vector of the same size as $\mathbf{y}$. Here $f(x)=\sqrt{x^2+\mu^2}$.
Taking the differential \begin{eqnarray} d\phi &=& (\mathbf{u}-\mathbf{u}_0):d\mathbf{u} + \alpha \mathbf{1}_N: f'(\mathbf{Du}) \odot d(\mathbf{Du})\\ &=& (\mathbf{u}-\mathbf{u}_0) + \alpha \mathbf{D}^T f'(\mathbf{Du}) : d\mathbf{u} \\ \end{eqnarray} where $f'(x)=\frac{x}{\sqrt{x^2+\mu^2}}$ and the colon denotes the inner product vector.
The gradient is the vector $$\mathbf{g} = (\mathbf{u}-\mathbf{u}_0) + \alpha \mathbf{D}^T f'(\mathbf{Du})$$ It follows \begin{eqnarray} d\mathbf{g} &=& d\mathbf{u} + \alpha \mathbf{D}^T \left[ f''(\mathbf{Du}) \odot d(\mathbf{Du}) \right] \\ &=& \left[ \mathbf{I} + \mathbf{D}^T \mathrm{Diag}(f''(\mathbf{Du})) \mathbf{D} \right] d\mathbf{u} \end{eqnarray} The bracket term is the Hessian.