Higher order partial derivative of vector function

Question

Let $$f(x_1,x_2)=\begin{bmatrix}x_1x_2\sin x_1\\x_2\cos x_1-x_1\end{bmatrix}$$ Now I need to find $\frac {\partial^2 f(x)}{\partial x^2}$. I have no idea how I should find this. The first order partial should be the jacobian but what is the 2nd order partial derivative? And how can I find a bilinear map $\frac {\partial^2 f(x)}{\partial x^2}(p,q)$? Any help is appreciated. Thank you.

Answer 1

To get the second partial derivative, you literally need to differentiate twice:

$$ f = \left[ \begin{array}{c} x_1 x_2 \sin x_1 \\ x_2 \cos x_1 - x_1 \end{array}\right] \implies \frac{\partial f}{\partial x_1} = \left[ \begin{array}{c} x_2 \sin x_1 + x_1 x_2 \cos x_1 \\ - x_2 \sin x_1 - 1 \end{array}\right]$$ $$ \implies \frac{\partial^2 f}{\partial x_1^2} = \left[ \begin{array}{c} 2 x_2 \cos x_1 - x_1 x_2 \sin x_1 \\ - x_2 \cos x_1 \end{array}\right].$$

To write the second derivative as a bilinear map $D^2 f: \mathbb R^2 \times \mathbb R^2 \to \mathbb R^2$, you first need to work out all four second derivatives. Then, if $v_i, w_i \in \mathbb R^2$ are a pair of directional vectors, the map $D^2 f$ is defined by: $$ D^2 f(v, w) = \sum_{i=1}^2 \sum_{j=1}^2 v_i w_j \frac{\partial^2 f}{\partial x_i \partial x_j} .$$

Answer 2

Observe $f: \mathbb{R}^2\rightarrow \mathbb{R}^2$ which means $Df: \mathbb{R}^2 \rightarrow L(\mathbb{R}^2; \mathbb{R}^2)$ and $D^2f:\mathbb{R}^2 \rightarrow L(\mathbb{R}^2; L(\mathbb{R}^2; \mathbb{R}^2))$ where $L(\mathbb{R}^2; \mathbb{R}^2)$ is the space of linear maps from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$; likewise, $L(\mathbb{R}^2; L^2(\mathbb{R}^2; \mathbb{R}^2))$ is the space of linear maps from $\mathbb{R}^2$ to the linear space of linear maps.

In short $Df$ maps $(x_1, x_2)$ to a matrix $Df(x_1, x_2)$ which in this case is \begin{align} Df(x_1, x_2)= \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{pmatrix} = \begin{pmatrix} x_2\sin x_1+ x_1x_2\cos x_1 & x_1\sin x_1\\ -x_2\sin x_1-1 & \cos x_1 \end{pmatrix}. \end{align} Alternative, we could say that $Df$ is a (1, 1)-tensor since it takes in a vector and a co-vector, i.e. $Df(x_1, x_2)[v^T, w]$. Hence, we can write $Df$ as \begin{align} Df =&\ \sum_{i, j}\frac{\partial f_i}{\partial x_j}\ e_i\otimes dx_j\\ =&\ (x_2\sin x_1+ x_1x_2\cos x_1)\ e_1\otimes dx_1+ (x_1\sin x_1)\ e_1\otimes dx_2\\ &-\ (x_2\sin x_1+1)\ e_2\otimes dx_1+ (\cos x_2)\ e_2\otimes dx_2. \end{align}

For the second derivative of $f$, $D^2f$, we get a $(1, 2)$-tensor where the "matrix" components are given by \begin{align} (D^2f)_{jk}^i = \frac{\partial^2 f_i}{\partial x_j\partial x_k} \end{align} which means \begin{align} D^2f =&\ \sum_{i, j , k}\frac{\partial^2 f_i}{\partial x_j\partial x_k}\ e_i\otimes dx_j\otimes dx_k\\ =&\ (2x_2\cos x_1-x_1x_2\sin x_1)\ e_1\otimes dx_1\otimes dx_1+(\sin x_1+x_1\cos x_1)\ e_1\otimes dx_1\otimes dx_2\\ &+(\sin x_1 +x_1\cos x_1)\ e_1\otimes dx_2\otimes dx_1 + (0)\ e_1\otimes dx_2\otimes dx_2\\ &-(x_2\cos x_1)\ e_2\otimes dx_1\otimes dx_1 - (\sin x_1)\ e_2\otimes dx_1\otimes dx_2\\ &-(\sin x_1)\ e_2\otimes dx_2\otimes dx_1+(0)\ e_2\otimes dx_2\otimes dx_2 \end{align}

Edit: One should note that \begin{align} L(V; W) \cong V^\ast\otimes W \end{align} which means \begin{align} L(\mathbb{R}^2; L(\mathbb{R}^2; \mathbb{R}^2))\cong (\mathbb{R}^2)^\ast\otimes (\mathbb{R}^2)^\ast\otimes \mathbb{R}^2. \end{align}