Calculating the Jacobian for a function containing a derivative

Question

I have the equation

$F(t) = \phi u + \frac{1}{2}\frac{d^2u}{dt^2} + u^3$

and broadly speaking, my task is to calculate the $\phi$ and $u(t)$ such that $F(t) = 0$.

I am testing out a new algorithm to do this (and so there are no in-built ways to implement it). This requires calculating the Jacobian of the function.

I am implementing this on MATLAB.

I am unable to figure out how exactly I take the partial derivative with $u$ for the term with the derivative.

Here's what I have done so far -

Assuming $\textbf{u}_3$ is the row vector form of $u$ which has been cubed elementwise, I calculated the Jacobian as -

$\textbf{J} = \phi\textbf{I} + \textbf{D} + \textbf{u}_3\textbf{I}$

where D is the appropriate finite difference matrix and I is the identity matrix.

My first question is - is this correct?

If yes, my second question is, can the same thing be done for a different differentiation technique (like spectral differentiation) by simply replacing D by the appropriate matrix?

Answer 1

There are some unknowns in what you are doing but for simplicity, suppose we want to find $u(t)$ as discrete times $t_1, t_2, \cdots, t_n$. Let $\textbf{F} = [F(t_1), F(t_2), \cdots, F(t_n)]^T$ and $\textbf{u} = [u(t_1), u(t_2), \cdots, u(t_n)]^T$ be column vectors representing $F$ and $u$ evaluated at the desired times. From your problem statement, you wish to find $\phi$ and $\textbf{u}$ such that $\textbf{F} = \textbf{0}$. We can write out the vector equation as

\begin{align} \textbf{F} &= \underbrace{(\phi I + D)}_{A}\textbf{u} + \left(\textbf{u} \odot \textbf{u} \odot \textbf{u}\right) \end{align}

where $D$ is your appropriate finite difference matrix and $\odot$ corresponds to the Hadamard product. The jacobian you are after is going to have partial derivatives of the form $\frac{\partial F_i}{ \partial u_j}$, where $F_i = F(t_i)$ and $u_j = u(t_j)$. We can look at the above vector system elementwise and see that for a fixed $i$, we have that

$$F_i = u_i^3 + \sum_{j} A_{ij} u_j$$

If you take the partial derivative of $F_i$ with respect to some $u_k$, you get that \begin{align} \frac{\partial F_i}{\partial u_k} &= \frac{\partial}{\partial u_k} \left\lbrace u_i^3 + \sum_j A_{ij} u_j \right\rbrace \\ &= 3 \delta_{ik} u_i^2 + A_{ik} \end{align}

where $\delta_{ij} = 1$ if $i = j$ and is $0$ otherwise. This implies that your jacobian portion $J = [\frac{\partial F_i}{\partial u_k}]$ (where $i$ corresponds to rows and $k$ to columns) has the matrix form of

$$J = A + 3 \text{diag}\left(\textbf{u} \odot \textbf{u}\right)$$