# Calculating the Jacobian for a function containing a derivative

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?

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

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