1. ODEs, Adjoint forms and matrix representation

Hi MHF members,

I was looking up in the books I have but I could not go anywhere.
Consider the higher-order ode
$\displaystyle L[y]:=y^{(n)}+\sum_{k=0}^{n-1}p_{n-k}y^{(k)}=0$,
which has the matrix form
$\displaystyle \pmb{Y}^{\prime}+\pmb{A}\pmb{Y}=\pmb{0}$,
where
$\displaystyle \pmb{Y}=\left( \begin{array}{c} y \\ y^{\prime} \\ \vdots \\ y^{(n-1)} \\ \end{array} \right)$ and $\displaystyle \pmb{A}=\left( \begin{array}{cccc} & -1 & & \\ & & \ddots & \\ & & & -1 \\ p_{n} & p_{n-1} & \cdots & p_{1} \\ \end{array} \right)$.
The adjoint form of $\displaystyle L$ is given by
$\displaystyle L^{\ast}[z]=z^{(n)}+\sum_{k=0}^{n-1}(-1)^{n-k}[p_{n-k}z]^{(k)}=0$,
where the coefficients are assumed to be real and sufficiently smooth.
How the matrix form of the adjoint equation can be derived?
Have any of you met with this in a book/reference?

Thanks.
bkarpuz

2. Re: ODEs, Adjoint forms and matrix representation

Originally Posted by bkarpuz
Hi MHF
How the matrix form of the adjoint equation can be derived?
Actually, I can use the Leibnitz rule for higher-order derivatives of
the product term$\displaystyle p_{n-k}z$ for expansion, and then reverse the order of two sums and get
$\displaystyle L^{\ast}[z]=z^{(n)}+\sum_{k=0}^{n-1}\underbrace{\bigg[\sum_{\ell=k}^{n-1}(-1)^{n-\ell}\binom{\ell}{k}p_{n-\ell}^{(\ell-k)}\bigg]}_{\text{Coefficient}}z^{(k)}$.
Then write the matrix form of this equation. However, I was looking for a different way.
For instance, having the coloumns of the unknown both involving $\displaystyle p$ and $\displaystyle z$, if possible.