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**

Re: ODEs, Adjoint forms and matrix representation

Quote:

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.