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Math Help - ODEs, Adjoint forms and matrix representation

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    Senior Member bkarpuz's Avatar
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    Exclamation 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
    L[y]:=y^{(n)}+\sum_{k=0}^{n-1}p_{n-k}y^{(k)}=0,
    which has the matrix form
    \pmb{Y}^{\prime}+\pmb{A}\pmb{Y}=\pmb{0},
    where
    \pmb{Y}=\left( \begin{array}{c} y \\ y^{\prime} \\ \vdots \\ y^{(n-1)} \\ \end{array} \right) and \pmb{A}=\left( \begin{array}{cccc} & -1 &  &  \\ &  & \ddots &  \\ &  &  & -1 \\ p_{n} & p_{n-1} & \cdots & p_{1} \\ \end{array} \right).
    The adjoint form of L is given by
    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
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    Senior Member bkarpuz's Avatar
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    Re: ODEs, Adjoint forms and matrix representation

    Quote Originally Posted by bkarpuz View Post
    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 p_{n-k}z for expansion, and then reverse the order of two sums and get
    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 p and z, if possible.
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