Abstract It is shown that the strongly noncanonical fourth order operator Ly=r3(t)r2(t)r1(t)y′(t)′′′$$\begin{array}{} \displaystyle \mathcal {L}\,y=\left(r_3(t)\left(r_2(t)\left(r_1(t)y'(t)\right)'\right)'\right)' \end{array}$$ can be written in essentially unique canonical form as Ly=q4(t)q3(t)q2(t)q1(t)q0(t)y(t)′′′′.$$\begin{array}{} \displaystyle \mathcal {L}\,y =… Click to show full abstract
Abstract It is shown that the strongly noncanonical fourth order operator Ly=r3(t)r2(t)r1(t)y′(t)′′′$$\begin{array}{} \displaystyle \mathcal {L}\,y=\left(r_3(t)\left(r_2(t)\left(r_1(t)y'(t)\right)'\right)'\right)' \end{array}$$ can be written in essentially unique canonical form as Ly=q4(t)q3(t)q2(t)q1(t)q0(t)y(t)′′′′.$$\begin{array}{} \displaystyle \mathcal {L}\,y = q_4(t)\left(q_3(t)\left(q_2(t)\left(q_1(t)\left(q_0(t)y(t)\right)'\right)'\right)'\right)'. \end{array}$$ The canonical representation essentially simplifies examination of the fourth order strongly noncanonical equations r3(t)r2(t)r1(t)y′(t)′′′+p(t)y(τ(t))=0.$$\begin{array}{} \displaystyle \left(r_3(t)\left(r_2(t)\left(r_1(t)y'(t)\right)'\right)'\right)'+p(t)y(\tau(t))=0. \end{array}$$
               
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