Bretherton equation

The Bretherton equation derives from the Lagrangian density:^[6]

${\displaystyle {\mathcal {L}}={\tfrac {1}{2}}\left(u_{t}\right)^{2}+{\tfrac {1}{2}}\left(u_{x}\right)^{2}-{\tfrac {1}{2}}\left(u_{xx}\right)^{2}-{\tfrac {1}{2}}u^{2}+{\tfrac {1}{p+1}}u^{p+1}}$

through the Euler–Lagrange equation:

${\displaystyle {\frac {\partial }{\partial t}}\left({\frac {\partial {\mathcal {L}}}{\partial u_{t}}}\right)+{\frac {\partial }{\partial x}}\left({\frac {\partial {\mathcal {L}}}{\partial u_{x}}}\right)-{\frac {\partial ^{2}}{\partial x^{2}}}\left({\frac {\partial {\mathcal {L}}}{\partial u_{xx}}}\right)-{\frac {\partial {\mathcal {L}}}{\partial u}}=0.}$

The equation can also be formulated as a Hamiltonian system:^[7]

${\displaystyle {\begin{aligned}u_{t}&-{\frac {\delta {H}}{\delta v}}=0,\\v_{t}&+{\frac {\delta {H}}{\delta u}}=0,\end{aligned}}}$

in terms of functional derivatives involving the Hamiltonian ${\displaystyle H:}$

${\displaystyle H(u,v)=\int {\mathcal {H}}(u,v;x,t)\;\mathrm {d} x}$   and   ${\displaystyle {\mathcal {H}}(u,v;x,t)={\tfrac {1}{2}}v^{2}-{\tfrac {1}{2}}\left(u_{x}\right)^{2}+{\tfrac {1}{2}}\left(u_{xx}\right)^{2}+{\tfrac {1}{2}}u^{2}-{\tfrac {1}{p+1}}u^{p+1}}$

with ${\displaystyle {\mathcal {H}}}$ the Hamiltonian density – consequently ${\displaystyle v=u_{t}.}$ The Hamiltonian ${\displaystyle H}$ is the total energy of the system, and is conserved over time.^[7][8]