Brief notes on the derivation of the KdV equation

where θ 2 (ξ, τ) is an arbitrary function independent of z. This process can be repeated in anarbitrary number of times. However, for the present purpose, we do not need higher orderapproximations.We now calculate η n ’s using both (11) and (12). For the zero-th order term η 0 , (11) andthe result Φ 0 = θ 0 (ξ, τ) implyη 0 = ∂Φ 0∂ξ = ∂θ 0∂ξ .We will use this and previous results for Φ 1 and Φ 2 and their derivatives in what follows.Note that each z appearing in Φ n ’s and their derivatives will be replaced by αη since z = αηin both (11) and (12). By substituting the full power series for both η and Φ into (11) andretaining only terms up to α, we obtainη 0 + α η 1 −(η 0 + α ∂θ )1+ α ∂θ 0∂ξ ∂τ + α 2⇒ η 1 − ∂θ 1∂ξ + ∂θ 0∂τ + 1 2 η2 0 = 0.( ) 2 ∂θ0= 0∂ξDifferentiating this wrt ξ and noting that ∂ 2 θ 0 /∂τ∂ξ = ∂η 0 /∂τ we obtain∂η 1∂ξ − ∂2 θ 1∂ξ + ∂η 02 ∂τ + η ∂η 00∂ξ= 0. (21)This equation is nonlinear and involves both η 0 and η 1 . It also contains an undeterminedfunction θ 1 (ξ, τ).A similar nonlinear equation involving both η 0 and η 1 and θ 1 (ξ, τ) can be derived bysubstituting the full power series for both η and Φ into (12) and retaining terms up to α 2 .By so doing we obtainα ∂Φ 1∂z + ∂Φ (α2 2∂z = α − ∂η 0∂ξ − α ∂η 1∂ξ + α ∂η 0∂τ + α ∂η 0∂ξ)∂Φ 0∂ξ⇒ − (1 + α η 0 ) ∂η ( )0 1∂ξ − α ∂ 3 η 03 ∂ξ + ∂2 θ 1= − ∂η 03 ∂ξ 2 ∂ξ − α ∂η 1∂ξ + α ∂η 0∂τ + α ∂η 0∂ξ η 0∂η 0⇒ − η 0∂ξ − 1 ∂ 3 η 03 ∂ξ − ∂2 θ 13 ∂ξ = −∂η 12 ∂ξ + ∂η 0∂τ + η ∂η 00∂ξAdding (21) and (22) we obtain⇒ ∂η 0∂τ + 2η ∂η 00∂ξ + 1 ∂ 3 η 03 ∂ξ − ∂η 13 ∂ξ + ∂2 θ 1= 0. (22)∂ξ2 2 ∂η 0∂τ + 3η ∂η 00∂ξ + 1 3∂ 3 η 0= 0. (23)∂ξ3 This equation governs the evolution of the zero-th order approximation η 0 to the surfaceelevation η.6