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## Since ψ0(x) breaks the

Since ψ0(x) breaks the continuous translation symmetry L is singular and the eigenvector associated with the 0 eigenvalue is the translation mode ∂xψ0. Note: • if ψ0 did not break the translation symmetry, ∂xψ0 would vanish and not represent an eigenvector and there would be no 0 eigenvalue associated with the translation symmetry. L is self-adjoint ⇒ ∂xψ0 is also its left 0-eigenvector. Project (39) on ∂xψ0 Thus Notes: v1 � ∞ −∞ 0 = v1 � +∞ −∞ ∂xψ0 � −v1∂xψ0 − λ1∂λ∂ψ ˆ � � � V (λ; ψ) � dx λ=0,ψ=ψ0 � ∞ (∂xψ0) 2 dx = −λ1∂λ −∞ � +∞ −∞ ∂ψV (0, ψ0) ∂xψ0 dx � �� � ∂xV (0;ψ0(x)) (∂xψ0) 2 dx = −λ1 ∂λ [V (λ; ψ0(x))]| λ=0 | +∞ −∞ ≈ − V (λ1; ψ0)| +∞ −∞ • the lhs of the equation can be read as the amount of work performed by the friction � +∞ � β −∞ dx � dx dt dt dt � �� � dx • the rhs of the equation can be read as the difference in potential energy between initial and final state • the perturbation method does not rely on the existence of a potential ⇒ works also when there are multiple coupled components ψj(x, t) satisfying nonlinear PDEs that cannot be derived from a potential. 4.2 Interaction between Fronts Consider fronts of the nonlinear diffusion equation Notes: ∂tψ = ∂ 2 x ψ − ψ + cψ3 − ψ 5 • the coefficients of ∂ 2 x ψ, ψ, and of ψ5 can be chosen to have magnitude 1 by rescaling of space, time and ψ. 90

• the coefficient of ψ is chosen negative: ψ = 0 is linearly stable • the coefficient of ψ 5 is chosen negative: saturation at large values of ψ Homogeneous stationary states: linearly stable linearly unstable V Figure 18: Potential with minima at ψ0 and ±ψ0. ψ = 0 or ψ 2 0 = c + √ c 2 − 4 2 ψ 2 u = c − √ c 2 − 4 2 Consider fronts that connect ψ = 0 with ψ = ψ0 ψ ψ R L L x x ψ L m Figure 19: Front positions. Goal: evolution equations for xL and xR, which describe the interaction between the two fronts. We need separation of time scales • individual fronts should move slowly • interaction between the fronts weak: for xR − xL large the fronts deform each other only weakly 91 x R ψ ψ R

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