Views
5 years ago

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Ansatz: with

Ansatz: with Notes: ψ = ψL + ψR −ψ0 +ǫψ1 + . . . ���� subtract common part ψL = ψF (x − xL(T)) ψR = ψF (xR(T) − x) T = ǫt c = c0 + ǫc1 c0 = 4 √ 3 ψF(ζ) = ψ0 � 1 2 (1 + tanhζ) ψ0 = 3 1 4 • c0 can be determined as the point where the potential V (ψ) = + 1 the same value at the two minimal ψ = 0 and ψ = ψ0 (cf. Sec.4.1.1). Denote Insert expansion ψ ′ L ≡ dψF(ζ) dζ � � � � ζ=x−xL ψ ′ R ≡ dψF(ζ) dζ � � � � ζ=xR−x 0 = ǫ {ψ ′ L ∂TxL − ψ ′ R ∂TxR} + ψ ′′ L + ψ′′ R + ǫψ′′ 1 − ψL − ψR + ψ0 − ǫψ1 + (c0 + ǫc1) {ψL + ψR − ψ0 + ǫψ1} 3 − {ψL + ψR − ψ0 + ǫψ1} 5 For x < xm we have xR − x ≫ 1: �� 1 ψR − ψ0 = ψ0 2 (1 + tanh (xR � − x)) − 1 �� 1 = ψ0 = ψ0 exR−x x−xR + e + exR−x x−xR − e exR−x x−xR + e 2 � � 1 √ − 1 1 + e−2(xR−x) Analogously for x > xm and x − xL ≫ 1: − 1 → − 1 −2(xR−x) ψ0e 2 ψL − ψ → − 1 −2(x−xL) ψ0e 2 Consider expansion separately for x < xm and x > xm. For x < xm: and � 2ψ2 − 1 4cψ4 + 1 6ψ6 has for xR − x → ∞ {ψL + (ψR − ψ0) + ǫψ1} 3 = ψ 3 L + 3ψ 2 L (ψR − ψ0) + +3ǫψ1ψ 2 L + O � (ψR − ψ0) 2 , ǫ (ψR − ψ0),ǫ 2� {ψL + (ψR − ψ0) + ǫψ1} 5 = ψ 5 L + 5ψ 4 L (ψR − ψ0) + +5ǫψ1ψ 4 L + O � (ψR − ψ0) 2 , ǫ (ψR − ψ0),ǫ 2� 92

Using that ψL,R satisfy the O(ǫ 0 ) equations ψ ′′ L − ψL + c0ψ 3 L − ψ5 L = 0 ψ′′ R − ψR + c0ψ 3 R − ψ5 R we get for x < xm −ǫ � ψ ′′ 1 − ψ1 + 3c0ψ 2 Lψ1 − 5ψ 4 Lψ1 � � �� � with For x > xm with LLψ1 = 0 = ψ ′′ R + (ψR − ψ0) � −1 + 3c0ψ 2 L − 5ψ4 � L + +ǫ � c1ψ 3 L + ∂TxLψ ′ L + ∂TxRψ ′ R LL = ∂ 2 x − 1 + 3c0ψ 2 L − 5ψ 4 L −ǫLRψ1 = ψ ′′ L + (ψL − ψ0) � −1 + 3c0ψ 2 R − 5ψ4 � R + +ǫ � c1ψ 3 R + ∂TxLψ ′ L + ∂TxRψ ′ � R LR = ∂ 2 x − 1 + 3c0ψ 2 R − 5ψ 4 R To obtain evolution equations for xL and xR we need two solvability conditions Translation symmetry: Note: • single front: ∂xψL,R is a 0-eigenvector • two interacting fronts: there is only one exactly vanishing eigenvalue with the eigenvector arising from the double-front solution ∂xψL+R, for which the xR − xL is not growing or shrinking • the double-front solution is stationary for c slightly different than c0 due to the interaction between the two fronts. How do we get a second solvability condition? We want the perturbation expansion to remain well-ordered in the limit xR − xL → ∞, i.e. if the fronts are infinitely far apart, i.e. ψ1 has to remain small compared to ψL + ψR − ψ0 • for any finite L: only 1 translation mode • for L = ∞: 2 independent fronts ⇒ expect 2 translation modes LL∂xψL = 0 LR∂xψR = 0 93 �

Physics 106P: Lecture 5 Notes - The Burns Home Page
Page 2 Lecture Notes in Computer Science 4475 Commenced ...
SPH4U: Lecture 7 Notes - The Burns Home Page
SPH4UI Lecture 1 Notes - The Burns Home Page
Physics 106P: Lecture 5 Notes - The Burns Home Page
Introduction to Computational Neuroscience - ESAM Home Page ...
Interdisciplinary Nonlinear Dynamics (438) - ESAM Home Page
lecture notes 8 - Personal Home Pages (at UEL)
SPH4U: Lecture 8 Notes - The Burns Home Page
Week 5 Lecture Notes - Leeward CC eMedia Server home page
Ancient Civilizations Mr. Hanover Lecture Notes ... - LS Home Page
Lecture Notes - Link home page - Carnegie Mellon University