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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Project in the two

Project in the two domains x < xmand x > xm separately onto the translation modes ∂xψL,R, respectively. x < xm: � xm −ǫ � xm � xm ψ −∞ ′ LLLψ1dx = ǫ∂TxL ψ −∞ ′2 Ldx + ǫ∂TxR −∞ � xm + ψ −∞ ′ � L −1 + 3c0ψ 2 L − 5ψ4 L Integrate lhs by parts � xm −∞ ψ ′ LLLψ1dx = ψ ′ Lψ ′ 1| xm −∞ − ψ ′′ L � xm ψ −∞ ′′ Lψ ′ 1dx � �� � ψ1| xm ∞ −R xm −∞ ∂2 x (∂xψL)ψ1dx ψ ′ Lψ′ � xm Rdx + −∞ ψ ′ L ψ′′ R � (ψR − ψ0)dx + ǫc1 dx + � xm −∞ ψ ′ L ψ3 L dx � xm + ψ −∞ ′ � L −ψ1 + 3c0ψ 2 Lψ1 − 5ψ 4 � Lψ1 dx ψ ′ L and ψ′′ L are exponentially small at xm and x → −∞ ⇒ boundary terms are exponentially small and can be ignored at this order since they are already multiplied by ǫ. The remainder of the lhs is LL∂xψL = 0 ⇒ obtain one solvability condition. To estimate and evaluate the integrals rewrite in terms of We get then ψL = ψ0 s = e x−xL and L = xR − xL dx = 1 s ds � � 1 1 + 2 s − 1 s s + 1 s � � xm . . .dx = −∞ = ψ0 � e L 2 0 . . . 1 s ds � 1 1 + s 2 2 + s2 − 1 1 + s2 = ψ0 s √ 1 + s2 ψR − ψ0 → − 1 2 ψ0e −2xR 2(x−xL) 1 2xL e e = − 2 ψ0e −2L s 2 ψ ′ R → −ψ0s ds dx e−2L = −ψ0s 2 e −2L ψ ′′ R → −2ψ0s 2 e −2L ∂sψL = ψ0 � 0 = ǫ∂TxLψ 2 0 1 √ 1 + s 2 − � e L 2 +ψ 2 0 0 +ψ 2 � L e 2 0 0 1 +ǫc1 � e L 2 0 4 ψ4 L x=−∞ � �� � ψ4 0 +h.o.t. s 2 √ 1 + s 2 3 s 2 (1 + s 2 ) 3 � 1 = ψ0 √ 1 + s2 3 ⇒ ψ′ L = ψ0 √ 1 + s2 3 1 s ds + ǫ∂TxRψ 2 � L e 2 0 0 s √ 1 + s 2 3 s � 2 −s � e −2L1 ds + s s √ 1 + s2 3 � 2 −2s � e −2L1 ds + (40) s s √ 1 + s2 3 � −1 + 3c0ψ 2 s 0 2 1 + s2 − 5ψ4 s 0 4 (1 + s2 ) 2 � 2 −s 2 e−2L1 s ds � �x=xm 94

Analogously for x > xm: 0 = ǫ∂TxL � ∞ + xm � ∞ ψ xm ′ Lψ′ Rdx + ǫ∂TxR xm ψ ′ � R −1 + 3c0ψ 2 R − 5ψ 4 R Rewrite these integrals in terms of u = e xR−x dx = − 1 du u du dx = −u ψR u = ψ0 √ 1 + u2 Since ψ ′ R = −dψR dx ψL − ψ0 → − 1 2 ψ0u 2 e −2L � ∞ . . .dx → xm � 0 e x R −xm = udψR du � ∞ ψ ′2 � ∞ Rdx + ψ xm ′ Rψ′′ Ldx + � � ∞ (ψL − ψ0) dx + ǫc1 ψ (41) ′ Rψ 3 Rdx ψ ′ L → ψ0u 2 e −2L u = ψ0 √ 1 + u2 3 � . . . − 1 � � L e 2 du = u 0 xm 1 ∂uψR = ψ0 √ 1 + u2 3 ψ ′′ L → −2ψ0u 2 e −2L . . . 1 u du each integral in the expression for x > xm has a corresponding integral for x < xm and their magnitudes are the same. Add (41) and (40) 0 = ǫ (∂TxL − ∂TxR)ψ 2 0 � e L 2 +2e −2L ψ 2 0 0 +2e −2L ψ 2 � L e 2 0 0 1 +ǫc1 2 ψ4 0 ⎧ ⎨ ⎩ � e L 2 0 s 2 (1 + s 2 ) 3 s √ 1 + s2 3 � 2 −2s � 1 ds + s s √ 1 + s2 3 � −1 + 3c0ψ 2 0 For large s all integrands decay at least as 1 s � L e 2 1 ds − s 0 s 2 s √ 1 + s2 3 � 2 −s � e −2L1 s 4 1 + s2 − 5ψ4 0 (1 + s2 ) 2 � −s 2 2 s ds ⎫ ⎬ ⎭ + 1 ds + s ⇒ the integrals are at most O(ln s) = O(L): • We can therefore ignore the second integral with respect to the first integral in the first term • ǫ must be exponentially small in L to balance the terms 95

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