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2d2u. = hr0(xi. +. X2). aM2. dt2 V°dx2 2Mu0 8x ' iS^P = [e0 - 2JB{T,q)]<p(x,t) - J^B^q)^'^ dt dx2 , Kxi ±X2)ro du{x, t) H ^ <p{x, t). w0 dx (9.44) (9.45) ...

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X(t) (0 < t < T) (0 < t < T) (0 < x < X{T)) Since 1 2d2u ( a2\ du qu = q — re thus from the discussion of the properties of the free boundary in §6.5, ...

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... to the conditions that u{0,t) = 0, u(L,t) = 0 u(x,0) = sin(7:x/L) +sin(2irx/L) ut(x,0) = 0. 2.2 Find a separable solution of the equation d2u _ 2d2u ...

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Thus, integrating (10.4) over the whole u plane, π 4 D2 = π 4D20 + ∫ |(u)|2d2u (10.5) in which D20 is the zero (unscattered) order, which is the integrated ...

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(2.9.115) Integrating in t ∈ [0,t] gives D2u2L2 + 7 8u10L10 + 25 2 ∥ ∥ |u|2Du ∥ ∥2L2 + 5Re(|u|2u2,(Du)2 ) +5Im ( |u|2D2u,Du ) – 35 8Im(|u|6u, ...

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Since ukl(t) -* u(t) in L2, we have v + u(t)2/2 = d~2d2u(t), and (9.4.47) follows. Since E23)(uk(t)) = E23)(<pk) (see (9.4.4)), the inequality (9.4.45) ...

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