证明
Lemma 6 ([The free cap edge]). Retain the model geometry and flow coordinates…Lemma 5 ([High derivative equation]). Let l≥8 . Suppose that the ordinary…观点
We record the energy argument, which will also be used on the pulse. If v satisfies v_ θθ -P_1v_ θξ -S_1v_ ξξ =b^θv_θ+b^ξv_ξ+R, with the above coefficient bounds and bounded b^i , use the energy density and flux e=(v_θ^2+S_1v_ξ^2)/(2), j=(P_1v_θ^2)/(2)+S_1v_ξv_θ. One has j≤(P_1+√S_1)e . Integrate on intervals whose endpoints move inward at speed L/4 . Multiplication by v_θ and integration by parts, with the nonpositive boundary loss discarded, gives E'(θ) ≤C E(θ) +√2 E(θ)\, R_ L^2 , E(θ)=∫eξ. Thus √ E is controlled by its initial value and the time integral of R_2 , with a Gronwall factor. One can justify division at zero energy by first using √ E+η and then letting η0 . Apply this estimate to v=_ξ^jz , j≥8 , obtained by differentiating. The top second-jet terms give the principal coefficients in. If a second-jet argument receives j-1 differentiations, it gives v_ξ or v_θ ; put these terms, together with the direct first-jet linearization, into b^i_iv . Their coefficients use only bounded low derivatives. Put V=(z_ξ,z_θ) . Every remaining solution factor is a spatial derivative of V of order at most j-1 , already controlled by the induction. A factor of order j-1 requires at least j-2 of the differentiations, since a second-jet argument is already one spatial derivative of V . There cannot be two such factors, because 2(j-2)>j . Bound that factor in L^2 , if it occurs. All other factors have order at most j-2 , so the one-dimensional Sobolev inequality and the available H^ j-1 bound for V control them in supremum norm. This gives the L^2 bound for the remainder in. The same moving interval is used at every derivative order; its length has a fixed positive lower bound, so the one-dimensional Sobolev constants are uniform. The initial first-jet bounds through spatial order seven follow from the C^8 class bound. For each larger order, the initial energy on the cut θ=-r is bounded by the cap estimate. The metric derivatives are bounded on -r≤θ≤-δ/τ , so closes the spatial induction there. The intervals start as [-Lr,Lr] and lose at most Lr/4 at each end, leaving room for Sobolev estimates on ξ≤Lr/2 . Reverse time and start at θ=r to obtain the same bounds at the upper pulse edge.