Errata. Updated as of June 13, (η(u)φt. + q(u)φ x + εη(u)φ xx. 2 for x < 3t/2+3x 1 2x 2, u(x, t) = 1 for x 3t/2+3x 1 2x 2.

Transcription

1 Errata H. Holden and N. H. Risebro, Front Tracking for Hyperbolic Conservation Laws Applied Mathematical Sciences, volume 3, Springer Verlag, New York, 00 Updated as of June 3, 007 Changes appear in red. Chapter Page 7, line 7 should read: L p loc U = {f f Lp K for every compact set K U}. Page, line after equation.8 should read:... the velocity v is given... Page 9, line 6+ should read:... Roždestvenskiĭ and... Chapter Page 6, last line of equation.0 should read: ηuφt + quφ x + εηuφ xx dx dt,.0 Page 39, line four above figure should read:... depicts this solution in the x, t plane... Page 39, equation.4 should read: { for x < 3t/+3x x, ux, t = for x 3t/+3x x..4 Page 44, line in epigraph should read: Henrik Ibsen, Peer Gynt 867 Pages 46-47, from line 6+ p. 46 to line 7+ p. 47 the text should be replaced by: Returning now to conservation laws and.49, we must make a smart choice of a test function φx, t, y, s. Let ψx, t be a test function that has support in t > ɛ 0. We then define x + y φx, t, y, s = ψ, t + s ω ε0 t sω ε x y, where ε 0 and ε are small positive numbers. In this case φ t + φ s = ψ x + y, t + s ω ε0 t sω ε x y, t We are grateful to F. Gossler, O. Sete, and M. Rejske. Beware! Here ψ x+y t, t+s means the partial derivative of ψ with respect to the second variable, and this derivative is evaluated at x+y, t+s.

2 and 3 Introduce φ x + φ y = ψ x + y, t + s ω ε0 t sω ε x y. x K = ux, t vx, t ψ t + qu, vψ x and new variables with Jacobian equal 4 ξ = x + y, σ = t + s, η = x y, τ = t s.50a with inverse with Jacobian equal 4 x = ξ + η, t = σ + τ, y = ξ η, s = σ τ..50b We can write.49 as Kξ, η, σ, τω ε0 τω ε η4 dξdηdσdτ 0..50c Heuristically, we see that.50c converges to Kξ, 0, σ, 0 dξdσ = ux, t vx, t ψ t + qu, vψ x dt dx 0.50d as ε 0 and ε tend to zero. Write {, if σ + τ 0 and σ τ 0, χσ, τ =..50e 0, otherwise. To show rigorously that.50c converges to.50d, we consider Kξ, η, σ, τω ε0 τω ε η dξdηdσdτ Kξ, 0, σ, 0 dξdσ 4 = Kξ, η, σ, τχσ, τω ε0 τω ε η dξdηdσdτ Kξ, 0, σ, 0χσ, 0ω ε0 τω ε η dξdηdσdτ = Kξ, η, σ, τχσ, τ Kξ, 0, σ, 0χσ, 0 ω ε0 τω ε η dξdηdσdτ.50f = Kξ, η, σ, τ Kξ, 0, σ, 0 χσ, τω ε0 τω ε η dξdηdσdτ + Kξ, 0, σ, 0 χσ, τ χσ, 0 ω ε0 τω ε η dξdηdσdτ. 3 As in the previous equation, ψ x+y x, t+s to the first variable, and this derivative is evaluated at means the partial derivative of ψ with respect x+y, t+s.

3 Thus Kξ, η, σ, τω ε0 τω ε η dξdηdσdτ Kξ, 0, σ, 0 dξdσ Kξ, η, σ, τ Kξ, 0, σ, 0 χσ, τω ε0 τω ε η dξdηdσdτ + Kξ, 0, σ, 0 χσ, τ χσ, 0 ω ε0 τω ε η dξdηdσdτ = A ε,ε0 + B ε,ε0..50g The term A ε,ε0 can be estimated as follows: A ε,ε0 = Kξ, η, σ, τ Kξ, 0, σ, 0 χσ, τω ε0 τω ε η dξdηdσdτ Kξ, η, σ, τ Kξ, 0, σ, 0 ω ε0 τω ε η dξdηdσdτ uξ C + η, σ + τ uξ, σ + vξ + η, σ + τ vξ, σ ω ε0 τω ε η dξdηdσdτ. Here we used the Lipschitz continuity of q as well as the boundedness of the derivatives of the test function ψ. The constants are absorbed in the constant C. Furthermore, A ε,ε0 C ε 0 ε ω supp ψ 0 as ε, ε 0 0. suppω ε0 ω ε uξ + η, σ + τ uξ, σ + vξ + η, σ + τ vξ, σ dξdσdηdτ Here we used the following argument: For every Lebesgue point ξ, σ we have that uξ + η, σ + τ uξ, σ dηdτ suppω ε0 ω ε suppω ε0 ω ε Using 0 as ε, ε 0 0. χ supp ψ ess sup supp ψ uξ + η, σ + τ uξ, σ dηdτ suppω ε0 ωε as a dominating function, we infer that uξ + η, σ + τ uξ, σ dξdσdηdτ ε 0 ε supp ψ suppω ε0 ω ε 0 as ε, ε 0 0. The argument concerning the function v is similar. Thus we have proved that A ε,ε0 0 as ε, ε

4 Next we study the term B ε,ε0. We find B ε,ε0 = Kξ, 0, σ, 0 χσ, τ χσ, 0 ω ε0 τω ε η dξdηdσdτ ω ε Kξ, 0, σ, 0 χσ, τ χσ, 0 dτdξdσ 0 supp ψ ω ε sup 0, σ, 0 dξ 0 σ supp ψ,σ Kξ, χσ, τ χσ, 0 dτdσ ε 0 ω sup 0, σ, 0 dξ σ supp ψ,σ Kξ, 0 as ε h Thus we conclude that the following key estimate ux, t vx, t ψ t + qu, vψ x dt dx 0.50i holds for any two weak solutions u and v and any nonnegative test function ψ with support in t > ɛ. Page 53, line 0 should read:... unique solution of the conservation law. Page 59, line 8+ should read: Chapter 3 Page 66, equation 3. should read: Page 69, line 6 should read: L t x = t σ η l η r q l q r 0, F U; j = f ũ j 0, t. 3. ux, t + t ux, t + t x fux, t fux x, t Page 75, line should read: Lipschitz continuity of the flux implies, for any fixed... Page 77, line should read: = U n j λ F U n j p,..., U n j+q F U n j p,..., U n j+q, Page 78, line + should read: G = Gu 0,..., u p+q+ and F = F u,..., u p+q+. Page 78, lines 6 7+ should read: i Gu 0,..., u p+q+ and i F u,..., u p+q+. 4

5 We set i F = 0 if i = 0. Page 78, line 8+ should read: Gu 0,..., u p+q+ = u j λ.... Page 78, equation 3.3 should read: p+q+ i= Page 78, equation 3.34 should read: p+q+ i F u,..., u = f u, 3.3 i Gu,..., u = Page 78, equation 3.35 should read: p+q+ i j i Gu,..., u = p+q+ p+q+ [ i jδ i,j δ i,j = ] λi j i F u,..., u i F u,..., u p+q+ = λ i + i i F u,..., u Pages 78 9, equation 3.36 should read: p+q+ i,k=0 i k i,kgu,..., u = λf u p+q+ = λ i k i,k F u,..., u i,kf u,..., u i,k=0 = λ p+q+ i,k=0 i + k + i k i,kf u,..., u = Page 79, lines 6 7+ should read: Gux p+ x, t..., ux, t,..., ux + q x, t. Set u i = ux + i p + x, t for i = 0,..., p + q +. 5

6 Page 79, lines 8 7+ should read: Gu 0,..., u p+q+ p+q+ = Gu j,..., u j + i Gu j,..., u j u i u j + p+q+ i,k=0 i,kgu j,..., u j u i u j u k u j + O x 3 p+q+ = ux, t + u x x, t x i j i Gu j,..., u j + p+q+ u xxx, t x i j i Gu j,..., u j + p+q+ u xx, t x i jk j i,kgu j,..., u j + O x 3 i,k=0 p+q+ = ux, t + u x x, t x i j i Gu j,..., u j + p+q+ x i j [ i Gu j,..., u j u x x, t] x x u xx, t + O x 3. p+q+ i,k i j i jk j i,kgu j,..., u j Page 80, the first line of equation 3.37 should read: Gux p+ x, t,..., ux + q x, t = ux, t tfux, t x Page 80, line 9+ should read: ux, t + t = ux, t tfu x + t Page 80, equation 3.38 should read: L t = t λ [ p+q+ i= [f u u x ] i j i Gu,..., u λ f u x + O t 3. u x ] =: t λ [βuu x] x + O t. Page 80, lines 6 7 should read: p+q+ λf u = i j i Gu,..., u = p+q+ i j i Gu,..., u i Gu,..., u. 6 x

7 Page 80, lines 3 4 should read: λ f u = p+q+ p+q+ Page 8, line + should read:... then Gu 0,..., u p+q+ = u k by... i j i Gu,..., u i j i Gu,..., u. p+q+ i Gu,..., u Chapter 4 Page 7, line in epigraph should read:... 3rd... Page 5, the last line of equation 4.0 should read: u 0 v 0 + t+ t min{t.v. u 0, T.V. v 0 } max{ f f Lip, g g Lip }. Page 7, the first line after equation 4.5 should read: In Figure Page 33, line + should read: + u n j x, t n + k φx, t n dx 0. R Page 33, line 3+ should read: tn+ un+/ i y, t k φ t + q g δ u n+/ i y, t, kφ y dt dy R t n+/ Page 33, lines + should read: + u η x, y, 0 k φx, y, 0 dx dy R N u η x, y, t n/ + k u η x, y, t n/ k n= R Page 34, lines 3 4+ should read ux, y, T k φx, y, T dx dy R + ux, y, 0 k φx, y, 0 dx dy 0, R Page 35, the last two line of equation 4.4 should read: ux, T k φx, T dx... dx m R m + u 0 x k φx, 0 dx dx m. R m 7

8 Page 48, equation 4.68 should read: u n u n L loc R m C n n t /m Page 48, equation 4.69 should read: u n u n φ C r n n t φ + max φ xj, 4.69 B j r Page 48, the second line below equation 4.69 should read:... the case n = n +, and... Page 49, the last line of the proof of Lemma 4.9 should read: Choosing ε = n n t proves... Page 5, line 9+ should read: + f ε m B r Page 53, equation 4.76 should read: R m t 0 y r ε δ x y dy dx ε uφ t + fu φ+εu φ dt dx = Page 53, equation 4.79 should read: t m + u tφ t + f t u t φ+εχ m+ u t φ dt dx = 0, 4.79 R m 0 Page 54, the last line of equation 4.8 should read: T sign u k ϕgx, t, u dx dx m dt, R m Page 56, line 6 should read: T.V. u n+ e γ t T.V. Page 57, the last displayed equation should read: t 0 u n+/ e γ t T.V. u n. u k φs +sign u k guφ ds+uφ s=t s=0 = 0. Page 58, line + should read: tn+/ t n u t k ϕ t +sign u t k gu t ϕ dx dt Page 58, line 6+ should read: + χ t sign u t k gu t ϕ dx dt Chapter 5 8

9 Page 65, line 8 should read: In Chapter we proved... Page 68, equation 5.7 should read: h q Page 73, line 4 should read: t ux, t = q + q h + = 0, 5.7 h x { u l u r for x < λ j u l t, for x λ j u l t, Page 76, equation 5.4 should read: h w ξ = := q ξ = R ξ; u l Page 76, equation 5.47 should read: u,ɛ = 9 v l + h l +ξ 7 v l h l + ξ v l + h l +ξ h,ɛ = R q,ɛ u l :=,ɛ hl + ɛ 3 vl + ɛ 3 hl + ɛ Page 79, two lines above equation 5.63 should read:... a unique differentiable function... Page 87, lines 4 5+ should read: the eigenvalues of the Jacobian of B, and... Page 89, add the following to the statement of Theorem 5.4: Assume that λ j r j =. Page 90, line 5 in epigraph should read:... Gegenstand... Page 95, equation 5.0 should read: r j u l l k u l r j u l + l k u l u 0 =

10 Page 97, the last equation of the proof of Lemma 5.8 should read: W ɛ,...,ɛ k+,0,...,0 u l = W k+,ɛk+ Wɛ,...,ɛ k u l = u l + + k ɛ i r i u l + i= k ɛ i D ri r i u l i= k ɛ i ɛ j D ri r j u l + ɛ k+ r k+ Wɛ,...,ɛ k,0,...,0u l i,j= i<j + ɛ k+d rk+ r k+ Wɛ,...,ɛ k,0,...,0u l + O ɛ 3 k+ = u l + ɛ i r i u l + + k+ i,j= i<j i= k+ ɛ i D ri r i u l i= ɛ i ɛ j D ri r j u l + O ɛ 3 Page 98, the line after equation 5.33 should read: using Page 98, the line before equation 5.34 should read: see from 5.39 that... Chapter 6 Page, lines 5 6 should read:... the speed of γ j is greater than the speed of γ i for j > i, we... Page 3, line 4+ should read:... if u L = u R = 0, then... Appendix A Page 95, line 0+ should read:... to be an u 3 M such that... Page 97, line should read: The Arzelà... Page 99, line should read: Appendix B sup t [0,T ] Page 306, line 8 should read: B u η x, t ux, t dx 0 as η 0, t f Lip u δ s us ds + δ t πt s Page 306, line 6 should read:... integrable measure dµs = f Lip / πt s. 0 0

11 Page 306, line 4 should read: = t δ exp t f Lip π, Page 307, line should read: The Arzelà... Page 33, lines 9 0+ should read: +µ sign η U U φ dx dt+µ U sign ηuφ dx dt. The third and the fourth integrals tend to zero as η 0 since f is Lipschitz and x sign ηx tends weakly to zero, and... Appendix C Page 38, the solution of Exercise.a should read: ux, y = y+ y x. Page 33, the first line should read:... proof of the... Page 33, the solution of Exercise.3b should read: 0 for x, ux, 0 = 3 x t for < x < t 3 +, for x t 3 +. Page 37, the equation at bottom should read: f EO u, v = fu ū + fv ū fū. Page 339, the first line of the solution of Exercise 4.8 should read:... heat equation u t = ε u... Page 348, the first line of the solution of Exercise A. should read:... both are nondecreasing... Page 348, the last line should read: References v δ v x+η Page 357, reference [07] should read:... J. Nečas, M. Rokyta, and M. Růžička. Page 358, reference [6] should read: B. L. Roždestvenskiĭ and... Page 358, reference [7] should read: Continuous Glimm functionals... Index Page 36, line 4+ should read: Arzelà Ascoli x v δ y vy dy > ηɛ,