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Lecture notes on the Stefan problem
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1 Lecture notes on the Stefan problem Daniele Andreucci Dipartimento di Metodi e Modelli Matematici Università di Roma La Sapienza via Antonio Scarpa Roma, Italy
3 Introduction These notes cover a portion of a course of the Doctorate Modelli e metodi matematici per la tecnologia e la società, given in 21 in Rome. The title of the course was Evolution equations and free boundary problems and its topics included, essentially, an introduction to Stefan and Hele-Shaw problems. Here only the material concerning the Stefan problem is partially reproduced. The present notes assume the reader has some knowledge of the elementary theory of L p and Sobolev spaces, as well as of the basic results of existence and regularity of solutions to smooth parabolic equations. The bibliography is minimal; only books and articles quoted in the text are referenced. See [12], [18] and [2] for further references. I thank the audience of the course for many stimulating comments and questions, and prof. R. Ricci for interesting discussions on the subject of these notes. Rome, January 22 Revised version: Rome, January 24 iii
5 Notation The notation employed here is essentially standard. Appendix D contains a list of the main symbols used in the text. Constants denoted by γ, C, C δ,...may change fom line to line. v
7 Contents Introduction Notation iii v Chapter 1. The classical formulation 1 1. The Stefan condition 1 2. The free boundary problem 4 3. Explicit examples of solutions 5 4. Basic estimates 7 5. Existence and uniqueness of the solution 8 6. Qualitative behaviour of the solution Regularity of the free boundary 15 Chapter 2. Weak formulation of the Stefan problem From the energy balance to the weak formulation Comparing the weak and the classical formulations 2 3. Definition of weak solution Uniqueness of the weak solution Existence of weak solutions A comparison result 37 Appendix A. Maximum principles for parabolic equations The weak maximum principle The strong maximum principle Hopf s lemma (the boundary point principle) Maximum principle for weak solutions 48 Appendix B. A theorem by Kruzhkov Mollifying kernels The main estimate 53 m+α m+α, Appendix C. The spaces H 2 (Q T ) Comments 55 Appendix D. Symbols used in text 57 Bibliography 59 vii
9 CHAPTER 1 The classical formulation In this chapter we consider the formulation of the Stefan problem as a classical initial boundary value problem for a parabolic partial differential equation. A portion of the boundary of the domain is a priori unknown (the free boundary), and therefore two boundary conditions must be prescribed on it, instead than only one, to obtain a well posed problem. 1. The Stefan condition The Stefan problem ([17]) is probably the simplest mathematical model of a phenomenon of change of phase. When a change of phase takes place, a latent heat is either absorbed or released, while the temperature of the material changing its phase remains constant. In the following we denote by L > the latent heat per unit of volume (p.u.v.), and neglect for the sake of simplicity any volume change in the material undergoing the change of phase. We also assume the critical temperature of change of phase to be a constant, θ. To be specific, consider at time t = t a domain A divided by the plane x 1 = s into two subdomains. At time t = t the sub-domain A 1 = A {x 1 < s } is filled by water, while A 2 = A {x 1 > s } is filled by ice. In the terminology of problems of change of phase, A 1 is the liquid phase and A 2 is the solid phase. The surface separating the two phases is referred to as the interface. Assume also the setting is plane symmetric, that is the temperature θ is a function of x 1 only, besides the time t, and the interface is a plane at all times. Denote by x 1 = s(t) the position of the interface at time t. Note that, due to the natural assumption that temperature is continuous, θ(s(t)+, t) = θ(s(t), t) = θ, for all t. (1.1) Assume ice is changing its phase, that is the interface is advancing into the solid phase. Due to the symmetry assumption we stipulate, we may confine ourselves to consider any portion D, say a disk, of the interface at time t. At a later time t 1 > t the interface occupies a position s(t 1 ) > s(t ) = s. The cylinder D (s(t 1 ), s(t )) has been melted over the time interval (t, t 1 ) (see Figure 1). The change of phase has therefore absorbed a quantity of heat volume of the melted cylinder latent heat p.u.v. = area(d)(s(t 1 ) s(t ))L. (1.2) The heat must be provided by diffusion, as we assume that no heat source or sink is present. We adopt for heat diffusion Fourier s law heat flux = k i Dθ, (1.3) where k 1 > is the diffusivity coefficient in water, and k 2 > is the diffusivity coefficient in ice (in principle k 1 k 2 ). Thus, the quantity of heat in (1.2) must 1
10 2 DANIELE ANDREUCCI water ice s(t ) s(t 1 ) e 1 Figure 1. Melting ice equal t 1 [ k 1 Dθ(s(t), t) e 1 k 2 Dθ(s(t)+, t) ( e 1 )] dx 2 dx 3 dt = t D(t) area(d) t 1 t [ k 1 θ x1 (s(t), t) + k 2 θ x1 (s(t)+, t)] dt. (1.4) Equating the two quantities, dividing the equation by t 1 t and letting t 1 t, we finally find k 1 θ x1 (s(t), t) + k 2 θ x1 (s(t)+, t) = Lṡ(t), (1.5) where we have substituted t with the general time t, as the same procedure can be obviously carried out at any time. This is called the Stefan condition on the free boundary. We stress the fact that the Stefan condition is merely a law of energetical balance. Several remarks are in order. Remark 1.1. Note that, although we did not assume anything on the values of θ(x 1, t) inside each one of the two phases, on physical grounds we should expect θ θ, in water, i.e., in A 1 ; θ θ, in ice, i.e., in A 2. (1.6) The equality θ θ in either one of the two phases (or in both) can not be ruled out in the model. Rather, it corresponds to the case when a whole phase is at critical temperature. Diffusion of heat, according to (1.3), can not take place in that phase, as θ x1 there. Thus (1.5) reduces to, e.g., if the solid phase is at constant temperature, k 1 θ x1 (s(t), t) = Lṡ(t). (1.7) Note that if θ > θ in water, (1.7) predicts that ṡ(t) >. In other words, melting of ice is predicted by the model, instead of solidification of water. This is consistent with obvious physical considerations. Problems where one of the two phases is everywhere at the critical temperature
11 1. THE STEFAN CONDITION 3 are usually referred to (somehow misleadingly) as one phase problems, while the general case where (1.5) is prescribed is the two phases problem. In the latter case, the sign of ṡ(t), and therefore the physical behaviour of the system water/ice predicted by the mathematical model, depends on the relative magnitude of the two heat fluxes at the interface. Remark 1.2. On the interface, which is also known as the free boundary two conditions are therefore prescribed: (1.1) and (1.5). In the case of a one phase problem, this fact has the following meaningful interpretation in terms of the general theory of parabolic PDE: The boundary of the domain where the heat equation (a parabolic equation based on (1.3)) is posed, contains an a priori unknown portion, corresponding to the interface separating the liquid phase from the solid one. Clearly, if only the Dirichlet boundary condition (1.1) was imposed on it, we could choose arbitrarily this part of the boundary, and solve the corresponding initial value boundary problem. Apparently the solution would not, in general, satisfy (1.7). A similar remark applies to solutions found imposing just (1.7) (where now the functional form of the arbitrarily given boundary x 1 = s(t) is explicitly taken into account). It is therefore evident that on the free boundary both conditions (1.1) and (1.7) should be prescribed in order to have a well posed problem. (Or, anyway, in more general free boundary problems, two different boundary conditions are required.) Incidentally, this circle of ideas provides the basic ingredient of a possible proof of the existence of solutions: we assign arbitrarily a candidate free boundary s and consider the solution θ to the problem, say, corresponding to the data (1.1). Then we define a transformed boundary s exploiting (1.7), i.e., k 1 θ x1 (s (t), t) = Lṡ (t). A fixed point of this transform corresponds to a solution of the complete problem. Remark 1.3. The meaning of conditions (1.1) and (1.5) in the context of two phases problems is probably better understood in terms of the weak formulation of the Stefan problem, which is discussed below in Chapter 2. We remark here that, actually, one phase problems are just two phases problems with one phase at constant temperature, so that the discussion in Chapter 2 applies to them too. Remark 1.4. Problems where the temperature restriction (1.6) is not fulfilled, are sometimes called undercooled Stefan problems. We do not treat them here, albeit their mathematical and physical interest (see however Subsection 2.4 of Chapter 2); let us only recall that they are, in some sense, ill posed Exercises Write the analogs of Stefan condition (1.5) in the cases of cylindrical and spherical symmetry in R Note that if we assume, in (1.5), θ < θ in A 1 and θ θ in A 2, we find ṡ(t) <. This appears to be inconsistent with physical intuition: a phase of ice at sub critical temperature should grow into a phase of water at identically critical temperature. Indeed, (1.5) is not a suitable model for the physical setting considered here. In other words, the Stefan condition in the form given above keeps memory of which side of the interface is occupied by which phase. Find out where we implicitly took into account this piece of information and write the Stefan condition when ice and water switch places.
12 4 DANIELE ANDREUCCI 1.3. Prove that Stefan condition (1.5) does not change its form if bounded volumetric heat sources are present (i.e., if the heat equation is not homogeneous). 2. The free boundary problem Keeping the plane symmetry setting considered above, we may of course assume the problem is one dimensional. Denoting by x the space variable, the complete two phases problem can be written as c 1 θ t k 1 θ xx =, in Q 1, (2.1) c 2 θ t k 2 θ xx =, in Q 2, (2.2) k 1 θ x (, t) = h 1 (t), < t < T, (2.3) k 2 θ x (d, t) = h 2 (t), < t < T, (2.4) θ(x, ) = Θ(x), < x < d, (2.5) k 1 θ x (s(t), t) + k 2 θ x (s(t)+, t) = Lṡ(t), < t < T, (2.6) θ(s(t), t) = θ(s(t)+, t) = θ, < t < T, (2.7) s() = b. (2.8) Here < b < d, T and c 1, c 2, k 1, k 2 are given positive numbers. The c i represent the thermal capacities in the two phases. The liquid phase occupies at the initial time t = the interval (, b), while the solid phase occupies (b, d). The problem is posed in the time interval (, T). Moreover we have set Q 1 = {(x, t) < x < s(t), < t < T }, Q 2 = {(x, t) s(t) < x < d, < t < T }. We are assuming that < s(t) < d for all < t < T. If the free boundary hits one of the two fixed boundaries x = and x = d, say at time t, of course the formulation above should be changed. In practice, one of the two phases disappears at t = t. We leave to the reader the simple task of writing the mathematical model for t > t. One could impose other types of boundary data, instead of (2.3), (2.4), e.g., Dirichlet data. If we are to attach the physical meaning of a change of phase model to the problem above, the data must satisfy suitable conditions. At any rate Θ(x) θ, < x < b; Θ(x) θ, b < x < d. Essentially, we need θ θ in Q 1 and θ θ in Q 2. Actually, we will deal mainly with the one phase version of (2.1) (2.8) where the solid phase is at constant temperature. Namely, after adimensionalization, we look at u t u xx =, in Q s,t, (2.9) u x (, t) = h(t) >, < t < T, (2.1) u(x, ) = u (x), < x < b, (2.11) u x (s(t), t) = ṡ(t), < t < T, (2.12) u(s(t), t) =, < t < T, (2.13) s() = b. (2.14)
13 3. EXPLICIT EXAMPLES OF SOLUTIONS 5 (We have kept the old names for all variables excepting the unknown u.) Here we denote for each positive function s C([, T]), such s() = b, Q s,t = {(x, t) < x < s(t), < t < T }. We regard the rescaled temperature u as a function defined in Q s,t. The solid phase therefore does not appear explicitly in the problem. As a matter of fact we assume it to be unbounded in the positive x direction (i.e., d = + ), so that no upper limit has to be imposed on the growth of the free boundary s. The sign restrictions in (2.1) and in (2.11) are imposed so that u > in Q s,t, see Proposition 4.1 below. Definition 2.1. A solution to problem (2.9) (2.14) is a pair (u, s) with s C 1 ((, T]) C([, T]), s() = b, s(t) >, t T ; u C(Q s,t ) C 2,1 (Q s,t ), u x C(Q s,t {t = }), and such that (2.9) (2.14) are satisfied in a classical pointwise sense. We prove a theorem of existence and uniqueness of solutions to (2.9) (2.14), under the assumptions h C([, T]), h(t) >, t T ; (2.15) u C([, b]), u (x) H(b x), x b. (2.16) We also study some qualitative behaviour of the solution. The adimensionalization of the problem does not play a substantial role in the mathematical theory we develop here. For further reading on the one-phase Stefan problem, we refer the reader to [5], [3]; we employ in this chapter the techniques found there, with some changes. Remark 2.1. The free boundary problem (2.9) (2.14) is strongly non linear, in spite of the linearity of the PDE and of the boundary conditions there. Indeed, recall that the free boundary s itself is an unknown of the problem; its dependence on the data is not linear (as, e.g., the explicit examples of Section 3 show) Exercises Prove that the change of variables θ αu + θ, x βξ, t γτ, allows one to write the one phase problem in the adimensionalized form (2.9) (2.14), for a suitable choice of the constants α, β, γ. Also note that adimensionalizing the complete two phases problem similarly is in general impossible. 3. Explicit examples of solutions Example 1. An explicit solution of the heat equation (2.9) is given by ( ) x v(x, t) = erf 2 = 2 x/2 t π t v(x, ) = 1, x >. e z2 dz, x,t >,
14 6 DANIELE ANDREUCCI Here erf denotes the well known error function. Fix C > arbitrarily, and set, for a α > to be chosen presently, { ( )} x u(x, t) = C erf α erf 2. t Define also s(t) = 2α t; note that s() =. Thus u > in Q s,t, and (2.9) as well as (2.13) are satisfied. By direct calculation u x (x, t) = C πt e x2 4t. Hence, u x (s(t), t) = u x (2α t, t) = C πt e α2 = ṡ(t) = α t, if and only if C = παe α2. Note that on the fixed boundary x = we may select either one of the conditions u x (, t) = C πt, or u(, t) = C erf α. We have proven that, when α is chosen as above, u(x, t) = 2αe α2 α x/2 t e z2 dz (3.1) solves the problem sketched in Figure 2. Note however that u is not continuous at (, ); the notion of solution in this connection should be suitably redefined. t s(t) = 2α t u x = C/ πt or u = C erf α u t u xx = u = u x = ṡ(t) = α/ t x Figure 2. The Stefan problem solved by u in (3.1) Example 2. It is obvious by direct inspection that the function u(x, t) = e t x 1, (3.2) solves the Stefan problem in Figure 3, corresponding to the free boundary s(t) = t. Note that we are forced to prescribe an exponentially increasing flux on x = in order to obtain a linear growth for s(t).
15 4. BASIC ESTIMATES 7 t s(t) = t u x = e t or u = e t 1 u t u xx = u = u x = ṡ(t) = 1 x Figure 3. The Stefan problem solved by u in (3.2) 3.1. Exercises Convince yourself that the solutions corresponding to u x (, t) = 2C/ πt, in the case of Example 1, and to u x (, t) = 2e t in the case of Example 2, can not be obtained by linearity from the ones given above. 4. Basic estimates Proposition 4.1. If (u, s) is a solution to (2.9) (2.14), then u(x, t) >, in Q s,t ; (4.1) ṡ(t) >, for all t >. (4.2) Proof. By virtue of the weak maximum principle, u must attain its maximum on the parabolic boundary of Q s,t, i.e., on Q s,t {(x, t) t = T, < x < s(t)}. The data h being positive, the maximum is attained on t = or on x = s(t). Therefore u (remember that u ). If we had u( x, t) = in some ( x, t) Q s,t, invoking the strong maximum principle we would obtain u in Q s,t {t < t}. This is again inconsistent with h >. Thus (4.1) is proven. Then, the value u = attained on the free boundary is a minimum for u. Recalling the parabolic version of Hopf s lemma, we infer ṡ(t) = u x (s(t), t) >, for all t >. Remark 4.1. The proof of estimate (4.1) does not make use of the Stefan condition (2.12). In the same spirit, we consider in the following results solutions to the initial value boundary problem obtained removing Stefan condition from the formulation. The rationale for this approach is that we want to apply those results to approximating solutions constructed according to the ideas of Remark 1.2. Lemma 4.1. Let u be a solution of (2.9), (2.1), (2.11), (2.13), where s is assumed to be a positive non decreasing Lipschitz continuous function in [, T], such that s() = b. Let (2.15) and (2.16) be in force. Then u x is continuous up to all the points of the boundary of Q s,t of the form (, t), (s(t), t), with T t >.
16 8 DANIELE ANDREUCCI The regularity of u x up to x = is classical; the proof of this result will be completed in Section 7, see Lemma 7.1. Proposition 4.2. Let u, s be as in Lemma 4.1. Then where M = max( h, H). < u(x, t) M(s(t) x), in Q s,t, (4.3) Proof. Define v(x, t) = M(s(t) x). It follows immediately v t v xx = Mṡ(t), in Q s,t ; v(s(t), t) = u(s(t), t) =, t T, v x (, t) = M h = u x (, t), < t < T, v(x, ) = M(b x) u (x) = u(x, ), x b. Therefore, taking into account the results of Appendix A, v(x, t) u(x, t), in Q s,t. Corollary 4.1. If u, s are as in Proposition 4.2, then > u x (s(t), t) M, < t < T, (4.4) where M is the constant defined in Proposition 4.2. Proof. A trivial consequence of Proposition 4.1 and of Proposition 4.2, as well as of Lemma 4.1. We also keep in mind Remark 4.1, and of course make use of Hopf s lemma for the strict inequality in (4.4). Remark 4.2. We stress the fact that the barrier construction of Proposition 4.2 is made possible by the fact that ṡ. In turn, this is for solutions of the Stefan problem a consequence of the positivity of u. Thus, such a barrier function, and in general any similar barrier function, does not exist in the case of the undercooled Stefan problem Exercises Note that the function v defined in the proof of Proposition 4.2 is just Lipschitz continuous in t. In spite of this fact, one may apply to v u the weak maximum principle in the form given in Section 4 of Appendix A. Carry out the proof in detail. We prove here 5. Existence and uniqueness of the solution Theorem 5.1. Assume (2.15), (2.16). Then there exists a unique solution to (2.9) (2.14). Let u, s be as in Proposition 4.2. Moreover assume that s Σ := {σ Lip([, T]) σ M, σ() = b}. Here M is the constant defined in Proposition 4.2. The set Σ is a convex compact subset of the Banach space C([, T]), equipped with the max norm. A useful property of all s Σ is b + Mt s(t) b, t T.
17 Define the transform T (s) by 5. EXISTENCE AND UNIQUENESS OF THE SOLUTION 9 t T (s)(t) = b u x (s(τ), τ) dτ, T t. Note that, as a consequence of Lemma 4.1 and of Corollary 4.1, and T (s) Lip([, T]) C 1 ((, T]), M d dt T (s)(t) = u x(s(t), t), t >. Then T : Σ Σ. Also note that a fixed point of T corresponds to a solution of our Stefan problem. We use the divergence theorem to transform the boundary flux integral defining T (s). Namely we write = t s(τ) (u τ u xx ) dxdτ = b t u (x) dx u(s(τ), τ)ṡ(τ) dτ + s(t) u(x, t) dx t t u x (s(τ), τ) dτ h(τ) dτ. Therefore t T (s)(t) = b b b + u x (s(τ), τ) dτ = t u (x) dx + h(τ) dτ s(t) u(x, t) dx =: F(t) s(t) u(x, t) dx. (5.1) Note that this equality allows us to express T (s) in terms of more regular functions than the flux u x (s(t), t), which appeared in its original definition. We are now in a position to prove that T is continuous in the max norm. Let s 1, s 2 Σ. Let us define α(t) = min(s 1 (t), s 2 (t)), β(t) = max(s 1 (t), s 2 (t)), i = 1, if β(t) = s 1 (t), i = 2, otherwise. (The number i is a function of time; this will not have any specific relevance.) Let us also define v(x, t) = u 1 (x, t) u 2 (x, t). Then v satisfies v t v xx =, in Q α,t, (5.2) v x (, t) =, < t < T, (5.3) v(x, ) =, < x < b, (5.4) v(α(t), t) = u i (α(t), t) M(β(t) α(t)), < t < T. (5.5)
18 1 DANIELE ANDREUCCI Therefore, we may invoke the maximum principle to obtain where we also denote On the other hand, we have v,t := max Q α,t v M s 1 s 2,t, (5.6) s 1 s 2,t = max τ t s 1(τ) s 2 (τ). T (s 1 )(t) T (s 2 )(t) = Therefore s 2 (t) s 1 (t) u 2 (x, t) dx u 1 (x, t) dx = α(t) v(x, t) dx + ( 1) i β(t) α(t) u i (x, t) dx. T (s 1 )(t) T (s 2 )(t) α(t) v,t + M(β(t) α(t)) 2 (b + MT)M s 1 s 2,t + M s 1 s 2 2,t, (5.7) and the continuity of T : Σ Σ is proven. By Schauder s theorem, it follows that a fixed point of T exists, and thus existence of a solution. Uniqueness might be proven invoking the monotone dependence result given below (see Theorem 6.1). However, mainly with the purpose of elucidating the role played in the theory of free boundary problems by local integral estimates, we proceed to give a direct proof of uniqueness of solutions. More explicitly, we prove the contractive character (for small t) of T, thereby obtaining existence and uniqueness of a fixed point Local estimates vs the maximum principle. Estimates like (5.6), obtained through the maximum principle, have the advantage of providing an immediate sup estimate of the solution in the whole domain of definition. However, the bound they give may be too rough, at least in some regions of the domain. Consider for example, in the setting above, a point (b/2, ε), with ε 1. The maximum principle predicts for v(b/2, ε) a bound of order M s 1 s 2,ε, that is, obviously, the same bound satisfied by the boundary data for v. On the other hand, taking into account (5.3), (5.4) one might expect v(b/2, ε) to be much smaller than v(α(ε), ε). This is indeed the case, as we show below. In order to do so, we exploit local integral estimates of the solution, that is, estimates involving only values of v in the region of interest, in this case, away from the boundary. We aim at proving that T is a contraction, that is, T (s 1 ) T (s 2 ),t d s 1 s 2,t, (5.8) with d < 1 for small enough t. A quick glance at (5.7) shows that (5.8) does not, indeed, follow from there (unless bm < 1). This failure is due only to the term
19 5. EXISTENCE AND UNIQUENESS OF THE SOLUTION 11 originating from the estimate of α(t) v(x, t) dx. Thus, a better estimate of this integral is needed. A key step in any local estimation is a good choice of cut off functions. These are, typically, non negative smooth functions equal to 1 in the region we want to single out, and identically vanishing away from it. Let b/2 > δ >, and define the cut off function ζ(x), such that ζ(x) = 1, x b 2δ, ζ(x) =, b δ x, 2 δ ζ x(x). Multiply (5.2) by vζ 2 and integrate by parts in Q b,t = [, b] [, t]. We get = (v τ vζ 2 v xx vζ 2 ) dxdτ Q b,t = 1 2 b v(x, t) 2 ζ(x) 2 dx + vxζ 2 2 dxdτ + 2 Q b,t whence (using the inequality 2ab εa 2 + b 2 /ε, ε > ) Q b,t vv x ζζ x dxdτ, (5.9) 1 b v(x, t) 2 ζ(x) 2 dx + v 2 xζ 2 2 dxdτ 2 v x v ζ x ζ dxdτ Q b,t Q b,t 2 v 2 ζ 2 1 x dxdτ + v 2 xζ 2 2 dxdτ. Q b,t Q b,t By absorbing the last integral into the left hand side, we find b v(x, t) 2 ζ(x) 2 dx + vxζ 2 2 dxdτ 4 Q b,t Let us now go back to t = 4 T (s 1 )(t) T (s 2 )(t) = = α(t) b 2δ b δ b 2δ Q b,t v 2 ζ 2 x dxdτ v 2 ζ 2 x dxdτ 16 δ 2 tδ v 2,t = 16 δ t v 2,t. (5.1) v(x, t) dx + ( 1) i v(x, t) dx + α(t) b 2δ β(t) α(t) u i (x, t) dx v(x, t) dx + ( 1) i β(t) α(t) u i (x, t) dx.
20 12 DANIELE ANDREUCCI Use Hölder s inequality and (5.1) to bound the integral over (, b 2δ) (recall that ζ 1 there), and find T (s 1 )(t) T (s 2 )(t) ( b 2δ v(x, t) 2 dx) 1/2 b + (α(t) b + 2δ) v,t + M(β(t) α(t)) 2 b 4 δ t v,t + (Mt + 2δ) v,t + M(β(t) α(t)) 2. Take now δ = t (t is fixed in this argument), and apply again the sup estimate (5.6), finally obtaining T (s 1 ) T (s 2 ),t { 4M b 4 t + M(Mt + 2 t) + M 2 t } s 1 s 2,t. Clearly, for t = t (M, b), T is a contractive mapping Exercises In Subsection 5.1 we proved existence and uniqueness of a fixed point of T in a small time interval (locally in time). Show how the argument can be completed to give existence and uniqueness of a fixed point in any time interval (global existence) Why uniqueness of solutions (u, s) with s Σ is equivalent to uniqueness of solutions in the class of Definition 2.1, without further restrictions? 5.3. Give an interpretation of (5.1) as an energetical balance Prove that T has the property (see also Figure 4) s 1 (t) < s 2 (t), < t < T = T (s 1 )(t) > T (s 2 )(t), < t < T Extend the proof of existence and uniqueness of solutions to the case when h. What happens if h, u? 5.6. To carry out rigorously the calculations in (5.9) actually we need an approximation procedure: i.e., we need first perform integration in a smaller 2- dimensional domain, bounded away from the boundaries x =, t =. Recognize the need of this approach, and go over the (easy) details. 6. Qualitative behaviour of the solution Theorem 6.1. (Monotone dependence) Let (u i, s i ) be solutions of (2.9) (2.14), i = 1, 2, respectively corresponding to data h = h i, b = b i, u = u i. Assume both sets of data satisfy (2.15), (2.16). If h 1 (t) h 2 (t), < t < T ; b 1 b 2 ; u 1 (x) u 2 (x), < x < b 1 ; (6.1) then s 1 (t) s 2 (t), < t < T. (6.2) Proof. 1) Let us assume first b 1 < b 2. Reasoning by contradiction, assume t = inf{t s 1 (t) = s 2 (t)} (, T). Then the function v = u 2 u 1 is strictly positive in { < t < t, < x < s 1 (t)},
21 6. QUALITATIVE BEHAVIOUR OF THE SOLUTION 13 t T (s) = s x = σ(t) x = T (σ)(t) b x Figure 4. Behaviour of the transform T by virtue of the strong maximum principle and Hopf s lemma. Indeed, v(s 1 (t), t) >, < t < t. Then v attains a minimum at (s 1 ( t), t) = (s 2 ( t), t), where Thus, due to Hopf s lemma, But we compute v(s 1 ( t), t) =. v x (s 1 ( t), t) <. v x (s 1 ( t), t) = u 2x (s 2 ( t), t) u 1x (s 1 ( t), t) = ṡ 2 ( t) + ṡ 1 ( t). Hence ṡ 2 ( t) > ṡ 1 ( t), which is not consistent with the definition of t. 2) Assume now b 1 = b 2. Let us extend the data u 2 to zero over (b, b + δ), where < δ < 1 is arbitrary. Let (u δ, s δ ) be the solution of problem (2.9) (2.14) corresponding to the data h = h 2, b = b 2 + δ, u = u 2. Then, by the first part of the proof, s 2 < s δ, s 1 < s δ, and for all < t < T s δ (t) s 2 (t) = δ s 2 (t) [u δ u 2 ](x, t) dx s δ (t) s 2 (t) u δ (x, t) dx δ. Therefore s δ s 2 + δ, so that s 1 < s δ s 2 + δ. On letting δ we recover s 1 s 2. Let us investigate the behaviour of the solution of the Stefan problem for large times. In doing so, we of course assume that T =. Note that the result of existence and uniqueness applies over each finite time interval; a standard extension technique allows us to prove existence and uniqueness of a solution defined for all positive times. Owing to Proposition 4.2, we have s(t) S, < t < = u(x, t) MS, in Q s,. (6.3)
22 14 DANIELE ANDREUCCI Moreover, s being monotonic, certainly there exists s = lim t s(t). (6.4) Theorem 6.2. Let (u, s) be the solution of Theorem 5.1. Then Proof. 1) Assume first b s = lim s(t) = b + t u (x) dx + h(t) dt. (6.5) h(t) dt = +. (6.6) We only need show s is unbounded. Let us recall that, for all t >, b s(t) = b + t u (x) dx + h(τ) dτ s(t) u(x, t) dx. (6.7) From (6.3), it follows that, if s is bounded over (, ), then u is also bounded over (, ). This is clearly inconsistent with (6.7), when we keep in mind (6.6). 2) Assume h(t) dt < +. (6.8) The balance law (6.7), together with u >, immediately yields b s(t) < b + u (x) dx + h(t) dt < +. (6.9) Then we have s(t) s < ; it is only left to identify s as the quantity indicated above. Owing to (6.7) again, we only need show lim t s(t) u(x, t) dx =. (6.1) On multiplying (2.9) by u and integrating by parts in Q s,t, we get 1 2 s(t) u(x, t) 2 dx + u 2 x dxdτ = Q s,t 1 b t u (x) 2 dx + u(, τ)h(τ) dτ. (6.11) 2 Recalling (6.3) and (6.9), the last integral in (6.11) is majorised by Thus a consequence of (6.11) is Ms h(τ) dτ <. Q s, u x (x, t) 2 dxdt < +. (6.12)
23 Elementary calculus then shows that u(x, t) 2 dxdt = Q s, 7. REGULARITY OF THE FREE BOUNDARY 15 Q s, s Q s, [ s(t) x s(t) x u ξ (ξ, t) dξ] 2 dxdt u ξ (ξ, t) 2 dξ dxdt s 2 Then there exists a sequence {t n }, t n, such that But the function s(t n ) t u(x, t n ) 2 dx. s(t) u(x, t) 2 dx, Q s, u ξ (ξ, t) 2 dξ dt <. when we take into account (6.11), is easily seen to have limit as t, so that this limit is. By Hölder s inequality, s(t) u(x, t) dx [ s(t) 1/2 s u(x, t) dx] 2, t, completing the proof of (6.1) Exercises Find conditions ensuring that the inequality in (6.2) is strict Discuss the necessity of assumptions (2.15), (2.16) in Theorem Regularity of the free boundary The approach in this Section is taken from [16], and provides an example of bootstrap argument, i.e., of an inductive proof where any given smoothness of the solution allows us to prove even more regularity for it. Our first result will become the first step in the induction procedure, and is however required to prove Lemma 4.1. Lemma 7.1. Let u C 2,1 (Q s,t ) C(Q s,t ), where s Lip([, T]), and s(t) > for t T. Assume u fulfils u t u xx =, in Q s,t, (7.1) u(s(t), t) =, t T. (7.2) Then for each small enough ε >, u x is continuous in P ε, where P ε = {(x, t) ε < x < s(t), ε < t < T }.
24 16 DANIELE ANDREUCCI Proof. The proof is based on standard local regularity estimates for solutions of parabolic equations. Introduce the following change of variables { y = x s(t), v(y, τ) = u(ys(τ), τ). τ = t ; The set Q s,t is mapped onto R = (, 1) (, T), where v solves v τ 1 s(τ) 2v yy ṡ(τ) s(τ) yv y =, in R, (7.3) v(1, τ) =, τ T. (7.4) More explicitly, (7.3) is solved a.e. in R, as v is locally a Sobolev function in R. Classical results, see [11] Chapter IV, Section 1, imply that for any fixed ε >, v τ, v y, v yy L q ((ε, 1) (ε, T)), for all q > 1. Then we use the embedding Lemma 3.3 of [11] Chapter II, to infer that, for q > 3, v y H α, α 2 ([ε, 1] [ε, T]), (7.5) where α = 1 3/q (see also Remark 11.2 of [11], p. 218; the space H α, α 2 is defined in Appendix C). Since u x (x, t) = 1 s(t) v ( x y s(t), t), the result follows. Our next result implies that the free boundary in the Stefan problem (2.9) (2.14) is of class C (, T). Theorem 7.1. Assume u and s are as in Lemma 7.1, and moreover u x (s(t), t) = cṡ(t), < t < T, (7.6) where c is a given constant. Then s C (, T). Proof. For v defined as in the proof of Lemma 7.1, we rewrite (7.6) as ṡ(τ) = 1 cs(τ) v y(1, τ), < τ < T. (7.7) Choose α (, 1). Then (7.5) and (7.7) yield at once ṡ H α, α 2 ([ε, 1] [ε, T]), for each fixed ε >. (7.8) Next we make use of the following classical result: If the coefficients in (7.3) (i.e., ṡ), are of class m+α m+α, H 2 ([ε, 1] [ε, T]), then v is of class H 2+m+α,2+m+α 2 ([2ε, 1] [2ε, T]). Here m is any integer. We prove by induction that for all m (7.9) m+α m+α, ṡ H 2 ([ε, 1] [ε, T]), for each fixed ε >. (7.1) We already know this is the case when m =, from (7.8). Assume then (7.1) is in force for a given m. Then, owing to (7.9), v H 2+m+α,2+m+α 2 ([ε, 1] [ε, T]), for each fixed ε >. (7.11)
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So it's been over two months since my last post, and I hope to write about what has been going on lately soon. In the meantime, I have dipped my toes back into the writing world by writing a piece for Homeschoolers Anonymous.They are having a homeschooled LGBTQ week, so be sure to check out the stories of other people who grew up homeschooled and queer, or grew up opposing gay rights and later became allies. If you read the posts prior to this week, Homeschoolers Anonymous is giving a voice to people who grew up homeschooled and have something to say about it. Some of the stories are heartbreaking, others are inspiring, but all need to be told, and all need to be heard.Here … [Read more...]
Guest Post at Homeschooler's Anonymous
Oblivious to Privilege:Part Two
Part Two of Haley's Guest Post...The experience of living in two genders and perceived to have different sexual orientations, has made me aware of privilege and how oblivious I once was to it. I don't share my experiences as some guilt trip to people who possess privilege or to paint a picture that my life is really hard. My life is really great. I am living freely and openly. I wake up and I look forward to what the day will hold. Honestly, my life has never been better. However, reflecting on privilege and how my perceptions of it evolved as I evolved, can be illuminating. In fact while I've lost some privileges I've gained a new one. It is a privilege to grow in empathy and see so m … [Read more...]
Oblivious to Privilege: Part One
A guest post from Haley...I've lived with society perceiving me in different ways over my lifetime.At one time I was perceived to be a straight white cisgender male (now obviously I was really a closeted transgender woman but the world didn't know that.) In the year before I gender transitioned, I pushed gender boundaries, my style was fairly genderqueer and many people perceived me to be a gay male. Then during the first six months of living as a woman as the hormones started and a learning curve happened, I was regularly visibly identified to be a transgender woman. And for the last six months as my transition has progressed I "pass" and the world perceives me to be a white woman, … [Read more...]
Malpractice on the Air
It was almost midnight and I was driving home after my usual shift. Yawning, I pushed the radio button hoping to catch a peppy song to keep me awake, but the station that usually plays pop music had one of those call-in counseling shows on. "Hello Brooke, so you are 19 and pregnant?""Yes, unexpectedly so.""Wow, you're going to be a mom.""Yeah, I'm excited, but nervous, because I don't have a lot of support ya know?" I fumbled with the dial, trying to change the channel before I got sucked in, but it was too late. "Don't you have family?" The host asked."Well, they don't really want to be a part of my life.""Oh, that's sad, why not?""Well, they th … [Read more...]
On Being Evangelized
I can usually spot them when they come in. The "superchristians". The excessively patronizing men make the order while the women stand quietly in their long skirts behind them. The group of middle aged women who come in for their bible study, order each of their drinks individually, and never ever tip.Or the conservatively fashionable woman who takes forever to make up her mind what she wants, and then when she decides on an herbal tea for it's health benefits, she plucks up her courage and shakily makes her evangelistic comment about how amazing it is that god has made tea.One time I spotted a middle aged couple sitting rigidly next to each other across from their long-haired adult … [Read more...]
February in photos
February has seriously been the longest month ever. Of course, I'm not including all the kids, because I don't really put pictures of them here. And this should include pictures of Haley at school and me at work. And maybe a few of me laying around in my pajama's aching and sick, because that's how I spent two weeks of this month. (It felt much longer.) Who else is ready for spring? … [Read more...]
Every once in a while when I get a bunch of small posts that aren't really big enough to be a post of their own, I do a hodge-podge post. So here's the latest. On the drive home from dropping people off in the morning. Ms Drama: Mom do you looove the hair on your chin?Me: Um, I don't know? (What a weird question!)Ms Drama: You know, because it helps you win when you play video games. This is what comes of saying things like "I barely squeaked by there! Won by the hair on my chinny-chin-chin!" **************************************** So I was at work doing dishes and my co-worker was on break. So I decided to call Haley and chat while … [Read more...]

subset: commoncrawl


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The East Baton Rouge Parish Registrar of Voters started work Thursday validating the thousands and thousands of signatures submitted by the hopeful organizers of the proposed city of St. George.
The petition is more than 11,000 pages, and organizers estimate they contain about 18,000 signatures. They need the signatures of 25 percent of registered voters in the proposed boundaries in order to qualify for an election.
The exact number needed is still unclear because the boundaries of the proposed city have several split precincts that need to be evaluated in order to find the true city population.
Estimates for the required number have ranged from 16,500 to 18,500.
Organizers gathered the signatures over the course of more than a year. They were submitted to the Registrar's Office on Oct. 20, with the hope the petition would be validated in time for the December or March election.
However, registrar officials had not been able to start the process until this week because they say they've been inundated with fall election work.
The process to validate the petition will take at least a month, Registrar Elaine Lamb has said.
That means the petition is unlikely to be verified in time for the March election — and that St. George supporters probably will have to wait for the May 2 ballot.
If after verifying the petition St. George organizers come up short on signatures, they will have 60 days to make the difference. They have continued gathering additional signatures since submitting the petition to the Registrar's Office on Oct. 20.
Only registered voters within the proposed boundaries of the new city will have an opportunity to vote on its incorporation.
If successful, St. George will be about 80 square miles with a population of about 107,000 people.
St. George's budget would be carved out of funds that currently make up the East Baton Rouge city-parish budget.
Mary Olive Pierson, an attorney representing the city-parish on St. George issues, has said she will file a suit challenging the validity of the petition if it's successfully verified.
Follow Rebekah Allen on Twitter, @rebekahallen. For more coverage of city-parish government, follow City Hall Buzz blog at http://blogs. theadvocate.com/cityhallbuzz.


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The Ministry of Education has just launched its website for Shaping Education in Christchurch, New Zealand, following the earthquakes that have devastated the area over the last year. The website is a really positive step, however it does need creative and innovative thinking to lead to the changes needed. So let everyone know that they should have a say and that they should check out the links that are provided on the site. Read blog post >
Tag Archives: innovation in education
Impact Collaboration
Dr Cheryl Doig is currently curating a book on collaboration, with input from around the globe.
Click here to find out more information.


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Compete in swimming matches across the United States! You can play as an athlete from North America, Europe, Asia, or Australia. Swim fast to win gold medals in 100-, 200-, and 400-meter competitions. You must make perfect turns to always finish in first place. Become a swimming pro!
Swimming Pro is one of our selected Sports Games.
Report game as broken
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There wasn't a match played this past weekend, so there are no stats for it. But makeups are pointless now because there's no way the match would alter anyone's fates. So this post is to officially congratulate our two champions.
In the PS3 League, our champions of Season 03 are the DNUTs!
The DNUTs finished the season controlling 9 games, while Reality and the Boston Teabags each controlled only 5.
And in the X360 League, our champions are Boneless Chicken!
BC finished with 6 games under their control, while the Warriors only had 4.
I'm still working on a header for the champions, but I'm not great at these things, so if anyone wants to make a header for these champions, go ahead.
Information will be posted about Season 04 and my plans to finish 2013. Stay tuned.
Unprofessional League Gaming Essentials


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UK Delivery
Standard delivery within 3 working days
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Buy online and collect from our West London store
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Yamaha Recording Custom Gadd Style 5pc Shell Pack - Solid Black Finish
Yamaha has reintroduced its iconic Recording Custom Drum Series. Re-designed in collaboration with acclaimed drummer and Yamaha artist Steve Gadd, the Recording Custom Series has become known as one of the most recorded drum kits in history. Today it returns, packed with a number of new features that build upon the original spec.
- Shell: 100% Birch 6ply (with inner Dark Brown paint)
- Bearing Edge: 30 degrees
- Lug: Newly designed Weighted High-tention Lug (One-piece) TT/FT
- Hoop: Triple Flange Hoop (Steel 1.6mm)
- BD Hoop: Wood Hoop
- TT/FT Head: Top: Remo US Coated Ambassador, Bottom: Remo US Clear Ambassador
- BD Head:Front :Remo Smooth White PS3 with Yamaha Logo , Batter: Remo PS3 Coated
- Tom Mount: Y.E.S.S. System FT
- Bracket: Open type
- BD Leg: Convertible type
Weighted Lug
The revamped modern design retains the traditional essence of the Recording Custom. In order to max-imum the attenuating properties of the birch shell Yamaha have increased the weight of the lug, further enhanc-ing the artist' s expressive power and reducing undesirable noise from the shell without using a mute. This ensures optimum sustain for the core sound, producing a crisp, articulate tone. This feature in harmony with the thin ply bass drum shell delivers a very impressive sounding bass drum.
100% North American Birch 6 Ply Shell
Yamaha was the first company to use an All Birch shell design for Recording Custom with its familiar punchy sound. Birch reduces the undesirable sounds and prevents interference when the drums are next to each other. The new Design features all 6 Ply 6mm thick for all the Drums including the Bass drum which supports a focused deeper tone.
30 Degree Bearing Edge
While testing many combinations of shells Steve Gadd opted for 30 degree bearing edges, to deliver a sharp response. The shells High Gloss lacquer Yamaha introduced High Gloss Lacquer with the Recording Custom series in 1978, thanks to our specialist know how from our famous Piano craftsman.
High Gloss
Lacquer finishes create a very durable yet thin finish to maximise the shells resonance and tone. Colours include Solid Black, Surf Green and Classic Walnut along with Real Wood finish which was one of the finishes offered when it was first launched in 1975 finish.
Lifted 20" BD
This lifted bass drum produces a more punchy sound and accentuates the sound produced in the core of the shell. By lifting the bass drum off the floor we reduced interference and maximize the resonance of the shell. This also increases the length of the pedal beater rod, centering it precisely to ensure absolute accuracy during performances.
Steve Gadd Quotes:
'For the past 3 years I've been meeting with the Research and Development guys at Yamaha with the aim of bringing the Recording Custom Series up to date. I told Yamaha what I thought throughout the process, the result is that everything is a step up. They still feel like the Recording Custom, when you take them out of the box they don't feel new, they feel comfortable and I think that's good. The Bass Drum has a nice fat sound, the drum responds to different heads on it, and my goal was to have a simple Snare Drum design that could respond to a wide range of tuning. I think we did a great job, the new Recording Customs sound great and I think you'll like them'.


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This article, Yin and Yang: The Body and Eyes, is the property of iSavage.
This article, Yin and Yang: The Body and Eyes, is the property of Kaitan.
"I heard this girl has the Rinnegan here in the Land of Water, this will propel my power exponentially, I just need to find her."
Sayo, on the other hand, was sitting peacefully, seiza-style, in front of the simple grave in Kirigakure marked with a single, chilling inscription. All it said was Koten (古典; ; ; ). Age Eight. Sayo was decked out in full shinobi gear. Koten, I've got another assassination mission... this'll be a long one. I'll be back to talk to you in a month or so. I'll miss you. Rocking back to her heels and suddenly straightening, she exhaled slowly before leaping off towards the north.
"I hope I succeed, if I do, I can go through with my plan, though since she has the rinnegan, it won't be an easy fight, I might have to use that technique of the going gets tough." Yoshio stops to look at his map to see where he was going.
Sayo continued to run north, heading for a minor nation north of the Land of Thunder. It had a rich shipping businessman, one who had taken over a portion of Gato Shipping, who was threatening to cut off trade with Kiri. His kind should die anyway.
Yoshio then gets to this place where there were ships ever where, he then gets to this hotel where he stops to rest for the day. "Guess I will stop here, maybe I will run into her here." Yoshio says.
Sayo continued running, her hair flowing behind her in the wind as she ran across the ocean waves, the shore a distant memory as she continued north, to the Land of Thunder and beyond, to a place, a mission, and somewhere to come home from. She would complete her latest solo mission, and then come back to her team. They missed her when she went on missions, she knew that. Donning her Shark ANBU mask, she continued to run, breathing easily.
Yoshio then goes to the tavern to get something to eat, when he then sees a woman sitting at a table by herself. He goes over there and spends time with her, then they go to his hotel room where they sleep together. Then he wakes up in the morning and leaves to hotel room to go and search for this woman.
Outside of a walled compound, Sayo snorted at the opulence of the man's estate. The man had no sense of shame, and his guards weren't even ninja-trained. Simply walking around the compound, she came across a servant washing laundry in the nearby stream, and slid into the compound through an open gate, masking her whispering footsteps with chakra. Slipping on a Transformation of a nondescript servant in the same garb as the woman, she continued shuffling forward, her head down, holding a basket she had purloined from the same unobservant servant.
Yoshio leaves the town and goes on the search for Sayo. About an hour after he left, he realizes he left his first aid kit there so he heads back to go and get it.
Sayo, watching the progress of servants within the castle, allowed herself to fall into the flow and continue, peeling off at random intervals to map the interior better. Snorting slightly at her own stupidity, she turned and walked up the stairs. The man's room was the only one on the third floor, and that would make his death so much easier.
Yoshio gets a lead that the woman he is looking for is on a mission, but doesn't know the details, so he keeps on looking for leads on the girl to find out where she exactly is.
Slipping into the man's room, Sayo shuffled in slowly, her head down and holding a basket of laundry. The master rolled from the bed, shirtless, his swag belly wobbling as he approached Sayo in a predatory fashion. "Well well, aren't you a pretty one. I haven't seen you about the castle before."
Sayo's tremble was interpreted as nervousness and fear by the shipping magnate, when in reality she was violently reasserting her self-control as his hand reached for her chest and the clasp of her uniform.
Moments later, Sayo stood, half-naked and breathing heavily, the man in front of her slashed into a bloody mess. It seems her self-control wasn't so good after all.
Yoshio gets yet another lead that she was on a mission in the Land of Thunder, so thats where he goes even though it was about a days travel from where he was. "Man, the Land of Thunder, that will take a day, but its for a good cause so."
Suddenly feeling inexplicably tired, Sayo had only enough control to lock the door before the trembling got to her, voices invading her head, the memory of pain, of needles, of cold, sterile rooms, the flash of glasses... the world swirled around her as she sank to her knees, hands gripping the sides of her head, in a futile attempt to calm the blood roaring in her ears, the war-drums pounding in her skull. This was not a good time to have a panic attack! was her last conscious, surprisingly rational thought before she fell onto her back, her eyes closing as she slipped into darkness. The only sign of life was her bare chest rising and falling with each shallow, quick breath.
Yoshio then thinks to himself about what he is doing, "This girl has the rinnegan, and I heard that she went through experiments to get it too, like me with wood release, Danzo made me go through all that stuff. Well, I will see when I get to her."
About 2 hours later, Yoshio sets up for camp and thinks about his thoughts on going on with his plan. "Maybe I shouldn't even try to get the rinnegan, I won't be able to do anything because they will hunt me if I kill this girl and take her rinnegan. Maybe I should help her."
Sayo felt like she was clawing her way up through layers of soft, inky darkness, and as it slid over her skin she would lose her purchase and fall, before catching herself and climbing once more. Eventually, by a monumental effort, she broke free of the last layer of suffocating softness and rose to consciousness, opening her eyes. Her hair was plastered slickly to her face by her sweat, her heartbeat was still faster than normal as she placed a hand on her chest, her muscles feeling numb and her skin feeling foreign. Her fingers were clumsier than usual as she involuntarily shuddered before dispelling the transformation, and felt the familiar weight of her shinobi life envelop her, giving her the composure to once more stand up straight. Turning to the mangled corpse, she burned the man's body with a simple Phoenix Sage Fire that also incidentally began setting the surroundings on fire. Leaping out the window with a barely concealed grimace, she buckled as she hit the ground, before staggering away as her vision began to swim. Collapsing once more under a fir tree, safely out of sight and sense from the compound, Sayo fell back into the layers of flowing softness, and this time there was no climbing out as the hands reached out to claim her.
Yoshio wakes up 2 hours later to get the the Land of Thunder faster, and he did, within 3 hours her was there. He asks around if people have see Sayo, but to no avail. So he just casually goes around the land and then sees this compound, he rushes over there and sees this girl under a fir tree. At the moment, he didn't realize it was Sayo, "Why is this girl just under this fir tree, maybe I should take her back into town and get a hotel for her. (Yawn).... and me too." Yoshio picks up the girl and takes her back into town where he got 2 rooms, 1 for him and one for Sayo. He lays her on her bed then goes to is room, and just collapses on the bed and falls asleep.
Sayo struggled to wakefulness, panting, the sheets tangled around her legs. She lay there, marshaling her thoughts, before a horrifying thought came to her mind. Simply, she hadn't fallen asleep in a bed. Not moving a single muscle, she extended her chakra throughout her body, finding no repression seals or other limiters. With a sudden violent movement of her body, she was standing on the floor, finding herself in a rather boring hotel room. How... what had happened? Going back to the bed, she began to search it for clues, finding a strange scent underneath her own smell, and this one smelled of... a forest? This couldn't be good. Sayo once more bashed her way through a window, six robotic arms unfolding from her shoulders as she traveled away from the building in a strange insectoid fashion, switching between pairs of robotic limbs in an unusual way of rooftop travel.
"Umm, excuse me, but how come you just didn't jump out the window like normal people, instead you just had to use the rinnegan to turn your body into a nasty looking robot insect, what ever it is. And you didn't even think about telling the person who got you into that hotel room safe and sound thank you." Yoshio says, "So where are you going?" He asks Sayo.
"Back," muttered Sayo in a low voice. "Back where I am supposed to be, back where people are looking for me. So go **** yourself with a tree." With this inventively vile insult, she loped off at an angle, aiming to pass him and go towards the harbor. She now recognized herself as being in the Moon Quarter, and there were ships nearby that would take her back to the Land of Thunder and then the Land of Water.
"Hey that wasn't nice." Yoshio says as he then uses Water Release: Water Encampment Wall stopping her right in her tracks, "Why are you so pissed, all I did was help."
Sayo once more activated Chakra Propulsion, flying high over the wall to sail over the rooftops and out of sight, leaving only a cracked set of roof tiles behind, dashing towards the harbor hand over hand over hand.
Yoshio chases after the girl, "Wait, where are you going, I just want to talk there is no need for this." Yoshio says
Her only response was to pick up her speed even more, bounding over the rooftops faster than even many taijutsu specialists could move, her arms a windmill of perfectly timed mechanical running. Hitting the quay, she darted off to the end, and flung herself off of the jetty, falling dozens of feet to the water below.


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|Tangible Fixed Assets
|£ 350,000
|Shareholder Funds
|£ 302,782
|Profit Loss Account Reserve
|£ 200,867
|Revaluation Reserve
|£ 94,915
|Called Up Share Capital
|£ 7,000
|Net Assets Liabilities Including Pension Asset Liability
|£ 302,782
|Total Assets Less Current Liabilities
|£ 356,578
|Net Current Assets Liabilities
|£ 6,578
|Creditors Due Within One Year
|£ 18,997
|Cash Bank In Hand
|£ 25,575
|Share Capital Allotted Called Up Paid
|£ 6,900
|Number Shares Allotted
|£ 6,900
|Tangible Fixed Assets Cost Or Valuation
|£ 350,000
|Creditors Due After One Year
|£ 53,796
Information about the Private Limited Company BOWERS & LEACH (EXETER) LIMITED has been prepared for information purposes only. It is not intended to be nor does it constitute legal advice. This is public information provided by the official company register.


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Android 4.3 was officially unveiled and released today to the Android Open Source Project. In a surprisingly timely fashion, Google also released both the factory images and OTAs to the Nexus 4, 7, 10, and the Galaxy Nexus.
Android 4.3 was officially unveiled and released today to the Android Open Source Project. In a surprisingly timely fashion, Google also released both the factory images and OTAs to the Nexus 4, 7, 10, and the Galaxy Nexus.
Give the community enough time and almost any device can be cracked open, no matter how determined a carrier or OEM is to keep it locked down. The Verizon Galaxy S4 has proven a tough nut to crack, but a new root method is much less convoluted than previous ones. Just flash a kernel, run some tools, flash again, and you're done! Well, it's a little more involved than that, but not much.Read More
If you've been paying attention to TV Guide's official app, you know that it's needed some attention for a while. Its UI through version 2.x was an outdated pastiche of Gingerbread tabs and gradated iconography desperately in need of a redesign (and support for 4.0+). Today, TV Guide has fulfilled that need (for the most part), bringing to the Play Store TV Guide Mobile version 3.0. The update also brought "many cool new features" to the app, which we'll discuss momentarily.Read More
On this site, you'll read a lot of Android news (which is why we put it in our name). You will also hear a lot of tech news. You will maybe hear a tiny bit of movie news. Today, though, you're getting a triple-whammy of all three. The Sundance Film Festival, which actually began yesterday, has released the 2013 version of its mobile app to help you keep track of everything going on at the event.Read More
In the ongoing effort to make apps better on Android, Google has released another new guide for the developers who may be looking for guidance on just how to build a great tablet-oriented app. The piece has some fairly detailed information, including how much to adjust padding of UI elements and how to target different screen sizes and resolutions. There is also more broad guidance on how to make the most of larger screen real-estate.Read More
Sony's PlayStation Mobile for Android just went live for certified devices last night. But what if you don't have a certified device, which so far includes only a number of Sony phones and tablets as well as the HTC One series? Heck, I have a One series device - the EVO LTE - but turns out it's not certified. That's just silly.
Not a problem - if you have root on your phone or tablet, getting the PSM (PlayStation Mobile) apk to work is a matter of pushing a few files it relies on to your /system directory and rebooting.Read More
Walt Disney World Resort might be home to the Most Magical Place on Earth®, but with 30,000 acres and millions of visitors every year, it's awfully easy to have a bad day there. To that end, Disney has released its very own guide app, covering all four theme parks and the various satellite tourist areas around Disney World. The most useful feature is probably the detailed maps, complete with annotations for rides, shops, restaurants, et cetera, and GPS tracking to help you find your way around.Read More
Need some reading for the weekend? Google has you covered. To celebrate the shipping of the first batch of Nexus 7 pre-orders, Google has posted up the full, official, exhaustive changelog for Android 4.1 Jelly Bean.
Covering everything from Accessibility to Widgets, the changelog delivers comprehensive explanations of each and every change users can expect to find in Android's latest iteration, as well as changes to Google apps and the "new Google experience on Android," including Google Search and Google Now.Read More
Perhaps the most popular (and complete) free repair manual in existence, iFixit, launched an official app for Android recently, bringing detailed step-by-step repair instructions and (of course) the saucy teardown images we've come to know and love from the service's online counterpart to your Android devices.
For those who don't know, iFixit provides users with incredibly detailed repair guides for a huge variety of things from laptops to mobile devices, game consoles, and even cars, including great imagery and nice explanations for why hardware is the way it is.Read More
Sure, Hulu may still be struggling with supporting even a majority of Android devices, but who needs that mess when the CW has an app? Well, yes, people who want to watch shows by any of the other networks, but now all of the CW's top shows are available on (some of) your phones and tablets in their full-episode glory.
It's not quite the complete library experience we've all been hoping for that some shows occasionally get on Hulu or Netflix, but all of the 12 shows available come with 5 episodes each which, we assume, will be the five most recent episodes unless a show is on hiatus, as is the typical custom with most major streaming sources.Read More


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This article is not considered Canon in the world of LOST
of LOST, and is listed here for informational and reference purposes.
|What's Your Story?
|What's Your Story?
|Fire + Water
|Hurley, Libby
Libby and Hurley do laundry inside the hatch and discuss their past lives. Libby reveals that she was a shrink dealing with marriage and family counseling, and Hurley reveals that he is a lottery winner, and is shocked that she believes him.
Background InformationEdit
- This scene was meant to be in the episode Fire + Water.
- HURLEY: Oh, uh... I think these are yours.
- LIBBY: Sure they're not yours?
- HURLEY: So, what do you do? You know, before.
- LIBBY: I'm a shrink. I specialize in marriage and family counseling.
- HURLEY: So, you make hyper kids hit stuff with nerf bats?
- LIBBY: Thats EXACTLY what I do.
- HURLEY: So, are you married?
- LIBBY: Nope. But I was. Several times.
- HURLEY: How many is several?
- LIBBY: Well, if the annulments count, then three...Hey. At least now I know what NOT to do.
- LIBBY: OK, now it's your turn. What's the Hurley story?
- HURLEY: Uh, I've had a bunch of jobs. Mostly...dealing with chicken. And, and I kind of won the lottery.
- LIBBY: Really? How much did you win?
- HURLEY: 114 million. But now I'm worth like, 156 million, because of like, investments and stuff.
- LIBBY: Oh, that's amazing.
- HURLEY: Uh... you believe me?
- LIBBY: Yeah. Why would anyone lie about something like that?
- HURLEY: Libby, would you like to... take a walk sometime? With me I mean?
- LIBBY: Promise not to step on my foot?
- HURLEY: Well, yeah. Of course not.
- LIBBY: Then, I'd love to.


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Description: "She," is the great mythic creation of the 19th century, while "King Solomon's Mines" and "Allan Quatermain" are surging tales of adventure, full of sensational fights, blood-curdling perils and extraordinary escapes.
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List price: $17.95Publisher: Dover Publications, IncorporatedPublication date: 6/1/1951Binding: PaperbackPages: 656Size: 5.50" wide x 8.00" long x 1.25" tall Weight: 1.342Language: English
Sir Henry Rider Haggard (1856-1925) is best remembered for his 34 adventure fantasy novels set in exotic locations. As a child, Haggard, whose father was an English barrister, was considered dim-witted and was inclined to daydreaming. His parents ended his formal education when he was seventeen, and he was sent to work in South Africa, where his imagination was inspired by the people, animals, and jungle. He became close friends with authors Rudyard Kipling and Andrew Lang. Haggard's most popular books are King Solomon's Mines (1886) and She (1887). He also wrote short stories, as well as nonfiction on topics such as gardening, English farming, and rural life, interests which led to duties on government commissions concerned with land maintenance. For his literary contributions and his government service, Haggard was knighted in 1912. Several of Haggard's novels have been filmed. She was filmed in 1965, starring Ursula Andress. King Solomon's Mines was filmed with Stewart Granger and Deborah Kerr in 1950, and again with Richard Chamberlain and Sharon Stone in 1985. Also, the novel Allan Quatermain was filmed as Allan Quatermain and the Lost City of Gold with Richard Chamberlain and Sharon Stone in 1986.
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Prominent Designer M232; postmarked 1980; Very Maxwell - Misses' Coat, Jacket, Pants and Skirt.
With Vera Maxwell's super four-part wardrobe, you get lots of different looks for minimal time, work. Belt the coat and jacket or let them swing over the skirt or pants as well as other partners. Coat and jacket are cut the easy way with flip collar (all in one) and soft yoke detail. Skirt and pants have elastic waists for instant fit. Have fun with fabrics, play solids against checks, plaids, textures in knits, cotton, sheer wool.
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Jamal Crawford Dazzles In the Seattle Pro-Am
Jamal Crawford hails from Seattle, WA, which is one of the most underrated areas for NBA talent. The Seattle Pro-Am is one of the best summer amateur leagues in the country, and year after year top-flight NBA players make their mark in the league.
Crawford, who is one of the most exciting players in the NBA put on a show dropping 51 points in one of his most recent games. Take a look at Crawford do work in the video below.
S/O to @Ballislife for the footage.
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On-line literary magazine The Millions has a provocative interview with an artist who is also a copyright lawyer that is worth a read. It grows out of a recently mounted art show at New York's Drawing Center called "Day Job" which, according to the author,
"neatly upended the timeless lament of every struggling writer, artist and musician – My day job is robbing me of the time to do my REAL work! In a clever, counter-intuitive twist, the curator of the show asked a dozen artists to produce a work that illustrated how their day jobs enrich their art."
I thought this was a fascinating question, and one well worth posing here. How does your day job shape, inform, and even enrich your writing?