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9 Mar 2004: Sm =. 0. (ni=1. ψi(0) exp (2Viy))1/2. dy. (2.24). Recall thatn. ... 1 = 2/H0. (3.24)Hence from (3.16), by forming an expansion for ϕi, we find that.

9 Mar 2004: These substitutions transform (2.24) into. uN = I(hN) = Fr(hN) h1/2N. ... Appendix. The functions I and χA.1. The function I(hN). From the definitions (2.19), (2.24) and (2.25) we obtain.

9 Feb 2004: H 0 1!5 2F P12 f P1 1! 4.24! FIG. 2. ... On applicationthe boundary conditions4.23! and 4.24!, we find that. a 5 4.523 102 3, F P15 2 8.42 for Fr5 11.9.

9 Mar 2004: 0.9993 4.81 0 1.52 – –S3 0 45.8 79.80 0.574 1.375 1.0192 0.9996 19.24 0 1.36 – –S7 0 45.2 ... 19.24 0.0282 1.37 550 0.100R10 0.0500 46.3 79.97 0.579 1.390 1.0107 1.0007 9.80 0.0551 0.99 450

9 Mar 2004: Received 24 April 2000 and in revised form 7 December 2000).

9 Mar 2004: 15 P 24.2 P25.2 L25.2 P25.2 P30.2 L. T (C) T (C).

9 Mar 2004: Indeed, for most of the experimentsdiscussed by TML λ was in the range 0.02–0.24. ... tr dr rN. drNdthN. (2.24). We eliminate h/t with the aid of (2.12) and, after some rearrangement, we againobtain the result (2.22).

9 Mar 2004: which indicates that, at least initially,. s = (ΓT 4/3/L)t. (2.24). ... 24 H. E. Huppert. surface. The magma flows in conduits or dykes in a manner controlled by elastic, fluid.

9 Feb 2004: 5H1 2(H0U1)′=0, (24)3U1 H ′1 H ′0ϕ1. 1. 2H0ϕ. ′1=0, (25)ϕ1=2720. ... 4y2)], (38). and then subsequent substitution into (24) yields a second-order differential equation forH1(y), namely(1.

9 Mar 2004: salinity of the ice) is very small; k #1 0 24 [Gross et al., 1987].