# 14 the initial data splits into two waves having half

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Unformatted text preview: ing in the positive and negative x directions with speeds a and ;a, respectively. u(x,0) -1 u(x,1/2a) 1 x u(x,1/a) -1 1 -1 x 1 x u(x,3/2a) 1 x -1 Figure 10.1.4: Solution of Example 10.1.3 at t = 0 (upper left), 1=2a (upper right), 1=a (lower left), and 3=2a (lower right). 10.1.2 Rankine-Hugoniot Conditions For simplicity, let us neglect b(x t u) in (10.1.1a) and consider the integral form of the conservation law d Z udx = ;f (u)j = ;f (u( t)) + f (u( t)) (10.1.10) dt which states that the rate of change of u within the interval x is equal to the change in its ux through the boundaries x = , . If f and u are smooth functions, then (10.1.10) can be written as Z ut + f (u)x]dx = 0: 10.1. Conservation Laws 9 If this result is to hold for all \control volumes&quot; ( ), the integrand must vanish, and, hence, (10.1.1a) and (10.1.10) are equivalent. To further simplify matters, let con ne our attention to the scalar conservation law ut + f (u)x = 0 (u a(u) = dfdu ) and (10.1.11b) ut + a(u)ux = 0: with (10.1.11a) (10.1.11c) The characteristic equation is dx = = a(u): (10.1.12a) dt The scalar equation (10.1.11c) is already in the canonical form (10.1.8a). We calculate the directional derivative on the characteristic as du = u dt + u dx = u + a(u)u = 0: (10.1.12b) x x dt t dt t Thus, in this homogeneous scalar case, u(x t) is constant along the characteristic curve (10.1.9a). For an initial value problem for (10.1.11a) on ;1 &lt; x &lt; 1, t &gt; 0, the solution would have to satisfy the initial condition u(x 0) = u0(x) ; 1 &lt; x &lt; 1: (10.1.13) Since u is constant along characteristic curves, it must have the same value that it had initially. Thus, u = u0(x0) u0 along the characteristic that passes through (x0 0). From 0 (10.1.12a), we see that this characteristic satis es the ordinary initial value problem dx = a(u0) t&gt;0 x(0) = x0 : (10.1.14) 0 dt Integrating, we determine that the characteristic is the straight line x = x0 + a(u0)t: 0 (10.1.15) This procedure can be repeated to trace other characteristics and thereby construct the soluti...
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