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Unformatted text preview: ing in the positive and negative x directions with speeds a and ;a,
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
d Z udx = ;f (u)j = ;f (u( t)) + f (u( t))
which states that the rate of change of u within the interval
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
ut + f (u)x]dx = 0: 10.1. Conservation Laws 9 If this result is to hold for all \control volumes" ( ), 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
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):
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:
Thus, in this homogeneous scalar case, u(x t) is constant along the characteristic curve
For an initial value problem for (10.1.11a) on ;1 < x < 1, t > 0, the solution
would have to satisfy the initial condition
u(x 0) = u0(x) ; 1 < x < 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
(10.1.12a), we see that this characteristic satis es the ordinary initial value problem
dx = a(u0)
x(0) = x0 :
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
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- Spring '14