Differential Equations Lecture Work Solutions 273

Differential Equations Lecture Work Solutions 273 - 7. ut +...

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Unformatted text preview: 7. ut + uux = 0 u (x, 0) = 2 for x < 0 1 for 0 < x < 1 0 for x > 1 First find the shock characteristic for those with speed u = 2 and u = 1 [q ] = 1 22 1 3 u = (22 − 12 ) = 21 2 2 [u] = 2 − 1 = 1 Thus dxs 3 = dt 2 and the characteristic through x = 0 is then xs = 3 t 2 Similarly for the shock characteristic for those with speed u = 1 and u = 0 [q ] = 12 u 2 1 0 = 12 1 (1 − 02 ) = 2 2 [u] = 1 − 0 = 1 Thus 1 dxs = dt 2 and the characteristic through x = 1 is then xs = 1 t+1 2 Now these two shock charateristic going to intersect. The point of intersection is found by equating xs in both, i.e. 1 3 t+1= t 2 2 3 The solution is t = 1 and xs = . Now the speeds are u = 2 and u = 0 2 [q ] = 12 u 2 2 0 = 12 (2 − 02 ) = 2 2 273 ...
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