Differential Equations Lecture Work Solutions 268

Differential Equations Lecture Work Solutions 268 - x is:...

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3. u t +4 uu x =0 o r u t +(2 u 2 ) x =0 u ( x, 0) = 3 x< 1 2 x> 1 Shock again The shock characteristic is obtained by solving: dx s dt = 2 · 3 2 2 · 2 2 3 2 =10 x s =10 t + x s (0) | {z } =1 x s =10 t +1 Now we solve the ODE for u : du dt =0 u ( x, t )= u ( x 0 , 0) away from shock
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Unformatted text preview: x is: dx dt = 4 u = 4 u ( x , 0) x = 4 u ( x , 0)) t + x If x &lt; 1 x = x 12 t x &lt; 1 + 12 t x &lt; 1 x = x 8 t x &gt; 1 + 8 t 268...
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