Differential Equations Lecture Work Solutions 260

Differential Equations Lecture Work Solutions 260 - ct 1 To...

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8. b. Breaking time can be found by looking at the time that u t or u x tend to u t = cu + f 0 ± u t c (1 e ct ) ue ct ² e ct u x = f 0 ± 1+ u x c (1 e ct ) ² e ct Solve for u t u t = cu uf 0 1+ f 0 c (1 e ct ) and for u x u x = f 0 e ct 1+ f 0 c (1 e ct ) The smallest t for which the denominator is zero ( u x and u t →∞ )is 1+ f 0 c (1 e ct )=0 Since c, t are positive, the factor (1 e ct ) 0andso f 0 ( ξ ) < c Another way is by looking at the family of characteristics x ( t )= 1 c f ( ξ ) e ct 1 + ξ or F ( x, t, ξ )= x ( t )+ 1 c f ( ξ ) e
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Unformatted text preview: ct 1 To Fnd the envelope means to Fnd a solution for F = 0 , and F = 0 (i.e. the equation of characteristics is satisFed for all . Now F = f ( ) c e ct 1 1 = 0 Since t &gt; 0, this last condition gives the same result as before f ( ) &lt; c 260...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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