Differential Equations Lecture Work Solutions 260

# Differential Equations Lecture Work Solutions 260 - ct −...

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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 − ξ To Fnd the envelope means to Fnd a solution for F = 0 , and ∂F ∂ξ = 0 (i.e. the equation of characteristics is satisFed for all ξ . Now ∂F ∂ξ = f ( ξ ) c e − ct − 1 − 1 = 0 Since t > 0, this last condition gives the same result as before f ( ξ ) < − c 260...
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