Differential Equations Lecture Work Solutions 284

Differential Equations Lecture Work Solutions 284 - 5 4 3 t...

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Unformatted text preview: 5 4 3 t x+ct>0 2 u(0,t)=e^(−t) 1 x+ct<0 0 u(x,0)=sin x and ut (x,0)=0 −1 −5 −4 −3 −2 −1 0 x 1 2 3 4 Figure 56: Domain for problem 2 2. utt − c2 uxx = 0 x<0 u(x, 0) = sin x ut (x, 0) = 0 x<0 x<0 u(0, t) = e−t t>0 u(x, t) = F (x − ct) + G(x + ct) F (x) = 1 2 G(x) = 1 2 sin x since f = sin x, g = 0 sin x From boundary condition u(0, t) = F (− ct) + G(ct) = e−t If x + ct < 0 no influence of boundary at x = 0 ⇒ u(x, t) = 1 2 sin (x − ct) + 1 2 sin (x + ct) = sin x cos ct . . . after some trigonometric manipulation If x + ct > 0 then the argument of G is positive and thus G(ct) = e−t − F (−ct) or G(z ) = e−z/c − F (−z ) ⇒ G(x + ct) = e− x + ct c − F (−(x + ct)) 284 5 ...
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