Differential Equations Lecture Work Solutions 282

# Differential Equations Lecture Work Solutions 282 - 5 4...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5 4 x−ct<0 t 3 u(0,t)=h(t) 2 P(x,t) x−ct>0 1 D(0,t − x /c) 0 C(x−ct,0) B(ct−x,0) A(x+ct,0) u(x,0)=0 and ut (x,0)=0 −1 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Figure 55: Domain for problem 1 1. utt − c2 uxx = 0 u(x, 0) = 0 ut (x, 0) = 0 u(0, t) = h(t) Solution u(x, t) = F (x − ct) + G(x + ct) (*) F (x) = 1 1 f (x) − 2 2c G(x) = 1 1 f (x) + 2 2c x 0 g (τ ) dτ = 0 (#) since both f (x), g (x) are zero. Thus for x − ct > 0 the solution is zero (No inﬂuence of boundary at x = 0) For x − ct < 0 u(0, t) = h(t) ⇓ F (− ct) + G(ct) = h(t) u(x, t) = F (x − ct) + G(x + ct) 282 x 0 g (τ ) dτ = 0 ...
View Full Document

## 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.

Ask a homework question - tutors are online