PDE 83215 Solutions to exercises 5 1. (a) The general solution of utt ...

3. Let us write u(x, t) in the formThenand∞∑( ) nπxu(x, t) = f n (t) sinn=1L∞∑( ) nπxu t (x, t) = f n(t) ′ sinn=1L∞∑ ∞∑( ) ( )nπx mπxu t (x, t) 2 = f n(t)f ′ m(t) ′ sin sinn=1 m=1L Lwhen we integrate this expression over [0, L] the only terms that will contribute arethose with m = n. Using the principle that the average of sin 2 is 1 2we have∫ L0u t (x, t) 2 dx = L 2∞∑f n(t) ′ 2n=1Similarlyand∞∑( )nπ nπxu x (x, t) =n=1L f n(t) cosL∞∑ ∞∑u x (x, t) 2 nmπ 2( ) ( )nπx mπx=fn=1 m=1L 2 n (t)f m (t) cos cosL L∫ LPutting everything together we haveE = ρ 2∫ LNow we use the form of f n (t):Differentiating gives00u x (x, t) 2 dx = L 2(u2t + c 2 u 2 x)dx =ρL4∞∑n=1n 2 π 2L 2 f n(t) 2(∞∑f n(t) ′ 2 + c2 n 2 π 2 )fn=1L 2 n (t) 2( ) ( )nπctnπctf n (t) = α n cos + β n sinLLf ′ n(t) = nπcL( ( nπct−α n sinL) ( )) nπct+ β n cosLand thusf n(t) ′ 2 + c2 n 2 π 2fL 2 n (t) 2 = c2 n 2 π 2 ( )α2L 2 n + βn2The energy is thus writtenE = ρc2 π 24L∞∑n ( )2 αn 2 + βn2n=17