Let u(x, t) be the solution of

u_tt − u_xx = 0, 0 < x < 2, t > 0

u(0, t) = 0 = u(2, t), ∀ t > 0, 

u(x, 0) = sin (πx) + 2 sin(2πx), 0 ≤ x ≤ 2,

u_t(x, 0) = 0, 0 ≤ x ≤ 2.

Which of the following is true?

Explanation:

Given 

utt − uxx = 0, 0 < x < 2, t > 0

u(0, t) = 0 = u(2, t), ∀ t > 0, 

u(x, 0) = sin(πx) + 2sin(2πx), 0 ≤ x ≤ 2,

ut(x, 0) = 0, 0 ≤ x ≤ 2.

which is a wave equation of finite length. So solution is

u(x, t) = \(\sum D_n \sin({n\pi x\over l})\cos({n\pi ct\over l})\) where

\(D_n=\frac2l\int_0^lf(x)\sin({n\pi x\over l})dx\)

Here c = 1, l = 2, f(x) = sin(πx) + 2sin(2πx)

So, \(D_n=\frac22\int_0^2(\sin(\pi x)+2\sin (2\pi x))\sin({n\pi x\over 2})dx\)

and u(x, t) = \(\sum D_n \sin({n\pi x\over 2})\cos({n\pi ct\over 2})\)

u(x, 0) = \(\sum D_n \sin({n\pi x\over 2})\) = sin(πx) + 2sin(2πx)

Comparing we get

D₂ = 1, D₄ = 2, D_n = 0 for other natural number n

Hence we get

u(x, t) = sin(πx) cos(πt) + 2sin(2πx)cos(2πt)

Then u(1, 1) = 0

u(1/2, 1) = -1

u(1/2, 2) = 1

u(1/2, 1/2) = 0

Option (3) is correct, other are false.