Solving 1D Wave Equation w/ Initial Values

[stunner5000pt]

given the initial boundary value problem
$u_{tt} - u_{xx} = 0$, 0<x<1, t>0,
$u(x,0) =1$, $0\leq x \leq 1$,
$u_{t}(x,0) = \sin(\pi x)$, $0\leq x \leq 1$,
$u_{x} (0,t) = 0$, $t \geq 0$,
$u_{x} (1,t) = 0$.

find u(0.5,1). Where u is d'Alembert's solution for the 1D wave equation
well f(x) = 0, and g(x) = sin (pi x)
but i need to even extend g(x) since the wave is not fixed at its endpoints. WOuld the even extension of sin (pi x) simply be a shift to the right by pi/2 that is $\sin(\pi x - \frac{\pi}{2})$


Another question i have is
Find u(x,t) corresponding to initial values f(x) = sin x, g(x) = 0, when u(0,t) =0 and [tex] u_{x} (\frac{\pi}{2} ,t) = [/itex] , c=1.
Nowi know i have to extend f as an evenfunction about x = pi/2 only (right?). But how would one go aboutenxtending sine as an evne function? do i make sine into the cosine function and then shift that by pi/2 to the right? that is cos (x - pi/2).

Is this correct? Please help!

 

[lippyka]

Yes, your idea for the first problem is correct. The even extension of sin(πx) would be sin(πx-π/2). For the second problem, you are also correct that you need to extend f(x) as an even function about x=π/2. To do this, you can use the identity sin(x−π/2)=cos(x) and then extend the cosine function as an even function about x=π/2. So, the even extension of sin(x) should be cos(x-π/2).