given the initial boundary value problem(adsbygoogle = window.adsbygoogle || []).push({});

[itex] u_{tt} - u_{xx} = 0 [/itex], 0<x<1, t>0,

[itex] u(x,0) =1 [/itex], [itex] 0\leq x \leq 1 [/itex],

[itex] u_{t}(x,0) = \sin(\pi x) [/itex], [itex] 0\leq x \leq 1 [/itex],

[itex] u_{x} (0,t) = 0 [/itex], [itex] t \geq 0 [/itex],

[itex] u_{x} (1,t) = 0 [/itex].

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 [itex] \sin(\pi x - \frac{\pi}{2}) [/itex]

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!

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# Homework Help: PDE question

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