Solve the Neumann problem for the wave equation on the half line $-\infty < x <0 $.
$u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad -\infty < x < 0,\quad t>0$
$u(x,0)=0, u_t (x,0)=0 \quad (For \quad All \quad -\infty < x < 0)$
$u_x(0,t)=0 \quad \quad \quad \quad \quad(For \quad All \quad t > 0)$
What i tried
Im familiar with the derivation of the case when the wave equation of the line is $0 < x <\infty $.
$u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad 0 < x <\infty,\quad t>0$
$u(x,0)=0, u_t (x,0)=0 \quad (For \quad All \quad 0 < x <\infty)$
$u_x(0,t)=0 \quad \quad \quad \quad \quad(For \quad All \quad t > 0)$
Which after using the method of even extensions and reducing the problem to the wave equation on the whole line
$u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad -\infty < x <\infty,\quad t>0$
$u(x,0)=0, u_t (x,0)=0 \quad $
gives the solution of $$u(x,t) = \frac{1}{2} \left[ \phi(x+ct) + \phi(x-ct) \right] + \frac{1}{2c} \int_{x-ct}^{x+ct} \psi(y) \, dy \,\,\,\,\,\, \mathrm{for}\,\,\, x>ct$$ $$u(x,t) = \frac{1}{2} \left[ \phi(x+ct) + \phi(ct-x) \right] + \frac{1}{2c} \int_{0}^{ct-x} \psi(y) \, dy + \frac{1}{2c} \int_{0}^{ct+x} \psi(y) \, dy \,\,\,\,\,\, \mathrm {for} \,\,\,0<x<ct$$
Im thinking of the way to modify this solution such that it solves the original question when the wave equation on the half line range from $-\infty < x <0 $ an i got this solution as my answer
$$u(x,t) = \frac{1}{2} \left[ \phi(x+ct) + \phi(x-ct) \right] + \frac{1}{2c} \int_{x-ct}^{x+ct} \psi(y) \, dy \,\,\,\,\,\, \mathrm{for}\,\,\, x>-ct$$ $$u(x,t) = \frac{1}{2} \left[ \phi(-x-ct) + \phi(x-ct) \right] + \frac{1}{2c} \int_{0}^{x-ct} \psi(y) \, dy + \frac{1}{2c} \int_{0}^{-ct-x} \psi(y) \, dy \,\,\,\,\,\, \mathrm {for} \,\,\,-\infty<x<0$$
Am i correct. Could anybody explain. Thanks