Consider the wave equation and take Dirichlet boundary conditions:
$$v_{tt} - c^2v_{xx} = =0 \ \ \ \ \ \ ( 0 < x < \infty, \ \ -\infty < t < \infty) $$
$$ v(x,0) = \phi (x) , \ \ v_t(x,0) = \psi(x) \ \ \ \ \ \ for \ \ t = 0 $$
$$ v(0,t) = 0 \ \ \ \ \ \ for \ \ x = 0 $$
Convert the given problem into whole line by taking an odd extension of $\phi$ and $\psi$:
$$ \phi_{odd} = \left\{ \begin{array}{l l} \phi(x) & \quad \text{if x > 0}\\ -\phi(-x) & \quad \text{if x <0}\\ 0 & \quad \text{if x =0} \end{array} \right.$$
and
$$ \psi_{odd} = \left\{ \begin{array}{l l} \psi(x) & \quad \text{if x > 0}\\ -\psi(-x) & \quad \text{if x <0}\\ 0 & \quad \text{if x =0} \end{array} \right.$$
$$u_{tt} - c^2u_{xx} = =0 \ \ \ \ \ \ ( -\infty < x < \infty, \ \ -\infty < t < \infty) $$
$$ u(x,0) = \phi_{odd} (x) , \ \ u_t(x,0) = \psi_{odd}(x) \ \ \ \ \ \ for \ \ t = 0 $$
$$ u(0,t) = 0 \ \ \ \ \ \ for \ \ x = 0 $$
We know that the solution of the wave equation on the whole line is
$$u(x,t) = \frac{1}{2} \{\phi_{odd}(x+ct) + \phi_{odd}(x-ct) \} + \frac{1}{2c} \int \limits_{x-ct}^{x+ct} \psi_{odd}(y) dy.$$
I want to show that
$$v(x,t) = \frac{1}{2} \{\phi(x+ct) - \phi(ct-x) \} + \frac{1}{2c} \int \limits_{ct-x}^{x+ct} \psi(y) dy.$$
Please help me show that $\phi_{odd}(ct+x) = \phi(ct+x)$ and $\phi_{odd}(x-ct) = -\phi(ct-x)$.
Thank you
In this application of the method of rotations, we are looking at $\phi$ across the two intervals $x>c|t|$ and $0<x<c|t|$ which are the areas above and below the curve $x=t$ respectively in the first quadrant. The difference in formula is established by this. Note that in the region $x>c|t|$, everything is positive, hence $\phi_{\text{odd}}(ct+x)=\phi(ct+x)$. For the region $0<x<c|t|$, we have \begin{equation*} \phi_{\text{odd}}(x-ct)=\phi(-(ct-x))=-\phi(ct-x) \end{equation*} by the properties of odd functions. To get the formula for $v(x,t)$, replace $y$ with $-y$.