Suppose we are solving the wave equation in one dimension as given by:
$$
u_{tt} – c^2 u_{xx} = 0 , \quad x \in \mathbb{R}, t>0 \\
u(x,0) = \phi(x), \qquad x \in \mathbb{R} \\
u_t(x,0) = c \phi'(x), \quad x \in \mathbb{R}\\
$$
d'Alembert's solution gives us that $$ u(x,t) = \frac{1}{2} \left( \phi(x+ct) + \phi(x-ct) \right) + \frac{1}{2c}\int_{x-ct}^{x+ct} c\phi'(y) dy$$
When we evaluate the integral as $\frac{1}{2}( \phi(x +ct) – \phi(x-ct))$, the solution is then $$ u(x,t) = \phi(x+ct)$$
Which seems counterintuitive to me. Interpretting the d'Alembert solution as a left and right moving wave, we get that the left moving wave doesn't exist?
[This question arises from a problem to find the solution to the wave equation (in one dimension) on two media seperated at the origin, with the inital displacement only in $x>0$, which would mean the wave only travels away from the boundary, so no question at all?]
Best Answer
The boundary condition $u_t(x,0) = c \phi' (x)$ is very special, causing exactly the cancellation you observe. Changing this boundary condition to $u_t(x,0) = - c \phi'(x)$ cancels the other wave. (I recall; I haven't checked whether the other boundary condition must be tweaked also.)