Proving the Semigroup Property using D'Alembert So I am currently trying to prove the SemiGroup Property for the Wave Equation using D'Alembert. The way I am approaching it is as follows:
Suppose $u$ solves the PDE:
$$u_{tt}=c^2u_{xx}, \>\>\>\>\>\>u(x,0)=\phi(x), \>\>\>\>u_t(x,0)=\psi(x)$$
Set times $0<t_2<t_1$. Now consider the Initial Data:
$$\phi(x) = u(x,t_1), \>\>\>\>\psi(x) = u_t(x,t_1)$$
I want to show:
$$u(x,t_2) = \frac{1}{2}[u(x+c(t_2-t_1),t_1)+u(x-c(t_2-t_1),t_1)]+\frac{1}{2c}\int_{x-c(t_2-t_1)}^{x+c(t_2-t_1)}u_t(\tau,t_1)d\tau$$
Now I got:
$$u(x,t_2) = \frac{1}{2}[\phi(x+ct_2)+\phi(x-ct_2)]+\frac{1}{2c}\int_{x-ct_2}^{x+ct_2}\psi(\tau)d\tau$$
By D'Alembert. Now, I get:
$$\phi(x+ct_2) = u(x+ct_2,t_1) = \frac{1}{2}[\phi(x+c(t_2+t_1))+\phi(x+c(t_2-t_1))]+\frac{1}{2c}\int_{x+c(t_2-t_1)}^{x+c(t_2+t_1)}\psi(\tau)d\tau$$
$$\phi(x-ct_2) = u(x-ct_2,t_1) = \frac{1}{2}[\phi(x-c(t_2-t_1))+\phi(x-c(t_2+t_1))]+\frac{1}{2c}\int_{x-c(t_2+t_1)}^{x-c(t_2-t_1)}\psi(\tau)d\tau$$
$$\psi(\tau) = u_t(\tau,t_1) = \frac{c}{2}[\phi_t(x+ct_1)-\phi_t(x-ct_1)]+\frac{1}{2c}\left(c\psi(x+ct_1)+c\psi(x-ct_1)+\int_{x-ct_1}^{x+ct_1}\psi_t(\tau)d\tau\right)$$
How am I supposed to conclude anything? Is there an easier way to approach this?? I can't even tell whats happening because theres so many $\psi(x)$ and $\phi(x)$ everywhere. 
 A: Here is my approach: First by D'Alembert
$$u(x,t) = \frac 1 2 (\phi (x+ct) + \phi (x-ct)) + \frac 1 {2c} \int\limits_{x-ct}^{x+ct}\psi (\tau) d \tau$$
So of course:
$$u(x,t_2) = \frac 1 2 (\phi (x+ct_2) + \phi (x-ct_2)) + \frac 1 {2c} \int\limits_{x-ct_2}^{x+ct_2}\psi (\tau) d \tau$$
and
$$u(x +c(t_2-t_1),t_1) = \frac 1 2 (\phi(x+ct_2) + \phi (x+ ct_2 - 2ct_1)) + \frac 1 {2c} \int_{x+ct_2-2ct_1}^{x+ct_2}\psi (\tau) d \tau  \quad (*)$$
and
$$u(x -c(t_2-t_1),t_1) = \frac 1 2 (\phi(x- ct_2 + 2ct_1) + \phi (x- ct_2)) + \frac 1 {2c} \int\limits^{x-ct_2+2ct_1}_{x-ct_2}\psi (\tau) d \tau \quad (**)$$
Furthermore
$$u_t(\tau,t_1) = \frac{d}{dt}u(x,t) \Big|_{x=\tau,t=t_1} = \frac c 2(\phi ' (\tau +ct_1)-\phi ' (\tau - c t_1)) + \frac 1 {2c} (c \psi (\tau + ct_1) + c \psi (\tau - ct_1))$$
Note that then 
$$ \int\limits_{x-ct_2+ct_1}^{x+ct_2-ct_1} u_t (\tau,t_1) d \tau =\int\limits_{x-ct_2+ct_1}^{x+ct_2-ct_1} \left( \frac c 2(\phi ' (\tau + ct_1) - \phi ' (\tau - ct_1)) + \frac 1 {2c} (c \psi (\tau + ct_1) + c \psi (\tau -ct_1)) \right)d\tau \\ = \frac c 2 \int\limits_{x-ct_2 + 2ct_1}^{x+ct_2}\phi ' (\tau) d \tau - \frac c 2 \int\limits_{x-ct_2}^{x+ct_2-2ct_1}\phi ' (\tau) d \tau  +  \frac 1 2 \int\limits_{x-ct_2 + 2t_1}^{x+ct_2}\psi (\tau) d \tau +\frac 1 2 \int\limits_{x-ct_2}^{x+ct_2-2ct_1} \psi (\tau) d \tau \\ = \frac c 2 \left( \phi (x+ct_2) - \phi (x-ct_2 + 2ct_1) - \phi (x + ct_2 - 2ct_1) + \phi (x-ct_2) \right) \\ \quad +  \frac 1 2 \left ( \int\limits_{x-ct_2 + 2t_1}^{x+ct_2}\psi (\tau) d \tau + \int\limits_{x-ct_2}^{x+ct_2-2ct_1} \psi (\tau) d \tau \right) \quad (***)$$
So if we use $(*),(**),(***)$ in we get
$$\frac 1 2 (u(x +c(t_2-t_1),t_1)+ u(x -c(t_2-t_1),t_1)) + \frac 1 {2c} \int\limits_{x -ct_2 +ct_1}^{x + ct_2 - ct_1} u_t (\tau,t_1)d\tau \\ = \frac 1 2 \left(  \frac 1 2 (\phi(x+ct_2) + \phi (x+ ct_2 - 2ct_1)) + \frac 1 2 (\phi(x- ct_2 + 2ct_1) + \phi (x- ct_2)) \\ \quad + \frac 1 c \left(\frac c 2 \left( \phi (x+ct_2) - \phi (x-ct_2 + 2ct_1) - \phi (x + ct_2 - 2ct_1) + \phi (x-ct_2) \right)\right)\\ \quad + \frac 1 {2c} \int_{x+ct_2-2ct_1}^{x+ct_2}\psi (\tau) d \tau  + \frac 1 {2c} \int\limits^{x-ct_2+2ct_1}_{x-ct_2}\psi (\tau) d \tau \\ \quad +  \frac 1 c \left( \frac 1 2  \left( \int\limits_{x-ct_2 + 2t_1}^{x+ct_2}\psi (\tau) d \tau + \int\limits_{x-ct_2}^{x+ct_2-2ct_1} \psi (\tau) d \tau \right) \right) \right) \\ = \frac 1 2 (\phi (x+ct_2) + \phi (x-ct_2)) + \frac 1 {2c} \int\limits_{x-ct_2}^{x+ct_2}\psi (\tau) d \tau = u(x,t_2)$$
I don't see an easier way.
