# wave equation with moving end

Solve the wave equation with moving end: \begin{aligned} u_{t t} &=c^{2} u_{x x} \text { for } x>0, t>0 \\ u(x, 0) &=\varphi(x), u_{t}(x, 0)=\psi(x) \text { for } x \geq 0 \\ u(0, t) &=f(t) \text { for } t \geq 0 \end{aligned}

Suppose $$x_{0} \geq c t_{0}$$. Now $$\left[x_{0}-c t_{0}, x_{0}+c t_{0}\right]$$ lies entirely on the nonnegative part of the $$x$$ - axis, and not enough time has passed for the initial displacement $$f(t)$$ at time zero to reach any point in this interval. In this case, $$u\left(x_{0}, t_{0}\right)$$ is not influenced by $$f$$ and the d'Alembert formula holds. $$u\left(x_{0}, t_{0}\right)=F\left(x_{0}-c t_{0}\right)+B\left(x_{0}+c t_{0}\right) \text { for } x_{0} \geq c t_{0}$$ $$F$$ and $$B$$ are the forward and backward waves.

$$\psi(x)$$ are not defined for $$x_{0}-c t_{0}; hence $$F\left(x-c t_{0}\right)$$ is not defined. However, putting $$x=0$$ yields $$u\left(0, t_{0}\right)=f\left(t_{0}\right)=F\left(-c t_{0}\right)+B\left(c t_{0}\right) .$$

This suggests that we can extend $$F$$ to this negative value by defining $$F\left(-c t_{0}\right)=f\left(t_{0}\right)-B\left(c t_{0}\right) .$$ Both function values on the right are well defined. Further, since $$t_{0}$$ can be any positive number, we can think of $$c t_{0}$$ as any positive number and use this equation as a model to define $$F(-x)=f\left(\frac{x}{c}\right)-B(x)$$ for any positive number $$x$$. This extends $$F$$ to negative values. For simplicity, we are using the same symbol $$F$$ for the extended function. Now put, for $$x_{0}-c t_{0}<0$$ \begin{aligned} F\left(x_{0}-c t_{0}\right) &=F\left(-\left(c t_{0}-x_{0}\right)\right) \\ &=f\left(\frac{c t_{0}-x_{0}}{c}\right)-B\left(c t_{0}-x_{0}\right) \end{aligned} or $$F\left(x_{0}-c t_{0}\right)=f\left(t_{0}-\frac{x_{0}}{c}\right)-B\left(c t_{0}-x_{0}\right) .$$ Substituting this into d'Alembert's solution, we have $$u\left(x_{0}, t_{0}\right)=f\left(t_{0}-\frac{x_{0}}{c}\right)-B\left(c t_{0}-x_{0}\right)+B\left(x_{0}+c t_{0}\right) \text { for } x_{0}-c t_{0}<0$$ In view of the definition of the backward wave $$B$$, this equation can be written \begin{aligned} u\left(x_{0}, t_{0}\right) &=f\left(t_{0}-\frac{x_{0}}{c}\right)+\frac{1}{2}\left(\varphi\left(x_{0}+c t_{0}\right)-\varphi\left(c t_{0}-x_{0}\right)\right) \\ &+\frac{1}{2 c} \int_{c t_{0}-x_{0}}^{x_{0}+c t_{0}} \psi(s) d s \text { for } x_{0} We have used the zero subscript to discuss the solution at a particular point and maintain $$x$$ and $$t$$ as variables. However, we now drop the subscript and write the solution at any $$(x, t)$$ with $$x \geq 0, t \geq 0$$ : \bbox[5px, border:2px solid black] {\begin{aligned} u(x, t) &=\frac{1}{2}(\varphi(x-c t)+\varphi(x+c t)) \\ &+\frac{1}{2 c} \int_{x-c t}^{x+c t} \psi(s) d s \text { for } x \geq c t\\ u(x, t) &=f\left(t-\frac{x}{c}\right)+\frac{1}{2}(\varphi(x+c t)-\varphi(c t-x)) \\ &+\frac{1}{2 c} \int_{c t-x}^{x+c t} \psi(s) d s \text { for } x

I didn't understand the bolded lines, like why we split the domain in $$x< ct \& x\geq ct$$? And what was mean by, and not enough time has passed for the initial displacement $$f(t)$$ at time zero to reach any point in this interval. In this case, $$u\left(x_{0}, t_{0}\right)$$ is not influenced by $$f$$ and the d'Alembert formula holds.

I copy the whole page of Beginning partial differential equations by Peter V. O'Neil, page: 134, section: 4.5

Any help will be appreciated, Thanks in advance.

For ease of explanation let us consider a vibrating string obeying the PDE in your question. At time $$t=0$$ the string is in position $$\varphi(x)$$ and is released with initial velocity $$\psi(x).$$ You can interpret the condtion $$u(0,t) = f(t)$$ as disturbing the left endpoint of the string over time. Since the wave travels with speed $$c,$$ this disturbance propagates with the same speed.
Now, suppose you start from the initial conditions and you let time pass by. Consider a point $$x_0$$ of the spatial domain at time $$t_0$$. The condition $$x_0>ct_0$$ means that the disturbance that you caused at time $$t=0$$ has not reached this far. This point does not know yet that there was a disturbance. Therefore it obeys the same PDE (located at this point, as if it has started at time $$t=t_0$$) but with $$u(x_0,t_0) =0.$$ Hence you can write the solution by d'Alemberts formula.
If the point $$(x_0, t_0)$$ is such that $$x_0 < ct_0,$$ then that point has seen your disturbance that has travelled along the string.