Suppose a wave is travelling from right to left on x-axis. It satisfies $\Box \phi (x,t) =0$. Now I want this wave to get totally reflected at x=0. So what condition do I have to apply on $\phi (x,t)$ so that I get 100% reflection. The answer to this condition is supposed to be $\phi (x=0,t)=0$. But I can't figure why is it so? Nor can I come up with any argument to refute conditions like $\partial _x \phi (x=0,t)=0$, $\partial _t \phi (x=0,t)=0$, $\partial _{xt} \phi (x=0,t)=0$ etc.


An argument explaining why vanishing boundary conditions give rise to the total reflection of the wavepacket from the boundaries is as follows.

This argument is actually used to prove the existence of the solution under Dirichlet boundary conditions.

First consider the solution of $\square \phi(x,t)=0$ on the spatial segment $[-L,L]$ with periodic boundary conditions and for given initial conditions satisfying the same constraint.

This problem can be completely solved (proving an existence and uniqueness theorem if $\phi$ is supposed to be a bit more regular than $C^2([-L,L]\times \mathbb{R})$) using Fourier analysis, expanding the initial conditions and the solution in Fourier series. The solution turns out to be spatially periodic. As a matter of fact, the general solution is made of a couple of independent packets rigidly moving along the opposite directions: $\Phi(x,t)= f(x+t)+ g(x-t)$.

Now we come back to the problem we are actually interested in, i.e., the solution of $\square \phi=0$ on the spatial segment $[0,L]$ and assuming vanishing boundary conditions at $\pm L$, i.e. $\phi(0)=\phi(L) =0$ (obviously supposing that the initial conditions $\phi(x,0)$ and $\partial_t \phi(x,0)$ satisfy the same boundary constraint).

It is not difficult to prove that the solution can be obtained as follows.

(1) Consider the doubled spatial domain $[-L,L] = [-L,0] \cup [0,L]$,

(2) Define a new field $\Phi(x,t)$ on that enlarged domain which is the odd extension of $\phi$: $$\Phi(x,t) := \phi(x,t) \quad \mbox{if $x\in [0,L]$, }\qquad \Phi(x,t) := -\phi(-x,t) \quad \mbox{if $x\in [-L,0]$,}$$

(3) Analogously extend the initial conditions of $\phi$ to corresponding initial conditions of $\Phi$.

It is not difficult to see that the problem $\square \Phi(x,t)=0$ on $[-L,L] \times \mathbb{R}$ with the condition above is a problem with periodic boundary conditions over $[-L,L]$.

It therefore admits a unique solution as we said above. The half solution $\phi(x,t)=\Phi(t,x)|_{x \in [0,L]}$ is the solution of the original problem with vanishing boundary conditions on $[0,L]$.

It is now easy to figure out the form of the solution. Let us start with an initial packed $\phi(x,t)$ (and its time derivative) in $[0,L]$ concentrated far from the boundaries and moving towards $L$ (this is imposed by suitably fixing $\partial_t\phi(x,0))$.

Extend this packed on $[-L,0]$ as $-\psi(-x,0)$ and do the same for the other initial condition. Since we have this way reversed the sign of the time derivative of the initial datum in the segment $[-L,0]$, the packet in $[-L,0]$ will move towards $-L$ instead of $L$.

If thinking of the problem as a periodic problem on $[-L,L]$, everything is known: You have two packets, one in $[0,L]$ moving towards $L$ and another in $[-L,0]$ moving towards $-L$. The latter has a shape identical to the former up to the sign.

Since the extended problem is periodic in $[-L,L]$, when the packets reach the respective boundaries $\pm L$, they pass through them and pop out from the opposite side $\mp L$. In practice, restricting our attention to what happens in $[0,L]$ (where the original problem with vanishing boundary conditions is discussed), we see that, after the packet has reached the boundary at $L$, another packet appears around the same boundary moving towards $0$. This outgoing packet is identical to the ingoing one up to the sign. The net effect is a total reflecion of the ingoing packet from the boundary at $L$.


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