The Wave Equation — Question 5

Question 5 – Solution

Strategy. An odd or even spatial reflection enforces the corresponding boundary condition.

Step 1: Verify the two boundary models. Each translated profile solves $u_{tt}=u_{xx}$. Since $F$ is even and $F'$ is odd, $$\boxed{u_D(0,t)=F(t-a)-F(a-t)=0,\qquad (u_N)_x(0,t)=F'(t-a)+F'(a-t)=0.}$$ Thus $u_D$ satisfies the fixed-end condition and $u_N$ the free-end condition.

Step 2: Recover the actual initial data. Differentiating the time arguments carefully gives $$\begin{aligned} u_D(x,0)&=F(x-a)-F(x+a),& (u_D)_t(x,0)&=F'(x-a)+F'(x+a),\\ u_N(x,0)&=F(x-a)+F(x+a),& (u_N)_t(x,0)&=F'(x-a)-F'(x+a), \end{aligned}$$ where $F'(z)=-2ze^{-z^2}$. The incident term alone generally has neither zero boundary displacement nor zero boundary slope. Its reflected partner supplies the exact cancellation; small Gaussian tails are not identically zero.

Step 3: Inspect the collision instant. At $t=a$, the two profile arguments both equal $x$. Therefore $$\boxed{u_D(x,a)=0,\quad (u_D)_t(x,a)=2F'(x),\qquad u_N(x,a)=2F(x),\quad (u_N)_t(x,a)=0.}$$ The fixed-end string has zero displacement but nonzero velocity for $x>0$: it is not at rest. The free-end field has a doubled displacement profile at that instant and zero instantaneous velocity. Displacement and velocity must both be known to specify the state.

Step 4: Interpret the reflected sign. For $a=3$, the incoming peak at $t=1$ is near $x=2$. At $t=3$ the fixed-end displacement is exactly zero everywhere. At $t=5$ the reflected pulse is near $x=2$ with opposite sign. Indeed $u_D(x,5)=-u_D(x,1)$. A fixed end reverses displacement upon reflection; a free end preserves its sign. The plotted zero snapshot is an exact cancellation, not disappearance of the motion or its energy.