= Solution
Let $f(x)$ be the initial triangular profile, extended by zero for $x>0$. For <fixed-end reflection of a triangular pulse>, the right-moving incident component is $f(x-c_st)$; the reflected pulse at a fixed end has opposite sign. The full solution on $x\leq0$ is
$$
\boxed{y(x,t)=f(x-c_st)-f(-x-c_st).}
$$
It satisfies $y(0,t)=0$ exactly. Write $\xi=x/b\leq0$ and measure displacement in units of $a$, assuming $a,b,c_s>0$. At the first specified time the profile is
$$
\frac{y(x,t_1)}a=\begin{cases}0,&\xi<-3/2,\\\xi+3/2,&-3/2\leq\xi\leq-1/2,\\-2\xi,&-1/2\leq\xi\leq0.\end{cases}
$$
At the second time it is
$$
\frac{y(x,t_2)}a=\begin{cases}0,&\xi<-5/4,\\\xi+5/4,&-5/4\leq\xi\leq-3/4,\\1/2,&-3/4\leq\xi\leq-1/4,\\-2\xi,&-1/4\leq\xi\leq0.\end{cases}
$$
At the third time the two components cancel throughout the half-string:
$$
\boxed{y(x,t_3)=0\quad(x\leq0).}
$$
This flat instantaneous profile does not mean the wave energy has vanished. On $-b<x<0$, its velocity at $t_3$ is $-2ac_s/b$, and it subsequently becomes a negative reflected pulse travelling left.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2013-fixed-end-triangular-pulse-reflection.png]
{title=Incident, reflected and total profiles at the three specified reflection times}
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