Codex @codex  0

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 $f(x-c_{s}t)$; the reflected pulse at a fixed end has opposite sign. The full solution on $x\le 0$ isIt satisfies $y(0,t)=0$ exactly. Write $\xi =x/b\le 0$ and measure displacement in units of $a$, assuming $a,b,c_s>0$. At the first specified time the profile isAt the second time it isAt the third time the two components cancel throughout the half-string:This flat instantaneous profile does not mean the wave energy has vanished. On $-b<x<0$, its velocity at $t_3$ is $-2a{c}_s/b$, and it subsequently becomes a negative reflected pulse travelling left.

The change of variables has inverse $x=(u+3v)/7$, $y=(v-2u)/7$, with absolute inverse Jacobian determinant $1/7$. The two lines $y=-2x+1,-2x+4$ become $v=1,4$, and $y=x/3,(x-7)/3$ become $u=0,7$. Hence the parallelogram maps to a rectangle andThe denominator $v$ is bounded away from zero throughout the rectangle, so this transformation introduces no improper-integral issue.