Discussion
Let \(u(x, t)\) satisfy the wave equation
\(\dfrac{\partial^2u}{\partial t^2}=c^2\dfrac{\partial^2u}{\partial x^2}\),
\(-\infty \lt x \lt \infty\), \(t \gt 0\), \(c \gt 0\)
with
\[ u(x, 0) = \begin{cases} 1 & \text{if } |x| \lt 1 \\ 0 & \text{otherwise} \end{cases} \]
and
\[ \frac{\partial u}{\partial t}(x, 0) = 0. \]
By using D'Alembert's formula, the maximum value of \(u(0, t)\) for \(t \gt 0\) is
Answer: A
The maximum value is 1.
With zero initial velocity, D'Alembert's formula reduces to the average of two travelling copies of the initial shape:
\[ \begin{aligned} u(x, t) &= \tfrac{1}{2}[f(x - ct) + f(x + ct)] \\ u(0, t) &= \tfrac{1}{2}[f(-ct) + f(ct)] \end{aligned} \]
For \(0 \lt t \lt 1/c\) both \(|ct|\) and \(|-ct|\) are below 1, so both terms equal 1 and \(u(0, t) = 1\). For \(t \gt 1/c\) both copies have moved past the origin and \(u(0, t) = 0\).
So the largest value is 1. The value \(\frac{1}{2}\) is what a point outside the initial pulse sees when one half-pulse passes it, which is the usual trap; \(\frac{1}{4}\) and 0 are not the maximum at the origin.
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